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paul Offline OP
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A very informative video about a very important subject.

use pause and play to read every word throughout the video , then replay it again and read every word of it again.

this video contains videos its not just words scrolling.
then even if you never have posted here before join and leave a comment.

let us not speak softly now , because the hour is getting late


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the below link is to a google maps location above the gulf of mexico.

showing a massive area where methane hydrates have caused
enormous landslides due to massive methane releases.

unstable ground but its where the deep drilling is taking place

the below image shows the resulting formation of the
sea floor follwing a massive methane release.



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So? Just a bunch of random facts skewed in the direction of scare-mongering. Run for the hills guys, aircraft engines might explode! And watch out for the high pressure under the sea, it might shoot you like a gun! But at least you'll feel empowered knowing that "production casings are cemented to stop oil migrating to thief zones"

But the part in the middle with the sea erupting around the oil rig is pretty cool. Wouldn't want to be the guy standing on there going "oh crap we broke the sea".

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paul Offline OP
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Quote:
Just a bunch of random facts


I dont know if the facts you speak of are facts or not.
this guy might have just pieced this together the way you do math!

remember

3 cycles , with 3 forces in each cycle

and you quantify the entire 3 cycles with only 8 of the 9 forces.
it might be that he is cheating just like you.
I cant say because I dont know him , I just know you.

have you ever posted a topic?

or do you just troll around looking for some place to cheat.

Quote:
But at least you'll feel empowered knowing that "production casings are cemented to stop oil migrating to thief zones"


yes , I feel so empowered knowing that.
so do all the people who lost their livelyhood because the oil wells dont leak.

especially those who were earning $18,000 a week before BP
cut their livelyhood off , that are now getting a whopping
$1,500 a week compensation.

now their the ones who really feel empowered.

but I can just about guarantee you that the financial institutions that financed their boats and rigging are
also feeling empowered to take the boats away from them.

maybe BP should be forced to pay the bills that the fishermen can no longer pay.

maybe a economic safety net of 100 billion dollars per experimental deep well should be paid up front for any new wells and for any existing wells given that we now know the liabilities that these wells present to local economies , to ensure that any state , local government , industry , buisness , person , including wildlife will not undergo a similar experience in the future.

the reason I say this is that this single event will cause more damages than BP has assets to cover.

its not like the well is located off the coast of nigeria
BP could probably buy the entire nigerian coast line , this well and the current experimental wells are off the coast of the United States , this land is extremely expensive or was.

if / when property values reduce then BP is responcible for this reduction of property value and upon the sale of a property should be made to pay the difference in the properties value.

and should be made to provide continual payments above the 100 billion safety net for any reported property value reduction that is the result of this or any oil spill to be held along with the 100 billion per experimental well.

its endless.







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Originally Posted By: paul

and you quantify the entire 3 cycles with only 8 of the 9 forces.


You never showed it could work. Therefore you cannot know that it would work.

Quote:

or do you just troll around looking for some place to cheat.

I troll around looking for mistakes to fix. Misinformation is a burden on society.

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paul Offline OP
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your reply in the " Orion, Mission to Alpha Centauri " thread.

in message #35130

Originally Posted By: Kallog
Move the spaces in the list:


+1 accelerator.
-1 1st turn.

+1 2nd turn.
+2 accelerator.
-3 1st turn.

+3 2nd turn.
+3 accelerator.
-6 1st turn.

+6 2nd turn.
+4 accelerator.
-10 1st turn.


See how each group of 3 forces adds to zero? I've continued your pattern of the accelator's force increasing by 1 each cycle. But you can do that differently if you want, and they still add to zero.

LOL.

the only reason that the concept will work is because the two turnarounds cancel each other out.

they dont completely cancel each other out
because of friction as the mass passes through
each turnaround , however that friction is minimal
and can be considered negligible.

and the force used to accelerate the mass is the only remaining force that is applied to the pipe which will
cause the pipe to accelerate in the direction opposite of the direction that the mass is accelerated in.

but in order for you to cheat , you had to change the cycles.

the cycles begin with

acceleration
then the 1st turn
then 2nd turn.

using your numbers above and beginning correctly with
the acceleration force we can get a correct analysis.


+1 accelerator.
-1 1st turn.

+1 2nd turn.
----------------------- 1 complete cycle = +1
+2 accelerator.
-3 1st turn.

+3 2nd turn.
----------------------- 1 complete cycle = +2
+3 accelerator.
-6 1st turn.

+6 2nd turn.
----------------------- 1 complete cycle = +3
+4 accelerator.
-10 1st turn.

cheater

but dont feel alone because the DOE hires cheaters
just like you to determine if free energy funding applications
should be granted funding from the billions of tax payer money paid
out each year to large corporations.

large corporations that have no intent to save any energy.

Originally Posted By: Kallog
I troll around looking for mistakes to fix. Misinformation is a burden on society.


I dont see where you have fixed a mistake , only generated mistakes for others to fix.
just how much of a burden on society do you alone present

now we havent been able to get past your incorrect logic.

but by using 2 of these mass accelerators and turnaround
systems timed to fire so that when
1 mass is in the 1st turnaround
another mass is in the 2nd turnaround
there will be a smooth acceleration of
the pipe used in the example.

and the pipe would not jerk back and forth.








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Originally Posted By: paul

+1 accelerator.
-1 1st turn.


I've repeatedly corrected this consistent mistake of yours, but you haven't bothered to learn it. You don't have to believe me, just check with independent sources. There are free textbooks on Google books, there's a wealth of information in Wikipedia, there are forums dedicated to helping physics students with exactly this type of problem. And of course there's your own workshop where you can easily test this with a scale, a toy car and a stopwatch or video camera.

#1. These values need to be _impulses_ not forces. Forces obviously need not cancel each other out if you ignore the length of time they're applied for. I've explained this already.

#2. If the accelerator provides an impulse of +1, starting from stationary, then the 1st turn provides an impulse of -2, not -1. You've already agreed that the change in momentum would be -2 for an initial momentum of +1. Please stop perpetuating false information unless you can back it up with reasons.

#3. Impulses or forces that aren't applied simultaneously
won't cancel each other out in the short term. The pipe can jerk back and forth. If the timing isn't uniform (it isn't) then they may or may not cancel each other out in the long term either.

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paul Offline OP
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Quote:
#2. If the accelerator provides an impulse of +1, starting from stationary, then the 1st turn provides an impulse of -2, not -1. You've already agreed that the change in momentum would be -2 for an initial momentum of +1. Please stop perpetuating false information unless you can back it up with reasons.


first you put up your numbers that are obvious forces
because they do not include time.

then you want to tell me how impulse must be used ,then you still put up the same numbers.

impulse is force times time.

you cannot write impulse as +1 or -1

ie..
how to calculate impulse.

2N * 3 seconds = 6N-s

so if I apply 500N * 8 seconds the +impulse = 4000N-s

the two impulses in the two turns cancel each other out.
ie...
the -impulse in the 1st turn is 4000N-s
the +impulse in the 2nd turn is 4000N-s

the result for the 1st cycle is +4000N-s impulse.

its obvious that you were trying to pad the results by
leaving out important elements as you were trying to debunk the concept.

you cheat or try to but I catch you.

cheater.












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Originally Posted By: paul

so if I apply 500N * 8 seconds the +impulse = 4000N-s

the two impulses in the two turns cancel each other out.
ie...
the -impulse in the 1st turn is 4000N-s


Again, if the accelerator provides 4000Ns then the 1st turnaround provides -8000Ns the first time round.

Here's the sequence:
+4000Ns accelerator
-8000Ns 1st turn
+8000Ns 2nd turn
+4000Ns accelerator
-????Ns 1st turn

If I calculate the value of ???? you'll tell me I'm making it up, so you can figure it out.

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paul Offline OP
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Well Kallog

lets just face it , you have invented free energy , if you are correct.

by using a 500N force for 8 seconds you consume 4000N.

from the +4000N applied in the accelerator you somehow have magically turned that +4000N into 2 (8000N) or 16000N

all you need to do now is capture that 16000N at the turnarounds.

now each time you supply the 4000N your magic machine can capture more and more each cycle.

but it only provides 4000N ... LOL

so you were doubling the impulse in the turnarounds each cycle so lets calculate a few cycles using your logic.

----------- 1st cycle ------------

accelerator +4000N
1st turn -8000N
2nd turn +8000N

----------- 2nd cycle --------+16000N captured----

accelerator +4000N
1st turn -16000N
2nd turn +16000N

----------- 3rd cycle --------+32000N captured----

accelerator +4000N
1st turn -32000N
2nd turn +32000N

----------- 4th cycle --------+64000N captured----

accelerator +4000N
1st turn -64000N
2nd turn +64000N

----------- 5th cycle ---------+128000N captured----

accelerator +4000N
1st turn -128000N
2nd turn +128000N

------------------------------+256000N captured----

you have only used or consumed 20,000N

and your magic carpet has produced 496,000N

WOW


do you see where your mistake is?

or are you sudgesting that the above is possible?

the above really is possible except you can only use the force generated by the accelerator in the turns.

accelerator +4000N used as propulsion
1st turn -4000N captured to use in the next cycle - elastic collision.
2nd turn +4000N used as propulsion - non elastic collision

you can capture some or all of the 4000N in the 1st turn

so now we even have the power source to supply the initial 4000N used to accelerate the mass , and we have a excess 4000N after the first cycle.

the elastic collision in the 1st turn can be accomplished by
having the turn mounted to a shock absorbing mechanism perhaps
a large hydraulic piston that stores pressurized fluid in a accumulator.

also most of the impulse force can be captured by the piston
so any captured force would not subtract from the forward movement of the pipe.




















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Originally Posted By: paul

from the +4000N applied in the accelerator you somehow have magically turned that +4000N into 2 (8000N) or 16000N


It's only a coincidence that it's doubled the first couple of cycles. Just work it out yourself without copying me. I've turned the +4000Ns of momentum into -4000Ns of momuntum by the passive action of the turnaround. That required -8000Ns to accomplish. No need to argue this, just look it up. I posted a website recently with a ball bouncing off a wall. You agreed with that, so you also implicitly agree with this equivalent statement.



----------- 1st cycle ------------

accelerator +4000Ns. Mass has p=+4000Ns
1st turn -8000Ns. Mass has p=-4000Ns
2nd turn +8000Ns. Mass has p=+4000Ns

----------- 2nd cycle --------

accelerator +4000N. Mass has p=+8000Ns
1st turn -16000N. Mass has p=-8000Ns
2nd turn +16000N. Mass has p=+8000Ns

----------- 3rd cycle --------

accelerator +4000N. Mass has p=+12000Ns

Let's say we stop the machine here. And suppose that's done by stopping the moving mass. It doesn't have to, but I'll do it that way because it's easiest.

brake -12000Ns. Mass has p=0

---------- Shut down complete --------


Each transfer of momentum to the mass applies the same, but opposite to the pipe. So I'll list the velocities of the pipe assuming it has mass=1000kg.

----------- 1st cycle ------------
accelerator +4000Ns. Mass has p=+4000Ns. v = 0-4 = -4m/s
1st turn -8000Ns. Mass has p=-4000Ns. v = -4+8 = 4m/s
2nd turn +8000Ns. Mass has p=+4000Ns. v = 4-8 = -4m/s
----------- 2nd cycle --------
accelerator +4000N. Mass has p=+8000Ns. v = -4-4 = -8m/s
1st turn -16000N. Mass has p=-8000Ns. v = -8+16 = +8m/s
2nd turn +16000N. Mass has p=+8000Ns. v = 8-16 = -8m/s
----------- 3rd cycle --------
accelerator +4000N. Mass has p=+12000Ns. v = -8-4 = -12m/s
brake -12000Ns. Mass has p=0. v = -12+12 = 0m/s
---------- Shut down complete --------


So it's shaken back and forth, faster and faster, then stopped. Has it moved further than it's own length? To find that out you have to work out the time it spends travelling at each speed.



Quote:

do you see where your mistake is?

Yes. You're doubling the turnaround force/impulse each cycle. It isn't doubled. The impulse required to turn around is twice the magnitude of the momentum of the mass (as I said at the time). And that only increases by 4000Ns each cycle because that's what the accelerator adds to it.


Last edited by kallog; 07/03/10 03:32 AM.
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Quote:
----------- 3rd cycle --------

accelerator +4000N. Mass has p=+12000Ns

Let's say we stop the machine here. And suppose that's done by stopping the moving mass. It doesn't have to, but I'll do it that way because it's easiest.

brake -12000Ns. Mass has p=0

---------- Shut down complete --------


kallog , why would you waste the energy to accelerate the mass after you complete the two cycles then just stop the mass.

stopping the mass can be done in the accelerator.

you conviently used only 2 cycles that add to 12

then used only the next acceleration to MATCH that 12

LOL,you do try though.

complete the 3rd cycle and stop the mass in the
accelerator.

and see what the results are.


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Quote:

complete the 3rd cycle and stop the mass in the
accelerator.

and see what the results are.

Sure

1st number is the impulse applied to the mass
2nd number (p) is the momentum of the mass
3rd number (v) is the velocity of the pipe

Accelerator: +4000Ns. p=+4000Ns. v=-4m/s
1st turn: -8000Ns. p=-4000Ns. v=+4m/s
2nd turn: +8000Ns. p=+4000Ns. v=-4m/s
Accelerator: +4000Ns. p=+8000Ns. v=-8m/s
1st turn: -16000Ns. p=-8000Ns. v=+8m/s
2nd turn: +16000Ns. p=+8000Ns. v=-8m/s
Accelarator: +4000Ns. p=+12000Ns. v=-12m/s
1st turn: -24000Ns. p=-12000Ns. v=+12m/s
2nd turn: +24000Ns. p=+12000Ns. v=-12m/s
Brake: -12000Ns. p=0. v=0



Quote:

then used only the next acceleration to MATCH that 12

Yes. The brake's impulse does have to match the mass's momentum in order to stop it. If we had 4 cycles then the mass would get up to +16000Ns so the brake would have to apply -16000Ns. Of course you don't have to stop it, you could just let it keep freewheeling around with the accelerator switched off.

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I had to increase the distance to 160 meters instead of 152.4 meters that the 100 kg mass accelerates in , in order to use the 4000N.

we are still using a 500N force I presume.

using only 3 cycles we get

the 1st cycle the pipe moves 16 meters in 8 seconds.
the 2nd cycle the pipe moves 16 meters in 3.31 seconds.
the 3rd cycle the pipe moves 16 meters in 2.54 seconds.

16 * 3 = 48 meters in 13.85 seconds.

the pipe has a final velocity of 6.92 m/s
its mass as you used was 1000 kg

so its braking force is 6928.77N

braking the mass once acceleration is complete should be done
directly after the free float , perhaps the 2nd turnaround can be moved out of the way to allow the mass to be stopped linearly

so the stopping force of 6928.77N applied to stopping the mass will apply to the pipe and will not cause the pipe to slow but to increase velocity.

the pipe only moved 48 meters using a single mass.

but by using more of the accelerator and turnaround systems you can obtain much faster acceleration , and by timing them you can obtain a smoother acceleration that does not jerk back and forth.

but 48 meters distance in 14 seconds with a final velocity of apx 6.92 m/s for eternity as long as you dont run into something isnt bad.













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Originally Posted By: paul
I had to increase the distance to 160 meters instead of 152.4 meters that the 100 kg mass accelerates in , in order to use the 4000N.

the 1st cycle the pipe moves 16 meters in 8 seconds.
the 2nd cycle the pipe moves 16 meters in 3.31 seconds.
the 3rd cycle the pipe moves 16 meters in 2.54 seconds.


I've just put that 160m into a spreadsheet. Tho I didn't use any particular force. I suppose it come out as 500N. I also didn't apply a brake, but just truncated the calculations.

Some of my numbers match yours, but I'm not sure exactly which points in the cycle you're referring to, or if the distances are cumulative or any are backwards or what.

The key result is the displacement oscillates between +16m and -16m no matter how many cycles you do.



The equations I used were:
impulse = change in momentum
momentum = mass * velocity
velocity = change in displacement / change in time
For the acceleration stages I used the average velocity to calculate the time taken. That assumes constant acceleration.

Oh, those displacements are only for their stages. So you have to add them up to get the total displacement. Then that'll oscillate between -16m and 0.

Quote:

braking the mass once acceleration is complete should be done
directly after the free float , perhaps the 2nd turnaround can be moved out of the way to allow the mass to be stopped linearly

If you brake it before the 2nd turn, then yes you gain the impulse from the braking, but you also lose the impulse from the 2nd turn. It ends up the same either just before or just after the turn.

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I only used the acceleration stage in each cycle.

because the 2 turns cancel.

+16 m
-16 m
+16 m total displacement +16 m

then I used the final velocity of the pipe as the mass entered the accelerator.
and calculated the distance the pipe would again move in the next cycle.

+16 m
-16 m
+16 m total displacement +16 m * 2 = +32 m

then I used the final velocity of the pipe as the mass entered the accelerator.
and calculated the distance the pipe would again move in the next cycle.

+16 m
-16 m
+16 m total displacement +16 m * 3 = +48 m

I noticed in your spreadsheet for the first acceleration stage you are only
getting 8 m pipe displacement in the first 8 seconds.

this number should be 16 m vs 8 m

the accelerated mass is 100 kg
the pipe mass is 1000 kg as you used.

the 100 kg mass displaces 160 m
so the 1000 kg mass displaces 16 m

1000 kg /100 kg = 10 .... 160 m /10 = 16 m

you must have an error in your formula.


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Originally Posted By: paul

+16 m
-16 m
+16 m total displacement +16 m


You mean this?:
Pipe moves +16m while accelerating
Pipe moves -16m while floating
Pipe moves +16m while accelerating again??

That's more than 1 cycle so the next one will start at a different stage. Can you spell them all out step by step??


Quote:

I noticed in your spreadsheet for the first acceleration stage you are only
getting 8 m pipe displacement in the first 8 seconds.

Nope, 16m at 8s is the point selected on the graph. (or -16 as I have it).


Quote:

you must have an error in your formula.


It's fairly easy to check. I lifted the pipe velocities directly from my recent other message. Mass velocities are -10 times the pipe velocities. Everything else is determined by those and the 160m length.

Here's a clearer graph with accumulated displacement, so it represents the actual position of the pipe at any time. I also turned off the accelerator after a while to see what would happen.

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we need to get a few things straight before we continue.

1) we need to stop using different signs because
you use (-) and I use (+) for the same thing.

lets just use (+) for pipe movement in the direction that the accelerator moves the pipe the first time the mass is accelerated.


2) the 1st turn stops the pipe, it does not reverse the pipes direction.

3) the floating stage does not cause the pipe to accelerate.

4) the 2nd turn accelerates the pipe.

5) we can use 1 second for the time it takes for the mass to pass through each turn.

the mass has a final velocity of 40 m/s after it leaves the accelerator in the first cycle.

so we can use a 20 meter travel distance through each turnaround.

the mass will pass through the turnaround in 1 second.



your displacement line from 8 seconds shows the pipe moving backwards from -16 m to 0 m at 12 seconds.

the displacement line should advance from -16 m to -18 m from 8 seconds to 9 seconds because as the mass passes
through the 1st turn the pipe displaces -2 meters while the pipe is stopping.

a force of 4000N will accelerate a mass of
1000 kg from 4 m/s/s to 0 m/s/s in 1 second.
its acceleration is 4 m/s/s , its average velocity is 2 m/s
it will travel a distance of 2 m in 1 second.

then the displacement line should stay at -18 m through the free float for another 8 seconds.

the pipe then displaces from -18 m to -20 m from 17 seconds to 18 seconds as the mass passes through turn 2.

a force of 4000N will accelerate a mass of
1000 kg from 0 m/s/s to 4 m/s/s in 1 second.
its acceleration is 4 m/s/s , its average velocity is 2 m/s
it will travel a distance of 2 m in 1 second.

also your showing the pipe going the opposite direction
when it wont go the opposite direction given that only the momentum of 4000N from the mass is used.

the pipe velocity line should go from 4 m/s to 0 m/s at 9 seconds , what you are showing is the change in direction
of the momentum not the pipe.

you have the line labeled as pipe velocity not change in momentums direction.

so since I have added the pipe movement durring each of the 4 stages in a single cycle , I now get -20 meters displacement after the first cycle.

but I will use +20 for pipe displacement.

+16 m acceleration
+2 m 1st turn
0 m free float
+2 m 2nd turn

because it sounds better ! LOL

--------------------------

just add more to remove the stopping of the pipe and you get
continous forward movement.


.


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Originally Posted By: paul

lets just use (+) for pipe movement in the direction that the accelerator moves the pipe the first time the mass is accelerated.

OK, I'll update my spreadsheet.


Quote:

2) the 1st turn stops the pipe, it does not reverse the pipes direction.

No. Just apply conservation of momentum to the turnaround and it'll show that the pipe reverses:

Initial momentum of pipe = +4000Ns (it's just been accelerated)
Initial momentum of mass = -4000Ns (-40m/s * 100kg)
Total initial momentum of system = +4000 + -4000 = 0

final momentum of mass = +4000Ns (reversed direction)
final momentum of pipe = ???
Total final momentum of system = Total initial momentum of system (law of conservation of momentum).
+4000Ns + ??? = 0
??? = -4000Ns = final momentum of pipe

This is fundamental to many of your other points, so we have to sort it out before going further. If you don't agree, please tell me where in my calculation, not just the result.



Quote:

3) the floating stage does not cause the pipe to accelerate.

Agreed. My graph shows no acceleration during the floating stages, just constant velocity.

Quote:

4) the 2nd turn accelerates the pipe.

Agreed.

Quote:

5) we can use 1 second for the time it takes for the mass to pass through each turn.

It's much easier to use 0s. I've been doing that. But I'll see if I can update my spreadsheet to use 1s.

Quote:

the mass has a final velocity of 40 m/s after it leaves the accelerator in the first cycle.

Yep, but negative so that the pipe's opposite velocity complies with our sign convention.


Quote:

so we can use a 20 meter travel distance through each turnaround.


This makes things really really complicated. Say if the mass had got up to 1000m/s. Then it travels through a 20m turn in 1s, it had to slow down to average 20m/s. Then accelerate again at the end of the turn.

I don't think it's necessary to specify any length or time spent in the turns. They could nearly instant like a ball bearing bouncing off a steel plate.

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paul Offline OP
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Quote:
No. Just apply conservation of momentum to the turnaround and it'll show that the pipe reverses:


conservation of momentum is the philosophic or
theoretical side of physics , and if it cannot
be proved corect by the math side of physics then
it is a flawed law.

this is a simple example that is easily
calculated , you used question marks instead of
filling in the amounts using math.

I repeat a 4000N force can accelerate a 1000 kg mass
to 4 m/s/s in 1 second , that mass will move a distance of
2 meters.

also , there is no additional force applied to the pipe
in the direction that would reverse the pipes direction.

you do agree that the 100kg mass traveling at 40 m/s
passing through the 1st turn will apply a 4000N force
to the 1st turn , correct?

and that this 4000N force will apply to the pipe, correct?

so if you agree on the above then the pipe will slow
from 4 m/s velocity to 0 m/s velocity in 1 second , correct?

lets just use math wherever possible.

because the math side of physics is the side that is used to check the theoretical side.

the theoretical side is not used to check the math side.




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Originally Posted By: paul

conservation of momentum is the philosophic or


OK so we don't use momentum at all anymore in this discussion.

Quote:

I repeat a 4000N force can accelerate a 1000 kg mass
to 4 m/s/s in 1 second , that mass will move a distance of
2 meters.


I agree.

Quote:

also , there is no additional force applied to the pipe
in the direction that would reverse the pipes direction.


During that acceleration? Yes I agree to that too.

Quote:

you do agree that the 100kg mass traveling at 40 m/s
passing through the 1st turn will apply a 4000N force
to the 1st turn , correct?


It should be a negative force applied to the turn. And its magnitude is free to be whatever you choose, so yes I agree it can be -4000N.

Quote:

and that this 4000N force will apply to the pipe, correct?

Yes but in the negative direction.


Quote:

so if you agree on the above then the pipe will slow
from 4 m/s velocity to 0 m/s velocity in 1 second , correct?

Yes I agree on that too.


Do agree that the same 1st turn will also apply a 4000N force to the mass, in the opposite direction to its motion?

If you agree on that, then you should also agree that the mass will slow from -40m/s velocity to 0m/s velocity in 1 second, correct? Here's how:

F=ma
4000N = 100kg * 40m/s^2
So it slows from -40m/s to 0 in 1 second.

Now the pipe's stopped, and the mass has stopped too. So the turnaround isn't complete. The mass needs to get back up to 40m/s in the opposite direction. Doing that will start moving the stationary pipe too.

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Quote:
Do agree that the same 1st turn will also apply a 4000N force to the mass, in the opposite direction to its motion?

If you agree on that, then you should also agree that the mass will slow from -40m/s velocity to 0m/s velocity in 1 second, correct? Here's how:

F=ma
4000N = 100kg * 40m/s^2
So it slows from -40m/s to 0 in 1 second.

Now the pipe's stopped, and the mass has stopped too. So the turnaround isn't complete. The mass needs to get back up to 40m/s in the opposite direction. Doing that will start moving the stationary pipe too.


I cant agree that the 100 kg mass will stop.

theres no force that is applied to stop it.

in order for the pipe to apply a force to the mass
the pipe must move in that direction first.

as the 100 kg mass travels through the 1st turn
it is pressing against the pipe in the (-) direction.

the 100 kg mass never presses against the pipe
in the (+) direction as it passes through the 1st turn.

so the pipe never has a (+) force placed on it , that
would cause the pipe to move in the (+) direction other
than the friction between the mass and the 1st turn.

and these cancel out also.

ie..
the (+)friction through the 2nd half of the 1st turn cancels
out the (-) friction through 1st half of the 1st turn.

--------------

now as for the mass stopping , there is no force applied
to the mass that would stop the mass.

the mass is only being turned around by the 1st turn.

think of a pendulum , it swings on a string but it doesnt
stop as it reaches the bottom at the end of the
1st downward swing and it will only stop at the top of
the upward swing.

yet if you put the pendulum on a (non elastic) scale
that measures downward force and the mass of the ball
of the pendulum is 100 kg and the string is 4.9035 m long
it will swing to the bottom in 1 second.

the force applied is 980.7N = 100 kg * 9.807 m/s/s gc.
its acceleration is 9.807 m/s/s
its initial velocity is 0 m/s.
its average velocity is 4.9035 m/s.
its final velocity is 9.807 m/s.
at the bottom of the swing.

so the ball does not stop at the bottom of the swing , it
keeps going , just like the mass in the 1st turn keeps going.

the only reason the ball stops at the top of the up swing
is the 9.807 gc force pulling on it.

the force applied is 980.7N = 100 kg * 9.807 m/s/s gc.
its acceleration is 9.807 m/s/s
its initial velocity is 9.807 m/s.
its average velocity is 4.9035 m/s.
its final velocity is 0 m/s.
at the top of the swing.

but the (non elastic) scale that measures downward force will
read a force of 980.7 N as the ball swings past the bottom
on its way to the top of the up swing.

so after the mass passes through the 1st turn it still has
its velocity of 40 m/s assumming no friction.

--------------












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I noticed a mistake when I included a distance
ro the turnarounds.
Quote:

5) we can use 1 second for the time it takes for the mass to pass through each turn.

the mass has a final velocity of 40 m/s after it leaves the accelerator in the first cycle.

so we can use a 20 meter travel distance through each turnaround.

the mass will pass through the turnaround in 1 second.


make that 0.5 seconds , sorry.
or we can make the turnarounds 40 meters in lenght and keep the 1 second.




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We can safely assume friction is zero. In reality we can get arbitrarily close to zero and add a bit of energy every now and then to compensate. The mass could even be an electron travelling through a superconductor, where there's exactly zero friction.

Originally Posted By: paul

the 100 kg mass never presses against the pipe
in the (+) direction as it passes through the 1st turn.

so the pipe never has a (+) force placed on it , that
would cause the pipe to move in the (+) direction other
than the friction between the mass and the 1st turn.


Yes I agree. The _pipe_ never has a + force on it in the 1st turn. But the mass does. That's from Newton's 3rd law: action and reaction are equal and opposite. You used this law to find that both the pipe and mass experience a force in the accelerator.

When I say 'stop' I mean stop in the direction we're considering. Sure it keeps going at a speed of 40m/s, but as its velocity changes from backwards to forwards, the longitudinal component of it must obviously pass through 0.

It's the same with a pendulum. The _vertical_ component of velocity reaches 0 at the bottom of the swing. Sure it keeps moving, but it's stopped in the vertical direction. If you remove all forces at that moment, then it'll remain stopped in the vertical direction - and fly off in a straight line sideways.

If you're saying the mass doesn't experience a force in the turnaround, then you're saying Newton's 1st, 2nd and 3rd laws are all wrong.

The 1st law says it should keep going in a straight line, since there's no force acting on it. But it's turning.

The 2nd law says F=ma: zero force = 100kg * zero acceleration. But it's clearly accelerating because its direction is changing.

The 3rd law says if it applies -4000N to the pipe, then the pipe applies +4000N to the mass.

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Quote:
When I say 'stop' I mean stop in the direction we're considering. Sure it keeps going at a speed of 40m/s, but as its velocity changes from backwards to forwards, the longitudinal component of it must obviously pass through 0.


ok , are you still saying that the pipe will stop and then
move in the (+) direction?

because we need to clear that up first.

durring the time that the mass passes through the 1st turn.

I am saying that the pipe slows and stops.
and
I am saying that the mass still has a velocity of 40 m/s.
what are you saying?

please finalize your answers by simply stating them as I did above.

Quote:
If you're saying the mass doesn't experience a force in the turnaround, then you're saying Newton's 1st, 2nd and 3rd laws are all wrong.



if these laws are correct then the math will back them up.

if these laws are not correct then the math will not back them up , and they should be disposed or corrected.

we cannot anlalyze this concept using laws.


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Originally Posted By: paul

ok , are you still saying that the pipe will stop and then
move in the (+) direction?

No, we're both saying the accelerator causes the pipe to start moving in the + direction. I'm saying when the mass turns around, it applies enough force to first stop the pipe, then start it moving in the - direction.

Quote:

I am saying that the pipe slows and stops.
and
I am saying that the mass still has a velocity of 40 m/s.


That's a contradiction. The mass changes direction, therefore its velocity must change. Because its velocity changes it must have a force applied to it. It's impossible to change the direction of a moving object without applying a force.

I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.

I'm also saying the mass changes direction during the 1st turn, maintaining a _speed_ of 40m/s. Of course its velocity changes from -40m/s to +40m/s, so it's accelerating as it passes through the turn.

It doesn't matter that the mass moves sideways in the middle of the turn. It goes into the turn at -40m/s and comes out at +40m/s. Whatever happened during the turn caused an acceleration (here change in direction) and applied a force to the mass (by the 1st law), as well as a force to the pipe (by the 3rd law).

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Quote:

That's a contradiction. The mass changes direction, therefore its velocity must change. Because its velocity changes it must have a force applied to it. It's impossible to change the direction of a moving object without applying a force.

I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.


kallog , the mass is pressing against the pipe and that causes the mass to change direction.

through the 1st half of the 1st turn the mass is moving in the (-) direction , and pressing against the pipe
in the (-) direction.

then

through the 2nd half of the 1st turn the mass is moving in the (+) direction , and pressing against the pipe
in the (-) direction.

the mass never presses against the pipe in the (+) direction in the 1st turn.

so its not a contradiction.
the mass does change direction but the direction that its force is applied is always in the (-) direction as it passes through the 1st turn.

there is never any force applied to the pipe in the
(+) direction by the mass as the mass passes through the 1st turn.

so there is no available force to reverse the direction of the pipe.

the change in direction of the mass is not a force.
and a force is required to change the direction of the pipe.

think of a 360 degree range of motion.

the (+) direction is 360 degrees.
the (-) direction is 180 degrees.

the mass enters the 1st turn at 270 degrees.
it travels through the turn to 180 degrees.
then it travels through the turn to 90 degrees.


all the time it travels through the 1st turn it places
a force outward that has a direction that can be plotted from the center of the turn through the mass to the outermost part of the turn.

these forces are less at 270 degrees and greatest at 180 degrees and less at 90 degrees , and if you add up every force applied in the (-) direction they will all add up to -4000N.

never does the force applied to the turn transit into
the (+) side , which would be from
90 degrees to 360 degrees to 270 degrees.

momentum changes are not forces.
but momentum changes require forces.








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Originally Posted By: paul

the mass never presses against the pipe in the (+) direction in the 1st turn.


Paul, are you making an effort to misunderstand me? I never said that, I even clarified that fact when you mentioned it before. Please read my messages more carefully. I never said there is a force in the + direction applied to the pipe during the 1st turn.

We both agree there is a force in the - direction applied to the pipe. Correct?

Do you also agree that Newton's 3rd law says there must also be a force in the + direction applied to the _mass_? <<<---- Note! that says mass, not pipe.

Either there really is such a force, or Newton's 3rd law is wrong. Which way do you want to go?

Or try to visualize it. Imagine you're riding on that mass, will you feel anything as it goes through the turn? Could you black out from the g's? Could you get a sore ass from scraping against the turn? Anything at all? Or would you be unaware of the turnaround if you shut your eyes?

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Originally Posted By: kallog
Paul, are you making an effort to misunderstand me? I never said that, I even clarified that fact when you mentioned it before. Please read my messages more carefully. I never said there is a force in the + direction applied to the pipe during the 1st turn.


LOL , dont try to maneuver your way out of that one, you have always said through the entire discussion that the pipe would somehow move in the (+) direction because the mass goes through the 1st turn.

it shows clearly in your graphs that you put up
showing the (+) direction indicating that the pipe doesnt just stop.



as is shown in the above graph you posted , the pink line represents the pipe displacement , you always return the pipe to ZERO because of the 1st turn.

that reversal would require +8000N and you used the total change in the momentums direction +8000N to accomplish that reverse in the pipes direction.

kallog concerning the above graph, in order for the pipe to reverse direction you must first
apply a +4000N force to stop the pipe and then to get the pipe back to the ZERO line you must apply another force.

THERE IS NO OTHER FORCE!!!!

YOU CANNOT USE THE TOTAL CHANGE IN DIRECTION OF MOMENTUM AS A FORCE ie the 8000N you were using because the mass changes from +4000 to -4000 , you were adding up the
2 4000 to get 8000

Originally Posted By: kallog
I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.



you are using (-4000N) to slow and stop the pipe
then using (-4000N) to reverse direction.

but the imaginary force you use does not exist.

my entire post was to try and get you to realize that the
pipe will only stop , it will not reverse direction because
any available force that could reverse the pipe direction is used up stopping the pipe.

you have been claiming that the force used to change the momentum of the mass was adding the extra force to reverse the pipes direction.

thats extremely evident in this thread as is in others where this
has been discussed.

momentum is not a force and the change in momentum is not a force and that is what I have been trying to tell you all along.

so stop trying to put the blame on others , accept your mistakes and own up to them.

have you ever watched a nascar race , where cars travel
around a track , they use gas engines because the turns do not propel the cars around the track the way you think the mass is being propelled because it changes direction.

the nascar cars also change direction but they still have to use up gasoline to go around and around , the turns of the race are not like a magic carpet.

and the turns in this concept also are not like a magic carpet.









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Originally Posted By: paul

have always said through the entire discussion that the pipe would somehow move in the (+) direction because the mass goes through the 1st turn.


Yes, until just after I posted that graph, when we agreed on the opposite sign convention. Let's just stick to this way instead of wasting time:

Originally Posted By: paul

lets just use (+) for pipe movement in the direction that the accelerator moves the pipe the first time the mass is accelerated.





Quote:

you are using (-4000N) to slow and stop the pipe
then using (-4000N) to reverse direction.

but the imaginary force you use does not exist.

my entire post was to try and get you to realize that the
pipe will only stop , it will not reverse direction because
any available force that could reverse the pipe direction is used up stopping the pipe.


You assumed -4000N is only applied for 1s. Show me your calculations that say it must be applied for 1 second. Here's my calculation that shows it must be applied for 2 seconds:

4000N = 100kg * a
a = 40m/s^2
That says the velocity of the mass increases by 40m/s each second. Since it started at -40m/s, after 1s it's reached 0m/s. After 2s it's reached +40m/s. That's the velocity it must have after leaving the turn, so it took 2s.

Either agree or identify the exact point where you think I made a mistake.


Quote:

momentum is not a force and the change in momentum is not a


We're no longer using momentum at all. You can ignore my older posts where I used momentum. Because you don't trust the law of conservation of momentum, the entire concept of momentum has no use, and I can accept that doubt. We don't need momentum, it's only a convenience to make maths easier.

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I want to super-clarify our positions in case we're still talking at cross-purposes. Is this right?

We both agree:
- We're only talking about the 1st time through the 1st turn.
- The mass begins with a velocity of -40m/s
- The pipe begins with a velocity of +4m/s
- The mass finishes with a velocity of +40m/s

I say:
- The pipe finishes with a velocity of -4m/s

You say:
- The pipe finishes with a velocity of 0m/s

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Quote:
Here's my calculation that shows it must be applied for 2 seconds:

4000N = 100kg * a
a = 40m/s^2
That says the velocity of the mass increases by 40m/s each second. Since it started at -40m/s, after 1s it's reached 0m/s. After 2s it's reached +40m/s. That's the velocity it must have after leaving the turn, so it took 2s.

Either agree or identify the exact point where you think I made a mistake.


the lenght of the turnaround is what determines the time it takes for the mass to pass through it.

the mass is no longer being forcefully accelerated , it is only passing through the turnaround.

I realize there are forces inside the turnaround however.

if the turnaround is 40 meters in lenght the mass will pass
through the turnaround in 1 second because the mass has a velocity of 40 m/s.


if the turnaround is 20 meters in lenght the mass will pass
through the turnaround in 0.5 seconds because the mass has a velocity of 40 m/s.

I think we should use a turnaround lenght of 40 meters.

if that will assist our discussion.

and if we need to the lenght of the accelerator can be increased also.



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kallog
Quote:
I want to super-clarify our positions in case we're still talking at cross-purposes. Is this right?

Quote:
We both agree:
- We're only talking about the 1st time through the 1st turn.

yes
Quote:
- The mass begins with a velocity of -40m/s

yes
Quote:
- The pipe begins with a velocity of +4m/s

yes
Quote:
- The mass finishes with a velocity of +40m/s

yes
Quote:
I say:
- The pipe finishes with a velocity of -4m/s

yes, thats what you say , but I dont agree nor does physics math.
Originally Posted By: kallog
You say:
- The pipe finishes with a velocity of 0m/s

yes, thats what I say , and physics math backs me up.

the pipe stops !

the math

the action
100kg * 40 m/s = 4000N ---->mass
the reaction
1000kg * 4 m/s = 4000N ---->pipe

since the pipe has a velocity of 4 m/s and a mass of 1000 kg it will stop in 1 second if a force of 4000N is applied
for 1 second in the opposite direction that the pipe is moving in.

a=f/m
4m/s=4000N/1000kg

the pipe will stop in 1 second as the mass applies its
-4000N force to the pipe , and the pipe will continue to move in the (+) direction until it stops.

at this point the initial -4000N force has transfered to
the pipe as a -4000N force.

and the 100kg mass can now be described as having a +4000N force due to its direction and its mass * its velocity of 40 m/s.

call it what you will , say that it violates the laws of physics , conservation of momentum , conservation of energy
or conservation of laws if you choose , it does not matter , the above is what physics math shows will happen , it cannot happen any other way.

a 500N force applied for 8 seconds to a 100kg mass accelerated both the pipe and the mass , that same
force stopped the pipe and the mass still has a velocity
of 40 m/s , when the mass reaches the 2nd turn it will accelerate the pipe again to 4 m/s/s , when it exits the 2nd turn it will still have a velocity of 40 m/s.

1)free energy = true
2)conservation of momentum = false
3)conservation of energy = false

laws 1,2,3 gone...

but only the way they are used today , or should I say the
definitions that the laws have been rearanged to represent.














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Quote:

the mass is no longer being forcefully accelerated , it is only passing through the turnaround.

You really really need to understand the terminology. _All_ accelerations are forceful, and anything changing its direction of motion is being accelerated - with force. The speed can remain a constant 40m/s, but it is still accelerating. I can't believe you still haven't looked this up. Again I have a sudden realization that I'm talking in a foreign language to you.


Quote:

if the turnaround is 40 meters in lenght the mass will pass
through the turnaround in 1 second because the mass has a velocity of 40 m/s.


speed=length/time
40m/s = 40m / 1s
Indeed. So a 40m, 1s turnaround is possible. And it can only have one constant force being applied to the mass.

Quote:

if the turnaround is 20 meters in lenght the mass will pass
through the turnaround in 0.5 seconds because the mass has a velocity of 40 m/s.

40m/s = 20m / 0.5s
Yep, so a 20m, 0.5s turnaround is also possible.

Quote:

I think we should use a turnaround lenght of 40 meters.

OK. Let's stick to the 40m, 1s turnaround.

The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N

Again, please either agree or point out the exact part(s) of my calculation that you don't agree with. You didn't last time I asked, but we're not going to get anywhere unless we can isolate our disagreement to its exact root.

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Originally Posted By: paul

the action
100kg * 40 m/s = 4000N ---->mass

This isn't a valid equation. The dimensions on the left (mass * velocity) are different to those on the right (force). How about find a reference for this equation, or explain it in terms of something we know we agree on.


However, since you concluded the mass experiences a 4000N force (I assume in the +ve dirction), I'll continue..

force = mass * acceleration
4000N = 100kg * acceleartion
acceleration = 40m/s^2

That means after 1s in the turnaround, it's reached a velocity of 0m/s (in the longitudinal direction, of course it's still doing 40m/s speed, but now sideways).

0m/s in the longitudinal direction isn't turned around, so that 4000N force is incompatible with a 1s turnaround time.

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Quote:
The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N


momentum is not a force !

the masses velocity does not change.
-----
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
-----
its momentum changes , momentum is not a force
momentum is just mass multiplied by speed.

so you cannot say that a 8000N force is placed on the
pipe by the change in momentum of the mass.

if you are so certain that the pipe will experience a
-8000N force then where does the force come from , it must
come from somewhere.

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N

4 m/s/s * 1000 kg = 4000N

note: I added time as you require

where does the extra 4000N come from?

have you ever seen this happen in the real world , could
you give an example of this extra force in a turn?

does a ball dropped from 10 feet bounce to a height of 20 ft because of this principle?

because if that extra force is available in a turn then the ball bouncing off of the ground should also get a extra ummph because the earth is hard to move with a ball.


or does the earth also feel a force that is twice the
mass of the ball * its acceleration?

the reason I say twice the mass is because we know the velocity does not increase , so the mass of the ball
must increase.

am I right in assumming that that is what happens in the turn?

if we go in the middle of the turn and catch the mass we could have twice the mass? we could get rich by using gold masses.

at which point in the turn should we capture the gold before it has a chance to shrink its mass?

Quote:
That means after 1s in the turnaround, it's reached a velocity of 0m/s (in the longitudinal direction, of course it's still doing 40m/s speed, but now sideways).

0m/s in the longitudinal direction isn't turned around, so that 4000N force is incompatible with a 1s turnaround time.



the mass travels all the way through the 40 meter
turn in 1 second because it has a velocity of 40 m/s
40 meters /40 m/s =1s , so in 0.5 seconds the mass is half way through the turn.

its traveling longitudinal and pressing against the
turn.

the velocity/speed of the mass does not decrease.

again you are using the change in momentum but describing it as a change in velocity.

I thought we were going to leave momentum out of this
discussion , so why are you impersonating momentum with velocity?

momentum is not a force so lets not pretend that it is.
velocity is also not a force , btw.
and a change in velocity is also not a force.

just use the actual forces that would actually apply
a force to the pipe or the 100kg mass.







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Originally Posted By: paul

the masses velocity does not change.
-----
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s


Please please please look up 'velocity' on google or wikipedia or a text book, or anything. I thought I'd explained this before, but it seems you didn't look it up then either.

The _speed_ doesn't change, that remains at 40m/s throughout the turn. But the _velocity_ changes from "40m/s backwards" to "40m/s forwards". That's a huge change and can't just happen on a whim, it requires enough force.

Quote:

if you are so certain that the pipe will experience a
-8000N force then where does the force come from , it must
come from somewhere.

It comes from the mass pressing against the pipe as it turns around. If the mass didn't press against the pipe, it would be unable to turn around.


Quote:

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N

4 m/s/s * 1000 kg = 4000N

note: I added time as you require

4m/s/s is an acceleration not a velocity. If the pipe did accelerate at 4m/s/s then you'd be right, it'd need a 4000N force to do so.

Quote:

have you ever seen this happen in the real world , could
you give an example of this extra force in a turn?


Drive along, then brake hard. You feel the force of the seatbelt/floor/etc pushing your body backwards (of course you move forwards, but the backwards force stops you going through the windscreen).

As soon as you've stopped, go into reverse and floor it. Now you still feel a force pushing your body backwards (the same direction!).

That second application of force is an example of the extra 4000N I'm talking about. You wouldn't get it if you just stopped and stayed stopped, it was the direction-change that caused it.


Quote:

because if that extra force is available in a turn then the ball bouncing off of the ground should also get a extra ummph because the earth is hard to move with a ball.

It does get an extra oomph, enough to send it flying upwards again. Without that extra force it'd just stop when it hit the ground.

The extra oomph is powered by the stored elastic energy in the ball. If the ball is deflated it won't bounce because it can't store that elastic energy.


Quote:

velocity does not increase , so the mass of the ball

The velocity does increase. I hope by now you've looked up 'velocity', then you'll clearly see that it's increased even tho the speed has remained at 40m/s.

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Quote:

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N


This statement worries me. Are you sure you were thinking about what you said? Let me give you a clear example of why the whole concept is wrong.

A bungee jumper falls 50m before taking up the slack in the bungee cord and beginning to slow down. He feels some force, but it's comfortable.

A suicide jumper falls 50m and is caught by the ground. He feels a much higher force which is fatal.

Both jumpers were in free-fall for 50m. Therefore they both had the same speed at 50m down. Both were stopped by a force. But one experienced a much higher force than the other.

This really is common sense. It's why soft-soled sneakers are more comfortable for running in than Doc Martins.

My point here is that the force required to stop (or reverse) something isn't only determined by the thing's mass and velocity. It also depends how quickly it's stopped (or reversed).

You've said many times that force is mass * velocity. But clearly it depends on more than just mass and velocity.

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you didnt comment on my question , does the earth feel a force
that is twice the mass of the ball * its acceleration?

that is a very important question that you need to answer.

lets use the 100kg mass to figure this out.

the mass has fallen for a distance and now
has a acceleration of 40 m/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?

I recall earlier you said that you were treating
the turn as a spring.

in that case the force is absorbed by the spring but the pipe
also would be pushed by the force , then as the spring expands
the pipe would again be pushed by the force exerted by the spring.

is this how you are describing the turns still?

I would still like your answer to the force that the mass
would apply to the earth.












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Quote:
A bungee jumper falls 50m before taking up the slack in the bungee cord and beginning to slow down. He feels some force, but it's comfortable.

A suicide jumper falls 50m and is caught by the ground. He feels a much higher force which is fatal.



the bungee jumper feels a smaller force for a longer time.
the suicide jumper feels the total force all at one time.

both forces are the same exact force the the earth feels
and both forces are the same force felt by the two jumpers.

only the forces are spread out over time.


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Originally Posted By: paul

the mass has fallen for a distance and now
has a acceleration of 40 m/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


You can't know from just that information.

F=ma

What's the acceleration while it's applying a force to the earth? Who knows. Depends how soft it is. Depends how long it spends applying that force before completely stopping.

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Originally Posted By: paul

the bungee jumper feels a smaller force for a longer time.


Exactly. The force depends how long it takes to stop. Sudden hard stop = high force, slow cushioned stop = low force.


How about you either explain where you got

F = m v

from or stop using it. I've given you two very good reasons why it's wrong. Maybe you just used it by mistake once and are clinging to it to save face? Whatever the reason, back it up with something, or throw it out.

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the 100kg mass has fallen for a distance and now
has a acceleration of 40 m/s/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


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Originally Posted By: paul
the 100kg mass has fallen for a distance and now
has a acceleration of 40 m/s/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


You mean 40m/s/s is the acceleration it experiences while stopping, not while falling? Then it's easy:

F=ma
F=100kg * 40m/s^2
F=4000N

boom.

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No , I mean it has a acceleration of 40 m/s/s as it strikes
the earth.

we can say that the collision is inelastic so when the two
collide they do not deform , the 100kg mass transfers its force to the earth.

why didnt you say it strikes the earth with a 8000N force

its the same mass and the same acceleration as in the example.
the 100kg mass can not tell if its hitting the earth or being turned around in the pipe.


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Originally Posted By: paul

we can say that the collision is inelastic so when the two
collide they do not deform , the 100kg mass transfers its force to the earth.

That's not what 'inelastic' usually means. But if they really don't deform then the ball stops instantly. Stopping instantly applies an infinite force, which of course isn't possible. Perfectly undeformable bodies aren't possible either so it's OK.


Quote:

its the same mass and the same acceleration as in the example.

No. In the pipe the acceleration wasn't 40m/s^2. Just work through my calculations in bold in a recent message, I think that sums up my entire argument pretty clearly. Point to any specific lines you don't like.


I'm about to give up Paul. You used to be pretty competent with physics, but all your skills seem to have gone to the dogs with this tube thing, and getting progressively worse.

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I tried to use the velocity that the mass has as it strikes.

you demanded acceleration as it strikes.

then I used acceleration as it strikes.

then you say that is wrong , let me ask you the question
and you plug in whatever you require.

the 100kg mass is traveling at a speed of 40 m/s
the 100kg mass strikes the earth , what force will the mass apply to the earth?

4000N or 8000N


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Originally Posted By: paul

you demanded acceleration as it strikes.

then I used acceleration as it strikes.


You just changed the units. Doesn't mean it's the actual acceleration.


Quote:

the 100kg mass is traveling at a speed of 40 m/s
the 100kg mass strikes the earth , what force will the mass apply to the earth?

4000N or 8000N


We can't know without knowing its acceleration as it's stopping.

In the pipe case I worked out the acceleration using:

a = change in velocity / time taken
Right??

We both agreed that initial velocity is -40m/s and final velocity is +40m/s.

Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


What's wrong with that?
Don't like my 'change in velocity'? How would you do it?

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Quote:
We can't know without knowing its acceleration as it's stopping.


use 1 second to stop the mass.

Quote:
We both agreed that initial velocity is -40m/s and final velocity is +40m/s.

Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


What's wrong with that?
Don't like my 'change in velocity'? How would you do it?


nothing has a speed of 80 m/s

the mass has a speed of 40 m/s
the pipe has a speed of 4 m/s

you cant use 80 m/s as a speed or 80 m/s/s as a acceleration.
the distance the mass travels through the 1st turn is only 40 meters.

the turn does not stretch to 80 meters because the mass moves through it.

the mass will not increase its speed to 80 m/s in 1 second as it passes through the 1st turn.

you cannot use a change in direction as a force.





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Originally Posted By: paul
Quote:
We can't know without knowing its acceleration as it's stopping.


use 1 second to stop the mass.




OK so it's slightly different from the tube. Here the mass stops in 1s rather than reversing in 1s.

a=change in v / time
change in v = 0m/s - 40m/s
|change in v| = 40m/s

time = 1s

a = 40m/s / 1s
a = 40m/s^2

F=ma
F=100kg * 40m/s^2
F=4000N

Good enough? I got a bit sloppy with the + and - signs, so maybe it's -4000N but who cares.

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I think its amazing how we cannot seem to get a agreeement
on this , I think things have been made too confusing or too
hard to figure out using todays math.

are they teaching that momentum and a change in momentum is a force where you come from?

because you constantly use changes in direction as forces.

this is our main problem , I have never encountered momentum being used as a force so that is why I have trouble with your results.


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So you agree that a 4000N force will be placed on the earth
by the 100kg mass.

now suppose we move the earth and put the pipe there , the pipe would feel the same 4000N force?

if so then how could the mass moving through the turn apply a 8000N force to the pipe?

have you ever considered that the 4000N force is distributed evenly to the turn?



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Originally Posted By: paul

because you constantly use changes in direction as forces.

Not intentionally. Show me where I did and I'll correct my mistake.

Quote:

this is our main problem , I have never encountered momentum being used as a force so that is why I have trouble with your results.


Who said anything about momentum? We've dropped that concept.

I'm just using a couple of equations again and again for all sorts of purposes:

F=ma
a = (change in velocity) / (time taken)
(change in velocity) = (final velocity) - (initial velocity)

Well that's pretty much it really!!! Google them if you don't believe me.

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Originally Posted By: paul

now suppose we move the earth and put the pipe there , the pipe would feel the same 4000N force?

Of course.

Quote:

if so then how could the mass moving through the turn apply a 8000N force to the pipe?

Because the turning mass isn't just stopping, it's turning around. It's a whole different situation.

Quote:

have you ever considered that the 4000N force is distributed evenly to the turn?

Of course, that's been a fundamental assumption throughout. Sure it doesn't have to, but why complicate things with varying forces.

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Quote:
Who said anything about momentum? We've dropped that concept.

then how are you getting 8000N from 4000N if your not using
a change in momentum or a change in velocity to get the
8000N?


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Quote:
have you ever considered that the 4000N force is distributed evenly to the turn?

Of course, that's been a fundamental assumption throughout. Sure it doesn't have to, but why complicate things with varying forces.


so are you saying that the mass moving through the 1st turn only applies a force of 4000N to the pipe?

because that is where we dissagree , you say it has a force of 8000N applied I say it has a force of 4000N applied.

which of these do you agree on?


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Originally Posted By: paul
then how are you getting 8000N from 4000N if your not using
a change in momentum or a change in velocity to get the
8000N?


I'm using a change in velocity, not a change in momentum.

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Originally Posted By: paul

because that is where we dissagree , you say it has a force of 8000N applied I say it has a force of 4000N applied.


I say 8000N and I've shown my calculations of how I got that number.

Why do you say 4000N? Where's your calculations? Not the ones using F=mv.

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Quote:
Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


your using the change in velocity above to get the 80 m/s/s
acceleration.

F=ma = 80m/s/s * 100kg = 8000N

change in velocity is not a force

and change in velocity cannot be used to accelerate the pipe.

my calculations?

F=ma

the 100kg mass applies a force of 4000N for 1 second.
since force is determined by m*a
and acceleration is the change in velocity over time

so force = d/(dt)*mv

F=40m/(40m*1s)*mv
F=40m/40 * mv
F=1*mv
F=1*100kg*40m/s
F=1*4000
F=4000N

that 4000N force is Applied to the pipe
and it is applied to the mass.

so the mass still has +4000N
and the pipe has recieved -4000N

you were getting 8000N but the only way you can get
a 8000N force is to calculate the force by using a
200kg mass , if you keep the time and the velocity the same.

F=40m/(40m*1)*mv
F=40m/40 * mv
F=1*mv
F=1*200kg*40m/s
F=1*8000
F=8000N







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Originally Posted By: paul

your using the change in velocity above to get the 80 m/s/s
acceleration.

F=ma = 80m/s/s * 100kg = 8000N

change in velocity is not a force


Of course not, if it was, I'd have written:

F = change in velocity
F = 80m/s

which is nonsense.


Change in velocity becomes a force after you divide it by the time it took and multiply it by the mass who's velocity is changing. That's Newton's 2nd law again, F=ma.

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Can you give me an example of how you'd use F=ma?

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Paul, why is it so hard to understand?

before entering the turn the mass has a speed of +40m/s to the RIGHT, after the turn the mass has a speed of +40m/s to the LEFT.
Instead of LEFT and RIGHT we are using signs:
+40m/s (to the RIGHT), -40m/s (to the LEFT).

The difference is a velocity of 80m/s.
Because velocity has a direction, its a vector.

As kallog stated several times: depending on how much time you allow for this velocity change of 80m/s the force applied to the mass can be anything.
You chose 1 second, so the mass (100kg) will get a force of 8000N applied.

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I tell you what is so hard to understand , its the
extra 4000N he is claiming.

that is nonsence , you cannot get -8000N force from a -4000N force simply by turning the mass around , it will change its direction but it will not apply a force of -8000N to the pipe.


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Quote:
Can you give me an example of how you'd use F=ma?

Quote:
F=ma = 80m/s/s * 100kg = 8000N


change in velocity is not a force

and change in velocity cannot be used to accelerate the pipe.

my calculations?

F=ma

the 100kg mass applies a force of 4000N for 1 second.
since force is determined by m*a
and acceleration is the change in velocity over time

so force = d/(dt)*mv

F=40m/(40m*1s)*mv
F=40m/40 * mv
F=1*mv
F=1*100kg*40m/s
F=1*4000
F=4000N

that 4000N force is Applied to the pipe
and it is applied to the mass.

so the mass still has +4000N
and the pipe has recieved -4000N

you were getting 8000N but the only way you can get
a 8000N force is to calculate the force by using a
200kg mass , if you keep the time and the velocity the same.

F=40m/(40m*1)*mv
F=40m/40 * mv
F=1*mv
F=1*200kg*40m/s
F=1*8000
F=8000N

the mass did not increase to 200kg so your assumption that the mass would apply a -8000N force to the pipe is incorrect.

the above is a force calculation not a momentum or direction change equation , it would apply to the amount of force that the pipe will recieve as the mass passes through the turn , where a change of direction calculation can only calculate the amount of the change in direction.

a force is required to change the direction of a mass and
that force is 4000N supplied by the pipe.

stop using momentum calculations to make it appear that
the pipe will have a larger force applied to it.

use the same force calculation as above that calculates FORCE.



FORCE


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A Force of 4000N applied for 1 second on a 100kg mass with a velocity of 40m/s against its direction of movement, will result in the mass stopping.

But you want the mass to move again with 40m/s in the other direction.

The velocity difference between 40m/s to the RIGHT and 40m/s to the LEFT is 80m/s.

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not if the mass is turning around !
on a track.

just like a pendulum.

the 100kg mass will not just stop as it turns around.
but it does apply a 4000N force to the pipe as it turns around.

you must have just dropped in.

you should read up on the issue before replying.

yes , the velocity difference or the change in velocity
is 80 m/s , I never questioned that.
I questioned kallogs use of the 80m/s * 100kg to get
a force of 8000N , that he applied to the pipe to cause
the pipe to stop and then move in the other direction.

I agree the pipe just stops with the force of 4000N

but a change in velocity is like this
---------------- -40 m/s ------------>
<--------------- +40 m/s -------------

it turned around add them up the total change

IN VELOCITY / DIRECTION is 80 m/s , thats not a FORCE
its a change in direction.

or you can do it like this
--------------- 40 m/s North------------>
<-------------- 40 m/s South-------------

total change in velocity is 80 m/s
its still not a force , its a change in direction.

its like driving your car at 50 miles per hour
for 30 minutes to point A
and then driving it back at 50 miles per hour
for 30 minutes to point B

the total change in velocity durring the 1 hour
is 100 miles per hour , but you never drove your
car over 50 miles an hour.

the change in velocity of the car required no extra gasoline than it would if you just
kept driving north for an entire hour.

do you have the comprehension ability to comprehend that?

because kallog couldnt ever grasp that.

or he was just using that to try to make it appear that the pipe would just go forward
then backward , resulting in the pipe never moving further than 16 meters in any direction.




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Originally Posted By: paul

extra 4000N he is claiming.


Where did the first 4000N come from? It came from F=mv which we know is wrong.

If you really insist on 4000N, all you have to do is change the turnaround time from 1s to 2s and it's back to 4000N again. Would that make you happy?

Or we could use 0.1s and it becomes 80,000N. Or 10s and it's 800N.

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Originally Posted By: paul
not if the mass is turning around !
on a track.


We're treating it as a 1-dimensional turnaround. Like a spring or something. Alternatively it's the same as having a proper turning track but only considering the longitudinal component of the velocity. In both cases the velocity we use does indeed go to zero - the mass stops.

If you absolutely must analysise it in 2D, then do that. Of course then there'll be sideways forces too, which you must include. Just makes it harder. And it won't change the fact that the velocity of the 100kg mass changes by 80m/s during 1 second. So it'll end up exactly the same as my calculations.


Quote:

you should read up on the issue before replying.

He's hit the nail on the head.

Quote:

I questioned kallogs use of the 80m/s * 100kg to get
a force of 8000N , that he applied to the pipe to cause
the pipe to stop and then move in the other direction.

I didn't do that. Check my calculation again. I also divided it by the turnaround time. Then I got the 8000N.

Quote:

total change in velocity is 80 m/s
its still not a force , its a change in direction.

Who said it was a force? Only you. It's not a force, it's a change in velocity. It does cause a force to be applied, but we have to actually calculate that force, it's not 80N.


Quote:

the total change in velocity durring the 1 hour
is 100 miles per hour , but you never drove your
car over 50 miles an hour.

Sure. And I never said the mass travels over 40m/s at any point. Where did you get that idea from?

Quote:

the change in velocity of the car required no extra gasoline than it would if you just
kept driving north for an entire hour.

The mass's turnaround requires no burning of gas or use or electricity or anything. It's a passive device too. Again I never said we need to put energy into it to make it turn.

In the car example you could turn off the engine and let it cost around a 180deg curved track. In the absence of friction it'll get a 50mph change in velocity too. And of course it'll apply a force to the track, despite not burning any fuel and not losing any speed.

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Originally Posted By: paul

Quote:
F=ma = 80m/s/s * 100kg = 8000N


change in velocity is not a force

Correct. And 80m/s/s is not a change in velocity, it's an acceleration. The change in velocity is only 80m/s.


Quote:

so force = d/(dt)*mv
F=40m/(40m*1s)*mv
F=40m/40 * mv
F=1*mv
F=1*100kg*40m/s
F=1*4000
F=4000N


In this statement of Newton's 2nd law, d is not distance, is part of the derivative symbol.

F=d/dt (mv)
m is constant so
F=m d/dt v
F=m dv/dt
dv/dt is, by definition, equal to acceleration
F=ma
Then continue as I've already shown recently.



Here's a quote from the very same page you referenced. Didn't bother to read any of the text with that equation?

"
the use of the constant factor rule in differentiation allows the mass to move outside the derivative operator, and the equation becomes
F=m dv/dt
By substituting the definition of acceleration, the algebraic version of Newton's second law is derived:
F=ma
"

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the 1000kg pipe has a speed of 4m/s

in order to stop the pipe in 1 second you must apply a force
of 4000N for 1 second.

the turn is 40 meters long.
the half way point in the turn is 20 meters.

the 100kg mass travels to the 180 degree position in
0.5 seconds because it has a speed of 40 m/s.
the 100kg mass only applies its force to the pipe for 0.5 seconds
before it reaches the half way point.

40m/s is 20meters/half second
its velocity is 20 meters / half second

the below force calculation uses acceleration not velocity
acceleration is determined by distance over time

acceleration is (((( DISTANCE / ((((DISTANCE * TIME ))))
acceleration is 20 m / (20 m * 0.5 seconds)
acceleration is 20 m / 10
acceleration is 2 m/s/s

the pipe has a mass of 1000kg

2000N = 1000kg * 2m/s/s

the pipe slows down durring the first half of the 1st turn from
4 m/s to 2 m/s
it does not stop durring the first half of the 1st turn.

the end of the second half of the 1st turn
is when the pipe stops.


I think your problem is that you dont include time.
or when you do you dont take into consideration that
half of a second is a big difference when using a
time frame of 1 second.





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Originally Posted By: paul
the 1000kg pipe has a speed of 4m/s

in order to stop the pipe in 1 second you must apply a force
of 4000N for 1 second.

I agree.

Quote:

the 100kg mass only applies its force to the pipe for 0.5 seconds
before it reaches the half way point.

Yes. And it also applies a force to the pipe during the 2nd half of its travels out the other side.


Quote:

acceleration is (((( DISTANCE / ((((DISTANCE * TIME ))))

You love inventing dimensionally inconsistent equations to serve your pre-assumed result, don't you? That isn't acceleration.


Quote:

I think your problem is that you dont include time.
or when you do you dont take into consideration that
half of a second is a big difference when using a
time frame of 1 second.


How about trying to understand my calculation? - the one in bold I posted a couple of days ago is probably the most complete and up-to-date. Then you can actually identify what my problem is, rather than guessing.

I consistently do exactly that for you. See in this message I understood that you used a wrong formula for acceleration and pointed it out. You can try to defend it, but good luck finding any websites or books or experiments or people or anything that support the equation a = distance / (distance * time)

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Quote:
acceleration is (((( DISTANCE / ((((DISTANCE * TIME ))))

You love inventing dimensionally inconsistent equations to serve your pre-assumed result, don't you? That isn't acceleration.




I didnt make that up.



that is how you determine acceleration given that your
knowns are
the distance and the time that the distance was traveled

and the knowns for the force calculations above are

v=20 m/s
t=0.5 seconds
d=20 meters
m=100kg mass

use the above FORCE equation and you will
get 4000N

---------- WHERE IS THE FORCE EQUATION THAT YOU USE?-------------------
or are you using a MOMENTUM CHANGE equation
or a VELOCITY CHANGE equation to find for FORCE?

or is there one , I would like to see it.

--------------------------- distance ---------------------------



Displacement is not Distance as is evident above.
traveling a distance is going from point (a) to point (b) along a given path.

Displacement is the shortest distance between two points.

traveling that same distance in 1 second is going from
point (a) to point (b) in 1 second

if that distance is 40 meters and you travel that distance
in 1 second you have traveled 40 m/s

---------------------------- acceleration -----------------------

if you accelerate from 0 m/s/s to 40 m/s/s in 1 second
your acceleration is 40 m/s/s
and you traveled 20 meters in that 1 second.

when I use Distance / Distance*time it is acceleration

a 1kg mass with a acceleration of 1m/s/s requires a force of 1N
if that mass travels a distance of 0.5 meters in 1 second it has traveled 0.5m/s
that means that mass had a acceleration of 1m/s/s

lets check that with a physics math formula used for finding acceleration
a=f/m
1m/s/s=1N/1kg

lets use the formula D/dt to check the acceleration result

0.5 meters distance / 0.5 meters distance * 1 second = 1m/s/s acceleration








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Originally Posted By: paul



I didnt make that up.




Those formulas are correct, but as I said (and you ignored), you're misinterpreting them - grossly.

d isn't distance. It isn't any number, it's part of the differential operator d/dt. d/dt means "rate of change with respect to t of...". I'm not going to try to teach you calculus, but it's not needed. Those formulas above are identical to

F=ma
and
a=F/m

The proof is shown on the Wikipedia force page you linked to before.

You'll never understand things if you make an effort to ignore anything you don't understand. When I don't understand a word, I don't pretend it wasn't there, I go and look it up, or at least make an educated guess and keep in mind that I might've been wrong in case it leads to problems later.




Quote:

---------- WHERE IS THE FORMULA THAT YOU USE?-------------------

is there one , I would like to see it.


You've seen them many times. They're clearly spelt out in bold in post #35266.


I'm getting tired of this Paul. Maybe you're just playing the fool to be irritating, or maybe you're afraid to do anything correctly because you realize it could prove your idea wrong. If you post one more wrong equation I'm giving up.

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they seem to work well for me , whats the problem that you have with them.

how do you determine acceleration if you dont use
distance and time , where do you start?
suppose you dont know what the force is , or you dont know what the mass is?

what do you do then?

----------------

could you please put your equation up again , I cant find it.

and Im not playing a fool , Im genuinely concerned with how
you get 8000N from 4000N

forces dont just appear , they must come from somewhere
and I want to find out where your getting the extra
4000N from.










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Originally Posted By: paul
they seem to work well for me , whats the problem that you have with them.


They work well because you invent suitable units and meanings that lead to the 4000N force you somehow decided you should be getting. Doing it correctly, you have no choice about units, you have no choice what the answer's going to be, it just comes out one way, like it or not.


Quote:

how do you determine acceleration if you dont use
distance and time , where do you start?
suppose you dont know what the force is , or you dont know what the mass is?


If you know enough other variables, you don't need to know distance. In this case we know the time it takes, we know the velocity change it experiences, we know it's mass, so we can calculate the force as below.



Quote:

could you please put your equation up again , I cant find it.



The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N



Quote:

and Im not playing a fool , Im genuinely concerned with how
you get 8000N from 4000N


Maybe your stumbling block is assuming it somehow 'started' as 4000N then increased to 8000N. There was never any 4000N to begin with.

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I found your error...
you use the mass*acceleration to get the 8000N
then you use the same to get another 8000N to act against the pipe.

the mass and the pipe both experience the same 8000N
you cannot seperate this action and then get two seperate forces from it.

Quote:

The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N


the 8000N is divided between the mass and the pipe.
4000N goes to the pipe as I have always said.

you dont get your extra free force to move the
pipe in the opposite direction.

in order for you to apply a 8000N force to the mass and a 8000N force to the pipe , you would need 16,000N

so the pipe stops , and your bulls&^$t stops also.
im pretty much tired of your scammy ways kallog

end of our discussion.







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Originally Posted By: paul
I found your error...
you use the mass*acceleration to get the 8000N

Yes, this is a direct application of Newton's 2nd law: F=ma

Quote:

then you use the same to get another 8000N to act against the pipe.

I don't use F=ma for that, I use Newton's 3rd law:
3rd law on Wolfram Alpha

Quote:

the mass and the pipe both experience the same 8000N
you cannot seperate this action and then get two seperate forces from it.

In some fundamental way they're the same, that's why the have the same magnitude. But the direction is opposite. You push on your desk, the desk will experience a force and may move because of it. You hand will also experience a force, the same force, but in the opposite direction, you'll feel it pushing back at you. You don't feel nothing, and you don't feel the desk pulling your hand away from you.

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centripetal acceleration

centripetal acceleration calculator

radius of turn = 12.732 meters
velocity is the tangential velocity = 40 m/s
ac=v^2/r

ac=125.6676m/s/s

centripetal force

mass = 100kg
Fc=mac=m*(v^2/r)
Fc=100kg*125.6676 m/s/s = 12566.76N

Fc=((40m/s*40m/s)/12.732m radius) * 100kg = 12566.76N

so the outward force is the same as the inward force

so the pipe really does have a 12,566N force pushing
it in the opposite direction.

I find that hard to believe but I recall that that is
the exact principle that another invention I had opperated on.

so from a 4000N force we get an extra 12,566N force

amazing , I apologize to you , I was wrong.

from what I have read up on , the magnatude of the force
can be greater with a shorter turn and less with a larger turn.

interesting, I can see possibilities here.

anyway in 0.6368 seconds the pipe slows to 0m/s velocity.
while it is slowing it continues to move for 4 meters in the
(+) direction.
so now it has traveled +16 meters & +4 meters = +20 meters

after the 1 second has passed the pipe has reversed direction and is now moving in the opposite direction at a velocity of 8.566 m/s

it has traveled a distance of 2.283 meters in the opposite direction while the mass completes the 1st turn , it free floats the 160 meters for 18.678 seconds.

the pipe has a velocity of 8.566 m/s so it travels a distance of 8.566 m/s * 18.678 seconds = 159.995 meters in the opposite direction.

you would need to apply a force of 8566N to stop the pipe in 1 second.

the mass then enters the 2nd turn , having the same results as in the above so the pipe continues for a distance of 4 more meters in the (-) direction as it slows to 0m/s.

the mass applies a force of 12566.76N to the pipe and
the pipe slows to 0m/s

the pipe then moves in the (+) direction for
a distance of 2.283 meters before the mass enters the accelerator again.

the mass enters the accelerator at a velocity of 8.566 m/s
it accelerates the 100kg mass to 40.907 m/s in 6.4682 seconds

it now enters the 1st turn with a force of 4090.7N

so you were right again , it does go further backwards.

+16m
+4m
-----------> +20m
-2.283m
-159.996m
-4m
<----------- -166.273m

once again , I apologise.

you were right.

sorry.

but it will work and that is the important part.




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Originally Posted By: paul

amazing , I apologize to you , I was wrong.

Haha, amazing indeed! I have to frame that and hang it on my wall wink


Quote:

so the pipe really does have a 12,566N force pushing
it in the opposite direction.

Not exactly. That's the outward force. It's not always directed along the pipe, so the actual average longditudinal force during the turn will be different. Not sure exactly how to calculate it at the moment.


Quote:

but it will work and that is the important part.


Jumping the gun again? :P

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I noticed that I used the pipe velocity for the
mass velocity in my previous post as the mass moved
into the accelerator again , it was late
and it was friday.

so dont go framing up anything yet.

like you said that is the outward force,LOL.

I might have to figure out the total linear force
to see if I was right or if you were right.

LOL

I believe if I use a 180 degree turn I can divide the
1 second by 180 , since the mass is traveling in a circular path I can use the linear angle and the fc angle to find the amount of linear force for each degree of rotation , then add up the forces as I said before.

there will be 180 fc forces in the outward direction
and 180 linear forces in the linear direction.

starting at 270 degrees the linear force would be
very small , then at 180 degrees the linear force would
be very large then at 90 degrees the linear force would be very small again.

since the longitudinal forces outnumber the linear forces
2-1

and 12,000 / 3 = 4000

I seem to think I was right all along.

there are 8000N longitudinal
there are 4000N linear

so for now I recant my apology , LOL.

but save the frame you may need it later.












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Quote:

starting at 270 degrees the linear force would be
very small , then at 180 degrees the linear force would
be very large then at 90 degrees the linear force would be very small again.


You can quantify "very small" using trig as in this picture. I might have measured my angles from a different direction to you but it doesn't really matter.


Oh, what I call "longitudinal" I think is what you call "linear". The component of the force parallel to the direction the pipe moves.

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Originally Posted By: paul

I believe if I use a 180 degree turn I can divide the
1 second by 180 , since the mass is traveling in a circular path I can use the linear angle and the fc angle to find the amount of linear force for each degree of rotation , then add up the forces as I said before.

Sound like a plan. I've added them all up as you suggest using a spreadsheet instead of a questionable simplification.

I used the equation in my previous message:
F_longitudinal = F_outward * cos(theta)

You can see that with coarse 10 degree steps it isn't quite 8000N, but the smaller the steps the more it approaches 8000N.

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looks right to me , pretty good excellmanship also.
that is exactly what a spring would do , a spring
would be compressed by the mass slowing the pipe to a stop.

if the spring stops the mass in 0.5 seconds it requires 8000N
to do that , it then has 8000N stored force to push the mass
away again accelerating the pipe in the opposite direction.

if the spring stops the mass in 1 second , it requires 4000N
to do that , but that would only slow the pipe down , it then
has 4000N stored to push the mass away with stopping the pipe.

so if the turn is increased in lenght to 80 meters the
pipe would stop , and the mass would still have a speed of 40 m/s when it exits the turn.

correct?

I think you had also mentioned that earlier.
so you were right about that.




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Originally Posted By: paul

if the spring stops the mass in 0.5 seconds it requires 8000N
to do that , it then has 8000N stored force to push the mass
away again accelerating the pipe in the opposite direction.

Indeed. If 'opposite direction' means opposite to the direction the accelerator started moving the pipe in

Quote:

so if the turn is increased in lenght to 80 meters the
pipe would stop , and the mass would still have a speed of 40 m/s when it exits the turn.


Careful, all these variables aren't independent of each other. Since it's twice as long, the mass now spends 2s in the turnaround, so the overall effect is half the force applied for twice as long.

When I earlier worked out that the pipe would reverse direction, I was using a 1s turnaround. With a 2s turnaround it only needs half the force to achieve the same result.

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I think I will make a program to figure this out for me.

I tried several times but theres always something wrong.

so give me a few hours perhaps.


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well , I couldnt get the force to lower , so
I suppose that the design needs to change.
the lenght and time of the turn never affected the total force applied.

so it needs a opposing accelerator to work.

the two turns cancel each other out.
-8000N
+8000N

so the +4000N from the acceleration is all that is left.

if 2 accelerators are used
and they are timed so that one mass
enters its second turn as the other mass enters its first turn then these will cancel each other out.

so its now a matter of timming these two masses.

if they are in seperate loops then they should
syncronize with each other.

perhaps instead of accelerating the masses at 5m/s/s the
acceleration could be slowly increased from 0 m/s/s
this can be controlled by the amount of electricity used
to accelerate the masses in the accelerators.

this can help to time the masses also.

now if you had 120 accelerators , every second there would be 2 masses begining acceleration , and this would greatly smooth out the ride.

but we should start with two accelerators first.
we already know what happens when only 1 is used.









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Originally Posted By: paul
well , I couldnt get the force to lower , so
I suppose that the design needs to change.
the lenght and time of the turn never affected the total force applied.

Really? If the turnaround was longer, the radius would be larger, causing a reduced centrifugal force.

Quote:

the two turns cancel each other out.
-8000N
+8000N

If they're being used simultaneously.


Quote:

so its now a matter of timming these two masses.

Not sure where the 2nd acccelerator would go ?? But I don't think it's possible to keep them synchronized -

Suppose both masses enter and exit their opposite turnarounds at the same time and with the same speed.

The mass that went through the 1st turn floats to the 2nd turn. But the mass that's gone through the 2nd turn is then accelerated and reaches its 1st turn sooner. They're now out of sync.

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I let the program figure it out , I increased the
lenght of the turn , then got the radius for that lenght.

then used the v^2/r*mass to get the force.

then I used your excell results and divided the
result by 1.57956307021181 which was taken from
your 12k / 7k results.

it always showed a total of 12k for total force when
time was included.

but the cf did decrease for each meter of lenght I used.

cf * time = always 12,566
and the total linear force was always the same
12,566 / 1.5795 = always 7955.599

it probably will be extremely hard to get syncronization
because if you increased mass velocity you get increased force through the turns.

I think the answer is somewhere in numbers.










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Originally Posted By: paul

cf * time = always 12,566
and the total linear force was always the same
12,566 / 1.5795 = always 7955.599


Hehe, at risk of being an 'I told you so'. Remember I used to keep saying the impulse applied to the pipe is always of magnitude 8000Ns regardless of how long the mass spends turning around, or how much force it applies? That's exactly what your unchanging result is, that same old impulse.

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OK , rub it in...

LOL.

I still cant exactly see how 8000N can be directed in the
(-) direction from a 12000N force , that only leaves 4000N
dispersed in the lateral directions.

are you sure that you used the correct equation?


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Originally Posted By: paul
OK , rub it in...


Alright, if you insist, my favorite paragraph :P ..

as far as I am concerned I have won this discussion because
of my use of non - fictional forces and numbers , you have done nothing but try to use fictional forces and went as far as adding up those fictional forces to arrive at fictional numbers.



Quote:

I still cant exactly see how 8000N can be directed in the
(-) direction from a 12000N force , that only leaves 4000N
dispersed in the lateral directions.


I can't quite visualize it very well. Tho there isn't really 'more' longitudinal force than lateral. Tho the entirety of the average force is longitudinal, and none lateral.

If you took just one of the two quarter-turns, it'd have the same 8000N average force in both directions. So the total average force would be directed at 45 degrees. When you also average that with the other quarter-turn, the two opposite lateral components average to zero, leaving only the two 8000N longitudinal components, which average to 8000N.

If that makes any sense at all :P

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This might be illuminating somehow

pi/2 = 1.5708

pi/2 * 8000 = 12566

!

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when you think about it as being a spring it makes
complete sence , 4000 in 4000 out

I think I have found a way to send most of the
(-) force lateraly.

using springs that are compressed lateraly.

I will need to brush up on that first.

think of the turn being mounted on a moveable track.
the turn has 4 arms that extend backward at a 45 degree angle to 4 springs
the 4 springs are mounted lateraly , the mass enters the turn , the force pushes the mount backward this compresses the 4 springs , most of the force is absorbed
lateraly by the 4 springs.

the springs then push the turn forward again sending the mass forward , the turn might need to be just a half turn
or the turn can be removed.

and this is also done lateraly , so most of the force is absorbed and expended lateraly by the 4 springs slowing stopping and accelerating the mass.

what do you think?

oh and you can use the accelerator to accelerate the
mass from 40 m/s to 0 m/s and that is also in the
(+) direction , and its ready to go again.

I think weve done it.

and by the way , the mass can generate electricity
while the accelerator is braking it.

this way it wont cost anything to opperate it.

but that would only be in a perfect world scenario
there has got to be some losses somewhere.










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Originally Posted By: paul

the 4 springs , most of the force is absorbed
lateraly by the 4 springs.


It'll collapse unless the springs have enough stiffness in the longitudinal direction. They'll still carry the same load that way. They just might also carry a much higher load laterally as well.

There's no way you can arrange springs to make a component of the force dissapear. Imagine, if there was, you could lift a car with one hand. Just put this 'force redirecter' between you and the car, so its weight is sent sideways.

Quote:

oh and you can use the accelerator to accelerate the
mass from 40 m/s to 0 m/s and that is also in the
(+) direction , and its ready to go again.

Then the mass is stopped so the machine can no longer operate??

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Quote:
It'll collapse unless the springs have enough stiffness in the longitudinal direction. They'll still carry the same load that way. They just might also carry a much higher load laterally as well.


then you would want to use springs that have enough
stiffness.
they will carry the same load that the 100kg mass * its
speed places on it because it is what the force of the
load is being placed on.

the load begins at a 45 degree angle , and ends at 1 degree angle.

the brunt of the force that is applied is transmitted lateraly into the springs , not in the (-) direction.

this is a common way that engineers use structual design
to make car crashes less strenous on the driver.

springs are used throughout structual engineering and are
in use to reduce stresses.

there will be some force directed in the (-) direction
but most will be directed in the lateral or sideways directions.

like before when the turn was placing 8000N of the 12000N in the (-) direction because the directional forces applied to the turn were not being absorbed by springs and the forces were not being directed outwards.

using the springs directs most of the force outwards
the stress that the pipe feels is outwards.

Quote:
There's no way you can arrange springs to make a component of the force dissapear. Imagine, if there was, you could lift a car with one hand. Just put this 'force redirecter' between you and the car, so its weight is sent sideways.


I can lift a car with 1 hand , all I need is a long lever
and fulcrum or a set of pulleys , the springs do not make any forces dissapear , I know you know that , they do not go away they are stored by the springs , and while the forces are being held by the springs inside the pipe those forces are all sideways.


Quote:
Then the mass is stopped so the machine can no longer operate??


you stop the mass after the springs have tossed it
back into the accelerator.

this removes the free float side.

there are no more turns.

accelerate the mass with the accelerator.(-)40m/s
the springs accelerate the mass to ()0m/s
the springs accelerate the mass to (+)40m/s
the accelerator accelerates the mass to()0m/s

then the mass is setting where it began , it
is waiting to be accelerated again.

this way the mass never travels faster than
40m/s so the springs will always be stiff enough to
stop the mass.

so now its like this.

4000N (+) pipe , as the mass accelerates

low (-/+) force to pipe as springs reverse mass direction

4000N (+) pipe , as accelerator stops the mass.

repeat , each cycle adds acceleration to the pipe.

the magnatude of the (+) pipe velocity will prevent
the low (-) forces from stopping the pipe , the low
(-) forces will slow the pipe , however by directing
the brunt of the (-) forces outwards the (+) forces will
always win a battle between the forces.

may the force be with you.
live long and prosper.

fascinating!









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Originally Posted By: paul

the load begins at a 45 degree angle , and ends at 1 degree angle.

All you're doing is creating extra lateral forces. The average longitudinal force is still 8000N, and the average of the total lateral force of all 4 springs is still 0.

Quote:

this is a common way that engineers use structual design
to make car crashes less strenous on the driver.

No. They reduce the force by increasing the length of time the impulse is applied to the occupants for. You can do that if you want, but it's the same old choice between 8000N for 1s, or 4000N for 2s. Neither gives any advantage.


Quote:

springs are used throughout structual engineering and are
in use to reduce stresses.

Yes. But they also increase the length of time the parts are subject to that reduced stress.


Quote:

using the springs directs most of the force outwards
the stress that the pipe feels is outwards.

True, You're creating extra stress on the pipe walls without reducing the longitudinal force. It's effectively a lever mechanism.


Quote:

I can lift a car with 1 hand , all I need is a long lever
and fulcrum or a set of pulleys , the springs do not make

It's still transmitting force to the ground through the fulcrum/pulleys. The total weight force felt by the ground is no different.

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Just to be clear, this is what you're saying, right?




If you take the machine and put it on scales, with a static mass just sitting there, it'd weigh less?

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I'll upload a picture of what I'm talking about.
thats not exactly what I have in mind.

I havent tried to figure out the stresses and the direction
that they will be applied or the time , so I cant really say much more for now.

the picture will give a clearer understanding of the spring setup , and the force directions involved.


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Quote:
No. They reduce the force by increasing the length of time the impulse is applied to the occupants for. You can do that if you want, but it's the same old choice between 8000N for 1s, or 4000N for 2s. Neither gives any advantage.


since Im no longer turning the mass around , the only force
I must overcome is 4000N , the 100kg mass can be accelerated
to 40m/s velocity in 1 second and it will travel a distance of 20 meters in that 1 second.

so stopping the mass requires that exact same force applied in the opposite direction in that same time.

f=ma
4000N = 100/40
a=f/m
40=4000/100

in 1 second the mass stops.

it will not require 8000N
to stop the mass in 20 meters and 1 second

BTW , the springs in your picture arent lateral they are at
a 45 degree angle , so they would direct the force to a 45 degree angle , I said the springs would direct the force in a lateral direction ie...90 degrees from the direction that the 100kg mass is moving.







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Quote:
If you take the machine and put it on scales, with a static mass just sitting there, it'd weigh less?


wow , weve moved back to the earth?

good thing were not talking about what would happen on earth.

put the scales in space and see what the scales read.


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Originally Posted By: paul

since Im no longer turning the mass around , the only force
I must overcome is 4000N , the 100kg mass can be


Since when? If you mean the 2nd accelerator that stops it, then no, you are still turning it around. There's clearly no way you can bring it back to the starting point without turning it round.

Of course you could just stop it at the 1st turn. The pipe gets 4000N from the accelerator and -4000N from the stop. Is that what you're saying?


Quote:

BTW , the springs in your picture arent lateral they are at
a 45 degree angle , so they would direct the force to a 45 degree angle , I said the springs would direct the force in a lateral direction ie...90 degrees from the direction that the 100kg mass is moving.


Drawing a spring laterally is misleading. As i said it must have some stiffness in the longitudinal direction otherwise it'll simply shear off and do nothing.

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Originally Posted By: paul

put the scales in space and see what the scales read.


We can get to that later if you actually think it'll tell us anything useful.

Obviously if this is an actual machine then it can, in principal be transported to earth. And it can be weighed. I'm asking you what would happen if you did that. I explained this in my message, you're just being contrary for no reason other than to attempt to hide your mistake.

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the below is more like what I had in mind to direct the
force laterly.













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OK that's what I thought. But it's missing any support in the longitudinal direction. The whole thing will just fall off when the mass hits it. Sure it might absorb some of the impact, but you don't need any fancy sideways springs or linkages for that, just a big block of rubber or a single spring floating in space.

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Quote:
Since when? If you mean the 2nd accelerator that stops it, then no, you are still turning it around. There's clearly no way you can bring it back to the starting point without turning it round.

Of course you could just stop it at the 1st turn. The pipe gets 4000N from the accelerator and -4000N from the stop. Is that what you're saying?


yes , it just stops , its force is distributed laterly as in the images above.

then the mass is accelerated by the springs into the accelerator.




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Quote:
OK that's what I thought. But it's missing any support in the longitudinal direction.


I didnt draw what the mass that pushes the springs are sliding on.

the springs are attached to the pipe.
the arms push the sliding mass that compresses the springs.


I thought you could just imagine them sliding on a support.

think of a rod that passes through the springs and the masses.




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OK. But you forgot some forces on your diagrams:


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yes there will be some force placed in the (-) direction
but the brunt of the force will be directed outwards.

your arrows are too long , they should more clearly represent the force.

they should be much smaller.
and they should get smaller and smaller as the sliding mass
slides outward.


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Originally Posted By: paul

I thought you could just imagine them sliding on a support.


OK. Those 4 sliding supports together carry the entire load applied by the mass. All 4000N, and they apply that to the pipe, pushing it backwards.

There's no "law of conservation of magnitude of force". You can't convert a 4000N longitudinal force to 4 x 1000N lateral forces. Sure you can get 4 x 1000N lateral forces using that mechanism, or you can get 1,000,000N laterally if you want. Either way it doesn't reduce the 4000N longitudinal component.

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Originally Posted By: paul

they should be much smaller.
and they should get smaller and smaller as the sliding mass
slides outward.


If they do get smaller then this is an antigravity device.

But they don't. In total, they're always the same as the applied force regardless of increasing lateral forces.

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sure it does , its done all the time.

picture the sliding masses as cars.

the cars have weight because they are sitting on the earth
so this same setup would work on the earth.

if a force is applied to the center the force will extend to push the 4 cars away from the center , the 4 cars roll on the ground.

they still have weight.

but the force is directed away from the center.
because of the mechanism the force is not directed downwards.







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Originally Posted By: paul
they still have weight.


Yes. And they also each have 1/4 the downward force pushing them into the ground, additional to their own weight.

Come on, didn't you ever play with Meccanno as a kid? This is a bog-standard toggle mechanism. It's a common way of amplifying forces. Don't you think somebody would have noticed that it also does antigravity?

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forces are directional , the force that is applied in a direction cannot change direction by itself , so the cars feel a outward lateral force.

they do not feel any additional force.

in fact the force that the earth feels because of the
cars weight decreases because the cars are moving.

if a car is riding down a street made up of weight scales
the car would appear to weigh less and less as it passes over the weight scales as the car accelerated down the street , because the force has changed direction.

this is why a car driven off a cliff at high speeds does not go straight down as soon as it clears the cliff it
falls at a angle due to the force of gravity acting on it.

the force is mostly in the direction that the car is traveling at.

you seemed to understand the outward forces when we
were using the turn , in fact you believed that a
4000N force became 12,566N because the forces were
inward due to the acceleration towards the center of the turn that the 100kg mass experienced because it turned, but now you seem to think that forces can not be
directed outward as it was in the turn.

you need to get your story straight.

I can tell your getting stumped on this one , because
your reaching into the insult bag to try and cover
your losses.

Quote:
Come on, didn't you ever play with Meccanno as a kid? This is a bog-standard toggle mechanism. It's a common way of amplifying forces. Don't you think somebody would have noticed that it also does antigravity?


I never had any toys like that , just tools and wood
and metal , and such.

I used to want toys like that but they were too expensive.

and Im not sure where you get amplifying forces
from, the 4 springs dont amplify forces , they
only direct forces , I didnt include any mechanical advantages in the design.

heres one other point you seem to be overlooking
completely , even if only a portion of the force
is directed outward , then the final result would
be that the pipe would be capable of forward motion
each cycle and its displacement would continuously increase because thats the way forces work.

if you cant defeat the 4000N , then the pipe doesnt stop.
it only slows , then as the mass is tossed back into
the accelerator and slowed to a stop in the accelerator
by the acclerator the pipe accelerates more , then the accelerator accelerates the pipe even more by accelerating the mass again , it never stops, as long as the accelerator is switched on.

so unless you can figure out a real reason that the mass will direct 100% of its (-) force to the pipe
in the (-) direction , then the concept is 100% viable.

and it will work !!!














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Originally Posted By: paul

in fact the force that the earth feels because of the
cars weight decreases because the cars are moving.

Let's not get side-tracked into movement.


Quote:

were using the turn , in fact you believed that a
4000N force became 12,566N because the forces were

It didn't 'become'. 8000N was simply the average longitudinal component of the radial force.


Quote:

I can tell your getting stumped on this one , because
your reaching into the insult bag to try and cover
your losses.

I'm doing that because you're frustrating me. You're also being dishonest. Why didn't you show the downward forces in your diagrams? All those arrows but you neglected to include, or even mention the very ones that I had claimed would be there.


Quote:

and Im not sure where you get amplifying forces
from, the 4 springs dont amplify forces , they
only direct forces , I didnt include any mechanical advantages in the design.


Yep, your mechanism gives more mechanical advantage the further it's flattened down. Look up "toggle mechanism". With 4000N applied on top, the side forces can increase way beyond 1000N each. In theory, they become infinite when the linkages are horizontal. That's how vice-grips work.


Quote:

completely , even if only a portion of the force
is directed outward , then the final result would

Exactly all of the applied force is directed in the same direction it's applied in. There are _extra_ forces directed outward, but these don't take anything away from the longitudinal force.



If you still can't afford Meccanno, you can build this with biro springs and popsicle sticks. Test it by placing it on a scales with a weight on top. See that the weight weighs less with the device under it.

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I'm getting tired of analysing all your inventions only to have you tell me my methods aren't valid. You havn't once shown that any of my methods have been faulty. I was about to do it using Newton's 1st law, but I expect you'll say that might be wrong.

How about you analyse this one? Be sure to include references for your methods/equations/etc.

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"this is why a car driven off a cliff at high speeds does not go straight down as soon as it clears the cliff it
falls at a angle due to the force of gravity acting on it."

True, but not because the force of gravity is redirected or anything. A Car driving with fullspeed over the edge of a cliff will hit the ground at the exact same time as a car dropped from the same height.

A more practical example would be the bullet of a gun, if fired exactly parallel to the ground. The bullet will hit the ground just as soon as any other object falling from the same height.

Also a car driving on a flat street doesn't become lighter. The donward force of gravity is independent from any sideway forces.

Here is something you can try for yourself:
Get some weight (e.g. 1kg) and a string.
Put the weight on the middle of the string.
Attach one end of the string to a door/wall.
Now you can pull on the other end of the string.

The 1kg of the weight will always have a downward force of 9.8 N.

Nevertheless you can pull with all your strength, there will be always a slight tilt of the string. There is an exact relation between the tilting angle and strength you are pulling at, so that always exactly 9.8 N are applied in a vertical direction to support the weight.

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Quote:
If you still can't afford Meccanno, you can build this with biro springs and popsicle sticks. Test it by placing it on a scales with a weight on top. See that the weight weighs less with the device under it.


its not the weight kallog that Im directing , its the force
of the mass times its acceleration.
Im sure you understand that but you cant find a way to get
force to fit into your naysay.

so you use a example of a static weight sitting on a scale.

of course the mass would WEIGH the same sitting on a mechanism such as the 4 springs , but that has nothing to do with what Im talking about.


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Quote:
I'm getting tired of analysing all your inventions only to have you tell me my methods aren't valid. You havn't once shown that any of my methods have been faulty. I was about to do it using Newton's 1st law, but I expect you'll say that might be wrong.


Its really funny the way you wanted to use 8000N as a (-)
force in the turn , starting with a 4000N force.

then when the cards turn you no longer use outward force.

LOL.

and now you say that all the (-) force is still there
and its added to the force that has been sent outwards ,
how do you suppose you can apply a force in 1 direction
then divide that force in 4 directions and still
have the same amount of force in the (-) direction?

what you are sudgesting is free energy.

Im saying the 4000N is divided up in the 4 lateral directions and in the (-) direction.


because if the 4000N force can be reclaimed in the
(-) direction , then all the other force is free !!!

the trouble with that is that it cant be done.

just show me how you do that with a force diagram.









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shut up momos none of your replies had any meaning to what I
was saying.

your replies are out of context , stay out of this.

Quote:
the force that the earth feels because of the
cars weight decreases


Quote:
the car would appear to weigh less


you think Im wrong put your weight scale in your yard
run over it and see what the thing says.
LOL

Quote:
this is why a car driven off a cliff at high speeds does not go straight down as soon as it clears the cliff it
falls at a angle due to the force of gravity acting on it.



gravity is the only reason the car or a bullet or anything else falls , you think Im wrong , hold your computer out the window and drop it?

dont throw it downwards or horizontaly , just drop it.
or you can throw it upwards to make it last longer.





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Originally Posted By: paul

its not the weight kallog that Im directing , its the force
of the mass times its acceleration.
Im sure you understand that but you cant find a way to get
force to fit into your naysay.


I actually don't understand that at all. Until now I thought you were only redirecting forces. What exactly do you mean by "force of the mass times its acceleration"? Is that a cross product? Maybe you mean "the magnitude of the force times the magnitude of the accelearation"? That's a scalar quantity which doesn't have any direction to redirect. Can you be clearer?

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Originally Posted By: paul

Its really funny the way you wanted to use 8000N as a (-)
force in the turn , starting with a 4000N force.

then when the cards turn you no longer use outward force.


On the curved track, the radial forces nearly always had a component in the - direction. In this device the outward forces are always straight outward. They have no component in the - direction.


Quote:

then divide that force in 4 directions and still
have the same amount of force in the (-) direction?

what you are sudgesting is free energy.

It is free force, but not free energy (energy isn't force!). There's nothing wrong with free force, why should there be? Without it a lever couldn't work.

Quote:

Im saying the 4000N is divided up in the 4 lateral directions and in the (-) direction.

Based on what physical theory? You're assuming that the total magnitude of the force can't change. Show me a reference which supports that idea.


Quote:

just show me how you do that with a force diagram.

Noway!!!! Any hint of a valid force diagram will instantly destroy your concept. You can do that. Use a free body diagram.


There's my challenge to you. Draw a free-body diagram of it. Be sure to look up Wikipedia to find out the rules for free-body diagrams.


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I had to use a force field to redirect the force in the (-)
direction and when I did look who showed up !!!
it was pac man all along.

but hes just imaginary isnt he?



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Quote:
Yep, your mechanism gives more mechanical advantage the further it's flattened down.


it never is a mechanical advantage there is no
pivot or fulcrum.

it gives more force to the springs as it is flattened down but never more than the 4000N (-) force divided by 4 = 1000N each spring.

it starts off at 45 degrees , and the (-) force is greatest there , then at 0 degrees the (-) force is weakest.

and at 45 degrees the lateral force is weakest and strongest at 0 degrees.

but you cant treat the forces the same as you did with your excell calculations.
because the extra force you got in the excell calculations
was due to the the pipe supplying a force to turn the mass around 180 degrees.

the springs do not turn the mass.

the mass only presents a 4000N force to the center.

not 8000N.









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You didn't really look up free-body diagrams did you?

Where are the forces applied to the ramp? Where are the reaction forces?



Here (above) you've shown a single force on the right-hand ball. Under that force it'll accelerate towards the bottom-left. But the ramp's in the way. Where does it go without any other forces to redirect it?

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Originally Posted By: paul
it never is a mechanical advantage there is no
pivot or fulcrum.


You can look it up in the dictionary


Quote:

and at 45 degrees the lateral force is weakest and strongest at 0 degrees.

I agree with that part at least.



Quote:

the springs do not turn the mass.

the mass only presents a 4000N force to the center.

not 8000N.


If it's stopping in 1s then it's an average of 4000N.
If it's changing direction in 1s then it's an average of 8000N.
That's applied to both the 'center' and the pipe.
You keep changing the design so I forget if it's still reversing or maybe it's stopping now?


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I cant seem to find any use for a free body diagram
its more like filler for physics books , and lecture times.

were you going to use one to show that there really are no
forces involved?

just say 1-1=0 and 1-1=0 then your done.

all they do is show the exact reverse of a force.

you apply a force
at -0
and the same force is applied back
at +0

whats the point?


Quote:
If it's stopping in 1s then it's an average of 4000N.


no its not , use the force kallog.

F = ma

unless f=ma is wrong , are you saying that?


you must be trying to average something out again , trying to get an angle on it.

I suppose that since the 100kg mass was accelerated for
8 seconds then its average would be 500N right?

so I suppose you could say that the average could be only
2000N if you had an extra second , but you only have 1 to work with.

but if you cut the second in half you could get 2 1/2 seconds , this way you could get an average of 2000N
over the 2 1/2 seconds.

nope that wouldnt even be right you would still have
4000N because a Newton is measured in seconds not 1/2 seconds.

I guess I should elaborate on that 1N = 1kg meter/s^2
to avoid the comments from the peanut gallery.

a force that will accelerate a mass of 1 kg
at a rate of 1 meter per second per second.

if you cut the second in half it is only half the newtons.

Quote:
If it's changing direction in 1s then it's an average of 8000N.


it changes direction in 2 seconds not 1 second.
so its not 8000N its 4000N

I will re-post that part.

Originally Posted By: paul
since Im no longer turning the mass around , the only force
I must overcome is 4000N , the 100kg mass can be accelerated
to 40m/s velocity in 1 second and it will travel a distance of 20 meters in that 1 second.

so stopping the mass requires that exact same force applied in the opposite direction in that same time.

f=ma
4000N = 100/40
a=f/m
40=4000/100

in 1 second the mass stops.

it will not require 8000N
to stop the mass in 20 meters and 1 second







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Originally Posted By: paul
you apply a force
at -0
and the same force is applied back
at +0

whats the point?

You answered your own question. So are you going to do it or are you afraid it'll reveal what you're trying to sweep under the rug?


Quote:

I suppose that since the 100kg mass was accelerated for
8 seconds then its average would be 500N right?

Instead of supposing, why don't you actually work things out?

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Quote:
Instead of supposing, why don't you actually work things out?


I did you just forgot.

I was refering to you wanting to average out the 4000N
to fit into another scheme.


the free body diagram is crap.

it doesnt calculate anything.

you cant determine if anything can or cannot be accelerated
with it , so whats the point.

and your sudgestion that I use a nothing to prove something
just falls in line with what you are trying to accomplish.

heres a free body diagram , can you use it to
calculate anything?



according to the above free body diagram the block should fly upwards off of the incline , because the brunt of the
forces are in that direction.

and that is exactly why I use the word crap when describing such crap.

http://en.wikipedia.org/wiki/Free_body_diagram
All external contacts and constraints are left out and replaced with force arrows as described above.

Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the
ball applies a force to the table, and the table applies
an equal and opposite force to the ball. The FBD of
the ball only includes the force that the table causes on the ball
.


therefore a FBD is useless when trying to
calculate forces in nature as the ball would
apply a force to the table , like the block
above would provide a force downward opposite of F3,
and the block would provide a force downward opposite
of force F2 and the incline would provide a force
upward opposite force F1 , your sudgestion of using
a FBD is senceless and only shows your intent to
cloud the subject matter with crap.

of course using crap like that would have worked
out just fine you wouldnt it.



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Originally Posted By: paul




according to the above free body diagram the block should fly upwards off of the incline , because the brunt of the
forces are in that direction.


Yes, if you make up values for the forces. Work it out correctly and it'll tell you what actually happens. If F3 is zero, the diagram tells you it acceleartes down the slope, and how fast. If you pick another value F3 it tells you which direction it'll go and with what acceleration. If you want to hold it stationary the diagram tells you what value of F3 to use for that too. F1 and F2 are determined by the block's weight.


Quote:

therefore a FBD is useless when trying to
calculate forces in nature as the ball would
apply a force to the table , like the block
above would provide a force downward opposite of F3,

It's useless when you're trying to fool people. You can't easily hide force components with a free-body diagram. More relevent here, you can't easily hide them _from yourself_. If you do the diagram you'll discover your device is very different from what you've described.

I know it's a simple idea. That's why it's so powerful. It's often easy to misjudge how forces work, to mix up reactions with actions, to reverse directions, to omit force components, to guess wrong values for magnitudes, etc. But it's hard to make a mistake with a free body diagram.

By the way, it's not always trivially easy to do. The block picture above requires some trig to calculate F2. You do actually have to do maths sometimes!

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Originally Posted By: paul

Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the
ball applies a force to the table, and the table applies
an equal and opposite force to the ball. The FBD of
the ball only includes the force that the table causes on the ball
.


You've found a mistake in Wikipedia, the part you underlined is wrong. The rest of the paragraph is correct.

"The FBD of the (anything) only includes forces applied to the (thing)"

Kind of make sense really. The ball doesn't care what's happening to some bat on the other side of the world. It's only influenced by forces applied to itself.

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I dont think so , I have watched several lectures on the
free body diagrams and I never considered this type of diagram valuable as all forces are not included.

in the diagram I posted with the block and incline the incline
is showing as a force , but not the force that the block would
exert in the opposite direction.

I think its best to include all forces in our
discussion as you might try to slip something around
a corner to lean the discussion in your favor.


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Originally Posted By: paul

in the diagram I posted with the block and incline the incline
is showing as a force , but not the force that the block would
exert in the opposite direction.


That diagram would be used to analyse the behaviour of the block, not the incline. So it includes all the forces on the block, and none of the forces on the incline.

If you wanted to analyze the incline, you'd do a separate FBD for that.

If you wanted to analyze the complete system, you'd still have 2 seperate FBDs, but you'd find that some forces have the same magnitude in both diagrams, so you can connect them together with equations to describe the complete system thoroughly.

Come on, do FBDs for the components of your mechanism. Do each moving part with a seperate diagram. The diagrams you've shown are incomplete and ambiguous. FBDs are crystal clear because they're done with such simple rules.

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this thing is really pretty easy to figure out
the 4000 N force is applied to the center.

it is then divided by 4 because there are 4 arms extending
downwards from the center.

so each arm gets 1000N

at the end of each arm there is force at a 45 degree angle
this force continues straight but cannot apply itself in that
direction but the force can apply itself in a 45 degree angle
to the spring.

so the 4 1000N forces must pass through 2 45 degree angles before
they reach the springs.

since the original force was at 180 degrees the 180 degree force
is directed to the 90 degree angle , so the 4000N force is all
directed outward.

so the 1st 45 degree angle results in a force of 1000N at 45 degrees and
the 2nd 45 degree angle results in a force of 1000N at a 90 degree angle

these 4 1000N forces are lateral or at 90 degrees from the center.

the pipe wouldnt even feel any (-) force as the mass applies the 4000N force.

it seems to easy but this is the way forces work.

theres no need to draw a bunch of diagrams.

its a done deal , even it you do try and complicate it
its just too simple a machine.

its hard to complicate simplicity.



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Originally Posted By: paul

it is then divided by 4 because there are 4 arms extendig
downwards from the center.

so each arm gets 1000N


What physical theory are you using to find that?

Electric currents behave like that. Put 4000A into a conductor, split it 4 ways, and you'll find the total current in those 4 conductors is also 4000A. If it's symmetric (as yours is), then there'll be 1000A in each.

There's no such law for magnitudes of forces.


Quote:

it seems to easy but this is the way forces work.

Give me a reference.


A FBD would show what actually happens. I know you're afraid to draw one because you'll find you have to include forces that you pretended didn't exist.

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Quote:
What physical theory are you using to find that?


that doesnt need a theory its basic math.

4000N/4=1000N

no theory needed.


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Hey why don't we all put one of these on the front of our car? Then when we crash into a tree, we don't get hurt because hardly any force is transmitted through the device to the cabin.

Skydivers wouldn't need parachutes. Just tie these to their feet and when they land, they only get a fraction of the force from the ground.

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Quote:
Give me a reference.


heres a pretty good one , if someone sits on it
there weight is divided equally between the 4 legs.

you could put 4 weight scales under the 4 table legs
and each weight scale would read 1/4 a persons weight plus
1/4 the weight of the table.

but put the 4 scales and table in zero g
and someone sitting on the table will not
register any weight on the 4 scales

same same



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Originally Posted By: paul

no theory needed.


This equation works for static forces, right?

Put in on a scales and measure the weight of something sitting on the center.

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Originally Posted By: paul
heres a pretty good one , if someone sits on it
there weight is divided equally between the 4 legs.


Here all the forces are in the same direction.

It does indeed work when they're all in the same direction.

In fact you can see the legs are sloping a bit. Would you say the total vertical force on the floor is _less_ than the weight of somebody sitting on it? Because some of the force must be transmitted outward, right?

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Quote:
Hey why don't we all put one of these on the front of our car? Then when we crash into a tree, we don't get hurt because hardly any force is transmitted through the device to the cabin.

Skydivers wouldn't need parachutes. Just tie these to their feet and when they land, they only get a fraction of the force from the ground.


those arent bad ideas , only you would need them to be very large but they would work on the same principle.

you would need a much larger one for a car because we are
only talking about a 100kg mass impact and
thats only 220 lbs.


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Quote:
This equation works for static forces, right?

Put in on a scales and measure the weight of something sitting on the center.


well we could put a weight scale behind each spring.
then put the static 100kg mass on it , and each scale would read 25kg.

at least thats what should happen.

but thats because of gravity , in space zero g the mass wouldnt
show as having a weight.


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Originally Posted By: paul

you would need a much larger one for a car because we are
only talking about a 100kg mass impact and
thats only 220 lbs.


What difference does size make? You haven't specified a size for your device, and you say it'll work, therefore you think it'll work for any size.

Or do you mean materials might not be strong enough to support higher forces? Well a tree can survive a car hitting it, so we could even make them out of wood.

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You know the basic idea you're using is correct.

The only difference between your use of it and the correct way is you're using the _magnitudes_ of the forces. The correct way uses the complete forces - magnitude and direction.

Do the same maths with vector forces instead of scalars and it comes out right.

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Quote:
What difference does size make? You haven't specified a size for your device, and you say it'll work, therefore you think it'll work for any size.


I never got a chance , you imediately decided that
it wouldnt work , and from the information that I
posted , you said it wouldnt work so I think its you
that thinks it is a 1 size fits all.

of course its not , but telling you that wont do
much good either will it.


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Quote:
The only difference between your use of it and the correct way is you're using the _magnitudes_ of the forces. The correct way uses the complete forces - magnitude and direction.


I think you have that backwards , I've always used magnatudes and directions , you always just say that all the force is directed backwards.

a force cant travel through nothing , it must act on an object.
and the outward forces act on objects not nothings as your
backward forces act.

then the outward forces act on the pipe and then the springs expand tossing the mass back into the accelerator.







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Quote:
Do the same maths with vector forces instead of scalars and it comes out right.


I would kind of enjoy seeing how you would do that.

why dont you do it , and include all forces.

just use each force and the direction of each force.

we can add the resistance to the forces after
you have that done.

and its in space so you dont need to include any g forces or forces from mass due to g.

that is why we must start with the 4000N force and its direction because its in space in zero g and we cannot use static forces because there are none in zero g that would apply in this case.






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Quote:
Here all the forces are in the same direction.

It does indeed work when they're all in the same direction.

In fact you can see the legs are sloping a bit. Would you say the total vertical force on the floor is _less_ than the weight of somebody sitting on it? Because some of the force must be transmitted outward, right?


yes they are sloping and the force is directed through the slope.
inside the slope the direction of the force is downwards and lateral.

I picked this picture for that purpose.



I think if someone stood on this table and jumped upwards
when they landed on the table the outward forces applied to the legs
would bend the legs causing them to be sent outwards
from the center and the legs would break.

much in the same way that the 4 springs would get
the brunt of the 4000N force.




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OK I'll do it myself. Here's my FBD.

Can we just have 2 legs instead of 4? It can just have sliders to support it from collapsing into/out of the screen.



The variables F1, F2 and F3 are just the magnitudes of the forces, and they could be zero if need be. The arrows show their directions on each member.

Not sure how you want me to calculate the forces. I could treat it statically then it won't be deflecting, just statically supporting the 4000N. It might get a bit tricky to do dynamically - the 100kg mass would have to be included and it may not be a constant 4000N.

Notice that it's possible to choose values for F1, F2 and F3 so that all forces balance on all members, allowing it to support a static load if need be. If F3=0 then it can't support any static load because there's no way to choose F1 and F2 so the blocks have zero net force on them.

I think we really should sort out how it'll behave statically before trying to do it dynamically. Even tho it's not on earth, any real thing can, in principle have a static load applied to it. Maybe from gravity or maybe from a little propeller attached to it, or a big magnet, or whatever.

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Originally Posted By: paul

I think if someone stood on this table and jumped upwards
when they landed on the table the outward forces applied to the legs


Sure there'd be outward forces. I'm not disputing that.

But the downward forces wouldn't be in any way reduced. If the floor was just weak enough to break when someone jumped on a straight-legged table, then it'd equally break when they jumped on this sloping-legged table.

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so your saying that F1 stays constant with the force applied at the point where the 4000N is applied.

since the F1 force pushing on the block is
pushing downwards at a 45 degree angle shouldnt this force continue through the block , which will cause the F3 force
to be placed in the F3 direction , and shouldnt there also be a F3 force pushing down on the opposing side of the block?

the 2 F3 forces on the block causing a torque that
would balance

then the two F3 forces should balance.



in the above , the F1 force is directed through
the block causing torque t to be applied around
the center of mass of the block , this torque is translational to the downward force F3 which
results in the upward force F3 , the two forces total 0.

also you left out the opposing F2 force arrows , I drew
them the same size as we are not including the magnatude
of the forces only there direction.

so in the above the F3 forces can be striken , and unless we are
going to include frictional forces only the
F1 and F2 forces remain , heres where the magnatude of the
forces come in to decide if the block will move outward
I say it will move , how bout you?








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Paul, recommendation.

It has been decades since I studied statics or dynamics, but I will tell you this - you cannot solve this kind of problem consistently and correctly without free body diagrams. It is usually the first or second step to solving mechanics problems.

However, you cannot correctly set up free body diagrams without a clear understanding of vectors. I have a lot of difficulty following what you guys are arguing about, but I think kallog's FBD was correct the way it was. The red F2 is the opposing force to the component of F1 in the x direction. F3 is the opposing force for the component of F1 in the Y direction. The black F2 and F3 are not extra forces - they are the components of F1 in the x and y direction which means they can replace F1.

I admire that you're trying to wrestle this on your own, but you should focus on a few vector videos before trying to use free body diagrams.

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Originally Posted By: paul
so your saying that F1 stays constant with the force applied at the point where the 4000N is applied.

No. I just showed a 4000N input for clarity. F1 stays constant as long as the input stays at 4000N. Obviously if you increase the input force then F1 increases also. But then the diagram doesn't apply because it only says 4000N.

The equal F1 forces everywhere also assumes the diagonal linkages are massless, or it's a static analysis. I hope we can treat them as massless, if not the whole thing.

Quote:

since the F1 force pushing on the block is
pushing downwards at a 45 degree angle shouldnt this force continue through the block , which will cause the F3 force

It does. For a static analysis, the vector sum of F2 and F3 is a force antiparallel to F1 and with the same magnitude as F1.


Quote:

be a F3 force pushing down on the opposing side of the block?

There's nothing there to push the block down, so no such force can exist.

Quote:

also you left out the opposing F2 force arrows , I drew
them the same size as we are not including the magnatude

Again, there's nothing there to apply a force purely in that direction. The only things touching the block on that end are the base, which, being slippery can only apply an upward force (F3), and the diagonal linkage which is only capable of transmitting force parallel to its length because it's pin-jointed (F1).

The horizontal component of that F1 can balance F2. No need for an extra F2.

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Haha I might have got a bit carried away here, but I made a computer simulation. I had to adjust the spring constants to get approximately 2s turnaround time.



Here's the reaction force measured at the point shown on the diagram. It's the same as on the opposite side. All other vertical reaction forces are zero. The horizontal reaction forces at the ends of the springs depend on the angle on the diagonal members, but they came out at an average of 1500N each for the geometry shown. That kind of makes sense because they're a bit steeper than 45deg. I suppose at 45deg, there'd be 2000N on each spring.




The model assumes small displacements for everything, so it might be a bit inaccurate. However the larger you make the real thing, the closer it'll approximate the model.

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thats nice kalog.

but what does it mean?

does the 3200 mean 3200N?

you used fully constrained , does that mean that you
removed the springs?

and the force is just stopped there.

by the way the slides arent just laying against a floor
or something they are in space they must ride on some type
of track that would hold them from floating around.

so there is a torque from the 45 degree force placed on
the slider.

you said you made a program to get your results , did the
program take distance into consideration?

ie... how far did the slider move outward?

I cant see how you got 3200N from 1 leg if that is what
you are saying.

or trying to say.

Quote:
The model assumes small displacements for everything


if your working a 2 second time frame you should use a distance that would equal 40 meters in that 2 seconds
I would like to know how much your distance was.

or at least the displacement in each direction not both
together and then say it was 0 , as you mostly do.

the distance that the 100kg mass moved in the
(-) direction.


when you did get the 2 seconds what spring constants did
you use?

Quote:
No. I just showed a 4000N input for clarity. F1 stays constant as long as the input stays at 4000N.


the force F1 goes to zero in 1 second , did your program take that into consideration?

your diagram doesnt look like it would fit.

at 1 second your showing the highest vertical forces
as the 4000N has been reduced to zero , how do you account for that?














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Yea I suppose I didn't really explain it properly.

Originally Posted By: paul

does the 3200 mean 3200N?

Yes. That's the peak force, occurring in the middle of the motion.

Quote:

you used fully constrained , does that mean that you
removed the springs?

and the force is just stopped there.

The line segments closest to the ends are the springs. So the outside ends of the springs are fixed. That's also where the horizontal 1500N (average) reaction force occurs. The inside ends are free to move horizontally, and are connected to the blocks.

Quote:

by the way the slides arent just laying against a floor
or something they are in space they must ride on some type
of track that would hold them from floating around.

Yep, the 4 bottom elements are all prevented from moving vertically, but free to slide sideways.

Quote:

so there is a torque from the 45 degree force placed on
the slider.

There could be depending on the details of the connection. I assumed there's some slider material directly below the diagonal linkage's connection. That way no moments are applied.

Quote:

you said you made a program to get your results , did the
program take distance into consideration?

It's a general purpose program I used. No, that's one of the weaknesses. It assumes the digonal linkages have the same fixed angle when finding the forces, tho is does have them moving. The springs compressed a fairly spectacular amount, but that would just be because I made the dimensions quite small. If it were all scaled up a lot, there'd be hardly any movement of anything (relative to the size).

Quote:

I cant see how you got 3200N from 1 leg if that is what
you are saying.


Yes. 3200N on each leg. So 6400N total. Of course that's only the maximum and only occurs for an instant. Most of the 2s are spent at less force.

The average is 2000N/leg. Which is really quite close to the 4000N total I've been saying all along :P

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Quote:
Yea I suppose I didn't really explain it properly.


Originally Posted By: paul
does the 3200 mean 3200N?

Yes. That's the peak force, occurring in the middle of the motion.


your diagram should look more like a big "M" instead of
a mountain.

the largest forces your showing as resultant forces
dont have anything to result from.

the 4000N is zero at 1 second , you must have the program
wrong somewhere.

Quote:
That's also where the horizontal 1500N (average) reaction force occurs.


that 1500N average would also be wrong.

Quote:
Yep, the 4 bottom elements are all prevented from moving vertically, but free to slide sideways.


so the torque is present.
and there is a matching F3 force , that is striken then.

Quote:
It's a general purpose program I used. No, that's one of the weaknesses. It assumes the digonal linkages have the same fixed angle when finding the forces, tho is does have them moving. The springs compressed a fairly spectacular amount, but that would just be because I made the dimensions quite small. If it were all scaled up a lot, there'd be hardly any movement of anything (relative to the size).


whats the name of the program?
can it be downloaded?
I would like to plug in your numbers to see what results I
can get from it using different spring constants.

Quote:
Yes. 3200N on each leg. So 6400N total. Of course that's only the maximum and only occurs for an instant. Most of the 2s are spent at less force.


you must be using only 2 legs , thus the 6400N
that couldnt be the result from a force of 4000N
applied at a angle of 45 degrees.

Quote:
The average is 2000N/leg. Which is really quite close to the 4000N total I've been saying all along :P


we'll just have to wait and see , I think your program
doesnt include the proper elements or you wouldnt get
such large reaction forces when there are no action forces
present.









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Originally Posted By: paul

your diagram should look more like a big "M" instead of
a mountain.


Sink in the middle?? There's no joint that would allow that. Or do you mean the blocks tilt? I assumed the blocks were prevented from tilting. If they can then it's unstable because the springs will just pop them out of straightness as soon as any load is applied.

Quote:

the 4000N is zero at 1 second , you must have the program
wrong somewhere.


Not at all. 1 second is where the springs are the most compressed. Obviously they have quite a lot of force on them at that time.

But anyway, that simulation was a bit of a side-track. It won't prove anything because nobody's going to go through all the code checking it.

We should focus on the static analysis since we still disagree even on that.

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Originally Posted By: paul

so the torque is present.
and there is a matching F3 force , that is striken then.


The computer says there's no force there, It just needs to be restrained vertically to prevent instability. The FBD says there's no moment applied to the blocks, so there shouldn't be any anyway.

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Quote:

Not at all. 1 second is where the springs are the most compressed.


so the springs are at the point where you measured.
the legend of your diagram only shows the verticle reaction force.

there is no lateral force mentioned.

basically what your saying is that the springs are
compressed with 6400N , for 2 springs.

I dont know kallog it sounds suspicious to me.

Quote:
The computer says there's no force there, It just needs to be restrained vertically to prevent instability.


I think that without the math to examine , I will have
to decline your programs results.

and the computer program can be designed to neglect
forces that would be considered elements in a force diagram.

so I dont think the results are viable.

theres almost allways a force there , wherever there may be.






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Originally Posted By: paul

so the springs are at the point where you measured.
the legend of your diagram only shows the verticle reaction force.

Oh yea I didn't mention that. When I looked at the animation it sunk down then bounced up. The whole thing took about 2s and it was at the bottom of its deflection at 1s.

Quote:

basically what your saying is that the springs are
compressed with 6400N , for 2 springs.

No, that's the vertical force. The springs got an average of 1500N each.

Nonetheless, it's pretty amazing that the vertical force averaged to 4000N on both legs. I didn't have to tweak anything (except the springs to get the 2s you specified), despite the arbitrary angles of the legs and arbitrary overall dimensions.

Quote:

I think that without the math to examine , I will have
to decline your programs results.

Fair enough. Lets focus on the static. Can you respond to my last message about that?

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so from the 4000N longitudinal force you get

4000N longitudinal and 3000N lateral ???????

I think that you will find that the resultant
longitudinal forces will only be 1000N.





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Originally Posted By: paul
so from the 4000N longitudinal force you get

4000N longitudinal and 3000N lateral ???????

I think that you will find that the resultant
longitudinal forces will only be 1000N.


I didn't use a 4000N input. I used a 100kg mass bashing into it at 40m/s. But yea those were the average forces at the supports.

Go back to the FBD to see if it should be 1000N or not. No point just making up numbers because they feel good. Also no point making up a 'law of conservation of magnitude of force' when none exists. This mechanism has a mechanical advantage or disadvantage, so you can tune it to give whatever horizontal force you want. You can have 1000N if you want, but it won't detract from the 4000N pushing backwards.

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100kg mass bashing into it at 40m/s.
thats 4000N isnt it?
at least it is if the mechanism stops
the 100kg mass in 1 second.

whats the name of the program you used?

the reason I say 1000N longitudinal
is because you say 3000N lateral.
you cant have a reaction larger that the action.

for every action there is a equal and opposite reaction.


the cosine of 45 degrees is 0.7071

4000N * 0.7071 = 2828.4N
this should be the resultant lateral force that the
slider feels.

of course you would divide that by 4
when using 4 legs , you can use 1 to get the total.

pretty close to your 3000N !







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Originally Posted By: paul
100kg mass bashing into it at 40m/s.
thats 4000N isnt it?
at least it is if the mechanism stops
the 100kg mass in 1 second.

Yep, on average it is. It doesn't apply a constant 4000N.

Quote:

whats the name of the program you used?

LISA www.lisa-fet.com

Quote:

the reason I say 1000N longitudinal
is because you say 3000N lateral.
you cant have a reaction larger that the action.

The 'action' on the left spring is -1500N, the reaction applied by the spring is equal and opposite, +1500N. Similarly for the other spring.

Again, there's no reason to conserve magnitudes of forces. You can easily apply 4000N longitudinally and get 10,000N sideways. It's just mechanical advantage.

Quote:

for every action there is a equal and opposite reaction.

Yes, opposite, not sideways.

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Originally Posted By: paul

the cosine of 45 degrees is 0.7071

4000N * 0.7071 = 2828.4N
this should be the resultant lateral force that the
slider feels.


Can you go into more detail? You seem to be considering both legs together, as well as the input force. It's a bit confusing.

But I think the general idea of calculating components is right.

Hey by the way, do you tend to edit your messages, adding extra parts, and turning off the 'mark as edited' option? I'm not complaining, but I often find your messages are different after I've replied to them. Or maybe it's a bug in the forum?

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yes , opposite and opposing.
thats why you cant have the full 4000N , its not opposite.

the forces start at a 45 degree angle.
and the angle decreases.


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yes , its always a work in progress , you should
usualy wait a few minutes to reply.


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Originally Posted By: paul
yes , opposite and opposing.
thats why you cant have the full 4000N , its not opposite.


Word games mean nothing in physics. FBDs aren't ambiguous. You really should use them. They make it clear what all the forces are.

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if you can agree to the above , then I will add this
to my program to return the resultant forces using distances and time for each degree.

in the above none of the resultant forces are above the
4000N applied , so I trust this.

what I meant in saying you cant have the full 4000N
was that the longitudinal direction that you are claimimg that the 4000N force would be in , would not be 4000N at no time durring the cycle.

it will always be from 45 and decreasing as pictured above.






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Yea those diagrams perfectly correct in themselves.

But it's not 4000N along the diagonal member. It's 4000N longitudinally at the input, so that direction change has to be calculated in a similar way.

As well it's split between the two diagonal members, so they only get half each.

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I was just showing how the forces would direct through
the point that you used , of course the first angle
from the input point would also be at 45 degrees.

you didnt include that either did you?

yes I was intending to include 4 legs in the program
and using all elements involved , we can call the sliders
massless or we can give them a mass , also friction
in space would be the result of forces causing objects
to rub against each other , so there will be friction
should we use a friction coefficient also or can we
say that friction is negligible for now?

So if you can agree to this then I will put these
into the program Im building.

do you think you could help with a spring coefficient
to use?

I asked earlier but you didnt reply , what did you use
in the program you used?



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Originally Posted By: paul
I was just showing how the forces would direct through
the point that you used , of course the first angle
from the input point would also be at 45 degrees.


Yea I agree with what you did and that the first angle would be the same. Just not with using 4000N on the diagonals.

Quote:

do you think you could help with a spring coefficient
to use?


I used a spring constant of 70 N/m for each spring.

Everything massless would be good. And everything frictionless. Those two things can always be made arbitrarily small in real life.

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Im almost done with the program , but I have to say
its not looking good for your side , LOL.

well I am done with it mostly , the calculations are done
I just need to make a postable text file with it that
shows the results.

the 70N/m stops the mass to quickly using a 20 meter
stopping distance , it stopped in less than 1 second.

and in a 40 meter stopping distance it stops in 5 meters.

so I need to find a useable stopping distance and spring co efficient to use.

I want to make sure its all right before I post the
results anyway.


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Originally Posted By: paul

well I am done with it mostly , the calculations are done
I just need to make a postable text file with it that
shows the results.

Remember I won't be able to trust your computer results any more than you trusted mine. You'll have to describe the calculations in such a way that I can easily check and reproduce them.

What are the equations? And why not do the static calcs anyway? We never resolved even the static case, so I don't expect any dynamic analysis to have much purpose at this stage.

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good point

maybe I should include the equations in the readout that I make.

I havent determined the best arm lenght and its initial and
final angle yet , so when I do then I can include the
readout portion within the program.

Im using the spring co efficient as a resistive force.
and then I subtract that force from the 4000N before continueing
to the next set of calculations , so it should be pretty precise.

now the more sets of calculations that I allow the program to perform , the more precise the results are.

so it will be awhile , meanwhile I have recieved my new starter
for the hydrogen engine im building.
I purchased a old 15 hp engine frm the salvage yard and am
rebuilding it to use HHO only !!!

if it works well enought I will want to find a generator head
to attach to it , and then get my free energy.

it has a built in generator that charges batteries and powers lights and stuff on a riding lawn mower and this generator might just make all the HHO it needs to run.


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Originally Posted By: paul

I havent determined the best arm lenght and its initial and
final angle yet , so when I do then I can include the
readout portion within the program.


Don't bother optimizing it. To prove me wrong you only have to show that either of these claims is wrong, even slightly:

1. The average of the total force applied to the pipe is identical to the average force applied by the mass.

2. For a 100kg 40m/s mass, the average force applied by the mass is 8000Ns / turnaround time.


Quote:

Im using the spring co efficient as a resistive force.

It isn't a force. You have to multiply it by the amount of spring deflection to get the force.


Quote:

starter
for the hydrogen engine im building.

Cool. I'm glad you're actually trying it. I just hope that when it fails you don't make some excuse about too many strokes or not fully using the intake vacuum or whatever.

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Quote:

1. The average of the total force applied to the pipe is identical to the average force applied by the mass.

2. For a 100kg 40m/s mass, the average force applied by the mass is 8000Ns / turnaround time.



I suppose you mean the average force applied to the pipe in the
longitudinal direction in #1 above.
because alot of the force is being applied in the lateral direction.

and the force that is being applied in the longitudinal direction
is being applied as a torque or twisting force that stresses the
pipe along its sides , therefore that force can be dampened also
resulting in even less longitudinal force.

in #2 above it is 4000N that is required to stop the mass in 1 second as the mass stops , however it seems that the springs do not compress enought to return the mass at 40 m/s in the next second , so the total will be less than 8000N.

alot of the force is lost through the angles and in the twisting torque through the sides of the pipe.

but dont worry Im taking everything into consideration.

maybe I should make another drawing or two to explain this better.




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Originally Posted By: paul

I suppose you mean the average force applied to the pipe in the
longitudinal direction in #1 above.
because alot of the force is being applied in the lateral direction.

That would do too. I actually mean the entire total force. All lateral forces cancel out in the averaging operation, so only the longitudinal component contributes.

Quote:

pipe along its sides , therefore that force can be dampened also
resulting in even less longitudinal force.

Equations not words! If that's true, I'll admit I'm wrong, and you'll be confident that your machine will fly.

Quote:

m/s in the next second , so the total will be less than 8000N.

OK. I still stand by 2nd point, except I realize I used the wrong sign convention. Should be a negative in there. I'll add a more general #3 to allow for your new situation:

2. The average force applied by the mass is
F = -m * (v_2-v_1) / turnaround time.
Where v_1 is the initial velocity (-ve for backwards), and v_2 is the final velocity. This allows it to come out slower than it went in.


I've also made some unstated assumptions in how the averaging is done. But I don't think it'll become important unless you have masses inside the device. The averages should both be made over the same time period and it should be one in which all forces occur, even if that's longer than the turnaround time.

Quote:

maybe I should make another drawing or two to explain this better.


You should show some equations or even code. Actually I'm amazed you're programming a dynamic response solver. Engineering students do that in their final year. You really should do a FBD or any kind of static analysis. If there's mistakes in that then surely any work on the program will be a waste.

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Quote:
You should show some equations or even code


Im doing this in vb6 , because its much easier , do you have vb6 or higher?

if you have higher I could probably upgrade the project
to fit in with your compiler.

or just copy n paste the code and you can figure it out.

I have to warn you though its not pretty and I use a few tricks
to speed things up a bit.


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Oh ouch. I've had plenty of bad experiences with people trying to optimize VB6 code (don't do it!!!). I think I'll pass. At least until it's complete.

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not optimization , just tricks.

like passing or storing a variable or value in a lable or text box , versus working the code to pass the variable or value to a subroutine when needed.

this way the variable or value is there when I need to use it.
not stuck away in memory where I cant see its value.

its no way optimized , I dont optimize code I just clean it up
a bit.
shrink it where ever I can after it all works if Im going to
release it.

thats why I said its not pretty "code".

BTW , which compiler do you use?






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Originally Posted By: paul
like passing or storing a variable or value in a lable or text box , versus working the code to pass the variable or value to a subroutine when needed.


Cool, that's a good idea. Yea attach it in a private message if you can. I've got VB6.

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How's that program going Paul?

You seem to have falling into your old habit of mysteriously losing interest right before you're about to finally be proved right.

Can I assume from your silence that you discovered the program agrees with me?

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no


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I am concerned about the methane event Ryskin refers to and has theorised.

I have read his research and have many questions.

If soneone can help answer them please do so. Thanks.

Go to http://www.wna.org.au/Ryskin%20Theory.html to see my research on whether he is right and the impact on today's climate.

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Sorry, too many words for anyone to read.

Anthropomorphism isn't a good start to get any serious person's interest either.

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I suppose writing too much must mean that it lacks seriousness or perhaps it is yourself and this forum that is a waste of time - I will go to another forum than one that refers itself as science agogo. Of course it could not be that you are incapable of understanding it due to lack of intelligence. About as serious as the high school student you must be.

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