Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Methane Clathrates in Gulf and Worldwide

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

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.

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".

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.



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.

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


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

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

in message #35130

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.



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.

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.

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.

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.

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.

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.




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.

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.

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




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.

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.

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.


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.

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.

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??



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




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.

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.


.

OK, I'll update my spreadsheet.



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.




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


Agreed.


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.


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




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.

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.