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#35066 - 06/19/10 02:38 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100

Quote:

1. accelerate 20 masses applies 20 positive forces.

OK, impulse = +20
The pipe's moving forward with momentum +20

Quote:

2. turn 1 mass around 180 degrees applies a negative force.

-2 because it's reversing not just stopping.

Now we have to consider the actual possible layout. I suppose those 20 masses were spaced apart in their accelerator to begin with - or all lumped together. Either way, before any masses get to the 2nd u-bend, which I suppose is at the opposite end of the tube, all the other 19 have reached their first turn-around -

-2 * 19 = -38
Total = +20 - 2 - 2*19 = -20
Now the pipe's going backwards with momentum -20

Quote:

3. turn 1 mass around 180 degrees applies a positive force.

+2

Then the other 19 masses do their 2nd turn-around -

+2 * 19 = +38

Total = +20
Now the pipe's moving forward again with momentum +20


What next?





Edited by kallog (06/19/10 03:02 PM)

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#35068 - 06/19/10 03:59 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
What next?


LOL , ok...

Quote:
-2 because it's reversing not just stopping.


were not dealing with a free energy device here.
so you arent allowed to add free energy.

and the turnaround will not add any energy to the mass.

the turnaround does not supply a force to the mass it is like a curve in a road that your car travels on , I cant remember having to press the gas pedal to the floor board just to get through a curve , LOL.

I usually take my foot off the gas. LOL

the mass already has enought momentum to clear the turnaround.

and the most negative momentum you could possibly count
would be the mass * the masses velocity.
which is the masses momentum.

so whatever the single masses momentum is is the most
you can use as a negative momentum.

and the opposite occurs at the other end in the other turnaround , so once again I state that the two masses
momentums in the turnarounds cancel each other out.

the only remaining force is the force used to constantly accelerate the 20 masses.

we havent calculated the force to the pipe due to accelerating the 20 masses "constantly" so we cant use
your +20.

the number will be much greater than that.
to accelerate a 100 kg mass from 0 m/s to 5 m/s
requires more force that to accelerate that same mass
from 0 m/s to 1m/s.

until we know the amount of force applied to the pipe
due to accelerating the "20" masses constantly we will not be able to arrive at a close number.

note: constantly...

and by the time each of the "20 " masses gets to the end of the pipe and enters the turnaround its velocity might be 30-40 m/s faster than when it exited the turnaround at the other end.

so we cant determine the validity of the concept using your
assumptions.











_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35074 - 06/20/10 12:54 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
note: constantly...


constantly :
because
1) there are always 20 masses being accelerated.
2) there are always 20 masses free floating.
3) there are always 2 masses in the 2 turnarounds.

so using your math !!

-2 * 19 = -38
+2 * 19 = +38
+20 * 20000 = +400000
netural 20 * 0 = 0
Total = +400000 -38 +38 = +400000

see I can do the same stuff only in my favor as you do.

LOL
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35075 - 06/20/10 03:21 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
and the turnaround will not add any energy to the mass.


Correct. And of course we're talking about momentum, not energy, which I stated quite clearly.

Suppose a car has momentum of 1000. It crashes into a tree and stops. What's its new momentum? How much did it lose?

This point is crucial to it working or not. So we better clarify it before complicating things with constant accelerations.

Actually we can express it in an even clearer way. Suppose a car has velocity of 1000, it crashes into a tree and stops. What's its new velocity? What's the change in its velocity?

Sounds too simple doesn't it? But I'm curious what your answer will be.



Edited by kallog (06/20/10 03:29 AM)

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#35076 - 06/20/10 04:25 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
that all depends on your frame of reference and a tree is elastic to a point , then theres the wind resistance as the tree is approached by the car and the wind compresses and pushes back on the car with a momentum of -1000

however since the car is also pushing the compressed air
with a momentuum of +1000 the two momentums cancel each other out , the tree due to the elastisity absorbes the force of the "wind + car" system and the car pushes the tree towards the car at twice the cars momentum.

so after the tree impacts the car the car has a momentum
of car-1000 + tree+2000 = car - 1000.

the car has a momentum of -50,000

then the car will travel at the speed of light in the
opposite direction of its momentum - the doubled tree momentum that is squared and double squared just as the impact occurs due to the right angle acceleration of the wind and its angular velocity.

so
1000^2^2 w.T1-A1= 50,000

but only if the tree is elastic.
if the tree is not elastic then the results would be different.

which results in

-50,000 + +50,000 = -+50,000

depending on your frame of reference.

tree momentum durring collision is

T1(T/A^2) VC1-w = total tree momentum = +2000^3-%@.()

567,hd *(-) + a^3 / AT = wind angular velocity

so 50,000 - 567,hd = -1000

so although it appears that the tree is colliding with the tree , the compressed air between the "tree+car" system
is absorbing the cars momentum , the tree only looks as if the car is hitting the tree because the car is being pushed backwards by the elasticity of the "air+tree" system at the speed of light-c + 1000.





_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35077 - 06/20/10 05:43 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
that all depends on your frame of reference and a tree is elastic to a point , then theres the wind r


I assume all measurements are made from the same reference frame. It stopped from the point of view of whoever measured it as stopping. That same obsersver measured it having p=1000 before collision. And I'm asking for the final and change in momentum from the point of view of that same observer.

The elasticity of tree is also irrellivent because the car stopped. Any transient elastic behaviour was damped out causing the car to end up stopped.

Wind resistance is irrelivent, the car stopped, doesn't matter how.

Do you have an answer or do you give up?


Edited by kallog (06/20/10 05:46 AM)

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#35079 - 06/20/10 11:29 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
Suppose a car has momentum of 1000. It crashes into a tree and stops. What's its new momentum? How much did it lose?


its new momentum is 0 because it lost 1000

unless you want to get technical about it.
because its sitting on the earth , the earth is rotating around the sun at apx 66,000 mph
and who knows how fast our solar system is rotating around our galaxy , and who konws how fast our galaxy is rotating around our universe , etc...

but if its stopped on the earth then in reference to the earth its momentum is zero.
and it lost 1000.

Im not sure where your going with this unless your going to
try again to show that the concept is invalid using only 1 or 2 masses.

because if you stop one of the masses that has been accelerated at 5 m/s then you can subtract all the force that
was applied to accelerate the mass at the end by completely stopping the mass.

but if your going to stop 1 at one end then you must also stop 1 at the other end and this will
cancel out each -momentum with a +momentum.

even if you do that , there are still 20 masses being accelerated constantly at 5 m/s/s applying a force to the pipe.

but try as you may.



_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35080 - 06/21/10 05:05 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
but if its stopped on the earth then in reference to the earth its momentum is zero.
and it lost 1000.


Yes, same answer I got, funninly enough.

Here's where I'm going. Now the driver manages to restart his car and reverse away from the tree, he speeds up to a mometum of 1000 in the opposite direction and keeps going like that.

Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?



Edited by kallog (06/21/10 05:09 AM)

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#35082 - 06/21/10 12:07 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?

Q1:1000 ; the car was initaly traveling at 1000
Q2:1000 ; the car reaccelerated to 1000 after the collision
Q3:1000 ; after the collision the momentum goes to zero
Q4:1000 ; momentum at start is the same at end only opposite direction.


------ my turn ------

there are a total of 42 mases
20 being accelerated
20 free floating
2 in the turnarounds

Q1: how much force does each 100kg mass apply to the pipe in the opposite direction of the movement of the mass if accelerated for a distance of 500 ft at a constant acceleration of 5 m/s/s if initial velocity is zero.
Q2: how much force will be applied to the pipe in the opposite direction of mass movement by all 20 masses being accelerated at 5 m/s/s before the first mass enters the first turnaround.
Q3: what is the total force applied to the pipe by the 20 free floating masses.
Q4: how much force is required to accelerate 42 100kg
masses at 5m/s/s for a distance of 500 ft.









_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35083 - 06/21/10 01:59 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
Q1: What was his momentum before the collision
Q2: What is his momentum after driving away
Q3: What is the overall change in momentum of the car between the start and end of the story?
Q4: How much momentum did the car lose between the start and end?

Q1:1000 ; the car was initaly traveling at 1000
Q2:1000 ; the car reaccelerated to 1000 after the collision
Q3:1000 ; after the collision the momentum goes to zero
Q4:1000 ; momentum at start is the same at end only opposite direction.


Q1: Yes, 1000
Q2: No, -1000
Q3: Change in momentum = final momentum - initial momentum.
Q4: Negative of the answer to Q3

Do you want to reconsider Q3? That's crucial to the entire moving pipe machine.



Edited by kallog (06/21/10 02:05 PM)

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#35085 - 06/21/10 05:44 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
why dont you simply tell me what you think the answers are
then see if I agree to them.

meanwhile , I have calculated the braking force of the single mass that is accelerated at 5 m/s/s for a distance
of 152.4 meters = 500 ft.

I get the following , you can check the math for yourself.

500 kg-m/s/s or 500N
100 kg mass
a=f/m
5m/s/s = 500N/100kg

initial velocity = 0m/s
final velocity = 39.0385 m/s
average velocity = 19.51925 m/s
distance the mass traveled in 7.8077 seconds = 152.400448225 meters

so the single mass is traveling into the turnaround at
a velocity of 39.0385 m/s
F=ma
3903.85 N = 100 kg * 39.0385 m/s

the braking force required to stop the single mass is
3903.85 N

at any time durring acceleration the 20 masses being pushed by the 500 kg-m/s/s or 500N are pushing back on the pipe with the same force with a combined force of 500N * 20 = 10000N

we had used 500 kg as the mass of the pipe earlier in the discussion so the 10000N force applying to the 500kg mass
results in a acceleration of 20 m/s/s

a=f/m
20m/s/s = 10000N/500kg

so there is a constant force of 10000N applied to the pipe
at any moment durring acceleration of the 20 100 kg masses.

initial pipe velocity = 0m/s
initial mass velocity = 0m/s

given that all of the masses are distributed evenly
around in the system , by the time a single mass is accelerated between the first turnaround and the next turnaround
the pipe has a final velocity of 156.154 m/s
the pipes average velocity is 78.077 m/s
F=ma

78077N = 500 kg * 156.145


result

78077 N - 3903.85 N = 74173.15 N

we can keep dilly dallying around with examples but using the pipe example would be best.

to an observer outside the pipe the pipe has moved 609.6017929 meters after 7.8077 seconds.

just in case you are curious in 1 hour , 3600 seconds
the pipe will have traveled 129,600,000 meters
or 80,529.706 miles

its final velocity will be 72000 m/s or 44 miles / second

or 2684.323 mph

in 24 hours it will travel 39,158,514.937 miles
or 74649600000 meters

in 1 week 3657830400000 meters or 1,918,767,231.937 miles

and that is by using only 1 set of masses , a space ship could be fitted with hundreds of these tubes and each extra tube adds that much more acceleration , I need to set the program to calculate the speed and time to the nearest star.

what would the fastest acceleration be that humans could withstand?














_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35087 - 06/21/10 10:51 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
why dont you simply tell me what you think the answers are
then see if I agree to them.

I have told you and you don't agree. There really is no point going anywhere beyond this point until you understand how much momentum is transferred from an object that does a U-turn.

The answer to Q3 is -2000
We can verify this with:
Initial momentum + change in momentum = final momentum
1000 + -2000 = -1000


Quote:

we can keep dilly dallying around with examples but using the pipe example would be best.

I agree.


Acceleration of one mass:
500N for 7.8s = impulse of 3900 Ns -> gives the mass a momentum of 3900 kgm/s and the pipe -3900 kgm/s.


Quote:

Turnaround of one mass:
a velocity of 39.0385 m/s
F=ma
3903.85 N = 100 kg * 39.0385 m/s


This is the magnitude of the force that would deccelerate 100kg from 39.0m/s to 0 in 1 second. The mass then has to speed up again from 0 to 39m/s in the other direction, so this force is applied for 2 seconds.

Impulse applied to the mass = -3903.85N * 2s = -7808 Ns
This reduces the mass's momentum by 7808 kgm/s and increases the pipe's by 7808 kgm/s




Quote:

at any time durring acceleration the 20 masses being pushed by the 500 kg-m/s/s or 500N are pushing back on the pipe with the same force with a combined force of 500N * 20 = 10000N


So over 7.8s there's:
10000N pushing one way, impulse = 78000 Ns
and 20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.

The overall force on the pipe is continuously in the opposite direction to what the accelerations should cause. So the pipe's moving backwards.

What about the other U-bend bringing it back to the proper direction? I'm starting to get confused about where all the masses are. While that acceleration was occuring, there were 20 floaters, entering the 2nd u-bend one-by-one? These would go back into the accelerator at some speed - but that's impossible because we specified 20 masses accelerating from 0, not from 39m/s. I'm losing the ability to visualize it. Could you possibly draw a time-sequence of events?



Edited by kallog (06/21/10 10:55 PM)

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#35088 - 06/21/10 11:11 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul

78077 N - 3903.85 N = 74173.15 N


Even if you don't agree with my calculations, you should see a problem here. You've only accounted for 1 of the masses turning around but 20 accelerating. It takes 2s to do a turnaround so there are actually several masses in the u-bend at the same time, and their forces add together.

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#35089 - 06/22/10 01:07 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
This is the magnitude of the force that would deccelerate 100kg from 39.0m/s to 0 in 1 second. The mass then has to speed up again from 0 to 39m/s in the other direction, so this force is applied for 2 seconds.


I never said that the mass would be stopping , that was your
assumption.

Quote:
So over 7.8s there's:
10000N pushing one way, impulse = 78000 Ns
and 20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.


there is (((( ALWAYS )))) 1 mass in 1 turnaround and
there is (((( ALWAYS )))) 1 mass in the other turnaround.
Ive already said that Im not stopping the masses , That would be something a stupid person would do knowing that they would have to re-accelerate the masses using up precious energy.

you always seem to forget the other turnaround so your
Quote:
20 masses doing U-turns, which is 20 * 7808 Ns = 156200Ns in the opposite direction.

also has the other end attached to it.

and the result is :

+156200Ns plus -156200Ns = ZERO

they cancel each other out.

right !!!
yes right , youve agreed to it before already !!!

there is (((( ALWAYS )))) 20 masses being accelerated
and
there is (((( ALWAYS )))) 20 masses free floating

so the final result is: 20 * 3903.85N = 78077N

so at the end of the 7.8 seconds the pipe has a final pipe velocity of 156.154m/s in the opposite direction that the masses were accelerated.

its pretty clear that you are using math that is designed
to work only in your favor , by using the full stop on the masses because it would hopefully prove your point , but the only point you proved today is that you use these left handed
measures to accomplish a goal.

I guess you play around with the numbers until you see an opportunity , LOL , in fact ROFL...




_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35090 - 06/22/10 01:23 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
Even if you don't agree with my calculations, you should see a problem here. You've only accounted for 1 of the masses turning around but 20 accelerating.


I counted the braking force of 1 mass because it is the total force applied to 1 mass durring acceleration, the two masses in the turnarounds cancel each other out , why cant you understand that now ...

Originally Posted By: Kallog
Is this how it works?:

1. Accelerate the mass, starts moving the pipe.
2. Turn it around, applies a force the wrong way.3. Turn around again, applies the same force the opposite way.
4. Repeat.

The two turn-arounds cancel each other out. Leaving only the acceleration to propel the pipe?



_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35091 - 06/22/10 01:44 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul

I never said that the mass would be stopping , that was your
assumption.


A u-bend does stop the mass in 1 dimension, which is all we're considering, and all we need to consider. In order to turn it around. You have to decellerate it down to 0 then back up to full speed. This consumes _NO_ energy because it's an elastic 'collision'. Of course in the real 3D system the _speed_ remains constant throughout the turn, but the component of velocity in the only direction we're considering with must drop to zero and change sign.


Quote:

there is (((( ALWAYS )))) 1 mass in 1 turnaround and


That means we have to change the force it applies when going through:

In 7.8s 20 masses enter the turnaround.
Each mass spends 7.8/20 = 0.39s turning around.
Reversing the direction changes its momentum by 3900*2=7800 Ns
If the turning is done by applying a constant force between the mass and the tube, then that force is:
F = 7800Ns / 0.39s = 20,000N
So this force of magnitude 20,000N is applied continuously at the 1st turnaround.

Just as before, this is double, but opposite the constant 10,000N force applied by the accelerator.

And of course just as before I still havn't mentioned the 2nd turnaround which will help. But I don't know how you want it to work yet. See below..



Quote:

you always seem to forget the other turnaround so your
+156200Ns plus -156200Ns = ZERO
they cancel each other out.

I purposely omitted it because I didn't know what the setup should be. If there are masses going through the 2nd turn at the same speed as all the masses in the 1st turn (39m/s), then they're entering the accelerator already doing 39m/s and will overtake some of the masses starting at 0. But we calculated the forces at the 1st turn assuming all masses were accelerated from 0 to 39, not 39 to 39+39. So the force in the 1st turn now depends on which mass is going through it, and will always be equal or higher in magnitude than at the 2nd turn which takes some stragglers from the previous acceleration boost.

Quote:

there is (((( ALWAYS )))) 20 masses being accelerated
and
there is (((( ALWAYS )))) 20 masses free floating


I'm not sure this is physically possible. Could you produce an animation?

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#35092 - 06/22/10 02:16 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
the pipe is 152.4 meters long

152 / 20 = 7.6 meters

so the masses begin acceleration seperated by 7.6 meters each.

so we can have a turnaround radius of 4.83 meters.

this means we will need to add 9.66 meters to the lenght of the pipe because of the two turnarounds , however we still accelerate the masses a distance of 152.4 meters.

this way there is 1 mass in the middle of each turnaround when the acceleration begins.

so the turnarounds will need to be used to slightly accelerate the masses in the turnarounds bringing them up to a slight velocity before accelerating them at 5 m/s/s.

if we have to the pipe can always be lenghtened to avoid collisions between the masses.

Im only trying to show the concept not provide you with plans to build anything , so if you are considering collisions forget it.

the 20 masses in the free floating side are held in place by a opposing rail gun accelerator.

the opposing accelerator is removed after the masses are accelerated slightly as I mentioned earlier.

Quote:
A u-bend does stop the mass in 1 dimension, which is all we're considering, and all we need to consider. In order to turn it around. You have to decellerate it down to 0 then back up to full speed. This consumes _NO_ energy because it's an elastic 'collision'. Of course in the real 3D system the _speed_ remains constant throughout the turn, but the component of velocity in the only direction we're considering with must drop to zero and change sign.


No , you dont decelerate the mass .... the mass is only changing direction.

what you are talking about above is simple direction.

Quote:
That means we have to change the force it applies when going through:


No , we do not apply any force while the mass moves through the U turn.

the mass itself applies a force to the U turn and that force is its mass * velocity

100kg * 39.0385 m/s/s = 3903.85 N

which is exactly opposed at the other U turn..........
because its in zero g.

Quote:
Reversing the direction changes its momentum by 3900*2=7800 Ns


changes the momentum of the mass but not the momentum that is
either applied or subtracted from the pipes momentum.


knowing that the masses passing through the turnarounds cancel
each other out , why do you insist on discussing them?

where are you going with this , there is no gain or loss to pipe momentum because the masses pass through the turnarounds.

because we are not including any friction between the masses and the turnarounds.

its like you are arguing about two 5 lb weights tied to a string then flung over a clothesline.

they just sit there because they both present the same force to each other.













_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35093 - 06/22/10 02:59 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
the opposing accelerator is removed after the masses are accelerated slightly as I mentioned earlier.

So it accelerates them just enough to feed them into the main accelerator at the same rate other masses are leaving it, so as to maintain 20 in the main accelerator at all times?


Quote:

No , you dont decelerate the mass .... the mass is only changing direction.


Yes only changing direction, and that requires a force, which can be done just by letting it fly through a U-shaped pipe, or hitting a spring. No need to put any energy in.
Quote:

Originally Posted By: Kallog

To a physicist, 'acceleration' includes a change in speed or direction. Look up wikipedia to get a more thorough picture. This is the same acceleration in F=ma so it's crucial that we use a consistent definition.

I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.

picking at straws are we?



Quote:

No , we do not apply any force while the mass moves through the U turn.

We/mass/tube/etc. Somehow there's a force, as you say, and that's what I mean too. We don't need to fire up another rail gun or anything.


Quote:

the mass itself applies a force to the U turn and that force is its mass * velocity

I ignored this before because it didn't really seem to matter, but now we better straighten it out:
F=ma, F<>mv, p=mv.




Quote:

knowing that the masses passing through the turnarounds cancel
each other out , why do you insist on discussing them?

Because they don't. In this new situation with a slight acceleration provided at the turnaround before the accelerator, we have masses going very slowly through that, while _at the same time_ there are masses going at 39m/s through the opposite turnaround. Different speeds, different forces.

Sure the 39m/s masses eventually get to the other turnaround and apply the same but opposite force there. But by the time the last few of them are doing that, there'll be new, even faster masses that've been accelerated twice, going through the 1st turnaround, so they still don't cancel each other out.

This is why I keep asking for an animation. Even tho it's a bit of work to make, it'll clearly show what's happening without having to consider any forces or momentums. I'm sure you have the programming skills to generate such a thing automatically. I don't want to do it because I expect you'll say I've got it wrong and have to do it again.

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#35094 - 06/22/10 03:29 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
you realize Im sure that the acceleration does not have to be instantly at 5 m/s/s , that the acceleration could take place gradually to avoid collisions , however I assumed you would take that up.

lets drop all the confussing stuff.

we now have 1 mass and thats all.
we dont need a opposing rail gun.
we dont need to control the mass at all.
we just use the instant 5 m/s/s acceleration.

the mass is 100 kg
the pipes mass is 500 kg
the force supplied to the mass is 500kg-m/s/s or 500N
the pipes initial velocity is 0 m/s
the mass initial velocity is 0 m/s

we begin the mass acceleration
its initial velocity is 0 m/s
we accelerate it at 5m/s/s for a distance of 154.2 meters

accelerating this 1 mass results in a force of +3903.85N

to the pipe in the opposite direction.


the mass enters the first turnaround at a
velocity of 39.0385 m/s

it goes through the first turnaround
applies a force of -3903.85 N to the pipe.

it then free floats at a velocity of 39.0385 m/s
to the second turnaround.

it then goes through the second turnaround
and applies a force of +3903.85 N to the pipe.

the forces that apply to the pipe add up as follows

+3903.85 N acceleration
-3903.85 N 1st U turn
+3903.85 N 2nd U turn
-----------
+3903.85 N

its that simple.
the pipes initial velocity was 0 m/s
the pipe has a final velocity of 7.8077 m/s

the process repeats...

the pipes velocity increases every time the mass completes a cycle.










_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#35095 - 06/22/10 04:36 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul

we now have 1 mass and thats all.

Yay

Quote:

accelerating this 1 mass results in a force of +3903.85N

500N. That's the force supplied to the mass, and equally the pipe. That force is applied for 7.8s

Quote:

it goes through the first turnaround
applies a force of -3903.85 N to the pipe.

OK. Change in momentum = final momentum - initial momentum
= -39m/s*100kg - 39m/s*100kg
= -7800 kg m/s (Ns)
This means the -3900N force was applied for 2 seconds.

Quote:

it then goes through the second turnaround
and applies a force of +3903.85 N to the pipe.

OK. So again the force of +3900N is applied to the pipe for a time of 2s.



It's meaningless to add up forces while ignoring the times they're applied for. I'll include time below...


the forces add up as follows

+500 N acceleration for 7.8s. Impulse = 500*7.8 = 3900Ns
Pipe moves forwards with p=3900 kgm/s, v=7.8m/s

-3903.85 N 1st U turn for 2s. Impulse = -3900*2 = -7800Ns
Pipe reverses direction and has v=-7.8m/s

+3903.85 N 2nd U turn for 2s. Impulse = 3900*2 = 7800Ns
Pipe reverses direction again, v=7.8m/s

I'll do the next cycle. I suppose the accelerator applies the same impulse (change in momentum) each cycle.

So we go on with ..
Impulse = +3900 kg m/s, pipe speeds up to v=15.6m/s
Impulse = -15,600 kg m/s, pipe reverses to v=-15.6m/s
Impulse = +15,600 kg m/s, pipe reverses again to v=+15.6m/s

After 2 complete cycles it's still oscillating back and forth, but going faster and faster.

Maybe it's travelling a longer and longer distance each cycle too, or maybe it isn't. Havn't worked that out.

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