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#35096 - 06/22/10 04:41 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Where did 15,600 kg m/s come from?

That's the change in momentum of the mass as it goes through the 1st turnaround (for the 2nd time).

Speed on entering turnaround = -39m/s + -39m/s = -78m/s
Momentum entering turnaround = -78m/s * 100kg = -7800 kg m/s
Momentum exiting turnaround = 78m/s * 100kg = 7800 kgm/s
Change in momentum of mass through turnaround = 7800 - -7800 = 15,600 kg m/s
Change in momentum of pipe as mass turns around
= -15,600 kg m/s


Edited by kallog (06/22/10 04:46 AM)

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#35098 - 06/22/10 11:42 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
This means the -3900N force was applied for 2 seconds.


NO !!!!

the mass has a final velocity of 39.038 m/s when it enters the turnaround after being accelerated.

the turnaround would have to have a lenght of
39.038 meters * 2 = 78.076 meters
for the mass to consume 2 entire seconds while passing through it.

so your wrong.

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


and its even less meaningless to include erroneous times , Im not sure why your so concerned with time as the time is given in the acceleration of the mass.

ie...5 meters / SECOND / SECOND...

meaning each second..

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


maybe that is where your insertion of 2 seconds will come in !!!

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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#35100 - 06/22/10 12:57 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
NO !!!!


Yes. It's your design. You specified 3900N. That requires 2s, and 78m. If you want a shorter turnaround then you need to choose a higher force to get the job done faster. The value of this force is arbitrary, it doesn't change the outcome. That might sound strange, but any change in the force has to be compensated for by applying it for a different length of time so you end up with the same change in momentum.


Quote:

Im not sure why your so concerned with time as the time is given in the acceleration of the mass.


Have a free floating stationary block of 1kg. Push it with 10N for 1s. Then run round the other side and push it in the opposite direction with 10N for 1 hour. The block ends up stationary of course, because the two opposite forces balance, right?



Quote:

maybe that is where your insertion of 2 seconds will come in !!!

As I said before, there's no point going into other details unless you understand what that factor of 2 is for. Without it, the machine will indeed fly. Why don't you try to understand it? Don't have to listen to me, go check google or ask someone else. Or just look at this:
Answer to a physics problem.



Edited by kallog (06/22/10 01:00 PM)

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#35101 - 06/22/10 01:43 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
Yes. It's your design. You specified 3900N. That requires 2s, and 78m. If you want a shorter turnaround then you need to choose a higher force to get the job done faster. The value of this force is arbitrary, it doesn't change the outcome. That might sound strange, but any change in the force has to be compensated for by applying it for a different length of time so you end up with the same change in momentum.



ok , try this experiment out.
get a U shaped pipe , fire a riffle in one end of the pipe.
dont worry you have alot of time to walk out of the way before the bullet comes back towards you.
NO WAIT NOT REALLY , I mean DONT TRY IT !!!

LOL.

I must admit that your logic isnt logical.

the 2 seconds that you are using is a complete falsity.

suppose the mass were attached to a wheel , are you somehow sudgesting that a mass of 100 kg cannot have a velocity of
39.03 m/s given it is attached to the periphery of a wheel with a radius of 4 meters ? or 3 meters? or 2 meters?
or 1 meter?

that somehow the mass cannot travel that fast around a radius?

thats really going a bit to far kallog.

but you knowing that the requirement of 2 seconds in each turnaround would cause the concept to not work , so is that your scientific or your childish way of finding a solution.

to your problem.

when I say Your problem , I mean your inability to find a reason that the concept would not work.

because afterall that is what you are trying to do , you just cant do it because physics keeps getting in the way , spoiling your attempts.










_________________________
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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#35103 - 06/22/10 02:08 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
************** the second cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 39.0385 m/s
its final velocity is 52.193 m/s
its average velocity is 57.930 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.6309 seconds

its braking force is 5219.3 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 7.8077 m/s
final velocity is 10.4386 m/s
average velocity is 11.586088965 m/s

the final pipe velocity is 10.4386 m/s

the pipe has now moved a distance of 60.961 meters

the braking force of the pipe is 5219.3 N

**************** the third cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 52.193 m/s
its final velocity is 63.736 m/s
its average velocity is 66.0178799 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.3086 seconds

its braking force is 6373.6 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 10.4386 m/s
final velocity is 12.7472 m/s
average velocity is 13.203 m/s

the final pipe velocity is 12.7472 m/s

the pipe has now moved a distance of 91.443 meters

the braking force of the pipe is 6373.6 N


with each cycle the mass will be accelerated faster and faster
which in turn will accelerate the pipe faster and faster.

using only 1 mass is stupid but it seems to allow for your understanding
that the pipe will move as I stated , using a constant acceleration of masses
would be the ideal method to accomplish a smoother acceleration.

WOW , just think kallog in a few more cycles the mass will
require less than 2 seconds for acceleration , and your 4 second requirement for the mass to pass through the 2 turnarounds would make the concept , non working...

must be nice to just pick a time requirement out of thin air to use in your favor in a discussion.




_________________________
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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#35104 - 06/22/10 08:11 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
I found a flaw in my calculations , that gave incorrect numbers above.
the flaw is now corrected and I am posting the corrections.

************** the second cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 39.0385 m/s
its final velocity is 55.209 m/s
its average velocity is 47.12375 m/s

the time it takes the mass to accelerate the 152.4 meters is 3.2341 seconds

its braking force is 5520.9 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 7.8077 m/s
final velocity is 11.0418 m/s
average velocity is 9.42475 m/s

the final pipe velocity is 11.0418 m/s

the pipe has now moved a distance of 60.96067362 meters

the braking force of the pipe is 5520.9 N

**************** the third cycle ***********************

the 100 kg mass is accelerated at 5 m/s/s with the 500 N force.

for a distance of 152.4 meters.
-----------------------------------------
the mass...
-----------------------------------------
its initial velocity is 55.209 m/s
its final velocity is 67.617 m/s
its average velocity is 61.413 m/s

the time it takes the mass to accelerate the 152.4 meters is 2.4816 seconds

its braking force is 6761.7 N
-----------------------------------------
the pipe...
-----------------------------------------
mass = 500 kg
force is 500N
initial velocity is 11.0418 m/s
final velocity is 13.5234 m/s
average velocity is 12.2826 m/s

the final pipe velocity is 13.5234 m/s

the pipe has now moved a distance of 91.44117378 meters

the braking force of the pipe is 6761.7 N


with each cycle the mass will be accelerated faster and faster
which in turn will accelerate the pipe faster and faster.

using only 1 mass is stupid but it seems to allow for your understanding
that the pipe will move as I stated , using a constant acceleration of masses
would be the ideal method to accomplish a smoother acceleration.

WOW , just think kallog in a few more cycles the mass will
require less than 2 seconds for acceleration , and your 4 second requirement for the mass to pass through the 2 turnarounds would make the concept , non working...

must be nice to just pick a time requirement out of thin air to use in your favor in a discussion.

using 20 sets of tubes with 1 mass in each

in 1 year the pipe could travel a distance of
5.2 trillion miles.

using 40 sets

in 1 year the pipe could travel a distance of
10.43 trillion miles.

hey thats over 330,000 miles /second.

The speed of light is 186,000 miles per second

which can cut the 1 year travel time of
5.2 trillion miles in half to 6 months.













_________________________
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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#35105 - 06/22/10 09:52 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
that is the CHANGE IN MOMENTUM OF THE BALL kallog.

the ball still presents the same force to the wall.
the force that is its mass * velocity.
or your used to using p=mv
the momentum that is its mass * velocity

just like the mass still presents the same force to
the U turns.



its the elastisity of the ball that allows it to bounce
off the wall.

like compressing a spring.
the ball is compressed like a spring because of its
momentum.





_________________________
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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#35106 - 06/22/10 10:19 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
I must admit that given this new turn of events , I would have to say that I think we could travel to other solar systems using current energy technologies , that is if we
remain too stupid to use free energy sources.

_________________________
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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#35107 - 06/22/10 11:52 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Quote:

the 2 seconds that you are using is a complete falsity.


How long do you think it'll take? Here's a hint. F=3900N, m=100kg, initial speed = 39m/s, F=ma. Any ideas?

Or rather than wasting time, why don't we just change the arbitrary 3900N force that you chose. Make the turnaround force 100,000N and it'll take much less than 2s.



Quote:

because afterall that is what you are trying to do , you just cant do it because physics keeps getting in the way , spoiling your attempts.


I've noticed that every single insult you fire at me is actually describing yourself, not me. Where are your equations? I mean the ones using physics, not just made up. F=mv is just made up.

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#35109 - 06/23/10 02:33 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
kallog

Quote:
How long do you think it'll take? Here's a hint. F=3900N, m=100kg, initial speed = 39m/s, F=ma. Any ideas?


if the mass is traveling at a velocity of 39 m/s
and the U turn is 39 meters long , it will take
exactly 1 second.

if the mass is traveling at a velocity of 39 m/s
and the U turn is 19.5 meters long , it will take
exactly 1/2 second.

if the mass is traveling at a velocity of 39 m/s
and the U turn is 9.75 meters long , it will take
exactly 1/4 second.

picture a car on a hill beside a U shaped valley , the car coast down the hill but the car begins to coast up the other hill.

the other hill is the same size and shape of the first hill.

if resistance and friction are zero , the car could coast
up to the top of the other hill.

but will the car need assistance from the hill to coast
up to the other hill?
if so and given that the hill itself will not move and provide a force that would assist the car up the hill
how then can the hill assist the car?

the momentum of the car causes the car to go down the first hill and up the second hill.

the car lost all of its momentum in doing so.

just like the momentum of the mass causes the mass to go half way through the U turn and then through the other half of the U turn.

except the mass has a veocity before it enters the U turn.

now suppose the car was dropped from a
height of 500 ft, the cars momentum would
be enought to send the car down one hill
and up the other hill , and all the way
back up to the altitude it was first dropped from.

Quote:
F=mv is just made up.


nope , its used in hydraulics all the time when describing a fluid pressure that acts against a piston.
there is no time added because the time is itself added

via a=f/m

ie.. apply a force of 10 lb
against a 1 inch hydraulic piston and you get a force
of 100 lb from a opposing 10 sq inch piston.

the F=mv is used to calculate the force that the piston
itself can place on the fluid because of its
mass * velocity.

a lighter piston = less force
a heavier piston = more force

a lighter piston less force required to move the piston itself.
a heavier piston more force required to move the piston itself.

some people work with pistons that weigh several thousand pounds.

thus f=mv


_________________________
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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#35110 - 06/23/10 03:48 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

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

its final velocity is 52.193 m/s
its braking force is 5219.3 N


That force is chosen arbitrarily. It need not be 5219.3N. And in fact it cannot be now that you demand <<2s turnaround.

a=F/m
a = 5219.3N / 100kg
a = 52.2 m/s^2
At that acceleration/deceleration, how much time do you think it'll take to stop or turn around?

You must specify a higher braking force.

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#35111 - 06/23/10 04:03 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
that is the CHANGE IN MOMENTUM OF THE BALL

BINGO!


Quote:

the force that is its mass * velocity.

I don't care where you got this equation from or what other applications it's used in. IT IS AUTOMATICALLY WRONG. Why? Because the dimensions are inconsistent:
F = 100kg * 40m/s
F = 4000 kg m/s

Is "kg m/s" a unit of force? Can you convert it to newtons?
You can check with these google searches:
"4000 kg m/s in N"
"4000 kg m/s^2 in N"





Here's another reason it can't be right:
You claim:
F=ma
F=mv and
p=mv

Therefore:
F=p
ma=mv
a=v

Acceleration is always equal to velocity??



Quote:

or your used to using p=mv
the momentum that is its mass * velocity

At least we agree on something.

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#35112 - 06/23/10 04:06 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
remain too stupid to use free energy sources.


No, magic carpets are much better than free energy. Sadly whenever somebody flies a magic carpet, the CIA shoots them. That's why we find bodies dropping out of the sky from time to time.

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#35115 - 06/23/10 11:34 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:

a=F/m
a = 5219.3N / 100kg
a = 52.2 m/s^2

At that acceleration/deceleration, how much time do you think it'll take to stop or turn around?


if the turn around is 52 meters long it will take exactly
1 second to turn around because its velocity is 52m/s.

if the turnaround is shorter it will take less time.

if you apply a force of 5219.3N for 1 second you can stop it in 1 second and 1 meter distance.

applying less force will not stop it as fast and will also require more stopping distance.

applying more force will stop it faster and in less stopping distance.





_________________________
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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#35116 - 06/23/10 12:01 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

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

applying less force will not stop it as fast and will also require more stopping distance.

applying more force will stop it faster and in less stopping distance.


Yea. So if we need <<2s turnaround time we have to use a higher force. 5219N takes 2s and 104m to turn the mass around.

100,000N?

a=F/m
a=-100,000N / 100kg
a=-1000m/s^2

time = (change in v) /a
time to stop = -52.2m/s / (-1000m/s^2)
time to stop = 0.0522s

distance to stop = -1/2 a t^2
distance to stop = 1.36m

Equations from equations of motion

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#35117 - 06/23/10 12:06 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
Yea. So if we need <<2s turnaround time we have to use a higher force. 5219N takes 2s and 104m to turn the mass around.


Kallog , its not STOPPING !!!!

why cant you GRASP that ?

no force is REQUIRED to TURN IT AROUND
_________________________
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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#35118 - 06/23/10 01:23 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Suppose a car's coasting towards you at 100km/hr.

You can turn it around using no force. So you can reach out your hand, grab the wing mirror, and the car will spin round and head off in the opposite direction as you let go at the right time.

F=ma. Please look up Wikipedia or a high-school general science textbook instead of saying made up things.


Originally Posted By: paul

Quote:

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?




Edited by kallog (06/23/10 01:27 PM)

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#35119 - 06/23/10 01:50 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
I think it's starting to come together now.

The recurring problems you have would be consistent with treating velocity and momentum as scalars. Maybe you don't realize they're vectors? F=ma and p=mv are actually a vector equations. They can give wrong results if you only put the magnitudes of the quantities into them.


Can an object accelerate if its speed remains constant?

If an object reverses direction, must its momentum change?

If an object reverses direction, must its velocity change?

If an object reverses direction, must its speed change?

Is it possible to apply an unbalanced force to a moving mass without changing its speed?

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#35120 - 06/23/10 02:07 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Originally Posted By: paul
I have a clear understanding of that , and physicist also use decelerate to describe using a force to decelerate a object.


which object is being decelerated then kallog?

just tell me which object you think is being decelerated and what causes it to decelerate.

ie..where does the force come from that decelerates the object?

at what velocity does the object have the entire time it is decelerating.

what is the objects final velocity when it has finished decelerating?

if you are speaking about the mass as the object that is being decelerated.

1) how can a object declerate without changing its velocity.
2) there is no 2


the change in momentum you are talking about is nothing but
a change in the direction of the object.

that change in direction does not have anything to do with
the momentum of the pipe.


the force that the object presents to the pipe as it presses against the pipe while turning around is the only force that the object can possibly apply to the pipe while turning around.

and it will present a force that is exactly its
mass * its velocity.

How Force is related to Momentum

Momentum measures the 'motion content' of an object, and is based on the product of an object's mass and velocity. Momentum doubles, for example, when velocity doubles. Similarly, if two objects are moving with the same velocity, one with twice the mass of the other also has twice the momentum.

Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum. Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). Since acceleration is the change in velocity divided by time, you can connect the two concepts with the following relationship:

force = mass x (velocity / time) = (mass x velocity) / time = momentum / time

Multiplying both sides of this equation by time:

force x time = momentum



_________________________
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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#35121 - 06/23/10 02:41 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

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

just tell me which object you think is being decelerated and what causes it to decelerate.


I wouldn't use the term "decelerate" here because it's a bit vague. I would use "accelerate". The mass going through the u-bend is accelerating all the way through.



Quote:

at what velocity does the object have the entire time it is decelerating.

While it's turning around its velocity is continuously changing from, say 39m/s to the left, through 39m/s upward, till finally reaching 39m/s to the right. Its speed remains the same throughout.

Quote:

1) how can a object declerate without changing its velocity.

It can't. The velocity must change. Therefore, by definition, it is accelerating.


Quote:

the change in momentum you are talking about is nothing but
a change in the direction of the object.

Yes a change in direction. Which is a change in momentum. But not a change in the _magnitude_ of momentum.


Quote:

the force that the object presents to the pipe as it presses against the pipe while turning around is the only force that the object can possibly apply to the pipe while turning around.


Of course. That's the only force I've been talking about all the time, as well as its equal and opposite reaction. What force were you thinking of?


Most of what I just said will appear self-contradictory if you consider 'velocity' and 'speed' to be the same thing.

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