Manipulating Momentum

[Kirk]

Ovyyus - m1v1=m2v2
momentum is preserved

[ovyyus]

Yes, momentum is preserved. The dropped 1 lb weight transfers all it's momentum to the wheel and the 1/4 lb weight. Therefore all of the momentum of the dropped 1 lb weight isn't transferred just to the 1/4 lb weight.

Maybe I have it wrong. Is the 1/4 lb weight introduced after the dropped 1 lb weight has hit and transferred it's momentum to the wheel?

[nicbordeaux]

Nick, a 1 lb weight dropped 1 ft will hit a target at 8 fps. A 1 lb weight travelling at 8 fps has the same momentum as a 1/4 lb weight travelling at 32 fps. A 1/4 lb weight travelling at 32 fps can be thrown 16 ft high.

But how can we transfer the momentum of a 1 lb weight moving at 8 fps to a 1/4 lb weight so that it moves at 32 fps? Kirk's wheel lever proposes to transfer all of the momentum of the dropped 1 lb weight into wheel momentum and momentum of the 1/4 lb weight. The question is can the transaction deliver enough momentum to the 1/4 lb weight to throw it higher than the break-even point of 4 ft?

Bill, you have become a momentumist convert ;)

The hassle is the inertia or moment of inertia. The force required to set the device, whatever it may be, in to rotation. Stopping and starting doesn't seem too hot. But maybe Kirk has done a full 360° with no intervention, manual or divine.

I honestly wish Kirk every success, but still don't see it working. As to posible full transfer of momentum (= 100% efficiency transfer of energy) if we are talking impact, no two ways about it: suare imacts square. If one object is four times heavier, either it needs to be 4 x denser to be a square, or 4 times longer with an identical frontal area as the small square. An oblong. If impacting a lever, still a square weight impacting a square plate on lever.

If it could be done as described, somebody would have done it. Even if by accident.

[ovyyus]

Nick, I think you're making it overly complicated with issues of reset, weight shape, etc.

The action is simple. Transfer momentum of the dropped 1 lb weight to rotate the wheel mass. Transfer momentum of the rotating wheel mass to accelerate the 1/4 lb weight. How easy was that :)

[pequaide]

Nick: Stopping a large mass with a small mass has been done; you did it yourself with the bicycle wheel wrapped with a string. NASA has done it. I have done it thousands of times, with dozens of devices. I don’t know if it can be done with bouncing; but it has been done.

I do not feel comfortable with a bounce drive; so I modified the cause of the starting motion.

Kirk: You say you can throw it up 16 ft.

Okay let’s mount two pulleys 16 feet apart, one above the other. Let’s connect them with a light chain that has cups every foot along the chain. That would give us 16 of your ¼ pound masses placed in the 16 cups along the chain. The 1/4 pound masses are only on one side of course so this gives us 4 pounds of drive that can be dropped one foot; and only one ¼ pound mass is displaced 16 feet.

Now lets drop the cups on the chain one foot: to get 4 ft per second out of the bottom pulley the rate of acceleration must be 8 ft per sec per second. This is one fourth standard gravitational acceleration. This means that the bottom pulley can have a mass three times that of the accelerating mass on the chain. This would be 12 pounds; but there is a correction to be made here.
The bottom pulley is mounted in the center of a rim mass wheel that has a diameter 4 times that of the pulley. You strike the circumference of this wheel with a ¼ pound ball that is moving 16 ft /sec. this velocity will cost you 3.75 feet of drop to accelerate it from your 4 ft per second (on the chain) to 16 feet per second. That leaves you with 12.25 (¼ pound) weights on the chain to accelerate the bottom wheel. Three times this (12.25 * ¼) 3.06 lbs * 3 = 9.18 lb for the mass of the pulley.

This will give you a mass of 9.18 pounds for the rim of the pulley or 2.295 pounds for the rim of the wheel. So this is 2.295 pounds moving 16 feet per second. The chain (3.06 lbs) is still connected but it is moving only one fourth as fast, for .765 pounds moving 16 feet per second. This gives us a total of 3.06 pound moving 16 feet per second.

But I do not know what will happen when a 3.06 pound elastic bat (wheel) hits a ¼ pound elastic ball when both are moving 16 ft/sec. I know that a bat can stop in your hands when you hit a baseball just right.

[nicbordeaux]

Nick, I think you're making it overly complicated with issues of reset, weight shape, etc.

The action is simple. Transfer momentum of the dropped 1 lb weight to rotate the wheel mass. Transfer momentum of the rotating wheel mass to accelerate the 1/4 lb weight. How easy was that :)

As easy as flinging a ball off a wheel on a tether, except your driver has freefall gravity acceleration (if I get the idea right) before transfer. In which case you are looking at a wheel as light as possible. Not as heavy as possible. 100 % transfer is impossible in a collision scenario, otherwise the impact would be noiseless. It never is, is it ?

The fling stuff we played around with does the trick if (to be followed), it's just that we omitted something so patently obvious that it looks silly.

Anyway Bill, nice to see you playing the devil's advocate :)

I'll try to refrain from further comment on this topic and let it reach it's natural conclusion.

[Kirk]

Yes, momentum is preserved. The dropped 1 lb weight transfers all it's momentum to the wheel and the 1/4 lb weight. Therefore all of the momentum of the dropped 1 lb weight isn't transferred just to the 1/4 lb weight.

Maybe I have it wrong. Is the 1/4 lb weight introduced after the dropped 1 lb weight has hit and transferred it's momentum to the wheel?

the 1 pound weight transfers its momentum to the wheel. The wheel then transfers the momentum to the smaller mass

[Kirk]

Nick, a 1 lb weight dropped 1 ft will hit a target at 8 fps. A 1 lb weight travelling at 8 fps has the same momentum as a 1/4 lb weight travelling at 32 fps. A 1/4 lb weight travelling at 32 fps can be thrown 16 ft high.


If it could be done as described, somebody would have done it. Even if by accident.

and you are sure of this why? Just how many rotating machines fling mass?

[Kirk]

Nick, I think you're making it overly complicated with issues of reset, weight shape, etc.

The action is simple. Transfer momentum of the dropped 1 lb weight to rotate the wheel mass. Transfer momentum of the rotating wheel mass to accelerate the 1/4 lb weight. How easy was that :)

As easy as flinging a ball off a wheel on a tether, except your driver has freefall gravity acceleration (if I get the idea right) before transfer. In which case you are looking at a wheel as light as possible. Not as heavy as possible. 100 % transfer is impossible in a collision scenario, otherwise the impact would be noiseless. It never is, is it ?

The fling stuff we played around with does the trick if (to be followed), it's just that we omitted something so patently obvious that it looks silly.

Anyway Bill, nice to see you playing the devil's advocate :)

I'll try to refrain from further comment on this topic and let it reach it's natural conclusion.

wheel should have at least 100 times the momentum of the driving mass.

[Kirk]

Nick, a 1 lb weight dropped 1 ft will hit a target at 8 fps. A 1 lb weight travelling at 8 fps has the same momentum as a 1/4 lb weight travelling at 32 fps. A 1/4 lb weight travelling at 32 fps can be thrown 16 ft high.

But how can we transfer the momentum of a 1 lb weight moving at 8 fps to a 1/4 lb weight so that it moves at 32 fps? Kirk's wheel lever proposes to transfer all of the momentum of the dropped 1 lb weight into wheel momentum and momentum of the 1/4 lb weight. The question is can the transaction deliver enough momentum to the 1/4 lb weight to throw it higher than the break-even point of 4 ft?

Bill, you have become a momentumist convert ;)

The hassle is the inertia or moment of inertia. The force required to set the device, whatever it may be, in to rotation. Stopping and starting doesn't seem too hot. But maybe Kirk has done a full 360° with no intervention, manual or divine.

I honestly wish Kirk every success, but still don't see it working. As to posible full transfer of momentum (= 100% efficiency transfer of energy) if we are talking impact, no two ways about it: suare imacts square. If one object is four times heavier, either it needs to be 4 x denser to be a square, or 4 times longer with an identical frontal area as the small square. An oblong. If impacting a lever, still a square weight impacting a square plate on lever.

If it could be done as described, somebody would have done it. Even if by accident.

like an atomobile you start it first. Once spun up to operating speed you drop 1 pound ball then a quarter pounder ad infinitum. Where is this square stuff coming from? we are using tool steel balls with a mass ratio of 4:1 because the wheel is 4:1

Ebroider this and hang it on the wall:
m1v1=m2v2

[jim_mich]

Ebroider this and hang it on the wall:
m1v1=m2v2

Also:
1/2×m1×v1^2 ≠ 1/2×m2×v2^2

[Kirk]

In a closed system both energy and momentum are conserved, i.e., either one can be altered in its configuration but THE TOTAL IS CONSTANT,

According to Einstein, not only is momentum conserved -- and not only is energy conserved, BUT THE TOTAL OF MOMENTUM AND ENERGY IS CONSERVED. He stated that "whereas heretofore the conservation of momentum and the conservatrion of energy were considered two separate laws, they are now one: the conservation of momentum and energy.

but we see two equivalent momentums wth radically diferent ke
I notice Einstein said momentum and not ke

[Fletcher]

It's quite simple gentlemen & this is how you proceed to prove your theories ...

Use whatever mechanisms you want.

Measure the Input Energy [either Pe or Rotational &/or Linear Ke].

Measure the whole of system Pe before & after interaction has occurred.

If there is a gain Energy i.e. in whole of system Pe plus any Rotational/Linear Ke present, less your Input Energy [Pe or Ke], then you have proven that gravity is NOT conservative & momentumists RULE, ye-ha !


You will also have turned the Work Done Energy Equivalence axiom on its head !

This may be harder to empirically prove than you think because Joules [Nm] & Watts [J/s] are used to calculate capacity to do Work & Power - momentum is an all together unrelated unit of mv & is currently NOT the unit used for Work & Power etc in mechanics - so you'll have to come up with new units related to mv !

=> note from a die hard advocate of the current proven regime !

[pequaide]

I have stopped wheels that have mass ratios of over 40 to 1; so I see no reason why 100 to 1 would not work. Actually the velocities get to high; and air resistance becomes a big factor.

Nick: did you hear the weighted string unwinding from the bicycle rim? There might be a whirr from its high velocity but that has nothing to do with the momentum transfer. Your bicycle wheel was more massive than the small weigh; wasn’t it.

[Fletcher]

Taken from the thread called "Energy producing Experiments".

While I'm killing time for the next few minutes.

I might as well show how Kinematic equations can be used to find Energy [Capacity to do Work].

A weight of 50 newtons [50N's = 50/9.80665 = 5.01 kg, say 5 kg] is raised a height of 2 meters - it has gained additional Pe of 100 Joules i.e. [Pe = mgh = 5.01 x 9.81 x 2 = 100 Joules].

100 J was the Work Done to raise the mass to give it extra Energy of Position.

This is the Work it should be able to do returning to its original position.

N.B. Work Output = Work Input [assuming no losses] or Work Output < Work Input [with losses] - this is the Leverage Work Done equi-potentiality of F1 x D2 = F2 x D2.

Q. What is the Ke of a mass of m kg moving at v m/s ?

Ans. Let us assume that gravity accelerated it uniformly at a m/s^2 from zero velocity to v m/s^2 by being pushed by a constant Force of F Newtons.

The distance traveled during acceleration was s meters, then the Work Done i.e. the Ke, will be in Joules.

But v^2 = u^2 + 2as [u =0]
=> v^ = 2as
=> s = v^2/2a

N.B. => v^2 = 2as => v = sqrt [2as] = sqrt 40 = 6.32 m/s

BUT F = ma

SO ... Ke = Fs = ma x v^2/2a = 1/2mv^2 Joules

Thus Ke of 5 kg moving at 6.32 m/s after 2 meters

= 1/2 x 5 x 6.32^2

= 100 Joules.

Whilst waiting for an empirical demonstration & then retrofitting the math try the slam dunk shot by substituting your own units & values & reconciling the Kinematic Equations using Momentum to find the Ke & Capacity to do Work, assuming Uniform Acceleration from a Constant Force !