Manipulating Momentum

[ovyyus]

Duh

[ovyyus]

Kirk, is the attached diagram close to what you are describing?

The lever is assumed to be perfect and lossless. The red weight indicates your proposed momentum transfer gain over and above that normally expected (grey weight position).

[Kirk]

not sure what you mean by momentum gain. There is no gain in momentum as that would mean you added energy to the system and you dont. You put an mv in slow and take that same amount of mv out at a higher velocity but momentums are equivalent. Input 6 inches from axis or rotation taken out 24 inches from axis of rotation.
I dont think Besslers method was as efficient as I think I know what he was doing restricted to the confines of 1 wheel. Write more when I get back.

[ovyyus]

Sorry, I meant to say momentum transfer height gain.

Attached is a simpler version using steel balls dropped onto a rim anvil. The average weight pair COM has increased after the proposed transfer. Does this describe your idea?

[Kirk]

Draw a 4 inch circle. In the center of that draw a 1 inch circle. Place a dot on the 1 inch circle. Label that input. Put a dot on the rim. Label that output. 1 inch represents a foot. Drop a 1 pound ball 1 foot and hit the inner dot. Introduce a 1/4 pound ball to the dot on the rim. Watch it fly.

[ovyyus]

OK, 3rd time lucky. Is this it?

[Kirk]

you have it

[nicbordeaux]

Well, what are the considered opinions of our experts ? Jim, Bill, Fletcher, Wubbly and others ;)

[Bill_Mothershead]

Maybe the wheel is confusing people.

Certainly we have all seen the case where
TWO circus acrobats jump down onto one side of a lever
and the ONE acrobat on the other side of the lever gets thrown
high into the air.

If the distances on the lever were just right,
perhaps the little ball could suck up all the momentum
of the big heavy ball such that the big ball comes to rest.

[FunWithGravity2]

IMHO the issue with this and several other "momentum transfer" theories currently is the "complete" transfer. Why does everyone feel that their needs to be a complete transfer? Why does the device imparting the momentum need to stop ?


Dave

[Kirk]

Maybe the wheel is confusing people.

Certainly we have all seen the case where
TWO circus acrobats jump down onto one side of a lever
and the ONE acrobat on the other side of the lever gets thrown
high into the air.

If the distances on the lever were just right,
perhaps the little ball could suck up all the momentum
of the big heavy ball such that the big ball comes to rest.

It does. Just like a Pelton wheel does

[Grimer]

Yep, the Pelton wheel does indeed suck up the Jerk (change in acceleration) energy so anything based on that could open the way to PM.

[Kirk]

IMHO the issue with this and several other "momentum transfer" theories currently is the "complete" transfer. Why does everyone feel that their needs to be a complete transfer? Why does the device imparting the momentum need to stop ?


Dave

efficiency

[murilo]

OK, 3rd time lucky. Is this it?

If the distance of 1Lb to center is equal to 4X center to 1/4Lb the system will find equilibrium...
From this view, you can make counts for to establish non equilibrium at a side or to the other.
Best!
Others

[ovyyus]

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?