Manipulating Momentum

[Kirk]

You'll agree that the informs I got are too few (where are those hammers?)

Murilo,

Step 1: 1lb weight drops 1 ft and impacts flywheel (red dot input hammer)
Step 2: Flywheel is now rotating (1lb weight is no longer attached)
Step 3: 1/4lb weight introduced at flywheel rim
Step 4: Flywheel impacts 1/4lb weight at rim (red dot output hammer)
Step 5: 1/4lb weight is thrown 16ft high

Assuming no flywheel bearing and windage loss then no special input/output timing required. Assuming complete momentum transfer is another matter.

actually the red dot was moving 4 feet per second and then was struck

[murilo]

You'll agree that the informs I got are too few (where are those hammers?)

Murilo,

Step 1: 1lb weight drops 1 ft and impacts flywheel (red dot input hammer)
Step 2: Flywheel is now rotating (1lb weight is no longer attached)
Step 3: 1/4lb weight introduced at flywheel rim
Step 4: Flywheel impacts 1/4lb weight at rim (red dot output hammer)
Step 5: 1/4lb weight is thrown 16ft high

Assuming no flywheel bearing and windage loss then no special input/output timing required. Assuming complete momentum transfer is another matter.

Ovyyus, thanks!
Since it starts from stopped position, this design is somewhat we find at Cirque de Soleil - flying guys.
Well, maybe too much optimistic!
In a second shut the moving flywheel is going to met what I said before, about how hard to get straight moving or falling in turning wheels.

Welcome, Chris!
Unfortunately I deal to language barrier.

Best!
M.

[jim_mich]

Chris, welcome!

Is there an explanation for the WM2D lockup and "internal limit" message?

There is a limit to the number of steps (32767) that WM2D can store and index. If you want to continue longer then use the 'Start from here' feature. You will loose your original start positions and data but you can then run another 32767 steps.

[Gwheel]

Welcome to the forum Chris

WM2D has its problems and can not be fully trusted but it is good for quick ideas.

I have to get back to work.

Alan

Thanks Alan. Then I'll just use it for quick (and slow) ideas as well as configuration planning.

[Gwheel]

Jim_Mich

Thank you ! Great tip Jim. I had not used that selection, glad you brought it to my attention. I tried it once after reading your post and it seems to maintain the energy state which was in existence at the moment of termination.

[Kirk]

be sure to measure the coefficient of restitution of the driver and driving mass

[Kirk]

I measured some stainless balls and their COR was around .55
O had no idea a hard metal could be so poor.

establishing COR is very important in elastic conditions
alternatively use a cup as in a Pelton Wheel - loss to angular momentum if the ball spins, havent penciled it

[Grimer]

What was the driver and what was its COR?

[Kirk]

in the literature I have seen tool steel at .97 and I saw confirmation of the stainless at .55
My mesurements could have been .52-.57 so I took .55 as the right value.
Always pays to measure rather than wonder. using .55 material would lead to disappointment

[Tarsier79]

Gday Kirk

I have read momentum vs energy threads before, but from your explanation , this is the first time I understand why there is a possible "energy gain" to be had. WM2d doesn't conserve momentum when this is simmed. I am currently attempting a build to transfer momentum. Perhaps the conservation of energy trumps momentum? I do hope you will prove one way or the other.

Cheers

[Kirk]

never trust simulations

hopefully soon. waiting on the wheel

do the maths yourself and dont rely on a programmer

[pequaide]

One has to trump the other: But F = ma is an extremely tested formula. That needs no imaginary friends like 1/2mv² needs heat. In the formula; a = v/t; so Ft = mv.

[Tarsier79]

The problem may be the inertia of the wheel and the 4:1 leverage advantage needed for the smaller weight to be thrown. The heavier weight has to be closer to the axis, and so it is disadvantaged, even if it transfers all of its momentum to the wheel, it is at that smaller radius. Due to the weight distribution of a wheel, and its rotation, you are not only fighting against a leverage of 4:1(which is required), but also the inertial difference of the wheel compared with the two weights and their position. Because a wheel is suspended by its axle, the weight of the wheel that affects it most is furthest from the axle, or its inertial properties depend on the weight distribution between the axle and the rim.

I don't think I explained myself clearly, but I hope you get what I am trying to convey. Namely, I'm not sure if you are fighting 1 battle or 2.

[Kirk]

keep it simple Tarsier. Think of it as a lever. The distribution of forces is usually easier when you think in terms of a lever

[Tarsier79]

Speaking of leverage, I did a bit of a test (unsuccessful) using a 2:1 leverage ratio, which in a perfect world would give me, using your momentum theory, a gain of 2 also. However it looked closer to achieving the conservation of energy, rather than momentum.

Instead of P=MV, if P=MV/Leverage, then that would confirm the results of my test, and also bring the two mathematical formulas into alignment.