Hello Fellows

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Calloway
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Hello Fellows

Post by Calloway »

I have a question. If we have a wheel with weights evenly spaced around it's peripheral and we remove one weight from its configuration, which wheel setup supplies the most torque to the shaft? A wheel with 2,3,4,5,6,7 or 8 weights in the configuration? Or does it matter? I think this needs to be determined. We all have moved weights in and out and around the wheel in our experiments with no success. I think each weight with timing will have to be transferred to the shaft for any chance of the best torque and continuous rotation. We need the correct configuration that supplies the most torque. Your thoughts?

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agor95
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Re: Hello Fellows

Post by agor95 »

Hello Calloway

I will guess 5. At the right moment there is three on one side and two on the other.

Take the away on the two side and cancel out with the two side to see what torque is present.

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thx4
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Re: Hello Fellows

Post by thx4 »

Un seul poids suffit.
One weight is enough.
Last edited by thx4 on Fri Dec 06, 2024 5:00 pm, edited 1 time in total.
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Sam Peppiatt
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Re: Hello Fellows

Post by Sam Peppiatt »

Hi Calloway!!
Good to here from you. I don't know, seams like one would make the most difference----------Sam
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Re: Hello Fellows

Post by Kattla »

If they are equally spaced, then Thx is correct, 1 is the best option, since more (equally spaced) would make for a balanced wheel.

All 8 equally spaced would make for a better flywheel though.
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Re: Hello Fellows

Post by Robinhood46 »

The answer is two weights.
If you only have one weight evenly distributed, which means anywhere you like on the periphery, when you remove one weight you will have a perfectly balanced wheel. No torque whatsoever.

Any other number of weights than two, would mean you are only removing a fraction of the mass that is acting as a counter weight to the opposing side.

With two weights when you remove one, you remove 100% of the opposing mass to the other side and have the maximum amount of torque possible. The maximum amount of torque possible, is the weight of the removed mass at the distance from the axle.
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Re: Hello Fellows

Post by Robinhood46 »

Robinhood46 wrote: Fri Dec 06, 2024 6:59 pm The answer is two weights.
Are you sure, you bloody idiot?
I am having serious doubts since i posted that.

Surely the same logic would apply to 3 and 4 evenly distributed weights?
It is only when you go above 4 that the removal of 1 weight is less than 100% of the opposing weight.
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Tarsier79
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Re: Hello Fellows

Post by Tarsier79 »

Hi RH.

While your argument does come from sound reasoning, I think the answer is all of the above.

With one weight, you transition from maximum torque to no torque. Two weights act like you say. Both give you the same difference in torque.
If you have more weights, eg 4, you still get the same weight difference across the wheel, so the weight will apply the same torque to the wheel. With more weights, the wheel has more inertia. It will accelerate more slowly to a slower speed, but it will also be harder to stop. The wheel gains the same total KE regardless of how many weights it has. (obviously providing you remove/add the weight at the right spot)
Last edited by Tarsier79 on Fri Dec 06, 2024 7:44 pm, edited 1 time in total.
Robinhood46
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Re: Hello Fellows

Post by Robinhood46 »

I think it's the relation to the total mass, because of the % the wheel's mass changing, with the number of additional masses that threw me off with my initial answer.

It doesn't matter how many weights are on the wheel, because with the weight, or without the weight, will always be the difference of the mass of the weight. So you either have the torque of the mass of the weight, when it's not there, or you have a balanced wheel.

So i was talking nonsense. Well at least i think i was.
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Re: Hello Fellows

Post by andyblues »

funny i was pondering this question today and i did tests on a 8 compartment wheel removing one from one side creates 22.5 degrees of motion with a small amount of momentum but 5 weights and release one and you still get 22.5 degrees but more momentum i think this is because the weight is being moved from its heaviest point in the rim which makes me think this is the best place to exchange weight my thoughts are on removing from one side and transferring to the other except my wheel has one side rotating in the opposite direction which is still got loads of problems but shows a bit more potential and the possibility of creating a runner ,still scratching my head on this one but a good question i hope this helps Andy
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