Another idea to add to the mix

[rk17]

Hello to everybody on here, I've been a keen follower of the forum for a while so thought I'd jump in and offer up an idea - I'm sure it's been presented before in same shape or form, but anyway....

Imagine a wheel with weights at 12 o'clock (A) and 3 o'clock (B). When released, the wheel will naturally start turning clockwise. After half a revolution, the weights will now be at 6 o'clock and 9 o'clock. Ordinarily, this wheel would slow down pretty quickly at this point, maybe getting to, at most, 11 o'clock (B) and 8 o'clock (A). Yes?

My question is - if the weight that starts at 12 o'clock, A, were to disengage from the wheel at 6 o'clock, would the wheel keep turning with enough momentum to get weight B up to 12 o'clock? So effectively, weight A is falling from 12 o'clock through to 6 o'clock. Is this enough weight/force/momentum to propel weight B from 3 o'clock, through 6 and 9, and up to 12?

The reason I'm thinking about this is that it seems to me (a dabbler, not a scientist!!), that if weight A falls from 12 to 6 and then disengages from the wheel, it would have enough momentum to travel on a separate ramp back up to the 3/9 o'clock level. In addition, the momentum generated by its falling from 12 to 6 *could* be enough to move weight B (which does not disengage) from 3 o'clock through 6, 9 and up to 12. If this is so, then you're back where you started :-)

Apologies for the description if it's a bit verbose but I'm not too flash-hot with diagrams etc...hopefully if anyone understands what I mean they can dispel it, or even better, prove it! Like I said I'm sure the idea has probably been mooted before but hey, anything's worth a stab.

Thanks.

[DrWhat]

You mean like this?

I had time to kill :)

[rk17]

Hi, yep that's exactly what I was thinking of. I just wonder if it's workable although I won't be holding my breath.

[Fletcher]

You can draw a line from weight center to weight center - then take the mid point & connect it to the center of rotation - at that intersection is where the Center of Mass [CoM] is located - you'll see that it is at a different radius than the actual weights - the further apart the weights are i.e. further apart on the rim the closer the CoM is to the axle - the further apart the weights are also affects the Center of Gyration & because there is more twisting moment [than for close weights] more of the Kinetic Energy is rotational than translational i.e. total Ke = translational plus rotational - this doesn't effect the momentum though.

In the shots below I released the 12 o'cl yellow weight after 1 second - there was enough momemtum to get the blue weight over 12 o'cl i.e. it has residual velocity - there was insufficient momentum in the yellow weight to get it higher than 9- 3 o'cl which it would require to roll across to 3 o'cl to reset the mech - note that if I had used ramps that would have added more frictional losses - btw, no air drag was used in the model so it is practically at optimal settings.

N.B. RK17c is yellow weight at max height achieved.

[greendoor]

I believe that this basic principle could be modified to work. There are parameters that you can play with that change the maths.

You asked: "... if the weight that starts at 12 o'clock, A, were to disengage from the wheel at 6 o'clock, would the wheel keep turning with enough momentum to get weight B up to 12 o'clock?"

The big question is "How much momentum did we accumulate in that time", and "How much momentum do we need to return the flywheel and mass B to the top".

The answer to these questions are "it depends" - and so the success or failure depends on understand these parameters and engineering them to our advantage.

A HUGE variable is the mass of the flywheel. While it would seem that we want to minimise this, I believe we want to maximise it - for the following reason:

Momentum = Mass X Velocity - and it also equals Force X Time. (Specifically, an Impulse is Force applied for Time, which can result in accelerating a Mass up to Velocity, which = Momentum).

If you make the flywheel significantly more massive than the overbalancing weights, then you increase the inertia of the system. This results in the Force of gravity acting on the overbalance weight (A) being used to Accelerate the total flywheel & weight system. Force = Mass X Acceleration, which means Acceleration = Force/Mass.

What this mean - and you can see this in action - is that the Acceleration is much lower when you have a massive flywheel. You can watch the overbalance weight slowly, very slowly, build up speed, achieving maximum velocity at 6 o'clock (where you plan to disengage the weight). Most people dismiss slow velocity as having less energy, but Bessler warned that 'greed is an evil root' (translation thanks to John Collins).

If you calculate the Momentum gained in this extended period of Time - it is much greater than the momentum that could have been gained if the flywheel had zero mass and weight A could fall at maximum free-fall velocity. Much more.

Because Momentum = Force x Time. The force of gravity available never went away just because we were using it to accelerate a heavy flywheel. In fact, it was available for a much longer Time. We didn't waste it as heat or anything - we stored it in the flywheel. It's available to do work.

There is a catch if you believe that what I am proposing could be true - and this is where the nay-sayers don't get it and will attempt to prove it wrong.

If you believe that Momentum = Force X Time, then the reverse also applies when lifting weights back up. IF we lift a mass slowly, we have to oppose the force of gravity for a much longer Time also. So if you were hoping to use this surplus Momentum to lift mass B up to the top, you have to figure that because of the very slow flywheel velocity, the amount of Momentum you need to return it will be much greater as a result.

I am aware that this theory is contraversial ... the idea that Momentum available or required depends on Time. In most simple mechanical devices the Time factor nulls out. But if we purposefully engineer the Time for a Slow fall and a Fast rise, you have to factor this in.

Momentum is a conserved quantity. There are various ways of converting and transfering momentum between mass systems. A simple slingshot is an easy example to consider. We can slowly apply force to stretch an elastic material, and then suddenly release that energy in a very short period of time. We are taught to use Energy calculations for this, but I believe we can apply Impulse theory and relate it back to conserved Momentum. The Impulse applied is Force X Time. Relatively weak force applied for relatively long time. Then when it is released, we get a very strong force applied for a very short time. Impulse in = Impulse out. Momentum in = Momentum out. Sure - there are small heat & sound losses, but I believe this is relatively true.

IF you can accumulate an excess of Momentum as weights fall down, and then return them using less Momentum - you will obviously have a running wheel. I believe Bessler did it. Are we clever enough to engineer something that will work? Honestly - I don't believe most people would even consider what I am suggesting to be remotely true or possible. With that mindset, they will build all sorts of wheels that don't have a chance of succeeding. And maybe there are other valid methods that could work - but this theoretical principle seems the best chance I know of. (Credit to Pequaide for his concepts and experiments).

[Fletcher]

hmmm ... after reading greendoors comments I should add rk17 that the flywheel in the WM2D example I used had 0.0001 kg's of mass [virtually weightless] - this was to give maximum momentum & velocity to the weights - a heavier flywheel would slow everything down as greendoor says, & that means much less velocity if you disengage at or abouts 6 o'cl - naturally the velocity is required to get the weight back up a ramp to its highest vertical position possible.

If you're are wondering why the yellow mass didn't make it up to the horizontal axle line [3-9 o'cl] when there were virtually no losses in this simple approximation of your idea, it is because the blue weight went over the top [12 o'cl] with some velocity - velocity it didn't have before [it was stationary at the starting conditions] - that has to be bought & paid for & the yellow weight is the one that pays to give the blue one velocity aka energy - zero sum at best, IMO.

Obviously, you have to take everybody's opinions with a grain of salt & decide for yourself, or better still, build an experiment to prove the point.

[Jon J Hutton]

Once again guys, be careful when using wm2d. The bug is anytime you attach a round weight to a wheel and apply wind resistance you will get energy added into the system not taken away from. It is a bug that I talked to the designer about but so far has not been taken care of. Wm2d can only be trusted to simulate motion but not the physics of movement of how objects exchange forces. Sometimes it is too accurate especially when dealing with objects on an inclined plane. The only thing it is good for is taking away a headache from trying to picture in your mind too many moving parts.

JJH

[Fletcher]

Sound advice Jon - not to hard to build a table top sized wire frame [or lego] ferris wheel with battery powered electromagnetic release onto a suitably inclined ramp.

Why don't we have the balls further apart - say 12 o'cl & 5 o'cl ? - doesn't have to climb so high & then roll across eh ? - because there are no free lunches with gravity & even a near massless carrier wheel will not allow enough velocity to climb a ramp to more than the desired height to roll across & reset !

[pequaide]

You want the wheel to be heavy not light.

Construct a wheel that has a mass 20 times greater than that of the two added masses.

When the mass at 3 o’clock reaches 6 o’clock transfer all the linear Newtonian momentum of the wheel, and other mass, to the mass now at 6 o’clock: and disengage it. Then calculate how high this mass will rise.

The wheel will restart because there is now a stopped mass at 3 o’clock. Allow the wheel to reaccelerate and when the mass reaches 6 o’clock transfer all the linear Newtonian momentum of the wheel to the mass and disengage it. From its velocity calculate how high it will rise.

Formulas used: F = ma, mv, d = ½ v²/a

The wheel will accelerate just like a block on a frictionless plane being pulled by a mass draped over a pulley. The distribution of mass in the wheel is important. You get more momentum if the mass is concentrated near the circumference.

You can rearrange the distance formula (d = ½ v²/a) to find the wheel velocities at 6 o’clock.

[Fletcher]

You want the wheel to be heavy not light.

Construct a wheel that has a mass 20 times greater than that of the two added masses.

When the mass at 3 o’clock reaches 6 o’clock transfer all the linear Newtonian momentum of the wheel, and other mass, to the mass now at 6 o’clock: and disengage it. Then calculate how high this mass will rise.

RSK .. I guess you have to read pequaide's thread to see how to mechanically do that [or even part of it] - larger flywheel mass has more inertia to overcome therefore the mass arriving at the 6 o'cl release point will have less velocity by way of comparison to a light flywheel/carrier option.

[WaltzCee]

A descending mass beginning from 45 degrees could split in any proportion desired. That's not a mechanical obstacle. Masses at 12 and 3 are basically reduced to a single mass at 45 degrees.

The mass left dangling will sooner or later need to find it's way back to TDC. The mass that manages to get there by reason of momentum won't have enough energy to recover the dangling mass.

Essentially the momentum of the nondangling mass plus any gain from the dangling mass won't won't overcome that angle of the dangling mass. It has no hope of returning to TDC.

If you have no end of mass (like in the case of a water wheel) all this clarification is meaningless.

I hope this clears things up.

Hello rk17,

My boss has promised me a kegerator. Don't think poorly of me. I'm just a whore.




Walter

[rlortie]

Fletcher wrote in part;

larger flywheel mass has more inertia to overcome therefore the mass arriving at the 6 o'cl release point will have less velocity by way of comparison to a light flywheel/carrier option.

May I add that whats in the wheel can negate velocity as well! My present test wheel or drum is 69-3/4" in diameter made up of two MDF discs 1" thick.

I have just completed an experiment of an attempt to gain force from gravity to create angular motion by asymmetrical centripetal forces.

I am now tearing it down and in doing so, like a medical examiner doing an autopsy, I am weighing what I remove. I am now at 67.5 pounds and half completed. So I am looking at 135 pounds total in an attempt to guide 8 7 pound weights on a given vertical vortex path.

What have I learned in this attempt?: It did gain in angular momentum even though the 135 pounds is symmetrically balanced! Not enough rpm to build effective angular velocity although the AM was clearly present.

If I were to add an outside force, I may have what is commonly patented as a 'gravity amplifier' but we already have plenty of those.

Ralph

[WaltzCee]

reality's a bitch. You can put a pencil to it or you can experiment. Paper and pencil are cheap. So am I.

Please don't think poorly of me. I'm just trying to take delivery of a kegerator for Shelly and me. I suppose she might be the mother of my children. I hope so. With her looks and my mind they'll be the best ever.

I'm in love. Please spare me.




Walter

[rk17]

Hi all, good to get some feedback on the topic. As usual, the only real way to know will be to actually make it, so I'm going to have a go at that when I have a chance, probably a prettycrude model but hey. I appreciate you making the simulation Fletcher, but as pointed out, they shouldn't be relied upon 100%...! I agree greendoor with your statements regarding the mass of the wheel...well if I've any more ideas to add I know where to turn to!

Thanks.

[greendoor]

I also agree with Fletcher's observation that you need more velocity to return the mass upwards. This is why I believe a two-stage operation is required. It's fairly clear to me that by increasing the mass of the flywheel we slow down the ascent, and by extending the available Time, be accumulate more Momentum. But it's also very true that we can't use this Momentum directly, because it's too slow for the purpose.

IMO - unless we return the mass much quicker than it fell, we are going to burn up all that Momentum just overcoming the g-force for longer than is necessary.

This is where the second step is required. We have to convert the slow/heavy momentum into faster/lighter momentum. Which is completely legal, due to the Law of Conservation of Momentum. How to do it? Any way that works: catapult, slingshot, spring, yo-yo ... take your pick.

Because the bulk of the momentum is in the heavy flywheel, not so much in the overbalance weight, we have to extract it all from the flywheel - which means slowing it down (preferably to a dead stop). It's ok to do that, because the flywheel itself is balanced, and does not require reseting. If we can take all this momentum, convert it to faster equivalent momentum, and give THAT to the fallen mass, there seems to be no legal reason why this can't return the fallen mass - with SURPLUS.

I see very few wheel ideas that attempt to do this.