Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[stgpcm]

The top weight, all of it's force acting upon the wheel, can have it's torque calculated in the standard manner, 1 unit mass, acting on the wheel, calculate the component of the force due to gravity perpendicular to the line between the pivot and the center of mass of the weight = 1 . cos (60) = 1/2, multiplied by the distance from the pivot = 1 . 1/2, = 1/2 (or 0.5)

The method I believe you've been taking to caluculate the torque for the lower one, is to resolve it's interaction with the guide (which as the guide is vertical has no effect), and then use resolve that resultant with the wheel. This would produce a resultant of 1/2, multiplied but the distance 9/10, giving a CCW toque of 0.45. I may have misunderstood your method, but you won't work my simple example, so I can't tell.

Instead, as the weight is free to side on the wheel slot, you instead have to calculate a slightly more complex system -  force (f) applied to the weight by the slot (not the wheel, yet),  must have a vertical component sufficient to match the force due to gravity. (the horizontal component, as you rightly say, is leached away by the guides, and used to push the earth sideways, just a teeny bit). For this case,  f . sin(30) = 1, so f . 0.5 = 1, f = 2. If this force "f" is required by the slot to hold the weight against gravity, this must be the force the weight (due to gravity) is applying to the slot. So the torque is 2 . 9/10 = 18/10 or 1.8 CCW.

The system has a net torque of 1.3 CCW, so would rotate Counter Clockwise.

I hope this helps.

[Omnibus]

Not at all. The only force that causes torque is derivable solely from the unit force due to the given mass. Slots and guides cannot create mass to derive torque creating force from it.

The guide in this case is perpendicular, so the torque-creating force is of the magnitude and direction as it would be without a guide.

All this is in absence of forces of friction.

The above (considering the guides as some kind of mass generating entities) was a mistake @mondrasek was making but I think he understood it in the end.

[AquariuZ]

Question: Will you sell the energy yourself?
Initially that was what we had in mind, but this is no longer our goal. We have been advised to sell our systems to the current energy providers instead for the production of "beautiful energy" (electricity) as a substitute for power plants, wind turbines etc.

DAMMIT

I am going to call him again.  Selling is the same in this case as shelving.

GRRRRRRRRRRRRRRR %$$%**!

Edit: sorry here is the link with the newly posted information on Abelings website:
http://mooieenergie.nl/index.php/en/component/content/article/3-gewichtenergiecentrale/9-nieuwsbrief-mei-2009

Sjack, sjack, sjack. You know what´s going to happen with your invention don´t you?

[mondrasek]

Not at all. The only force that causes torque is derivable solely from the unit force due to the given mass. Slots and guides cannot create mass to derive torque creating force from it.

The guide in this case is perpendicular, so the torque-creating force is of the magnitude and direction as it would be without a guide.

All this is in absence of forces of friction.

The above (considering the guides as some kind of mass generating entities) was a mistake @mondrasek was making but I think he understood it in the end.

Not at all Omni.  You are still making the same mistake.  Even Eisenficker2000  confirmed the mistake in his (and your) original method and agreed that the way I explained and used is correct.  That was in this thread.  If you did not understand that you may re-read it.

The slots change the angle that the weights lean against the wheel.  So the angle of the slots may cause more of the weight to lean against the guides which are attached to the Earth.  The component that is pressing against the guides/Earth we agree can not cause rotation.  So the resultant force that *can* cause rotation is dependant on the angle of the slots.  You are not taking this into account.

Please realize that I will never agree with you that your current methods are correct.  That would be like asking me to agree that 1+1+1=2.  There is only one correct answer.  I am sorry that you have not been able to understand that the method you are employing is flawed.  And it is not just my opinion.  If you cannot trust that myself and Eisenficker2000 are now using the correct method, maybe you could discuss this with a local physics or mechanics instructor.

[stgpcm]

Yes this is without friction.

This is the answer found by people that build the wheel.

This is the answer wm2d gives you.

But you disagree with my static analysis?

OK, I'll do a dynamic analysis.

I will assume, as you believe, the system would rotate clockwise, I'll rotate the system one degree from angle a=29.5 to angle a=30.5. I will call the height of the pivot P

for the top weight, due to the inertialess catch (or the shape of the slot) the distance from the hub is constant. cos (a) is height/distance, so height = P + cos(a)/1

I'm adding, because the weight is above the Pivot

for the bottom weight, the vertical guide keeps the horizontal distance of the center of the mass from the hub a constant (in this case 0.45m), so the height of the mass is determined by tan(a) = 0.45/height, so height = P - 0.45/tan(a)

I'm (EDIT:typo) subtracting because the weight is below the Pivot

travelling from 29.5 to 30.5 degrees, the difference in height of the top weight would be END - START
(P + cos(30.5)) - (P + cos(29.5)),
= cos(30.5) - cos(29.5)

the difference in height of the lower weight would be
(P - 0.45/tan(30.5)) - (P - 0.45/tan(29.5)),
= 0.45 . (1/tan(29.5) - 1/tan(30.5))

at this point, I need to use my five figure tables, or a calculator. For speeds sake, I'll use a calculator

the difference in height for the top weight is

cos(30.5) - cos(29.5)
= 0.86162916044152574545106204815061 - 0.87035569593989968041595026212419
(quick check, the end is higher than the start... yes)
= -0.0087265354983739349648882139735844
(quick check, the object has got lower... yes)

the difference in height for the bottom weight is
0.45 . (1/tan(29.5) - 1/tan(30.5))
= 0.45 . (1/0.5657727781877700776025887010584 - 1/0.58904501642055107438483608526909 - )
= 0.45 . (1.7674940162428909073599175537677 - 1.6976631193260889100263622699454)
(quick check - START (P - 1/tan(29.5) is further below the pivot than END (P - 1/tan(20.5)... yes)
= 0.45 . 0.069830896916801997333555283822369
= 0.031423903612560898800099877720066
(quick check, the weight has gone up)

so turning clockwise the top weight has converted ~0.0087 potential energy into kinetic energy, to give to the system, and the bottom weight has taken ~0.0314 kinetic energy from the system to convert into potential energy. so turning the system clockwise has caused a LOSS of kinetic energy. So... the system would not actually accelerate clockwise, it would accelerate counter clockwise.