I want to know all of the maths concerning this scissor mechanism!

Maglatron

Jul 12, 2023
1,798
ok thats fine but for now I'm just trying to design a flywheel out of steel that has 10.8kgm^2 moment of inertia, in a ring format that is 10cm in width, as far as I can see there is nothing woowoo about that, if you can help that would be kind, thanks Mr Martaine2005

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Delta Prime

Jul 29, 2020
2,265
I'm not claiming overunity or free energy, and not selling either do the math yourself come to your own conclusion!
I never said that. I never even thought that.
Yes I do... But you hurt my feelings.
You'll have to wait I penciled you in, your fourth on my list....
1) Fermat’s Last Theorem: posed in 1637, solved in 1994 after a colossal effort requiring the mastery of two disparate areas of research: elliptic curves and modular forms (neither of which, at first glance, appear to relate to the problem per se).
2) Goldbach’s Conjecture: posed in 1742, still unsolved.
3) The Riemann Hypothesis: the only one of Hilbert’s problems from 1900 still unsolved
4) You ...

Maglatron

Jul 12, 2023
1,798
ok so do you want to help?

Alec_t

Jul 7, 2015
3,694
is there a way I can distribute the 60kg in a ring format that will have a thickness of 100mm if so can you enlighten me? cheers Mr Alec_t or perhaps provide the formula, thanks
There must be a way, but I'm not aware of the formula. Perhaps ChatGPT can come up with one? It would also have to take into account the masses of any spokes and hub used to support and rotate the ring. Those masses and their shapes would affect the moment of inertia of the flywheel.

Edit:
Think of the ring as a large disc with a smaller concentric disc chopped out of it.
If you ignore the masses of the hub and spokes then the formula would be :-
60kg = pi x (Ra^2 - Rb^2) x thickness x density, where Ra and Rb are the outer and inner radii (in metres) and thickness is also in metres.

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Maglatron

Jul 12, 2023
1,798
There must be a way, but I'm not aware of the formula. Perhaps ChatGPT can come up with one? It would also have to take into account the mass of any spokes and hub used to support and rotate the ring.
yep I agree!! I'll keep looking

Alec_t

Jul 7, 2015
3,694
See my edit to post #244.

Maglatron

Jul 12, 2023
1,798
thank you - struggling to get anywhere on chatgpt!! going for dinner I'll have a look after!

Maglatron

Jul 12, 2023
1,798
There must be a way, but I'm not aware of the formula. Perhaps ChatGPT can come up with one? It would also have to take into account the masses of any spokes and hub used to support and rotate the ring. Those masses and their shapes would affect the moment of inertia of the flywheel.

Edit:
Think of the ring as a large disc with a smaller concentric disc chopped out of it.
If you ignore the masses of the hub and spokes then the formula would be :-
60kg = pi x (Ra^2 - Rb^2) x thickness x density, where Ra and Rb are the outer and inner radii (in metres) and thickness is also in metres.
ok cool thats good how do I relate it to the desired moment of inertia of 10.8kgm^2? thanks
by the way It doesnt HAVE to be 60kg although would like it to be
I want it to be 100mm thick
Ra 0.5m outer
density 7850
need to work out inner diameter and weight to satisfy the inertia of 10.8kgm^2
I guess that the weight will determine the inner radius but not certain

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Jul 7, 2015
3,694

Maglatron

Jul 12, 2023
1,798
perhaps if a ring is just mr^2 then inertia10.8/ radius0.5 = mass =21.6kg but then that still needs the inner and outer edges and thickness plus I'm not too good with calculus there must be an easier way to calculate!! thanks though

Maglatron

Jul 12, 2023
1,798
I want to design a flywheel ring type and its inertia needs to be 10.8kgm^2 and the thickness should be 80mm to 100mm, also the hub is 100mm in diameter with a hole for a shaft equal to 35mm and should be 150mm thick in depth, the spokes go from the hub to the ring there will be 3 spokes and will be 50mm cylindrical and the flywheel should be made of steel with a density of 7850 the inner and outer diameters of the ring need to be calculated and the weight too! hope that sums it up

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Maglatron

Jul 12, 2023
1,798
maybe you might be able to understand it and give it a crack on my behalf chat gpt is just giving me ridiculus figures!

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Maglatron

Jul 12, 2023
1,798
im thinking if I do the inertia 10.8 / (Ri + Ro)^2 that might give me the mass, then I might be able to work out the thickness of the ring with volume and density! is this line of thinking correct??? or could it be 10.8 / 0.5 * (Ri + Ro)^2 for mass

edit:
so I found out that I = 1/2 * M * (Ri^2 + Ro^2)
the equation is M = 2I / (Ro^2 + Ri^2)

if
I = 10.8kgm^2
Ro = 0.5m
Ri = 0.45m
the equation would be
2 * 10.8 = 21.6
punching in the digits = 47.735kg

so now must find the thickness of the ring BUT definatly progress!

so to find the thickness we need to calculate the volume of the ring
V = M / density 47.735 / 7850 = 0.00608m^3
thickness = V /( Ro^2 - Ri^2) = 40.75mm which is good

I might mess with it a bit to get the thickness nearer to 100mm but I think that I dont need to but here goes
M = 2I / (Ro^2 + Ri^2) Ro = 0.4 Ri = 0.35
76.5kg
V = M / density 76.5 / 7850 = 0.00974^3
thickness 0.2597m or 26cm
again Ro = 0.45 Ri =0.4
59.586kg
volume 0.00759m^3
thickness 17.86cm
again
Ro = 0.48 Ri = 0.43
52kg
volume 0.00636m^3
thickness 14.56cm
again
Ro = 0.5 Ri 0.47
45.87kg
volume 0.00584m^3
thickness up again to 200mm thick
again
might just settle for the 14.56cm and make the hub 200mm normal to!! this is because I want the hub to be wider than the flywheel

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Maglatron

Jul 12, 2023
1,798
so did one more calculation with the outer radius of 0.55m and the inner radius of 0.5m the weight of the flywheel equals 39.095kg and that yields a thickness of 95mm so I think that will be the closest to the 100mm thickness I was after! now putting into solidworks!

Jul 12, 2023
1,798

Maglatron

Jul 12, 2023
1,798
put in key slot and filleted the sharp edges on the spokes

Jul 12, 2023
1,798

Maglatron

Jul 12, 2023
1,798
so what I need to do right now is to create a freewheel mechanism any suggestions people??

Maglatron

Jul 12, 2023
1,798
ok so back to the scissor mechanism, I have been giving this some thought, while I agree that at any given height of the blue circle at a stationary instant, you would feel the weight of the block at the top and nothing more at standstill rejecting mass of scissor, but when you go to lift the weight through the scissor at the blue point, it would be HARDER to lift than if you were strait lifting the block without the scissor - it just makes sence that that must be true, what are your thoughts please?

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Alec_t

Jul 7, 2015
3,694
Consider the energy used. Work = force x distance.

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