IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[PeterAX]

To Sam Peppiat.
=========================
Hi Sam,
--------------------------------------------
You really make me smile!:):) But how did you conclude that I am a girl?:)
No, I am not a girl. I am a boy. (And more precisely a "bigger" boy:), that is, a man.)
Anyway I appreciate your sense of humour!:) It's a pleasure for me to correspond with you!
---------------------------------------------
And what about your gravity wheel? Any progress? Ready to help, if necessary.
---------------------------------------------
Regards,

[PeterAX]

Asking my two simple questions for the 10th time.
------------------------------------------
QUESTION 1: V2 = ? (How many meters per second is V2 equal to?)
QUESTION 2: V3 = ? (How many meters per second is V3 equal to?)
------------------------------------------
Looking forward to your two answers for the 10th time.

[Sam Peppiatt]

Hi Peter AX,

What can I say; when I'm wrong I'm wrong.

The gravity wheel seams to have a mind of it's own. It refuses to cooperate with me.
As you may remember I'm working with pendulums. They have translating motion, called curvilinear translation. That is to say, they go around in a circle but don't rotate.

What I don't know is; what the centrifugal force would be on such a pend.
Would you consider trying to figure it out--------------------------------Sam

[WaltzCee]

Sam,
There are c.f. calculators online. You might search for one that suits you.

[Sam Peppiatt]

WaltzCee,

Yes, a good idea. I didn't see any thing for a body / weight translating in a circle. It's like a special case that they never consider.

Even if they did I probably wouldn't understand it------------------------Sam

[eccentrically1]

it's not a special case. the formula is the same no matter what calculator you use. F = mv2/r. So it's necessary to know the mass of the pendulum, its angular velocity, and length of radius.

here's one

https://www.omnicalculator.com/physics/ ... ugal-force


and an example here

https://www.omnicalculator.com/physics/ ... ugal-force

[Sam Peppiatt]

eccentrically,

Are you sure? I suspect that it might be different. Here's why. Example:
The connecting rod on a gasoline engine. If you squirt oil on the crank shaft, as it rotates, it will fly off on a tangent in every direction.

If you squirt oil onto the big end of the connecting rod, (which is translating not rotating), the oil just kind of slides back and forth, but doesn't fly off.

So, how can they be the same? Perhaps you can enlighten me----------------Sam

[eccentrically1]

I don't think it's a good example but the oil is in angular acceleration on the crankshaft so it would have that force: mv2/r.
The oil is only in linear acceleration on the rod so it experiences only linear force: ma. There isn't a change in direction as in the former case.

Anyway the two cases are different so you apply different formulas. But neither is special or not considered.

[Sam Peppiatt]

I stand corrected------------------------------------Sam

[Robinhood46]

Sam,
I think the difference in the behaviour of the oil rotating on the surface of a rotating surface, and rotating on the big end, is that the oil is always being forced away from the surface on the rotating surface because of the Cf.
On the big end, the oil is alternating between being forced against, laterally, away from and laterally (in the other direction) in relation to the surface.
So the oil is only really being thrown off the surface, when it is at the bottom. (in the case of a combustion engine, the right way up).

[Sam Peppiatt]

Robinhood46,

The test I did the rod was horizontal. I can't speak for the underneath side of it,
but on top the oil stayed on the rod just kind of shifting back and forth a little bit.

I thought it was different and special------------------------------------Sam

[PeterAX]

To Sam Peppiat.
=====================
Hi Sam,
Thank you for your post.
--------------------------------------
Well, it seems to me that your approach is perfectly correct in general. And may be only some small detail/component of the system has to be adjusted accordingly. Please give me some time to think over the problem. (And please send to me some pictures, if possible. Not pressing, only suggesting.)
--------------------------------------
It seems to me that this pendulum-related problem could be solved.
--------------------------------------
Looking forward to your answer.
Regards,

[PeterAX]

Asking my two simple questions for the 11th time.
------------------------------------------
QUESTION 1: V2 = ? (How many meters per second is V2 equal to?)
QUESTION 2: V3 = ? (How many meters per second is V3 equal to?)
------------------------------------------
Looking forward to your two answers for the 11th time.

[Sam Peppiatt]

PeterAX,

I suspect that eccentrically1, is exactly right, F=MA. It's as simple as that---------------------------Sam

[PeterAX]

To Sam Peppiat.
=================
Hi Sam,
Thank you for your post.
-------------------------------
I would suggest a somewhat "crazy" idea.:)
What would happen if we use two identical gravity wheels, (a) which are fixed to a common axis and (b) which work in opposite phase? That is, when the first gravity wheel's pendulum potential energy/centrifugal force is at its maximum the second gravity wheel's potential energy/centrigugal force is at its minimum. In other words, what would happen if we try to build a two-phase mechanical equivalent of a two-phase electric motor? Or to use three identical gravity wheels thus trying to build a three-phase mechanical equivalent of a three-phase electric motor? What do you think? Would this system work?
Looking forward to your answer.
Regards,