NEWS FLASH!! Fcdriver has a working wheel!!!!

[pequaide]

Videos of energy producing devices have already been posted; you will not be able to patent the concept.

There will be a plethora of working designs. The patents will be on those designs.

As far as I am concerned anyone that is still using angular momentum is not even in the game.

[Tarsier79]

Frank.

Show me how the angular momentum increases.

[Grimer]

LOL.

I'm sorry, Tarsier, - I can't explain it better than I have. If that doesn't satisfy you, sobeit.

[WaltzCee]

http://www.gazettejournal.net/index.php ... ntme/17448

[WaltzCee]

“It basically generates power more efficiently than solar power or windmills,� Driver said. He explained that where it would take 26 solar panels to produce 135 watts of energy for 8-10 hours, this invention would supply a similar amount of energy for 24 hours.

Commonly used solar modules often have efficiencies ranging from 11%-14% and produce about 10-13 watts per square foot. New modules are on the market that attain 20% efficiency. In general, the price per watt rises significantly with increasing panel efficiency.

I know there's a bit of confusion in the terms (power vs energy) but taking it all as is would make those panels 1/2 of a square foot.

pretty small panel.

[Grimer]

I don't think we need to worry about how much energy it produces compared to solar panels, Walter.
If it produces any energy at all on a continuous basis then it's the discovery of the millennium.

[Grimer]

I didn't follow the discussion, but invade anyway.

As far as I know the momentum ratio equals velocity ratio, thus:
When the velocity at 12-o' is 0 then via mgh=0.5*mv^2 one could determine relatively easy the expected velocity ratio.
Sqrt{ (E[11o']-E[12o']) / (E[5o']-E[12o']) }
If I didn't make a mistake then it should be something like, 1 : Sqrt{ [K+1-sqrt(0.75)] / [K+1+sqrt(0.75)] } -->for K=0, about 1 : 0.268
notes:
That K is used when the velocity is not 0 at 12-o'.
height(12-o')=>Cos(0)= + 1
height(11-o')=>Cos(30)= + sqrt(0.75)
height(5-o')=>Cos(150)= - sqrt(0.75)

Since you seem to be into the maths what would be useful is a formula which give the position of a pendulum bob after equal intervals of time so that one could say, for example,

At the start it's at 0.5°
After 1 seconds it will be at 3°
After 2 seconds it will be at 7°
After 3 seconds it will be at 12°
After 4 seconds it will be at 20°
After 5 seconds it will be at 32°
After 6 seconds it will be at 47°
After 7 seconds it will be at 67° ....and so on.

Then I could give a realistic visual picture of the change in speed of the bob as it orbits around the pivot rather than the crude guess I have used in the past.

[ME]

Perhaps I could... I guess slightly more complex, even more if you want some general solution.
For those things we have WM2D or other numerical integrating solvers, so why bother?

If you have an example, then I could construct a lookup table.

[Grimer]

A lookup table would be very useful.

How about length to centre of spherical bob = 20 inches

Diameter of bob = 2 inches

[ME]

How about: Radius, Start angle, start velocity (0 ?)
And then assume a point-like mass, and no air friction.

[Tarsier79]

Grimer.

So, another "Now I see" moment. You can't mathematically prove your theory. You can't properly explain your theory. You managed to describe it working on "3rd derivative energy", just as you have attributed it to every other device. I suspect when you revert to 3rd derivative mumbo jumbo it is to try to give the impression of not being stupid.

[justsomeone]

Didn't work. ;)

[rlortie]

A lookup table would be very useful.

How about length to centre of spherical bob = 20 inches

Diameter of bob = 2 inches

I fail to see of what value this is or what is to be gained in this discussion.

https://en.wikipedia.org/wiki/Pendulum "The period of swing of a simple gravity pendulum depends on its length."

Diameter of bob or its density appear to be irrelevant, not changing the period of the simple pendulum.

You are asking for a 20 inch radius or 40 inch diameter having a circumference of 125.644 inches. At a velocity of 20 RPM your pendulum is traveling 2,512.88" or 209.4 feet per minute.

Your visualization of "After 7 seconds it will be at 67°" seems to be a fairy tale or something from one of your derivatives!

[ME]

0th - position
1st - velocity: change in position
2nd - acceleration: change in velocity
3rd - jerk: change in acceleration
4th - shock: change in jerk

remember Taylor series?
xt=x0+v0*t+(1/2)*a0t^2+(1/6)*j0^3+(1/24)*s0^4

Mumbo-jumbo or not, at least I had a motive and an exercise to find me a method of extracting WM2D data (definitely needs some refinement).

A mass at a 20" radius (0.508 m) starting at the top.
As this would take an eternity to calculate, I just gave it a minimal push outside the significant numbers (who would notice anyway).
At least we know it should get a max. speed of 4.464 m/s when it reaches 6 o'-clock. That point is set as time-zero and angle-zero, so it's symmetrical around that point.
- btw: I have provided time by angle, and not angle by time.

disclaimer: I have the right to make errors.

---

[Grimer]

How do I turn that accurately into angle by time since I need to show the positions at equal time intervals?

I've clicked on the download button but it doesn't seem to work for me for some reason.