Chas Campbell free power motor

[hartiberlin]

So you agree Humbugger, that when I can do it with more than 27.7 degrees , that I have overunity
So Chas, go for 29 degrees !

[tinu]

Okay...I give up.  This is where the laughter breaks out as the arrogant professor tears out his hair and the plotting students roll on the floor at their successful crazy-making.

Stefan obviously is not stupid.  Everyone can clearly see the point I have made is correct.  Stefan pretends not to get it.  Humbugger goes insane being told he is wrong by an authority figure when everyone can plainly see he is right.

Humbugger, you are correct in your analysis, but I think yo should look closely at what stefan has said:

If you stop the wheel, when the next ball has come up at the left side,
it has enough time to run out to 4x.
Then always one ball is at the right side and it can then start the wheel again.

Stefan seems to be under the impression that you can stop the wheel and wait for the ball to go in.  Maybe this is possible with some type of mechanism to keep the wheel from rolling backwards.  With Chas's wheel, I think as soon as you have more force on the lift side than on the descending side, the wheel will start to move in the opposite direction.

Shrugged:

Yes, it is obvious I have failed to communicate well enough.  Please think about this one last time.

No matter the stopping and going.  No matter the rate of balls rolling along ramps.  No matter the number of extra balls waiting on ramps. 

If there is two feet of linear distance between each of the 12 cups on the outer wheel and if the fixed places where the ramps feed and retrieve the balls into and out of the wheel's cups are only one foot apart, then there will be some times when there are no balls in the outer wheel's cups

In other words, once again, when the cup with the ball in it gets to the exit ramp and moves out of the big wheel's cup and onto the ramp, at that very moment, no matter there may be 87 balls waiting and ready to get into the next available entry cup from the top ramp, the receiving cup isstill a whole foot away from being in position to receive a ball.  

In this case, the time of having a ball on the wheel and the time it would be empty would be equal.  For our torque add-up then, we would only count half of one ball on the right side of the wheel as an average.

Stefan keeps insisting there would always be at least one ball on the outer wheel.  He seems to think there will automatically be a cup in position to receive a new ball as soon as the ball is present at the feeding end of the upper ramp.  In reality, it can't get on the wheel until the next empty cup arrives.

Only a forward jerking infinite-speed motion could get the next empty cup in place to receive the new ball at the same time the old ball leaves its cup for the exit ramp.  Stopping the wheel does nothing.  I am already giving that the ramps always have a fresh ball in position whenever a cup comes by.

"Stefan seems to be under the impression that you can stop the wheel and wait for the ball to go in."

You would have an infinitely long wait on your hands if you stopped the wheel at the moment the lower ball leaves it cup.  The upper "feed" ramp may have a ball ready after only a moment, but the wheel will be even further out of position to receive it if it has been stopped or slowed.

Please tell me you get it now.  Please...even if you have to lie!   

Humbugger

Let people process, Hum.
You are right but the acceptance of bare truth takes some time.

I?ll try below put your idea in other wording:

The number of slots on the outer rim can not be larger than the number of slots of the inner rim.
If it is, the balls will deplete from the upper part and will accumulate in the lower part and in the final the wheel will inevitably come to a stop. Hence, same number of inner-outer slots, people!
This is the ?lack of information? some might have.

If you get it right, the equations are posted again bellow.
As a consequence, there is a maximum radius for the outer rim. It can not go to infinite because the distance between slots becomes too large. It further means that there would be times during which there is not at least one ball on the outer rim. So, if no balls on the outer rim, what angular momentum will make it turn?! It will stop and then start turning the other way, due to the balls on the inner rim.

So, it is not a matter of time, as Stefan implies.
It?s not, because you may have as many balls as you want. The balls will just wait in their line. (Just like in the movie). So, not the number of balls is the key for analyzing the machine but the number of slots (or cups, if you want). The supplementary balls, waiting in the line, will not affect the overall angular momentum until they occupy their slots.

Here is again the simple formula to play with:

Rint<Rext<Rint/sin(pi/n) in radians
Rint<Rext<Rint/sin(180/n) in degrees

where:
n ? total number of ball slots on each rim
Rint ? radius of the inner rim
Rext ? radius of the outer rim

Pick your n and Rint/Rex, compute the torque, mediate it and see it is always zero.

Tinu


Ok, here it is for your case, Stefan:

Angle a=30 degrees
if  m- mass of each ball, assume m*g=1, for simplicity
Right: 2 balls
Left: 5 balls (as per your picture)

Torque:
Right: Rext+Rext*cos(a)
Left: Rint+2*Rint*cos(a) +2*Rint*cos(2*a)

If Rint=1, Rext=Rmax theoretically possible for 12 total slots and 3 on the right = 2, you?ll have:

Right: 2+2*0.866 = 3.7321
Left: 1+2*0.866 + 2*0.5 = 3.7321

Great! The wheel stays!

Sorry Stefan, it doesn?t work; like I?ve told.
You wanted a proof and here it is.
I guess you?ve done a mistake but it?s your job to find it.
I suppose it?s in your understanding about Rmax. In your picture Rmax can not be larger than 2. In order to be 3.8637 (maximum possible), you have to redraw it. In that case the ball on the right will not be on horizontal but at 15 degrees above and the torque will be proportionally smaller.

I?m also going to bed,

Tinu

[Humbugger]

So you agree Humbugger, that when I can do it with more than 27.7 degrees , that I have overunity
So Chas, go for 29 degrees !

Yes, if he would just put exactly 360/29 (12.41379310...) cups on the wheel, I think it would work.  But only if the cups are those special ones I mentioned previously.  The Panacea brand by Campbell Corp. (only the Rev G or later will work in this application)

I heard they got a big grant and the cups will be ready in a couple of years.

Hum

[hartiberlin]

Tinu,
you did not analyze my last picture.
Things change, when you different setups.
I now have always only 1 ball at the right side, not 2.

I will try to calculate it tomorrow with your formulars
and then we will see, where the error is.

[hartiberlin]

So you agree Humbugger, that when I can do it with more than 27.7 degrees , that I have overunity
So Chas, go for 29 degrees !

Yes, if he would just put exactly 360/29 (12.41379310...) cups on the wheel, I think it would work.  But only if the cups are those special ones I mentioned previously.  The Panacea brand by Campbell Corp. (only the Rev G or later will work in this application)

I heard they got a big grant and the cups will be ready in a couple of years.

Hum

That is typically,
if you are caught by your own faults, the skeptics can only do
ridicule and never see their own mistakes...