NOTHING More than a wheel with swinging weights...

[nneba]

Hi,

I can appreciate what you are doing Raj, but when I say that gravity will have the same effect I mean that your weights will find their way to thier lowest point possible and end up at rest. I beleive that the reason for this is because the "seesaw" (levers) connect to the axel. It has been mentioned before that when the fulcrum is in the center of the wheel, power loss is inevitable. You had mentioned earlier something about an "orbital motion" this is the key...you must prevent any contact with the axel.

nneba

[shap-O-vert]

raj says:

Let us hear from you.

This is keeping me out of the workshop. But what the heck!

Perhaps raj, you could draw your approximation of the track of a single weight as the wheel turns (assume that it does). I say this because I can't see how there isn't a "double lift" on the upside created as a result of the lever being tilted by gravity when its fulcrum gets to the 8 o'clock (approx) position and the weight swings down accordingly. That is, when the lever gets to its tipping point.

What I can't see is how your flexible string facilitates an earlier "centering" of the weight.

Your estimate of the weight's track might help.

Regards

shap-O-vert.

[raj]

Thanks to you all.
One serious precision!
My concept is an empty-drum-like cylindrical wheel on horizontal axle (a bit like a lotto draw wheel) with eight cylindrical weights hanging/swinging, each on strings from eight points at 45 degrees interval on the inside of the outer rim of the wheel.
The weights hang/swing on both the strings and the flexible loops on the two pulleys, which guide/constrain the weights to move in an SMOOTH elliptical path/orbit round the axle of the wheel , but always a reasonable distance away from the axle.
Remember(1) We only have a wheel, weights on strings, flexible loops and pulleys and NOTHING else.
(2) the weights follow a SMOOTH elliptical curve.
Therefore there is NO LEVER or SEESAW (in the sense of the words) involved in my design.
I have made reference to seesaw only to explain how my concept works.
I have drawn the movement of the weights at 11.125 degrees interval to scale and measured/calculated the CCW and CW torques and found a continuous net positive torque CW in the range of 7 to 15 %.
I have built a crude physical model using a bicycle wheel to check the movements of the weights round the wheel axle.
The weights do follow a smooth elliptical path round the axle and stay always in a hanging/swinging position on the bicycle wheel.
There is absolutely NO double lifting of upward weight at anytine, anywhere on the wheel.
I hope you can all visualise my concept as it is.
Raj

[bluesgtr44]

Hey Raj, I worked up a quick 2 weight sim of this idea. I does present an eccentric pattern within the circle...but it is more illusion than a real redistribution of weight as it pertains to the circle. The setup for the two weight systems are the same and they are 180 degrees apart. Assuming you had 8 of these on the wheel, when you reach this point....you will be lifting 5 of those weighted sets with only 3 having the ability to drive on the wheel for about 30-45 degrees of the rotation. Because of the route that is being forced with the ropes and points....there is not reciprocation of this action to where there are 5 weights on the driving side of the wheel. I like this idea for a couple of reasons as far as Bessler is concerned. There is definitely a "connectedness" aspect about it and it does provide an offset path for the weights to follow even though it doesn't provide a constant imbalance. Thanks for sharing this!


Steve

[bluesgtr44]

I want to throw this one in real quick. This follows the path of the weights and definitely shows the eccentric path within the symmetrical circle. But, it's the tether points of the ropes where the weight is actually being applied to the wheel....not the points on the weights path. Still like this though, Raj!


Steve

[raj]

Steve!
Oh, THANK YOU ever so much for your excellent work done for me.
I wish I could do it myself.

Your SIM agrees 100 % with what I think my concept should work, EXCEPT for one tiny clarification needed from you.
It's the tether points of the ropes where the weight is actually being applied to the wheel...???
Can kindly clarify what does that mean in terms of torque producing effect on the wheel?

I have come to understand that no matter how a weight finds itself inside a wheel, it is the position of the weight with regards to the axle of the wheel that determines the torque producing effect by that weight.
Am I wrong in this view?

But when go back to the tether points of the weights, I must point out, as it's evident from the drawings, that the weights hang on three pivot points mostly, on two pivot points sometime, but NEVER on a single pivot point.
Would this matter with regards to your above assessment?

Assuming that the torque producing effect by the weights must be calculated on their positions inside the wheel, then even though, the five weights going up against three weights going down, I maintain that there will be continuous net positive CW torque, to help keep the in rotation.

The drawing below shows the eight weights in the same wheel position as in your two-weights sim.
Kindly check the torque calculation.
Do you agree that the three descending weights are producing more CW torque than the five ascending weights CCW torque.

Here are some precision about my drawings: Wheel outer rim diameter is 8 cm, wheel inner rim diameter is 6 cm, outer pulleys are fixed on inner rim, inner pulleys are fixed 0.4 cm from axle, strings are 4.4 cm, flexible loops circumference is 14.7 cm and weights axles diameter is 0.2 cm.
Thanks again for your very kind help.
Raj

[path_finder]

Dear raj,
There are my comments on your design (with the restriction I did forget perhaps an important point).

1. The idea to force the path of the weight by a rope is a good idea.
As you know the path obtained by a point of the rope around two other external points is an ellipse like shown earlier here:
http://www.besslerwheel.com/forum/download.php?id=6209
2. In your specific case the two poles of the ellipse are the two pulleys.
In the amended drawing below I show the points A and B and the corresponding ellipse (in blue) for the 3:00 position.
In fact and due to the attachment of the weights to the rim of the wheel, only a part of this ellipse is really used.
3. The key is certainly to use a rope with the same principle, with the exception the other end of the rope connected to a previous/next weight (instead to return back on the same weight)
4. This idea will be excellent if the ellipse could be fixed in the same position (this one in blue) for the whole rotation of the wheel.
Unfortunately it is not the case (it rotates WITH the wheel).
5. What is not clear for me is your drawing at 7:00 where the spring has been represented free.
IMHO the weight will continue to be sticked at the rim until 9:00 (the only position where it can overpass the vertical line).
This is the reason why on my drawing I made a copy of the 3:00 shape, made a rotation of 180 grades and paste it at 9:00, where the weight falls.
6. The property of an ellipse is: the sum of the distances from the point to the both poles is constant.
At 9:00 the fall of the weight will just change the position of the weight on the ellipse from one end of the arc to the other arc's end.
7. As explained here many times, any connecting point near the central axle must be absolutely avoided, reducing drastically the main torque

Your idea is good, but IMHO you should improve your design by fixing the two poles on the 3:00 radius by a permanent manner (whatever the rotation of the wheel).
It is just a variation of the ideal path (remember here:http://www.besslerwheel.com/forum/viewt ... 5325#55325)
As suggested many times an internal pendulum could be one solution despite the fact it is not stable enough
(like shown here:http://www.besslerwheel.com/forum/viewt ... 5811#55811)
I tried myself an elliptical cam (see here for memory her: http://www.besslerwheel.com/forum/viewt ... 7885#77885) but unsuccessfully (I don't know if because the poor level of the building -mainly the frictions-, or because the strings are connected to the cam instead to the rim-same comment as above-).

In any case I cannot give a definitive conclusion on your design. Even if it seems to work, in my experience only the building of the device can deliver a part of the solution.

[mickegg]

Hi raj

I like the idea.

Centrifugal force will want to throw all weights out to the maximumum position as shown by weights 2,3 and 4.

This will increase the negative torque.

Pairing the weights may be beneficial in this respect.

jim_mitch has a good understanding of these forces..if he's able to comment without giving away too much of his own research.

Regards

Mick

[raj]

A big thank to Path_Finder and Mick.
Your advices and suggestions have been gratefully received.

Path_Finder,
your suggestion for relocating the pulleys to a central position does not seem appropriate. If you read my posts from the beginning you will find me mentioning that I had tried something like that and it did not perform as I expected during a crude physical model testing.
About the weights clinging to the rim up to the 9 o'clock.
I am 100 % sure that do not happen. You see, even though the weights are tied by the strings to the outer rim of the wheel, the weights always rest and search/roll for the lowest points on the sets of flexible loops, until their tether(strings) stretch up completely, in whatever position the wheel may be.
By the way, the flexible loops pull/guide/constrain the weights to take the eccentric path round the axle, and the weights force the flexible loops to change shapes as the wheel would rotate.

Mick,
Your suggestion for pairing the weights look perfect, should centrifugal force comes into play.
I am a complete novice in this aspect. Rightly or wrongly, I have come to believe that centrifugal force is relative to the speed of rotation. I.e the higher the speed of rotation, the higher the CF. But I stand to corrected on this.
My proposed gravity wheel, in my view, if it works at all, will be a very slow moving wheel ( with big torque, perhaps?)
But if pairing of weights is necessary, then the pairs will be weights 1/5, 2/6,3/7 and 4/8 linked by strings.
I just checked the distance between the pairs of weight, in my last drawing above, and surprise, surprise they all have equal distance between them, that is between 1 and 5, 2 and 6, 3 and 7, and 4 and 8.

One correction to information given above: the diameter of the outer rim of the wheel is 16 cm and the diameter of the inner rim is 12 cm.
My apology for this mistake.
Raj

[jim_mich]

I am a complete novice in this aspect. Rightly or wrongly, I have come to believe that centrifugal force is relative to the speed of rotation. I.e the higher the speed of rotation, the higher the CF. But I stand to corrected on this.

CF increases by four times when the rotation speed increases by two times.
CF increases by nine times when the rotation speed increases by three times.
CF increases by sixteen times when the rotation speed increases by four times.

CF increases by two times when the radial distance increases by two times.

KE increases by four times when speed increases by two times.
KE increases by nine times when speed increases by three times.
KE increases by sixteen times when increases by four times.

CF = 0.000028403397 × Wt_Lbs × R_inches × RPM^2
CF = 0.000001118244 × Wt_kg × R_cm × RPM^2

The weight can be any desired units since the results units are the same as the weight force units. That is, if weight is pounds then the results are pounds of CF, and if weight is grams then results are grams of CF.

[bluesgtr44]

Hey Raj....Here's a quick pic of what I am talking about. It's not just about the position of the weight within the wheel...it's about where the weight is applying itself on the wheel. I don't have access to why WM2D files right now so I can't bring that up and show a better break down of that. You have a natural "sweet spot" in your set up at one point for a very short time, about 30 degrees, and this is when the outer point lets go of the weight and the inner point starts to pick it up.....the weight that is 180 degrees out from this is applying most all of it's weight on the outer point. Once the rotation continues though, you will reach the point where the force of both weights are on the same ascending side.


Steve

[raj]

Dear Jim! You are great! Thank you so much for your very enlightening information that I would use from now on, whenever I should have to calculate forces on my gravity wheel search trail.

Steve, I thank you again.
You have definitely help to understand my concept more now than when I first conceived my idea.
You have made my idea look much better TO ME with your sims.

Yes you are absolutely right to say that at a certain position of the wheel two 180-degree-opposite weights will find themselves on the same ascending side of the wheel.
I had known that ever since my first drawing of this new gravity wheel concept weeks ago.
Therefore it is plain that the wheel with only two weights WILL NOT work.
The wheel needs a minimum of eight weights to possibly (????) work.
But I found by measurement on drawings that even with five weights going up against three weights going down, there is a net positive CW torque. I do not think that it could ever be six weights against two weights.
Unfortunately I have to wait till next April when I shall be in a position to build a working wheel model to test my concept.
Its a long wait. That's life.
Raj

[raj]

For want of official records, my gravity wheel concept is now patent pending, re: Patent Application Number GB1017908.3

Raj

[path_finder]

Dear raj,
Good luck anyway.
I hope you are rich enough.

[raj]

Reviving this thread.

Raj