Mayday! Mayday!!!

[raj]

Now, Since my new design follow the same working principle as the simulation above, with weights, as prime movers acted upon by gravity, moving on elbowed rigid swings, as simulated above, around the axle of the main wheel, I have plenty to feel good about this new design.

Torque for neccessary rotation of wheel, is provided by weights in relation to their position and horizontal distance from the vertical line through the axle of the wheel.

The drawing below shows an accurate scaled positions of the 8 weights in the wheel, from 12 o'clock position turning 45 degrees counter-clockwise, in 3 stages: (1) starting position at 12 o'clock - weights in black circles (2) after 22.5 degrees turn counter-clockwise - weights in red circles. and (3) after 33.75 degrees - weights in large black dots -turn counter-clockwise.

The drawing is on graph sheet. Therefore working out the clockwise and counter-clockwise torque by the 8 weights from their positions in the drawing can be easily done by just looking closely to the drawing.

I have done my level best to explain my points and be open to your views.

Raj

[Ed]

Raj,

You have referred to path_finder's animation several times as a "simulation". Is that really true? I wasn't aware that he made videos of simulations, just animations of what he thinks will work. He also obviously does some nice builds, but has he actually built what is being represented in that animation?

[raj]

Ed,

Fair enough, ANIMATION!

Is my explanation of calculation of torque by the 8 weights in my drawing above farfetched?

If yes, kindly explain to me, how would YOU calculate the torque by these weights in the drawing?

Certainly not the same way as JB has explained?

Raj

[raj]

Look again at JB's drawing showing the weights' path and my drawing showing the weights' path around the axle.

[raj]

If motions of weights are going to be used as prime movers acted upon by gravity to rotate the wheel, then the wheel MUST be 100 % fully balanced with the centre of mass being on the axle of the wheel BEFORE the weights are added inside the wheel.

Thereafter, it is the NET torque by the weights of the wheel that is going to turn the wheel.

JB has explained his point by using the centre of mass in his argument.

Whilst, I am using net torque by the weights in my argument.

Raj

[raj]

Among other things on this forum, I have learnt how to find the centre of mass from a drawing. I think I learnt this from Tarsier79.

I have used this method to find the COM of my wheel drawing/design:
The red circles are COM (Mid point) of the four pairs of opposite weights.
The black circle is the COM (mid point) of the two pairs of red circles.

This black circle is the COM of the 8 weights, and the COM is on counter-clockwise side of the axle.

This proves that the wheel would turn counter-clockwise, as already shown by torque calculation.

Raj

[Tarsier79]

This proves that the wheel would turn counter-clockwise, as already shown by torque calculation.

Raj

Firstly, it looks to me like you have marked the COM slightly wrong, it should be slightly more to the right I think.

The COM is one way of calculating rotation, but it takes a little more than one drawing. Initially, what you need to do, is to redraw your wheel advanced and retarded by a certain number of degrees (for example 10), then recalculate the COM, and see if it has risen, or fallen. When the COM is at its lowest vertical point (regardless of horizontal displacement), that will be its resting point. If the COM doesn't fall, there won't be rotation. You might have to draw it a number of times to get the exact point of lowest COM.

IF you subscribe to this thinking: To reset, the COM will always fall just as much as it needs to rise. How is it possible then that "gravity only" will sustain a wheel?


Add: Also, I don't believe your other design PF is building will work. Gravity is not the prime mover Besslers wheel, it is just the environment enabling its operation.

[raj]

Tarsier79,
Thank you so much for your input,

I applaud your suggestion to check the COM at different angle of rotation, to ascertain genuine COM advantage.

I also agree that my COM point on my last drawing should be slightly to the right. But it would still be on the counter-clockwise side of the axle.

Please allow me to argue my point from a complete layman's perspective.

Bessler has said, from what I have learnt on this forum, that motion of weights inside his wheel could not find equilibrium that would stop the wheel from rotating continuously.

Now the key words in this Bessler's statement are '' motion of weights''.

My question here is what direction were bessler's weights moving inside his wheel?
Were these weights moving in the X-axis, Y-axis or Z-axis (direction)?
Or were they moving in the combined direction of 2 or 3 or nth axis?

Whatever direction Bessler's weights were moving inside the wheel, it could not been a horizontal path for a horizontal path would get them out and away from the wheel, with no turning back to the wheel.

Therefore it becomes obvious that the weights were moving in a circular path with the wheel. (Or separately?)

Whichever! Circular path means going up on one side and coming down on the other side: That is rising and falling of weights continuously.

Let's forget about the rising of weights for now and concentrate on the falling of weights.

Can weights fall without the help and action of GRAVITY? (in Vacuum?)

If the answer is ''YES'', then Bessler's wheel did not use gravity?

Raj

[johannesbender]

raj , the only reason I took interest
in responding to your design is because I am
playing with an idea that I saw in your design .

I did not want to share it as a visual image here
but I wil do so , I have drawn over it to show
what your mechanisms look like in regards
to what I am researching , even though mine
is different in function with my research, in its
application , I have used it here to show a point.

I am not out to tell you it wil not move or function
I merely exercise my opinion , if I am wrong or
not , I do not mind , opinions do not make a design fail or succeed .

you are free to dispute I do not mind .

now lets look at this picture of mine :
two arms pivoting at a point of convergence
where a weight is .

we wil start from where these arms are straightened such that A and B and the weight
forms a line .

this is at the top of your wheel.

now going to the left of your wheel there is
a squeeze on A and B , when we reduce the length between A and B the COM moves
along the yellow vector , in your case raj
the COM drops during the squeeze but
you are still lifting mass up in respect to A and B.

now on your right side of your wheel ,wich
can be seen by flipping my picture upside down
you are extending the distance between A and B
and this drops the COM but also raises mass
in respect to its orientation on the right .

the lifting of mass of each mechanism takes
energy/work wich you need to calculate along
with the total weight you lift on the right .

your real COM will not be at the centre of each
mechanism's centre weight , well depending on
your weight values , but it will still not be at
the positions you drew out , it shifts as the angle
of the arms close and open , okay it may be
to minuscule to be of importance to exactly
determine but anyway ..

then I also believe your total COM would be
very close to keel , I may be wrong but its my
opinion only , you have not shown weight values , only you would know .

take a look at the following drawing of mine also ,it may not be perfect but here I show that by using straight rods the work of the squeeze and
pull are taken away .
but in regards to weight relative to each wheel
we have 5 on top and 3 on bottom on one
wheel but 5 on bottom and 3 on top of the other.
the design is also flipped upside down compared to yours , ignore
the dotted crescent.

your wheel shares similar but different weight
positions in respect to each wheel , it is more
complicated than what you have shown by the
torques and COMs ..

good luck.
jb

http://en.m.wikipedia.org/wiki/Center_of_mass

[raj]

JB,
I sincerely appreciate your contribution.

I have followed Tarsier79's suggestion to draw two or more COM testing at different angle of rotation through 45 degrees because the overall weights positions reset then by design.

The following four drawings show the COM of the eight weights at four stages of rotation through 45 degrees:
1. starting position at 12 o'clock position.
2. at 11.25 degrees rotation ccw from starting position.
3. at 22.5 degrees rotation ccw from starting position.
4. at 33.75 degrees rotation from starting position.

In the next 11.25 degrees rotation, the overall position of weights would reset.

The pair of or single black circle(s) show the COM of the eight weights in eack drawing.

If you look closely, you will notice that the COM has moved up or down from one drawing to the next, BUT still the COM stays on the CCW side of the axle of the main wheel.

How do you interprete this falling and rising of COM on the same descending side of the wheel?

Moreover, these four drawings make it easy to calculate the NET torque by the 8 weights on the wheel.
Careful measurement would show a net CCW torque, BECAUSE there is always more weights going down than going up.

That is faster rise and slower fall of weights.

Raj

[johannesbender]

I am not going to dispute your calculations however take into account that your arms
arent crosses of equal length wich would
mean that they aren't balanced in the centre but
somewhere along a direction into the triangular
area , wich I have been trying to show you ,wich
could help you in being more accurate .

I am pretty sure you are correct to a certain acceptable degree of accuracy with the complete
COM though , since the point of balance would
be more close to the heavier side of the wheel as
you have been demonstrating .

I have no idea what wil happen but I do know
that being wrong teaches and if that's the case
I would rather be wrong here .

good luck and onward with your design.

edit : just to taunt you , just think how
well you could design your wheel if you could
pinpoint it to look something like this ,if
you knew your weight values etc.

[Tarsier79]

How do you interprete this falling and rising of COM on the same descending side of the wheel?

Exactly as you would in any other occasion. Disregard the horizontal displacement.


Here the COM is significantly to the RHS. This does not mean it will rotate CW.

Moreover, these four drawings make it easy to calculate the NET torque by the 8 weights on the wheel.
Careful measurement would show a net CCW torque, BECAUSE there is always more weights going down than going up.

In OB wheels, there is nearly always an imbalance between ascending and descending weights. The avalanche has double the weights descending, but will not rotate. The way you calculate torque will change depending on the mechanism. In this and your other design, you have to take into account gearing, ratios and leverage. EG. a mass might apply its weight 90% to the inner hub, which might be turning at double the speed, and 10% to the outer. Are you able to determine and calculate this?

[raj]

My dear Tarsier79,
Thank you again for your precise and constructive comments.

Let me recap on this, my new design.
1. A drumwheel (diameter 8 units) on axle
2. A smaller wheel (diameter 2 units) on hanging axle 2 units below axle of drumwheel inside drumwheel.
In other words, the top of circumference of the inner wheel is on the same level as axle of drumwheel.
3. The axles with sprockets of both wheels are identical and geared in a 1:1 ratio.
In other words, both wheels will rotate at the SAME speed. There is NO WAY one will rotating faster than the other.
4. The rigid arms connected to both wheels are IDENTICAL (4 units long).
5. Because of the 1:1 gearing of the wheels, all actions, forces and leverages acting upon and affecting the INNER wheel, will DIRECTLY act upon and affect the main wheel, as if all the weights on rigid elbowed arms are directly connected to only one, the MAIN wheel.
6. As a result, all torque and COM calculation could be calculated only with reference to the main wheel.

Raj

[raj]

Hello all forum members.

I am NOT CLAIMING that my design is a RUNNER.

I am ONLY trying to EVALUATE my design.

Raj

[path_finder]

Simply because we have not been able to find a suitable speed doubler mechanism

. After more than one year of research I have one such as required doubler.
Unfortunately I have been so much busy since few months for finalize the full wheel ( I still have heavy health problems, I made an important work in Cameroon and after the acquisition of an typical house in Normandy, I had to move all the stuff from the flat in the Paris area). But now I can again continue the work in Normandy.
This last week-end I finished the new slim doubler, and hope to get very soon the full wheel in oeration (all in wood, for this time).
I am very confident on this new design, because the three sub-assemblies have been separatively tested successfully (frame, doubler, primemover). Only the final building will tell us the truth.