question: evert rotor

[fredos]

Hello guys,

I have found this page relative to professor Evert which draws a lot of
designs of gravity motors.

the web page is:

http://www.evert.de/eft382e.htm

Personally, i would like to realize the design named:
Ring within wheel with fix supporting wheel

It seems to work if the contact between the disc and the external wheel has a lot of friction (tangential force creates torque on the external wheel!)
What i really think interresting in this design is that ,instead of making the weights going upward or downward like in overbalanced wheels, here, the gravity center of the disc is not moving ( but the rotation can be possible).

Does anyone on this forum tried to realize this design?

Thanks in advance for your responses

fredos

[jonnynet]

Oh great. I remember well on Evert's site. I studied many of his designs before I was a member of this forum. I always had the problem that the operation principles were a bit too complex for an assessment but this one looks very promising. If I understand right I can already see how it works!
The rotor with the small wheel shouldn't create a counter torque because it doesn't act as a lever...

[fredos]

Thanks for your response johnynet.

You seem to be the only person on the forum that has tried to study the evert rotors.
Personnaly, i have done some calculations on the design of the ring within wheel and , what is really strange, is that this design can theorically works if the system has a lot of friction. I am planning to realize this design so if some persons can prove me that this design has already been tested and doesnt work please let me know in order to save a lot of time.

Thanks for your responses guys

[jonnynet]

Referring to this design I have found out an interesting thing. I have build a simple (real) model where I believe it imitates the same principle and... oh wonder it worked. It was not in the form of a wheel but it represented the basic leverage.
I hope one of these days I can proof the function completely. Next I will add a drawing here and try to explain.

[fredos]

WOUAH! great job jonnynet

I am in holidays for 3 weeks but as soon as i am back at home, i will post my theoretical calculations on this motor. Normally, if we apply the normal laws of mecanics about the support reaction of the external wheel, we can show that the friction makes the external wheel turning which is unusual. Perhaps
it is the key of making a working devices.
My calculations explain that the momentum applied by the ring on the external wheel is in the form of:

M=m*g*cos(a)/(1+cos(a))

Where a is the angle between the horizontal contact of the ring and external wheel and the support of the ring (named SL by Evert).
m is the mass of the ring

If you take a ring of 10 kg, and an angle of 20° between the support and the horizontal line of the external wheel , we have a torque of :

T=m*g*Rw*cos(a)/(1+cos(a))

where Rw is the radius of the external wheel.

so : T=10*10*0.2*cos(20)/(1+cos(20))=14,53 N.m

let the external wheel to turn to 100 RPM and we have:

P=14,53*2*3.14*100/60=152 W

I dont know if this concept can work but i could be a solution of making
a gravity motor

[jim_mich]

You seem to be the only person on the forum that has tried to study the evert rotors.

Not true! We have discussed this many times before. Use the search button at the top of the page. Search for Evert and you will get 39 posts over a number of years. Many of us have studied it before and could find no evidence that it would work.

[jonnynet]

I may have understand his idea completely wrong but I will do what I have to do.

[murilo]

Jonny...
I loved this above response!!!
This is exactly the behavior one hope to get from all this forum members!!!
Thanks and best regards!
M.

[jonnynet]

Okay, drawing 'evertbas' shows the test I performed for Evert's design. A rod pinned to the wheel on the side where it shall to turn supported on the left by a static object (red circle). Would the rod to be supported on the right side - this is clear to everyone - the wheel wouldn't turn. But in this case it does.

And if we take a second smaller toothed wheel instead of the rod that interlocks with the outer one - it should be the same thing. The smaller wheel is also supported by a static roller or a counterweight and cannot fall down. Because the test was successful this gadgetry should be also successful and able to turn more than only 90 degrees.

[fredos]

Yes, jonny.

I think like you that a toothed wheel placed inside the toothed external wheel
can make a turning momentum.

Personnaly, i think also that the ring of the bessler rotor can be replaced by a weighted toothed disc.

The experience can be made by using a simple saucepan.

The main axis of the saucepan is fixed at the gravity center of the saucepan and the saucepan can freely turn around it.
You have just to put a weighted disc inside the saucepan and this disc is supported by the other fixed axis outside of the saucepan.

but my wife not allowed me to use the saucepan to do the experience.
lol!
perhaps jonny you can have more chance than me!

[jim_mich]

OK, here we go again... I'll try to paint a mental picture if I can so I don't need to paint and post a real picture.

Look at Jonneynet's picture 'evertbas.jpg'. If I understand Jonneynet correctly the red roller is free to roll but is hung out on an arm fixed to a rigid wheel axle that cannot swing. The outer wheel is free to rotate around the axle. The inner weighted circle rests upon the red roller and the inside of the outer circle. (This is difficult to model in WM2D because it requires the use of a flat sided polygon shaped like a donut with a very large hole. Somewhere in my 800+ files I have such a model)

Now the really important concept that you need to grasp is that in this type of situation the inner circle will always touch the outer circle at a point that is EXACTLY on a line draw through the centers of the two circles. This means that the force from the weight of the inner circle pushes straight outward along that line; that is it doesn't push downward at all. The same thing happens on the other side where the inner circle pushes straight inline against the red roller. You can use geometry to calculate exactly where the inner circle will rest. Because the forces don't push the outer wheel sideway, it won't turn. If you gave things a push start they would coast.

Now if the red roller was free to swing around the center axle then the inner circle would push it upward as the inner circle drops. Depending on how much the red roller weighs the inner circle would fall to straight down if the red roller were weightless. It would fall somewhere short of straight down if the red roller were heavy. In ether case once it reached that position it would just rest there and do nothing more.

[rlortie]

fredos,

I agree with Jim that this design as shown is not a winner.

but my wife not allowed me to use the saucepan to do the experience.

I purchase inexpensive aluminum pie plates AKA 'Pie Tins" for experiments of this nature. They are handy, light and easy to work with. sometimes you need to reinforce the axle hub with fender washers.

Better yet, buy the pie tin that already has a pie in it, My favorites are Pecan and Banana cream.

Ralph

[winkle]

if you put Ralph inside that wheel and hold a pie in front of him you might have a working wheel till the pie is gone
be sure to get the right flavor pie or you may wind up with another nonworking wheel

if you have any BBQ ribs i might take a turn in that wheel ;)

[jonnynet]

jim_mich, I've read your explanation carefully and now, after a detailed WM2D simulation I understand where the problem is.