right angles

[preoccupied]

I looked on PhysicsLab.org and found some information I tried to interpret for this. The math says it is not overbalanced but I will still look into new angles and distances. Maybe I will find something special.

35% less lever on one side does not overbalance.

In the beginning of the lift,
F=Cos30=.866 Sin60=.866 .5+.866=1.366, 1*.866*.866=.7499
T=F(.7499)*R(1)=.7499
( .7499 )
F=Cos45=.707 Sin45=.707 .707*1=.707, 7.07*7.07=.4998
T=F(.4998)*R(.65)=.3249
( .3249 )

At the end of the lift,
F =Cos20.7=.935 sin24.298=.411, 1*.935=.935, .935*.411=.3842
T=F(.3842)*R(1)=.3842
( .3842 )
F=Cos30=.866 sin105=.9659, .866*1=.866, .866*.9659=.8364
T=F(.8364)*R(.65) = .5437
( .5437 )

[path_finder]

In complement of my last assumption, hereafter a drawing showing how a balanced wheel can rotate under the displacement of its contact point.
Just a theoretical suggestion.

[path_finder]

Applying the above depicted principle, does have this assembly any chance to work?

The upper axle is fixed and linked to the ground.
The grey part is a pendulum supporting the curved rail where the central shaft (in light blue) has a mobile contact.
Any displacement of the wheel to the right will let rotate clockwise the main central shaft, therefore let rotate clockwise the main wheel thanks the unbalance.
The upper roller (in dark grey) is fixed, rotates on the inner rim of the wheel, and is intended to restore the top contact point at 12:00 of the main wheel thanks the belt.

This is just a first idea.
The motion could be complex, basically two pendula.
May be the ratio between the two pulleys must be adjusted.
May be a clutch is needed inside the upper shaft.

[preoccupied]

On the left side of the seesaw there are 15 axles taking turns turning 3 degrees each. On the right instead of starting level it starts already up 22.5 degrees and two axles turn from that position to the 45 degree position. The lever on the right is .85 inches and the lever on the left is 1 inch. The arms are 2 inch on the left and 1.202 on the right. The torque from 15 axles on the left calculated to be greater than the torque of two axles on the right and the distance traveled from two axles on the right is farther than the distance traveled from 15 axles on the left.

Math calculations did weight(1) X Cos from angle facing vertical piece X Sin from angle facing lever. That is Cos*Sin to get torque and the weight is 1 unit. I looked on physlab.org to try to figure out how to calculate this but I still don’t know if I did it correctly.

Torques
Cos30=.866 Sin60=.866, .866*.866=.7499 from when left side is level
Cos29.95=.866 and Sin57.05=.8391455, .866*.83912455=.7266606 From left side after 3 degree turn
Cos40.79=.75 Sin71.707=.9494, .9494*.75*.85=.6052 from beginning position on right at 22.5 degrees
Cos30=.866 sin105=.9659, .8364, .8364*R(.85)=.71094 ending position on right at 45 degrees

Distances
1.0409+.601=1.6419 is how high above the axle at the end of the lift on right side
.3252809+.91=1.2352 is how high above the axle when sitting at 22.5 degrees with the .85 inch lever.
1.2352 minus - 1.6419 distance above the axle after the lift = .4067
If two 22.5 degree turns on two different axles existed then .4067*2=.8134 distance pushed up. .78 is the distance traveled down from the left side and if .81 is the distance traveled up then we can create perpetual motion based on how I calculated it.