Calculating imperfect linkage angles

How would you calculate the changing angles and lifting limit of imperfect linkages?

(i don’t know the real name of the mechanism, but it’s a linkage that’s offset slightly to get varying angles as it rises)

Do you mean like a not parallel 4 bar or chain bar so something else entirely.
For chain bar you take the ratio say 4:3 so it is 1:3 more than it needs to be so it will be turned 1:3 of the angle raised.
4:3 ratio 180 degrees will angle shift 60 degrees.

cos B = (a2 + c2 -b2)/(2ac)

The capitol letter (angle) is opposite the lower case letter (side) on the triangle.
Just find the side lengths at the extremes, when the sides form a triangle.

The not parallel 4/6/8 bar

Is that the law of cosine?

Basically you have to solve the whole mess to get the resultant equation:
[ATTACH]6628[/ATTACH]
Anyone want to check my work?
*Supplementary

Let’s say on your average VEX 4-bar (in inches):
a = 3
b = 17
c = 17
And if you want to try a non-parallel lift, let’s plug in:
d = 4

Your equations now look like:

e = sqrt( 3^2+17^2-2*3*17cos( 90-B ) )
e = sqrt( 298-102cos( 90-B ) )
D = arccos( ( 4^2+e^2-17^2 ) )/( 2*4*e )
D = arccos( ( 25-102cos( 90-B ) )/( 8sqrt( 298-102cos( 90-B ) ) )
C = arcsin( 3sin( 90-B )/( e ) )
C = arcsin( 3sin( 90-B )/( sqrt( 298-102cos( 90-B ) ) ) )

A = 90-arcsin( 3sin( 90-B )/( sqrt( 298-102cos( 90-B ) ) ) )-arccos( ( 25-102cos( 90-B ) )/( 8sqrt( 298-102cos( 90-B ) ) )+B

And that’s the angle A as a function of parameter B and constants a, b, c, and d. Be cautious of the inverse trig functions; if you aren’t wary when using that equation you could get incorrect answers.
It’s a huge mess that I would avoid by using guess-and-check on the bar lengths.
Linkage Angles.JPG

Thanks!! That visual helped a lot