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)

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.

Thanks!! That visual helped a lot