# Torque in a 2-segment robot leg

I have a (hypothetical) robot with two legs, each with 2 segments, with a joint at the body of the robot and in the 'knee' of each leg, as in this (poorly-drawn) diagram: (Sorry about the quality and lighting and everything)

I want to know how much torque is applied at joint(s) B by the force of gravity, and therefore how much torque needs to be applied by a motor in the joint to exactly counteract the force of gravity and keep everything stationary.

It would be nice to get a general answer for the following ranges of the variables:

$$\hspace{1cm}90^\circ<A\leq 180^\circ, (270-A)^\circ\leq B \leq 180^\circ \hspace{1cm}0^\circ \leq C\leq90^\circ$$ $$0\ \text{kg}<M\leq100\ \text{kg}\hspace{1cm}0\ \text{m}<X,Y\leq 2\ \text{m}$$

The legs can be assumed to be weightless, and everything is at rest.

• I edited the question to reflect physically relevant values for the angles. In the most obvious cases, $C$ should be $\leq 90^\circ$, while $180-A+180-B$ should be $\leq 90^\circ$, so we get a convex shape.
– Danu
Oct 3, 2014 at 15:17

A quick diagram:

Key dimension here is the distance $d$. The weight of the robot $W = M \cdot g$ is carried equally by both legs, so we have a force $F$ along the lines $AC$ such that

$$F = \frac{W}{2 \cos\alpha}$$

This force results in a torque at point $B$ because $B$ is not on the line $AC$ - it is displaced by distance $d$, given by

$$d = Y \sin\beta$$

And finally, you combine to get

$$\Gamma_B = F\cdot d = \frac{M\cdot g\cdot Y}{2\ \cos \alpha \ \sin \beta}$$

I hope you can translate this to the dimensions / angles you were using in your diagram.

• I edited the question just now, I hope it didn't mess with (the relevance of) your answer. If it did, feel free to re-edit!
– Danu
Oct 3, 2014 at 15:18
• @Danu - pretty sure it's OK. I am leaving the translation from my frame of reference to that OP as an exercise... Oct 3, 2014 at 15:20