We’re going over the math for our climber gearbox and something about our physics doesn’t seem quite right to me. I’d love to get another set of eyes on it to confirm or deny my suspicions.
We’re planning a climb similar to Snow Problem’s, but using a sprocket driven arm instead of retracting a winch. We wanted to know how changing the position of the pivot on the arm (i.e. ratio of x/l in the drawing) would change the torque needed. When we drew it out however, we found that the maximum torque is only dependent on the distance between the arm pivot and robot CoM, not on the arm length. We are assuming quasi-static motion, so the robot CoM is always hanging below rung 2, and a massless arm.
If that’s right, we run into a problem when doing the chain calculations. With a full weight robot and a distance r of 0.7m, we get that the sprocket has to transfer 480 Nm of torque. Even with a large custom sprocket ~9" diam. sprocket, we’re still more than double the max working load of #35 chain and more than quadruple 25 chain. Not to mention that we’d need a ~500:1 ratio if we want to climb with 2 falcons without blowing breakers.
Something seems wrong here, but I’ve gone over the math a few times and I don’t see it. Maybe @snowproblemz did some calculations when designing their Ri3D climb? Thanks in advance for any help you can give.
Even if the CoM of the robot is directly below the rung, you’re clearly still doing work raising the robot. As the arm passes through the horizontal, the torque will be the robot weight times the arm length (mgx).
Your fallacy is thinking that \alpha will get to a right angle. In actuality, sin\alpha \le x/r.
You might save a bit of fabrication effort by using a bicycle crank.
Once the free end of the arm hits the higher rung, the arm is not free to rotate. So your chain driven system MUST rotate the body of the robot. So your math at the end is assuming this is the situation at all times.
Ever try to hold a 1kg mass on the end of a 2m long stick? Your hand, at roughly 0.5m from the end of the stick, needs to support ~10N of weight, or exert a force of 200N, or 45lbs. That is hard on the fingers!
Ok, that makes sense. We were assuming "plan for the worst case, which is \alpha=90^\circ. But now looking at it I agree that \alpha will never reach there, it’s limited at \arcsin(x/r). So we should be able to lower the torque on that sprocket by decreasing x fairly substantially.
I can’t speak to the calculations done by the team over the build, but a key reason we used a winch over a sprocket was that we were more confident in the winch’s ability to withstand the forces. Since the load on the winch was a tension through the rope, not a moment due to weight, it was easier to be sure that we wouldn’t blow out our motors.
I would also advise teams that from our experience, quasi-static motion is not necessarily a safe assumption, our design tended to swing quite a bit, contributing to our high bar climb not working very well.
Looking at your picture I would say the weight to the right would want to swing downward, which would drive the rotating arm upward in the non connected side, CoM wants to center below bar. Just my thoughts though
Why are you worried about that T_max? For T_max to equal mgr, you would need to have x=r and have your CoG inside Rung 2. The real worst case scenario would be when your arm of length x is horizontal. A horizontal arm makes r*sin(alpha)=x. Anything above or below horizontal would make rsin(alpha)<x.
So, correct me if I am wrong, but I think you should actually worry about T_max_real=mgx.
I don’t have a fancy tablet to draw on, so my yellow legal pad calcs will have to do.
