Weight guessing and arm construction

You can trade off weight for power, but you can also trade off power for speed. If you robot doesn't have to go fast, you can use a smaller motor, and add an extra stage in your geartrain.

You need to sit down with a calculator and a tablet of graph paper. Figure out the torque you need for each joint, and the speed the joint needs to turn, and then calculate how much power you need from the motor. Add an extra 50% or so to overcome friction.

You don't have to make your arm heavy to make it strong. Look at a construction crane. It is extremely strong, yet is very light. This is because it uses cross bracing and tension cables to give it strength. You need to do the same. A hollow tube is almost as strong as a solid rod. An L-shaped angle is stronger than a flat panel, and an I-beam is stronger still. If you need more strength, don't make your metal thicker, just add an extra angle or brace.

You should also consider "worst-case" vs "normal-case". The worst case is when your arm is fully extended horizontally, trying to lift the maximum payload. But do you really need to lift so much weight when it is fully extended? Can you design your robot's workflow so that situation is avoided? For instance, can heavier objects be located close to the robot, and lighter objects located further away?

If you seldom have to work with heavy loads at full horizontal extention, you can use a smaller motor, and handle the rare cases by over-volting your motor. For instance, if you have a 12 volt power supply, you can use a 6 volt motor, and use 50% duty cycle PWM to keep it to 6 volts 99% of the time. But when you really need some extra power, you can give it the full juice for a few seconds. You should only use this technique if the motor will have time to cool off between surges, so you don't burn it up.

-bob

I have too been thinking of building an arm for a mobile robot base (on wheels). I have been looking at different motor technologies for the arm, and have also noticed the problem of construction weight and torque. Its no good to have an arm that just have enough power to lift itself.

What kind of motors do people use in arms? I have looked at servos, but the bigger ones from e.g. hitec and futaba cost around $80 each. And I have still not understood how much they can hold. The torque is specified in oz-inch (or kg-cm metric), but what does this really mean? It is easy to misunderstand that if you build an arm that you translate the torque directly to the weight of the arm and what it is supposed to hold, but that cant be quite right, since I would assume the torque would be less the longer from the motor the weight is. The oz-inch implies to me that it would able to lift oz amount of weight of the arm was 1 inch long. Is this correct? (If you look at the metric measures they are in kg/cm which is always a higher number since 1 cm = 0.39 inches). Since the weight is distributed along the arm (can vary since joints are always heavier because of new servos), I guess you would have to consider the total weight as being the sum of weights in a ratio to the distance of the joint. It this correct?

In that case, if I ignore this distributed weight but look at a linear extension of the torque, and take an example servo like the Hitec HS-645MG which has 133.31 oz/inch (at 6v). Would a 10 inch arm (25.4 cm) be able to lift 13.3 oz? (0.37 kg).

Can anyone clarify? It seems to me that you need some really powerful servos to be able to make an arm that could lift e.g. 5 kg.

I have also learned that you need metal gears for heavy torque, as plastic can easily get stripped. Although people in the RC environment recommend servos with one plastic gear so that it gets stripped rather than having the whole motor ruined when the weight gets too high (too bad about those porcelain plates the robot was holding :-)

Most joints seem to benefit from the servo sitting inside the frame of the arm (e.g. elbow), but servos only have one fitting to fasten the frame too it. Are there any servos that have a run-through axle so that one could fasten the frame on each side of the servo? I would also assume that fastening the frame on only one side would easily add sideforce to the whole servo axle, maybe even break it?

Best regards, JC

Hi JC

If you want more power then you have to have more room and accept the extra weight. Choose your motor by referencing the max power figure which will be an rpm and a torque. eg: a 2" frame DC cheapy, 0.726 lb in at 4289 rpm (You do have the amps available we hope), calculate the max speed you require, say 60 rpm and hence the gear reduction 4289/60=71:1 multiply by the power

0.726 lb.in x 71=4.4 foot.lbs, reduce that by the efficiency of your chosen reduction gear system, guess 70% and you have 3 foot pounds at 1 rps. Then check the stall torque, say 4.27 lb.in (assuming you have tens of amps), and get 17.5 foot pounds. Consider whether your gear train/arm can handle that kind of power :o)

best regards

Robin G Hewitt