Max Load Capacity: Calculating Total Torque for Dual Servo Motors

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Quarck_boson
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The servo motor that I'm using says it has torque of 1.89 kg-cm. And the radius of the wheel I'm using is 3.5 cm.
So i wanted to know how much load can two of these motors can together carry?​
 

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I assume you put the wheel with a radius of 3.5 cm on the servo.
This gives the servo a bigger radius which will reduce the torque.
radius * torque = 1.8 kg-cm
Substituting
3.5cm *torque = 1.8
Solving
torge = 1.8/3.5 kg-cm
= .514 kg/cm
 
Always read the question billy_joule. There are two motors.
1.89 kg.cm divided by 3.5cm is 0.54 kg
With two servo motors the total will be 2 * 0.54 kg = 1.08 kg.
 
There are a few other considerations when you say you have a wheel.
  • Is the wheel running on the ground and the motor shaft carrying the gravity load. This axial load from gravity or the pressure angle of a gear train or the tension from a belt drive may exceed the allowable load for the motor bearings.
  • If the load you are carrying is vertical then you have safety factors to consider on top of the allowable load
  • Another point is the stall vs peak torque. If the motor is holding the load then you can only use the stall torque. If the load is for short term acceleration you can go close to the peak torque which is often from 3-8 times higher than the stall torque
    If you want to use the peak torque then the current rating of the drive has to match the peak current of the motor​
  • Finally you need to think about system stability. If the reflected inertia of the load Mr2 where M is the mass of the load and r is the wheel radius is > than about 4-5 times the rotary inertia of the servo motor "J" then the performance and accuracy of the system will be difficult to predict. 4 is an arbitrary number that varies depending on friction, mechanical system stiffness, feedback system and amplifier response time. Generally the lower the cost of your control system the lower the number should be. In the early days of servo system design it was said that J should always be greater than Mr2 now a very few systems can work with Mr2=50*J but this is very unlikely to work in your case