Force formula F=mgu m=mass g=gravity u=coefficient of friction (7kg)(9.81m/s^2)(0.5)=34N

Torque formula T=Fd t=torque f=force d=radius of pulley (34N)(4mm)=136Nmm add 50% for more torque, so it’s 206Nmm

  • i don’t know if I’m on the right track, can someone help me.
  • 1
    $\begingroup$ don't forget to accelerate your mass too. Just fending off gravity gets you moving at 0m/s aka not moving at all. plan out acceleration curves so you can make the moves you need to over the time you have. $\endgroup$
    – Abel
    Mar 4 at 2:02
  • $\begingroup$ I’m still confused. Do mind doing an example? $\endgroup$
    – Tyler Huor
    Mar 4 at 14:27
  • $\begingroup$ It's a pick and place cartesian x and y, so I think like 2 or 3 secs $\endgroup$
    – Tyler Huor
    Mar 4 at 14:28
  • $\begingroup$ 2 or 3 sec to travel the full range so how big is the range (distance in each axis)? $\endgroup$
    – Abel
    Mar 4 at 23:15
  • $\begingroup$ The X-axis is about 2 feet 5.5in and the Y-axis is about 1 feet 6.5in range. $\endgroup$
    – Tyler Huor
    Mar 5 at 4:38

1 Answer 1


Short version: You need beefier motors than what you got to move fast enough for pick and place at those speeds.

Motion requirements: A mass must move from any point in available range to another in S seconds.

Force = mass* acceleration

distanceUnderConstantAcceleration = 0.5accelerationtime^2

distanceUnderConstantVelocity = speed*time

Under normal operation, you accelerate to a max speed, run at that speed for a while, and then decelerate to a stop.

distance = 0.5accelerationacceltime^2 + speedtimeAtSpeed+speeddeceltime-0.5decelerationdeceltime^2

If you assume acceleration and deceleration happen at the same rates, it factors.

distance = 20.5acceleration*(timeToAccel)^2 + speed*remainingTime

There is a lot variable so to tease out rough numbers, start by pretending you spend all the time accelerating and decelerating (remainingTime=0). You can think of it as a need to cover half your range in half the time while accelerating as fast as possible.

For Y: 0.47m = acceleration*(3s/2)^2 or acceleration = .209m/s^2

similarly for the X axis: 0.75m = acceleration*(1.5s)^2 or acceleration = .334m/s^2

Force after friction considerations will usually be your 7kg times this for one axis and (7kg + mass of the other axis assembly) for the other.

For now just using 7 kg yields: Y: 1.47N X:2.34 N

Thing about accelerating though is that you impart a force on the other axis which adds to friction, just like gravity.

7kg*9.81m/s^2 = 68.67N

Combine via root mean square with the other axis and multiply by the coefficient of friction.

FrictionX = (coefficient)*sqrt((68.67N)^2 + (1.47N)^2)= 34.34N

FrictionY= 34.35N

(Not very much impact compared to just gravity unless you accelerate more)

37 N for X, 36 N for Y Minimum. On .04m pulleys, that's 20.9 and 20.4 oz-in. That might look friendly until you realize motor specs generally list stall torque and unloaded RPM.

A second implication is that at this target acceleration, you must be able to achieve a max velocity of acceleration*time. 0.314 and 0.5 m/s then yield 754 RPM and 1200 RPM @ .025m/rev. Usually one finds something slower than this with much higher torque and introduces the additional constant velocity segment.

At constant speed (assuming 0 time under acceleration), it would take .157 and .25 m/s or 377 and 600 RPM. Anything at or below or even slightly above would have insufficient time to accelerate.

Find motors, check their torque vs RPM curves to make sure you have excess of both torque and speed necessary.

Looking at the numbers, you may want to use a bigger pulley so you need less RPM. It does impact resolution with steppers, but you may be able to get finer stepping motors (.9 deg/ step instead of the usual 1.8) to make up for that.


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