verifying calculations for a motor spec

[tjump]

I'm throwing out some calculations to see if we are on target or have missed something.

We have a 10in arm that we are designing to lift 60lbs. The arm is driven by a 2in sprocket. The 2in sprocket is driven by a 1in sprocket (chain linking the two sprockets). The 1in sprocket is attached to the 3/8in output shaft of the motor.

What torque do we need out of the motor?

Using simple MA calculations and ignoring efficiency we have arrived at a value of approx. 160 in lbs.

Below is our calculation process

output lever has MA of 1:10 (radius of sprocket at 1" to length of arm at 10")
pulley system has MA of 2:1 (diameter of input sprocket at 1", diameter at output sprocket at 2in)
motor shaft to sprocket has MA of 3:8 (radius of motor shaft at 3/16in, radius of sprocket at 1/2in)

This gives us 800lbs at the edge of the motor shaft (3/16in radius) and thus at 1in out from the center of rotation of the motor shaft would be approximately 1/5 of this (3*5 = 15, ergo 15/16in) or 160 in lbs.

I would appreciate confirmation or corrections.

[ArcMan]

I'm throwing out some calculations to see if we are on target or have missed something.

We have a 10in arm that we are designing to lift 60lbs. The arm is driven by a 2in sprocket. The 2in sprocket is driven by a 1in sprocket (chain linking the two sprockets). The 1in sprocket is attached to the 3/8in output shaft of the motor.

What torque do we need out of the motor?

Using simple MA calculations and ignoring efficiency we have arrived at a value of approx. 160 in lbs.

Below is our calculation process

output lever has MA of 1:10 (radius of sprocket at 1" to length of arm at 10")
pulley system has MA of 2:1 (diameter of input sprocket at 1", diameter at output sprocket at 2in)
motor shaft to sprocket has MA of 3:8 (radius of motor shaft at 3/16in, radius of sprocket at 1/2in)

This gives us 800lbs at the edge of the motor shaft (3/16in radius) and thus at 1in out from the center of rotation of the motor shaft would be approximately 1/5 of this (3*5 = 15, ergo 15/16in) or 160 in lbs.

I would appreciate confirmation or corrections.

You're thinking into this too much.  The torque required by the motor is simply (60 * 10) * (1 / 2) = 300 lbf-in.

[Admin]

You should draw a FBD, it'll be obvious when you do.

Also, you can ignore all those gearing ratios and just use the motor shaft sprocket and main arm sprocket ratio.

The torque required by the motor is simply (60 * 10) * (1 / 2) = 300 lbf-in.

correct