Let's say we have a motor with a continuous power rating of 1/8 Hp at 1725 RPM, at 1.5A and 75 V. Using the conversion from HP to torque, which is 1 Hp = 550 Ft-Lb/Second, we can calculate that 1/8 Hp would be 68.75 Ft-Lb/Sec. So, this means a 1/8 Hp motor will be able to lift a 68.75 Lb load at the rate of 1 foot per second. This is the same as lifting a 1 Lb load at 68.75 Ft/Sec. Now, 1725 RPM = 28.75 Rev/Sec, which is the shaft speed for the rating point. What size spool would be needed to reel in a rope at the rate of 68.75 Ft/Sec?
Well 68.75 Ft/Sec / 28.75 Rev/Sec = 2.39 feet (circumference of spool). The radius of that spool (circumf. / (2*Pi) ) would be 0.38 feet, or 4.567". So, we have a torque of 1 Lb on a 4.567" lever arm, or 4.567 In-Lb.
So, the continuous rating point of the motor in question is 4.567 In-Lb. ! Ahh, that feels good to finally have a number to look at. This is the motor's Kt, or Torque Constant, when referred back to 1 Amp. So, the Kt = 4.567 In-Lb * (1 Amp/1.5 Amp) = 3.04 In-Lb/Amp. (In units of Oz-In, that torque would be 73 Oz-In., just multiply by 16.)
Now, this is continuous rated torque, so we have to select a peak value for the motor, based not only on the absolute maximum the motor can deliver, but on the servo amp's current capability, and other factors. To be conservative, let's select a 200% peak value, which would ocurr when this motor was getting about 3 Amps. The torque (ignoring armature reaction which should be minimal if the motor is designed for this current) will be essentially 200% of the cont. rating, or 9.1 In-Lb.
Let's assume a 2.5:1 belt reduction ratio from the motor to the leadscrew, and a 5 TPI leadscrew (equal to .200" lead per revolution). The 2.5:1 reduction multiplies torque by a factor of 2.5, so peak torque at the LEADSCREW will be 9.1 x 2.5 = 22.75 In-Lb. Now, to avoid a mess of geometry to calculate the pitch angle of the thread in the leadscrew, we take a shortcut and assume the leadscrew is just a rope wrapped around a pulley of some diameter to produce the equivalent motion (.200") for each revolution of the pulley. This is entirely equivalent, although it makes it impossible to figure out the efficiency of the screw. If you use a ballscrew, this should be good enough. With an Acme or other friction-producing screw, the friction under load can be substantial, but the calculation can be a bit messy.
So, to get an advance of .200" per revolution, the pulley would have to be .200/(2*Pi) = 0.0318" in radius. This is not practical, but it will produce the correct linear force from the ballscrew described! So, if torque 22.75 In-Lb is applied to a shaft, a force of 715 Lbs will be available at a radius of .0318". That approximates the linear force such a ballscrew would develop!
So, this system would deliver 715 Lbs. linear force to the table with 3 Amperes applied to the motor armature!
In this case, we know the motor is rated at 1725 RPM at 75 V. The motor speed constant Ke, expresses armature voltage at 1000 RPM, so this motor has a Ke = 75 * (1000 / 1725) = 43.48 V/KRPM.
Now, we know the leadscrew takes 5 turns/inch, so we can get leadscrew speed as IPM * 5 = RPM. So, if we want a 120 IPM rapid feed rate, that would be 600 LEADSCREW RPM. The motor has a belt reduction of 2.5, meaning the motor turns 2.5 times as fast as the leadscrew, so that would be 1500 MOTOR RPM. Motor voltage is Ke * (RPM/1000) = 43.48 * (1500/1000) = 65.22 Volts. You should add some factor for losses in the motor windings, amplifier and a bit for torque surges needed to keep the loop closed. At least a 1.25 factor would be a good idea, so 65.22 * 1.25 = 81.5 Volts. So, to reach the speed of 120 IPM, we should use a DC power supply of at least 81.5 Volts.