So if I have an average rpm of 100 rpm and the rates torque is .45 kgcm, does that mean this has a power output of 4000 Watts?

I know this answer is coming to you 3 years too late - and you probably figured this all out. Just in case - here is how you can calculate the power from the torque and RPM. A couple of quick notes: Torque is force times length - kg is not a force but a mass. If the "torque" is given in kg*cm - then this is not a proper torque number. It should be in Newton*cm. Newton is the unit for force and kg for mass. The force a mass experiences here on earth is the mass multiplied by the earth's acceleration of the mass in the earth's gravity. This acceleration is 9.81 m/s/s (meters per seconds squared). This unit is the same as N/kg (not that readily visible - but for now we'll just accept that as correct). Now we can calculate the proper torque by multiplying the two numbers: 4.5kg*cm * 9.81 N/kg = 44.2 N*cm. Power is work done per time - work, energy and torque are identical units of measure - force times length - the continuously delivered torque is the same as the work done by the motor. However, for the power to be correct we have to express work in N*m (SI units) - not N*cm. So we have translate N*cm into N*m. 100 cm = 1 m or 1cm = 0.01 m. We can just simply replace the cm with 0.01m and well get the torque (work) in N*m - 4.42N* 0.01m = 0.442 N*m. Next: power is Nm/s in SI units - that means we have to express 100 rpm in rps (revolutions per second = rev/sec). 1 min = 60 seconds. 100 rpm = 100 revs per 60 secs = 100/60 rps = 1.7 rps. Now we can calculate the power in N*m/s: 0.442N*m * 1.7 rps = 0.74N * m / s = 0.74W of mechanical power - which is even less power. Of course that's only correct if the 4.5kgcm number is correct for the motor. Some of the suggestions were to use the electrical power of the motor to estimate the mechanical output power. That too is a legit approach - however, that power setting given must be under the same conditions for which the mechanical output was given. E.g. the motor must run at 100 rpm and deliver the torque given - only then the current draw by the motor and the voltage accross the motor can be used to calculate the approximate mechanical power. For example - if the motor draws 1A at 12V when stalled out (motor is slowed down to a stop), then this is not the same condition as running the motor at 100 rpm. Sometimes the specifications given for those products can be a bit questionable.

Can this fit into a 1/4” metal rod? With a shaft collar or something?

If you are widening a coupler and you can, use a drill press or milling machine. Be sure the setup is as rigid as possible. Use a "reamer" 1/4" in diameter if you can get one. They are special purpose tooling but get used all the time for this exact job. They cost a few dollars. Use very low speed or even turn the chuck by hand. Do not use a hand dill at full speed with a twist bit. I would rather use 400 grit sand paper or fine abrasive padding than any sort of standard drill and bit.

Manufacturer	Greartisan
Part number	Greartisan
Item Weight	210 g
Item model number	Greartisan
Size	12 Volt
Color	50 RPM
Material	Metal
Voltage	12 Volts
Installation Method	Screw-In
Item Package Quantity	1
Speed	50 RPM
Cutting Diameter	37 Millimeters
Batteries Included?	No
Batteries Required?	No
Item diameter	37 Millimeters

Greartisan DC 12V 50RPM Gear Motor High Torque Electric Micro Speed Reduction Geared Motor Centric Output Shaft 37mm Diameter Gearbox

Details

Description

Specifications

Have a Question? See What Others Asked

Common Questions