Converting oz/in to Watts

[mikie]

Hey Guys,

I have tried looking everywhere to convert oz/in to watts and i can not find the multiplier for it.

Can anyone help.

I want to know what 200 watts gives me in oz/in

/Mikie

[ger21]

Watts = volts x amps. You can have 2 motors with the same power (watts), with a large difference in torque.

[OCNC]

Originally Posted by mikie Hey Guys, I have tried looking everywhere to convert oz/in to watts and i can not find the multiplier for it. Can anyone help. I want to know what 200 watts gives me in oz/in /Mikie

You're comparing apples and oranges. A watt is a power measurement and oz-in is an energy measurement in other words a watt is energy per time and oz-in is simply energy. Oz-in would convert to joules. Watts would convert to horsepower.

[Milo]

Watts is a measure of power (work) i.e. voltage x amperage, Torque x RPM. If a motor is generating 500 oz/inches with 0RPM then no work is being generated (500oz/in x 0RPM = 0). To convert torque to watts you will have to include the RPM in the conversion.

[Evodyne]

Originally Posted by ger21 Watts = volts x amps. You can have 2 motors with the same power (watts), with a large difference in torque.

Power is also speed x torque.

To get Watts, speed must be expressed in radians/second. To get this: 1 rev/min * 2PI radians/rev * 1 min/60 seconds = 0.10472 rad/sec. So 1500 rpm = 1500 * 0.10472 = 157.1 rad/sec.

Torque must be expressed in Newton-meters. To convert, note that 1 oz-inch = 0.0070616 Newton-meter. So 300 oz-in = 2.12 N-m.

The motor will be delivering 157.1 rad/sec * 2.12 N-m or 333 Watts.

746 Watts is 1 HP, so 333 Watts is 333/746 or 0.45 hp.

The motor will have zero power output when stalled (rad/sec=0) and at max no load speed (output torque = 0). Peak output power will be at 1/2 the max no load speed for the given supply voltage. The difference between the input power (V*I) and output power (speed*torque), at a given speed is, for the most part, what is turned into heat (I squared R losses). A very little amount of the loss is due to internal friction and viscous losses. The ratio of output power to input power is the motor's efficiency.

Hope I did my math right or I'll be totally embarrased!

Lance

[OCNC]

Originally Posted by Milo Watts is a measure of power (work) ...

Technically power is not work. Power is work per unit of time, ie. joules/sec. Work is simply energy, ie. joules.

[mikie]

Thanks guys,

that is quite a bit of information, i now have to rememebr all those years of Physics.

I will look p the specs on the units i bought and try to figure out what i can build with these units.

/M

[Switcher]

http://www.allegromicro.com/techpub2/compumot/a71.pdf

http://www.motionautomation.com/technica.htm

[mikie]

Thanks Switcher,

some great reference material there.

/M