Questions on Electric Motor 1

[VinceMash]

I am looking for information for the following motor:

http://www.aplussurplus.com/mirror/23238a.htm

I am looking for RPM on the output on the reducer.

The HP of the motor.

The torque on the output is 900 ncm = 79.65 in.lb

This is a german motor, I also do not know what the {i} on the reducer is, Md.max is the Torque (I was thinking that maybe the i means Ratio???)

Also is there any way to tell spin direction from the tag?

Thanks
Vince

 

[eemotor]

If the output due to the reducer is 79.65 in. lb. and if we assume constant HP output (rated output = 1/3 HP), the speed at the reducer output would be approximately 264 RPM based on:
T=(63025 x HP / N)

 

[VinceMash]

I know that the company selling this states that this is a 1/3HP motor but I don't know if I believe it, the motor only draws 1.4 Amps, is this possible?

Thanks

 

[Skogsgurra]

Hi Vince,

The name-plate data say 60 W output. That's a rather far cry from 1/3 HP - which is around 250 W. Input power is around 160 W. The motor speed is 1600 RPM at rated mechanical load. You will probably have 1800 RPM (two pole motor) at very light load.

The reducer has a 10:1 ratio (your guess is correct, "i" is used for gear ratio in Germany and elsewhere). The output shaft will run 160 RPM when fully loaded.

The Max torque number (900 Ncm) is what the shaft and the gears can take mechanically. Do not expect to get that torque out of the motor. 900 Ncm and 160 RPM would mean around 150 W out from the shaft. And you can not have that. Is this where the 1/3 HP comes from? Wrong interpretation if that is the case.

The insulation class is A, and that is rather poor. Do not run the motor hot! The built-in "Wärmewächter" will help you. A "Wärmewächter" is a thermal cut-off - like a Klixon - and it will shut off the supply if the motor gets too hot.

 

[Skogsgurra]

I forgot to mention that you will need a capacitor (20 microfarads at 200 V AC) to run the motor. It will run CW or CCW depending on how you connect the capacitor.

 

[VinceMash]

Thanks skogsgurra for the great info!

Two more questions:

What torque would you estimate I would get out of this motor?

The Motor has 3 leads and needs a capacitor to run, how would I need to hook this up? What would I vary to make the motor run CW as opposed to CCW


Thanks in advance

 

[jbartos]

Comment: Judging by the motor nameplate data:
V=115V
I=1.4A
P=60Watt ~ 1/(746(W/HP)/60W) ~ 1/12 HP
Speed=1600RPM

T=5260xHP/RPM in lb-ft
or
T=0.113x5260xHP/RPM in N-m
or
T=11.3x5260xHP/RPM in N-cm
=11.3x5260x(1/12)HP/1600RPM=3.1 N-cm
and the gear output will be
Tout=(10/1)x3.1N-cm=31 N-cm

 

[ScottyUK]

Hi skogsgurra,

Good post - but 1800rpm from a 2-pole (1 pole-pair) motor? 3600rpm from 2 poles at 60Hz, 1800rpm at 60Hz from 4 poles. This sounds like a 4-pole machine.




-----------------------------------

Ask a silly question and you are laughed at for a moment.

Don't ask the question and you might be laughed at for eternity.

 

[jbartos]

Correction (I beg your pardon)
T=(1/0.113)x5260xHP/RPM in N-m,
since 1 N-m=0.113 lb-ft
or
T=(100/0.113)x5260xHP/RPM in N-cm
=(100/0.113)x5260x(1/12)HP/1600RPM=242.44 N-cm
and the gear output will be
Tout=(10/1)x242.44N-cm=2424.4 N-cm
Apparently, the gear is supposed to be loaded 900 N-cm only.

 

[VinceMash]

Now I am really confused

so the output torque is 2424 Ncm...which converts to 214.52 lb.in. at 160 RPM...that really seems high for a 1/12 HP motor

 

[VinceMash]

I think I found the problem jbartos:

1 newton meter = 0.7376 foot pound-force

so T=[5252(1/12]/1600

T=.27354 Ft.Lb = 3.276 in.lb.

Tout= (10/1)3.276 = 32.76 in.lb. = 370 ncm


does this make sense?

 

[Skogsgurra]

Yes, Scotty. It is surely 4-pole. I do not know what made me say such a foolish thing. I didn't even think two pole pairs - a nice gesture from you to leave that way open...

 

[jbartos]

Thanks to VinceMash
Correction (I beg your pardon)
T=(1.3558)x5260xHP/RPM in N-m,
since lb-ft=1.3558 N-m
or
T=(100 x 1.3558)x5260xHP/RPM in N-cm
=(100 x 1.3558)x5260x(1/12)HP/1600RPM=37.14 N-cm
and the gear output will be
Tout=(10/1)x37.14N-cm=371.4 N-cm
Apparently, the gear is supposed to be loaded 900 N-cm only since some designs of motors allow the motor to have the motor starting torque of 250% or so of the motor full load torque.