Help selecting gearmotor

[jberge1]

Hey all,

I'm having trouble selecting a motor/gearbox combination to help me with a torque-testing machine I am trying to make. The motor distributor is recommending a motor that I think might be too big. What I want to do is apply a torque up to 2500 in lbs to a servo coupling, and turn it up to 500 RPM. There will be a gearbox on either side (hub city parallel shaft gearbox model 290) with a
Candy Controls Dynamics differential. The dynamics differential will have an input and output shaft that both turn the same direction and same speed. With a knob adjustment you can advance or retard the output shaft by up to 100 degrees. This will apply the torque to the coupling. Then since the output and input shafts continue to turn at 1:1, it maintains that torque while the whole thing is spinning.

I don't believe the gear motor itself needs to be capable of 2500 in lbs output of torque. That torque will be introduced into the gear train with or without the motor. But I do have quite a bit of inertia moving here (the two main gearboxes are about 12 feet apart.) And all that torsion would probably make the friction in the gearboxes resist the rotation by some amount.

Any suggestions on HP, gear reduction etc? I attached a screenshot of the concept.

 

[BrianE22]

We used to call that a "4-square". You only need to supply enough power and torque to overcome the losses in the gearboxes. Or if you need acceleration then you need the torque to do that. I assume the motor will be under speed control and the differential will be under torque control? You will need an estimate of the power you need for the electrical guys to suggest a motor type.

 

[jberge1]

Brian,
Finally someone gets it! Haha yes it's a Four-square. Not many people I've come across have heard of that.

I estimated a 30% combined loss from all of the gear boxes. I was planning on having an inline torque sensor from Magtrol monitor the torque. The motor would have a variable speed drive.

The big question I have at this point would be how do I estimate the power that the motor needs? Is there a formula based on inertia? Motor calculations is not my strong suit.

 

[jberge1]

I forgot to mention acceleration should be relatively slow so as to not spike the torque by any substantial amount.

 

[BrianE22]

If your gearboxes are 80% efficient then your motor/input gearbox torque will need to be 3 x .2 x 2500 = 1500 in-lbs. Output speed from the input gearbox would need to be 500 RPM. So gearbox output power = 11.9 HP. This ignores acceleration and any losses from the specimen.

You should check the torque required for a decent acceleration. You'll need to add the inertias of all the spinning items (shafts, gearboxes, specimen).

Add the acceleration torque and the "losses" torque and add a safety factor to get the actual required torque from the motor/input gearbox.

 

[jberge1]

Wow that was a great explanation. Thank you for helping me understand.

I calculated about 5HP needed to overcome the losses. For acceleration up to 500 RPM in about 10 seconds:

500 RPM *2 pi/10 seconds = 314 rad/sec^2 (fairly slow acceleration)

Solidworks tells me all that mass has a moment of inertia of 3 Kg*m^2 if I add the shafts end to end.

So torque = Inertia * Acceleration = 314*3 =~ 1000 Nm

Which I estimate would need 70HP+ 5HP for losses, totalling 75HP to get it from stopped to 500 rpm in 10 seconds.

Does that make sense?? If I got "only" a 30 HP motor and let it take much longer to accelerate, wouldn't I run the risk of stalling the motor and burning it out? 75 hp just seems excessively large to test some servo couplings.

 

[jberge1]

I found the error in my math. Should've divided 500 rpm by 60 first before getting angular velocity.
Can't thank you enough for helping me answer this.

 

[mikekilroy]

with a dyno and two motors, one drives, 1 loads, and both on same system, all you need to provide is losses; you only have one motor, so at the end of the day if you produce 2500##-in torque @ 500rpm, that is (2500/12)*500/5252=19.8hp or 19.8*.746kw/hp= 14.8kw

I understand the 4 square (never saw this before) is loading the backside of the coupling at upto 100 degrees position offset, producing the torque, at up to 500rpm at the coupling, but with a single motor, unless you have some type of perpetual motion machine attached and not shown, does it not require 2500#-in torque @ 500rpm?

Add to this your 30% losses for a required minimum 28.3hp. I will research 4 square to see what I am missing....

Since quick accel is not in your test requirement, then 0 additional torque is required for it - 10 sec is FOREVER to this kind of application and additional torque required for it with this system will by gut feel be in the 1-10#-in range: not worth calculating.

Assuming nothing, I would consider feeding this with a 3:1 belt drive with a 10hp 1800rpm motor & standard vfd drive.

http://www.KilroyWasHere<dot>com

 

[mikekilroy]

WOW! how cool! I found:

Several power re-circulating test rigs are available for low speed and high speed gear testing. Utilizing a 4-square kinematic mechanism, the load on the gears is obtained by applying a torque on the test gears against a pair of reversing gears. The motor driving the test rig has only to provide for the frictional losses in the kinematic mechanism. Low-speed (speeds up to 3000 rpm) test rigs are generally utilized for rotating bending fatigue tests and are configured for a 4 inch center distance.

So looks like Brian's answer is way to go.

I would call it (2500/12)*.7*500/5252= 13.9hp = 10.3kw

So I would use a 5hp, 1800rpm motor & vfd and call it a day! Thanks for showing this 4 square device!

http://www.KilroyWasHere<dot>com

 

[mikekilroy]

sure wish there was an edit button!

(2500/12)*.3*500/5252= 6hp
So I would use a 2hp, 1800rpm motor thru 3:1 belt & vfd

http://www.KilroyWasHere<dot>com