Swing Drive w/ ring & pinion

[somarp]

I need help – I am working on a hydraulic swing drive with ring and pinion. I am trying to determine what the final speed of the rotation will be.

The ring specs are:

12.5” pitch diameter
50 tooth

The pinion specs are:

3.5” pitch diamter
14 tooth


I am using a small Eaton gear motor to drive the hydraulic swing drive.

I have calculated the pump flow as follows

2450 (engine rpm) * 1.78 (pump disp) * .90 (Pvol eff.) * 1.116 (engine drive ration)/231 = 19 gpm

Then calculated the motor speed

231 * 19 gpm / 11 (motor disp) * .95 (Mvol eff) = 420 rpm

Then calculated the swing drive gearbox output speed

420 rpm / 25.78 (swing drive ratio) = 16.3 rpm

Now is where I am stuck – not sure how to calculate the final swing speed with consideration of the ring and pinion. I realize this may be an easy question for some, but I am not an engineer or anything.

Thanks.

 

[diamondjim]

16.3rpm times 3.5/12.5 = 4.564 rpm for the ring gear.
4.564 times pi times 12.5 divided by 12
yields 14.936 sfm (surface feet per minute).