Looking for a formula to convert RPM at motor to Hz at the VFD.

With the givin information I need to be able to determine the correct freq. for the VFD to run at to achieve a certain line speed. Here's the information:

Motor operating at 60 Hz has RPM of 1180
The Gear Box has a 159:1 ratio
The Drive sprocket on the gear box is a 13 tooth
The Driven sprocket has 19 a tooth
The net line speed for the sprocket pulling the 2 strand conveyor link off of the Driven is 67 Jobs per Hour

What is the freq needed for this line speed.

I keep tweaking the formula for each new freq but this results in a fractional number of poles for the original motor when compared to the actual conveyor found in the field. I think I just need a new perspective for this equation...

All the rest of your values are in units of rotation and time.

We have no idea how many revolutions of the output sprocket equals a "Job".

So you need to determine the ratio between a "Job" and the output sprocket.

There are 4.618 revolutions for each unit indexed..

67 "Jobs" x 4.618 revolutions = 309.406 revolutions per 60 minutes
= 5.1567 revolutions per minute.

If 60 Hz gives you 1180 RPM, through gearing ratios of 1:159 and 19:13, then 60 Hz gives you 10.8466 RPM on the output shaft.

(5.1567 / 10.8466) x 60 Hz means you need an output frequency of 28.525 Hz to get the desired line speed of 67 "Jobs" per hour.

It's late here and I'm running on a lot of coffee and not much sleep so check the math.

The VFD in the field is currently set at 55 Hz... hince the issue????