pic: Ira's Senior Project

[Andy Baker]

Originally Posted by Sylvia

My calculations were:

4200 RPM / 12.75 = 329.4117647 RPM output
329.4117647 x (2/1 gearbox to wheel ratio) = 658.8235294 = RPM wheels are driven at
658.8235294 RPM / 60 = 10.98039215 RPS (duh)
Circumference = d(pi) = 6(pi) = 19.44
10.98039215 RPS x 19.44” = 213.458823396 in/s
(213.458823396 in/s) / 12 = 17.77 ft/s

Sylvia,

All of this is correct. Good job on the calculations. Here is some advice:

1. Set yourself up a simple Excel file to do these calculations. Once you get your equations determined, then you can change a few variables (wheel dia., sprocket size, etc.) to get final speed results.

2. You don't need to track any decimal places past the tenths for this application.

3. Your 17.8 ft/sec makes sense, but it's too fast. Ira's setup was giving you about 5.8 ft/sec. If you increase the speed (by 2x) of the wheel by putting a larger sprocket on the gearbox, then you get 11.6 ft/sec. Also, if you increase the wheel size to 6", then that 11.6 is multiplied by 6/4, which results in 17.4 ft/sec.

What is the speed you want to go?

Keep it up,
Andy