Don Bruder wrote:
I've got a gearbox out of a piece of hardware. Opening it up, I find a 6
tooth gear fixed on the input shaft driving a 48 tooth gear. Concentric
with (and fixed to) the 48 tooth gear is a 10 tooth gear, which in turn
drives a 64 tooth gear, which is fixed to the output shaft of the
gearbox.

Am I straight out of my mind to figure that I need 512 turns of the
input shaft to get one turn on the output shaft?

My figuring:
Output shaft, with fixed 64 tooth gear, needs 6.4 turns of the 10 tooth
gear for one rev, which means the 48 tooth gear the 10 tooth is attached
to likewise rotate needs to 6.4 times. In turn, this means that the 6
tooth input shaft needs to turn 48 / 6 = 8 times for one turn of the 48
tooth gear, so to get 6.4 turns on the 48 tooth gear, the input has to
spin 8 times 6.4 = .... Uh-oh ... writing it out like that makes me
realize I misplaced a decimal point... I need 51.2 (instead of 512)
turns of the 6-tooth gear to get 6.4 turns of the 48/10 tooth pair to
get one turn of the 64 tooth gear on the output shaft, don't I?

I *KNEW* 512 seemed like an awful high number of input turns...

So correction: Am I right in figuring this as a 51.2:1 reduction
gear-train? Or have I bollixed things up hopelessly, and I'm still wrong
even after recovering my missing decimal?

6 10 60 1
-- x -- = ---- = ----
48 64 3072 51.2

Looks good to me.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/