Sanity-check on gearing question, please?
Correct!
Tim Wescott wrote:
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.
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