"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 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?


It's usually easier to do the arithmetic _backward_ up the gear train.

1 output turn causes 6.4 turns of the intermediate shaft.

1 intermediate shaft turn causes 8 turns of the input pinion.

Thus you get 51.2 turns of the input per turn of the output.

It's harder to lose track of the decimal point that way :-)

-- TP