In article ,
"tonyp" 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 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




I *THOUGHT* I *DID* do it backwards!

I got myself COMPLETELY lost trying to work it from input to output, and
switched around to figuring out what it would take on the input side of
each pair to get the desired rotation of the output side, starting from
the output shaft and working back to the input.

I still can't figure out where that darn decimal hid, though! Since I
managed to find it in the end, I guess it doesn't much matter

--
Don Bruder - - If your "From:" address isn't on my whitelist,
or the subject of the message doesn't contain the exact text "PopperAndShadow"
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