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Don Bruder
 
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Default Sanity-check on gearing question, please?


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?

--
Don Bruder - - If your "From:" address isn't on my whitelist,
or the subject of the message doesn't contain the exact text "PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more info
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Tim Wescott
 
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Default Sanity-check on gearing question, please?

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/
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RoyJ
 
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Default 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.

  #4   Report Post  
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Koz
 
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Default Sanity-check on gearing question, please?



RoyJ wrote:

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.

Just curious about something here. Along the same line as the "which
side of the car should the gas filler be on", 1:51.2 seems like a weird
and arbitrary gearing. I can assume that it was just conventiently
fitting gears that matched some existing enclosure casting or something
like that but is there some logical reason someone would need 1:51.2 vs
1:50, 1:60 or something more "round"? Assuming a 1725 RPM motor (no
slip) the output of a 1:51.2 vs a 1:50 would be less than 1 rpm (2.4%)
different .

There are a lot of "weird" ratio gearboxes out there and I'm just
curious if there was more of a purpose or in the pre-VFD days, you just
had a lot of selection for any possible strange ratio you might decide
that you need.

Koz

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tonyp
 
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Default Sanity-check on gearing question, please?


"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





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Leo Lichtman
 
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Default Sanity-check on gearing question, please?


"tonyp" wrote: It's usually easier to do the arithmetic _backward_ up the
gear train. (clip)
^^^^^^^^^^^^^^^^^^
I divide driven/driving. Driven=product of teeth in all the driven gears.
Driving=product of all the driving gears.


  #7   Report Post  
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Don Bruder
 
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Default Sanity-check on gearing question, please?

In article ,
Koz wrote:

RoyJ wrote:

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.

Just curious about something here. Along the same line as the "which
side of the car should the gas filler be on", 1:51.2 seems like a weird
and arbitrary gearing. I can assume that it was just conventiently
fitting gears that matched some existing enclosure casting or something
like that but is there some logical reason someone would need 1:51.2 vs
1:50, 1:60 or something more "round"? Assuming a 1725 RPM motor (no
slip) the output of a 1:51.2 vs a 1:50 would be less than 1 rpm (2.4%)
different .

There are a lot of "weird" ratio gearboxes out there and I'm just
curious if there was more of a purpose or in the pre-VFD days, you just
had a lot of selection for any possible strange ratio you might decide
that you need.


The gearset under discussion is being driven by a 7.5 degree stepper
motor scavenged out of the steering system of an old robot. (Old as in
designed somewhere in the neighborhood of 1978)

I haven't diddled around with the math yet, but I'm thinking it likely
that the gearing was chosen to make "X steps equals 1 degree" a true
statement to ease the "which way is it pointing/which way will it be
pointing if I step X times?" type doings involved.

Hmmm... Since I'm thinking about it, and I'm going to need it (or at
least have it come in handy) later anyway...

360 / 7.5 = 48 steps per rev, so 51.2 turns = 1 rev at the gearbox
output means 48 * 51.2 = 2457.6 steps, with the assembly mechanically
constrained to 180 degrees of rotation, so / 2 = 1228.8 steps for the
half-turn on the output shaft... 180 / 1228.8 steps makes it 15.36 steps
per degree = close enough to 16 steps per degreee of desired high-torque
rotation (with a little bit of error, of course - but in this particular
case, that little bit of inaccuracy would have been irrelevant, since it
would be constantly adjusting anyway due to positioning feedback and/or
recalculation - Hmmm... Unless they were working with a 0-15 count to
get 16 possible values to feed into a 16-to-4 demux, which would make it
even closer to perfect) on four stepper phases makes it 4 cycles through
the phases per degree of desired rotation, which would make the
processing *REAL* easy for an old, slow 8-bit processor using integer
arithmetic, and would get the wheel pointed in a reasonable facsimile of
the proper direction fairly rapidly, even at the lowly 1MHz the poor old
6808 processor driving the beast was set up to run.

Makes sense to me - How about youse-guys? Or have I shot myself in the
foot again and not noticed I'm bleeding yet?

(Warning: *ALL* of this math was done "on the fly" as I keyed up this
message, so I wouldn't be too terribly surprised to find myself
"arse-deep in 'error'-gators"... Do please "pull me out" if that's the
case, huh? )

--
Don Bruder - - If your "From:" address isn't on my whitelist,
or the subject of the message doesn't contain the exact text "PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more info
  #8   Report Post  
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Don Bruder
 
Posts: n/a
Default Sanity-check on gearing question, please?

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"
somewhere, any message sent to this address will go in the garbage without my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more info
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john
 
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Default Sanity-check on gearing question, please?



Koz wrote:



RoyJ wrote:

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.

Just curious about something here. Along the same line as the "which
side of the car should the gas filler be on", 1:51.2 seems like a weird
and arbitrary gearing. I can assume that it was just conventiently
fitting gears that matched some existing enclosure casting or something
like that but is there some logical reason someone would need 1:51.2 vs
1:50, 1:60 or something more "round"? Assuming a 1725 RPM motor (no
slip) the output of a 1:51.2 vs a 1:50 would be less than 1 rpm (2.4%)
different .

There are a lot of "weird" ratio gearboxes out there and I'm just
curious if there was more of a purpose or in the pre-VFD days, you just
had a lot of selection for any possible strange ratio you might decide
that you need.

Koz



1, 2, 4 , 8, 16, 32, 64, 128,256, 512, an even pole stepper motor
will work out in even steps per revolution using this ratio.



John

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Emmo
 
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Default Sanity-check on gearing question, please?

is this from a Hero?

"Don Bruder" wrote in message
...

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?

--
Don Bruder - - If your "From:" address isn't on my
whitelist,
or the subject of the message doesn't contain the exact text
"PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without
my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more
info





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Ron Bean
 
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Default Sanity-check on gearing question, please?


Koz writes:

Just curious about something here. Along the same line as the "which
side of the car should the gas filler be on", 1:51.2 seems like a weird
and arbitrary gearing. I can assume that it was just conventiently
fitting gears that matched some existing enclosure casting or something
like that but is there some logical reason someone would need 1:51.2 vs
1:50, 1:60 or something more "round"?


The explanation I've heard is that if the gear ratio is an exact
integer, the same teeth will engage each other for every revolution of
the gear. If one tooth has a flaw, it will quickly grind a matching flaw
into the corresponding tooth on the other gear, causing various bad
things to happen.

If the gear ratio is not an exact integer, the wear gets distributed
among all the teeth, and the gear lasts longer (at least in theory).

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Emmo
 
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Default Sanity-check on gearing question, please?

I've got one sitting in my spare bedroom. Love to get rid of it - I'm in
Austin TX...

"Don Bruder" wrote in message
...
In article ,
"Emmo" wrote:

is this from a Hero?



Aye. What makes ya ask?

--
Don Bruder - - If your "From:" address isn't on my
whitelist,
or the subject of the message doesn't contain the exact text
"PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without
my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more
info





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Don Bruder
 
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Default Sanity-check on gearing question, please?

In article ,
"Emmo" wrote:

I've got one sitting in my spare bedroom. Love to get rid of it - I'm in
Austin TX...


Don't suppose you've got the schematics? Particularly the section(s)
pertaining to the "Main Drive" board?

--
Don Bruder - - If your "From:" address isn't on my whitelist,
or the subject of the message doesn't contain the exact text "PopperAndShadow"
somewhere, any message sent to this address will go in the garbage without my
ever knowing it arrived. Sorry... http://www.sonic.net/~dakidd for more info
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