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Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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#1
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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 |
#2
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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/ |
#3
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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
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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 |
#5
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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 |
#6
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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
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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
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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 |
#9
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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 |
#10
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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 |
#11
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Sanity-check on gearing question, please?
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 |
#12
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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). |
#13
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Sanity-check on gearing question, please?
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#15
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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 |
#16
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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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