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Gear reduction question 
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#1
Mar1812, 10:35 AM

P: 159

hi everyone!
i was wondering how gear reduction is done besides connecting a small gear to a larger gear (radius speaking)? In my problem i have a gear  radius:5cm, angular velocity: 1000rpm I have a second gear  radius:5 cm, anguler velocity required: 1 rpm How do i make this happen? do i play with the numer of "teeths" on each gear, say that the first has 1 teeth and the second has 1000? Thanks 


#2
Mar1912, 05:57 PM

P: 343

The smallest pinions rarely ever have fewer than 12 teeth to avoid under cutting of the teeth during forming.
In order to achieve a particular gear ratio, it is necessary to find a number pair, a number quad (for two stage gearing), a number hex (for three stage gearing), etc. which in combination will give the required gear ratio. There are definite procedures for doing this, but most people simple start guessing. Only in rare circumstances is it required to hit a specific gear ratio with great accuracy. 


#3
Mar1912, 07:55 PM

P: 588

Here are a couple of pages that might help:
http://www.teamdavinci.com/understan..._reduction.htm http://www.cs.cmu.edu/~rapidproto/mechanisms/chpt7.html 


#4
Mar2012, 10:44 AM

Sci Advisor
PF Gold
P: 2,242

Gear reduction question
Keep in mind also that if the gears are directly meshing, they have the same tooth pitch. This means for a given diameter (5cm for example) the gears would have to have the same number of teeth. To accomplish gear reduction in this case you have to transition to a smaller pitch.



#5
Mar2012, 02:57 PM

P: 490

How much power do you have to transmit?



#6
Mar2212, 09:19 PM

P: 2

Gears should have an odd number of teeth as well to even out wearing.



#7
Mar2212, 09:29 PM

P: 343

"Gears should have an odd number of teeth as well to even out wearing."
Actually, this is not quite correct. The requirement is that in any individual mesh, the tooth numbers should be relatively prime. This means that there must be no common factor between the tooth numbers. If, for example one of the tooth numbers is a prime number such as 57, then the other tooth number could be any number and the condition of being relatively prime is still met since there would be no common factor between the prime number 57 and the second number which may be even or odd. 


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