What is the Calculation for Finding Gear Ratio in a Multi-Gear System?

  • Thread starter Thread starter TSN79
  • Start date Start date
  • Tags Tags
    Motor Rpm
AI Thread Summary
To find the gear ratio in a multi-gear system, the initial calculation involves the motor speed of 1450 rpm and the desired output speed of 80 rpm, yielding a ratio of 18.125. The first gear's teeth are calculated using its diameter of 95 mm and module of 5 mm, resulting in 19 teeth. The square root of the gear ratio, approximately 4.26, is then used to determine the number of teeth on the second gear, calculated as 81 teeth. The discussion emphasizes the importance of understanding each gear's parameters and calculations for accurate results. Clarification on the entire problem setup is encouraged for comprehensive understanding.
TSN79
Messages
422
Reaction score
0
A motor runs at 1450 rpm. Attached is a gear that has a gear module of 5 mm, and a mid tooth-diameter of 95 mm. This wheel is toothed up with a second wheel who's shaft drives a third wheel. This third wheel is then toothed up with the fourth and last wheel who's shaft runs let's say dryer, which is supposed to run at 80 rpm. This last gear has a module of 8 mm. The last gear's diameter must be less than 700 mm 'cause of space. I'm supposed to find the number of teeth and the mid tooth-diameter for each gear. Here what my book does to begin with:

The ratio is calculated as

\frac{1450}{80}=18,125

z_1=\frac{d_1}{m_1}=\frac{95}{5}=19 (the number of teeth on the first gear)

Now my book does something I don't get. It finds:

\sqrt{18,125}=4,26

This number is then used to find the number of teeth on the second wheel:

U=\frac{z_2}{z_1} \Rightarrow z_2=U\cdot z_1=4,26 \cdot 19=81

As I said earlier, I can't see why the square root of the ratio is calculated...
 
Last edited:
Engineering news on Phys.org
Obviously the diameter of the first gear is 95 mm. Is there anything else you have failed to mention? How about writing out the entire problem so we know exactly what we have to deal with?
 
I've updated the problem above to include the whole thing now...
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
I have an engine that uses a dry sump oiling system. The oil collection pan has three AN fittings to use for scavenging. Two of the fittings are approximately on the same level, the third is about 1/2 to 3/4 inch higher than the other two. The system ran for years with no problem using a three stage pump (one pressure and two scavenge stages). The two scavenge stages were connected at times to any two of the three AN fittings on the tank. Recently I tried an upgrade to a four stage pump...
Back
Top