# Involute gear ratios?

• posted

In the explanations (W.O.Davis, T.C.F.Stott) for constant angular velocity, appeal is made to a smooth belt coming off the base circle of one gear wheel and being pulled onto the base circle of the second gear wheel, and it is then said that provided the line of action between two mating teeth follows the path of the imaginary belt, then constant angular velocity will be imparted from one gear wheel to the other.

I have no dispute with that. It is a concise explanation of what is happening.

My question is this. The explanation given suggests that the velocity ratio between the two gears would be determined by the ratio of the diameters of the base circles taken with the imagined belt, but it is not, it is determined by the ratio of diameters of the pitch circles, which are bigger.

Can anybody out there solve this quandary?

• posted

That is because the velocity ratio between two gears is not directly determined by the ratio of pitch diameters. It is determined by the number of teeth on each gear. This number is directly proportional to the pitch diameter.

Paul K. Dickman

• posted

I thought the base circle reference explained constant angular velocity as the point of contact rolled across the two mating teeth, the same as the involute string unwinding. The base and pitch diameters are in equal proportion so you could also say the tooth tip speeds are equal, which is true but irrelevant.

jsw

• posted

Gear teeth engage at the pitch diameters, not at the base diameters. The belt analogy is valid only if a belt of some thickness (like a V belt) runs on base dia, or if a belt of infinitesimal thickness somehow runs on the pitch diameters. In the thick belt case, there is necessarily a thin part running at the pitch diameters; other parts of the belt compress or stretch as the belt goes round a pulley.

• posted

Oops.

Should have read the next two pages in the book before posing my question :-)

The base circles are in proportion to the pitch circles by a factor of the cosine of the pressure angle, and thus the two base circles are in the same proportion to each other as are the corresponding pitch circles.

• posted

Thanks Jim, as I subsequently found and posted.

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