(adsbygoogle = window.adsbygoogle || []).push({}); 1. This is a question that I've asked a few people and have not been able to get a solid explanation to:

The crankshaft gear (A) has 14 teeth. It drives gear (B) with 21 teeth which is on a shaft & on the other end of the shaft is gear (C) with 9 teeth. The distributor is driven off gear (C); calculate the number of teeth on the distributor gear (D).

2. In normal 4 stroke internal combustion engines, the distributor rotates at 1/2 the speed of the crankshaft. This means the distributor gear would have twice the number of teeth of the crankshaft gear (A) but in this case, not true due to the addition of gears (B) & (C).

3. My lecturer has given us a solution to this question, but it's long-winded & confusing & I'm in doubt of it because this is not his field of specialty.

Lecturer's solution:

Given gear (A) = 14 teeth gear (B) = 21 teeth gear (C) = 9 teeth, gear (D) = ?

B/A = 3/2 so 2 revolutions of B = 3 revolutions of A.

If gear A)makes 2 revolutions then gear C makes:

2/3 * 2 = 4/3

2 revolutions of Gear (B) = 2 revolutions of Gear (C) so:

4/3 * 9 = 12.

Answer: 12 teeth.

My solution:

Since the distributor gear (D) should have double the teeth of the crankshaft gear (A) & the ratio provided by gears (B) and (C) is 9/21 then:

28 * 9/21 = 12 teeth.

Answer: 12 teeth.

I don't know if either of these answers are correct or if the approach to the solutions are correct so any logical information will useful and much appreciated, thank you in advance!!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Calculating gear teeth & ratios.

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