Gear problem and number of revolutions

[donnagirl]

A gear, named p, and another gear, named q mesh smoothly so that they turn simultaneously with each other. Each gear rotates about its fixed center. If gear p has 17 teeth and gear q has 51 teeth, how many revolutions will gear q make if gear p makes 15 revolutions?

The answer should be 5 but how do I show this? Proportions gives me an answer of 45

 

[pappus]

A gear, named p, and another gear, named q mesh smoothly so that they turn simultaneously with each other. Each gear rotates about its fixed center. If gear p has 17 teeth and gear q has 51 teeth, how many revolutions will gear q make if gear p makes 15 revolutions?

The answer should be 5 but how do I show this? Proportions gives me an answer of 45

1. The number of revolutions is reciprocal to the number of teeth. The more teeth a gear has the less revolutions it makes.

2. Let r_q denotes the number of revolutions of gear q:

\(\displaystyle \displaystyle{\frac{r_q}{15} = \frac{\frac1{51}}{\frac1{17}} = \frac{17}{51} = \frac13}\)

3. Solve for \(\displaystyle r_q\)

 

[donnagirl]

But why does that proportion work? Shouldn't '15' go in the numerator as it is the number of revolutions that correspond to the gear with 17 teeth?

 

[srmichael]

But why does that proportion work? Shouldn't '15' go in the numerator as it is the number of revolutions that correspond to the gear with 17 teeth?

Pappus stated that the number of revolutions is reciprocal to the number of teeth. That is why in the proportion he set up he did \(\displaystyle \displaystyle \frac{\frac{1}{51}}{\frac{1}{17}}\) which, when you do this division becomes \(\displaystyle \displaystyle \frac{17}{51}\)

 

[donnagirl]

Ah yes of course. Thank you both!