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How to calculate mechanical advantage of a worm gear?

My textbook simply use the turn ratio as the mechanical advantage, but I'm not sure how that works.

My thinking: If the worm has a radius of $r$, and a turn ratio $n$ (turns/meter), the input distance is $2\pi r$, and the output distance is $1/n$ for each turn of the worm.

Therefore, mechanical advantage is the former divided by the latter which is $2\pi r \times n$. I'm not sure how this relates to the turn ratio. Is their definition of mechanical advantage different?

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Hmmm...the turn ratio gives the the ratio of torques without reference to the radius at which the worm is driven. However "mechanical advantage" usually means the ratio of forces for which we will need to know at what radius the worm is driven. You may be meant to assume that it is driven at it's own maximum radius. –  dmckee Apr 28 '13 at 14:19
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