Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.