# Lenth of line around an object?

by Physics is Phun
Tags: lenth, line, object
 P: 1,004 The area that a line of length L and width D is S = LD. The problem is you don't have the width of the line (its thickness), D. But you have Z, the number of times you wrap the line, so you can use it to find D = X/Z. Going back to the area S, that's also the area of the zone the line covers which is S = 2πYX. So finally you have: $$S = LD = L\frac{X}{Z} = 2\pi YX$$ $$L = 2\pi YZ$$ Grizzlycomet's method is of course shorter and easier, but this is a more general description of how to go about solving such problems when dealing with objects more complicated than cylinders.
 Sci Advisor PF Gold P: 2,226 Well the surface of a Cylinder is Euclidan, so imagine drawing the line your string makes and unwrapping the surface of the cyclinder, what would it look like? If you wnat the string to go from top to bottom (or bottom to top) and once around the cylinder it's lenght will be given by Pytahgoras's theorum where r is the radius and h the height: $$L^2 = 4\pi^2r^2 + h^2$$ For n twists it will be: $$L^2 = n^24\pi^2r^2 + {h^2}$$