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So here is the problem. My dad designed this umbrella that is to be wrapped with rope. The rope is 1 inch thick, and the umbrella is made of poles that arch away from a center beam extending upward from the ground. My question is how many linear feet of rope would I need if I were to wrap the poles, under and over, starting from the leftmost (or rightmost) pole and working towards the center.

http://i.imgur.com/eUeyv.jpg

Larger view of the image with measurements

The red arrows are the rope, and 1 foot is to be left at the end after wrapping the rest of the umbrella. I only included arrows on two of the drawings. How would you go about solving this? Let me know if you need any more information. Thanks!!

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The diagram doesn't have enough information to solve the problem.

The answer is $2\pi\cdot{R_i + R_o\over 2} N$ where $R_i$ and $R_o$ are the radii of the innermost and outermost circles of rope, and $N$ is the number of windings around. We have $R_i$, which is 1 foot (1 foot left unwrapped as shown in the upper left of the picture), or maybe 1 foot plus the thickness of the inner stem, and $R_o$, which is 14 feet (8+6 feet as shown in the upper right of the picture).

If the umbrella were flat, we would have $N$ equal to $R_o-R_i = 13$ feet divided by the thickness of the rope, which is 156 times around. But the ribs are curved, so they are more than 13 feet long. We need to know the length of the ribs, as measured with a flexible tape measure from the central stem to the end of one of the ribs. If this length is $L$, then $N= L\div 1\text{ inch}$, and you should be able to use the formula above. But there is nothing in the diagram to tell us what $L$ is.

Still we can be sure that the rope will have to go around considerably more than 156 times, so each umbrella will need considerably more than $2\pi\cdot7\frac12\text{ feet }\cdot156 = 7,351$ feet of rope.

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    Yes, and that's why the formula involves the length of the *average* pass, $2\pi\cdot\frac12(R_i+R_o)$, rather than the length of the longest pass, which is $2\pi\cdot R_o$, or the length of the shortest pass, which is $2\pi\cdot R_i$.2012-10-12