Well...back to the old drawing board. lol... well I'll keep thinking about it but this might be one of those problems that have no simple solution.

]]>(edit) whoops, messed up there, make that 543.226 inches. And I forgot to add the first quarter turn. But we don't need to worry about that much right now.

Well I finished a program and I think I got all the bugs out. It keeps giving me 542.667 but my formula gives me 543.226. Still I'm relying on the accuracy of the built in trig function of my C++ compiler and also the level of accuracy to which it can hold numbers.

Still, a difference of about 1.4 is enough to worry me.

I'm going to try writing a program to use numerical integration to find the area under one arch of cos x and compare it to the exact answer I get by integrating and see how much error there is.

(edit) did that and it came out pretty close. Either theres a minor bug or my formula is wrong. I suspect the latter. :-/

]]>Look at this little experment.

Say we divide a circle into n subsections, as shown. (illustrated as if n = 8) we then draw a radius to the edge of each subsection and then draw a perpendicular line at the point of tangency and extend it untill it intersects the path of the original radius. We know have n sides with a length of h where h/r = tan ( 2 pi /n ) if we sum all n sides we have n * tan ( 2 pi / n) (n * h)

If we take the limit as n approaches infinity, it ends up being 2 pi r which is what it would be if it were a circle. So the infinitsimal incorrect position of the line did not accumulate after an infinite summation and still remained at zero.

Ugh... it makes sense yet it doesn't make sense.

]]>In my summation system, finding an arc m at every point, we have to find the length of the string at that point, and rotate it an infinitsimal degree at the point of tangency, however, as the angle chances even by an infinitsimal, the point of rotation changes and the radius also changes. Do we need to take this ratio of change into account? Or will the infinitsimal difference in the radius and rotaton point tend to nothing? One thing calculus teaches us, is that an infinitsimal, times infinity does not always equal nothing.

]]>So we write a formula to find arc m at each value of k, then find how many rotations is required to real the string all the way in (easy stuff)

but first I need to convince myself that this type of summaton will work. I'll explain in a minute.

]]>Say you have a 36 inch string 1/8 inches thick, attached to a ball at one end, and attached to a poll with a radius of 1 inch at the other, the ball starts at the position shown and begins to rotate about the poll. As the string wraps around the poll the string adds to the radius of the poll. How far does the ball travel before hitting the stick?

For the simplicity of the problem.

Assume the location of the ball is at the very tip of the string, and the ball itself does not get in the way when the very end of the string hits the poll.

Assume the growth of the radius of the stick is at a constant (a spiral) and does not simply increase by the thickness of the string only after the rotation is complete. Instead, assume that in half a rotation the radius has grown by half the thickness of the string, in 5/16 rotations, the radius has grown by 5/16 the thickness of the string and well you get the idea.

Also note, the ball will travel 90 degree's before the string even begins to wrap.

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