Find the angle that lets you make a bendy pipe

Question

I have a pipe with $n$ segments of equal length. I move the farthest end nearer than the combined length to make the pipe bend. How do I find the angle between each segment with only the combined length (straight length), the distance between each end, and the number of segments?

$theta$ is the angle to solve for
$l$ is segment length
$n$ is the number of segments
$dx$ is how the distance between ends

With only two segments I would use the law of cosines:$dx^2=2*l^2 - 2*l^2*cos theta $

Ross Millikan · Answer

You have to assume all the angles between the segments are the same or the problem is underdetermined. In that case all the bends lie on the arc of a circle. A figure with four segments is below. $BG$ is your $dx$, $n=4$, and $l approx 3.6$ as drawn. All the triangles are isosceles. If angle $DAB=theta, GAB=4theta, GB=2rsin(2theta),l=BD=2rsin(theta/2)$. You need to use $l, GB$ to get $r, theta$

Find the angle that lets you make a bendy pipe

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