pole around corner with diameter considered

[galactus]

Hello everyone.

Has anyone ever derived a general formula for the pole of max diameter which can be moved around a corner when the diameter is considered?. I am sure everyone has seen the famous "pole around the corner" problem from calculus. The thing is, the diameter is neglected. That is unrealistic, but makes the problem easier.

The general formula when diameter is negligible is \(\displaystyle \L\\\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\), where a and b are the hallway widths.

This gets much more interesting and realistic when the diameter is considered.

 

[tkhunny]

No height? Even more interesting and more realistic.

 

[galactus]

Well, here's what I done with it.

In terms of theta, where x and y are the widths of the hallways and w is the diameter of the pipe.

\(\displaystyle \L\\L=ycsc({\theta})+xsec({\theta})-\frac{2w}{sin(2{\theta})}\)...[1]

Differentiate and get:

\(\displaystyle \L\\-ycsc({\theta})cot({\theta})+xsec({\theta})tan({\theta})-w(sec^{2}({\theta})-csc^{2}({\theta}))\)......[2]

Use technology, but set to 0 and solve for theta. Once we have theta, sub back into L and find the max length.

I will illustrate using y=8, x=4 and w=0.5(say, 6 inches in diameter):

Using those parameters in [2], setting to 0 and solving for theta, we find

\(\displaystyle \L\\{\theta}=0.910407119097\)

Sub into [1] and we find the max length is 15.62 feet.

This also works if the width is negligible as in the usual case where the diameter is not considered. \(\displaystyle \L\\\frac{2w}{sin(2{\theta})}\) would be 0.

Using the parameters from above, we find L=16.65. Which is what we get if we use the formula from the first post.

\(\displaystyle \L\\\left(8^{\frac{2}{3}}+4^{\frac{2}{3}})^{\frac{3}{2}}=16.65\)

I believe it works pretty good. Though, another method independent of theta would be nice to see, if practical.