An easier version of this puzzle is
here.
A large panel needs to be moved through a corridor, the panel is tall as the corridor. The corridor is A feet wide before a right angle turn, after the turn, it is B feet wide.
What is the maximum length of the panel that can pass through this corner.
Overhead view of the hallway:
+------------+---
| / |
| / |B ft
| / |
| /+------
| / |
| / |
| / |
| / |
| / |
| / |
| / |
|/ |
+<-A ft-->|
Let's call the portion of the panel from the inside corner to the wall of corridor A x, and the portion from the corner to the wall of B y. Then x + y = Length. Theta is the angle formed between the panel and the side wall of A, and also between the panel and a line parallel to width B.<o:p></o:p>
sin theta = A/x cos theta = B/y<o:p></o:p>
x = A/sin theta y = B/cos theta<o:p></o:p>
L = x + y = A/sin theta + B/cos theta<o:p></o:p>
To find the minimum length (or the max allowable), differentiate the above equation to get:<o:p></o:p>
- Acos theta/(sin theta)squared + Bsin theta/(cos theta)squared = 0<o:p></o:p>
or Acos theta/(sin theta)squared = Bsin theta/(cos theta)squared<o:p></o:p>
rewrite to A/B = (tan theta)cubed or tan theta = cube root of A / cube root of B<o:p></o:p>
So picture a right triangle with sides of cube root of A and cube root of B. The hypotenuse H equals the square root of the sum of the (cube root of A) squared and the (cube root of B) squared.<o:p></o:p>
Sin theta = (cube root of A) / H cos theta = (cube root of B) / H<o:p></o:p>
L = A/sin theta + B/cos theta = A/(cube root of A) / H + B/(cube root of B) / H<o:p></o:p>
Simplifying, min. L = (A to the 2/3 power plus B to the 2/3 power) raised to the 3/2 power.<o:p></o:p>
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Posted by lake
on 2004-10-26 16:48:45 |