4 ms·
Consider a box l x b x h let x = width of one corridor. let y = width of the perpendicular corridor. We will disregard the height of the box here since I ass
by niyazpk 17y ago
Consider a box l x b x h
let x = width of one corridor.
let y = width of the perpendicular corridor.
We will disregard the height of the box here since I assume that you can find out the correct way to orient the box.
The box will fit throught the hallway if:
(sqrt(x^2+y^2) - l/2) >= b
I will post the reasoning after I get out of my office.
BTW, thanks for posting this intersting problem.
- NoBSWebDesign 17y agoI really am curious as to how you arrived at this equation. It is a vast over-simplification of the problem. The equation I came up with is as follows: y1 = [l/(2b(1/tan(a)^2 - cos(a))] * [x - l/cos(a) + b/sin(a)] + l/cos(a) + b/sin(a) where you plugin b, l, and x based on your parameters, and find the maximum value for y1 when a goes from 0 to 90 degrees. That maximum value should be less than y for the box to be able to fit around the corner.