STEM Quiz #4

Hard to believe it’s been more than a year! Anyways, I have a new problem, inspired by a piece of cast iron cookware that I want to store safely. A correct solution isn’t particularly difficult, but there appear to be plenty of inefficient ways to approach the problem.

A part’s cross section is the union of a circle of diameter D, and a rectangle of size L-by-W which shares its center with the circle. L≥D≥W. What is the smallest breadth B of a box with width D which can contain the part? (This is a 2-D problem; if you want to got 3-d, assume this is a really long or really short cylinder.)

  1. (really easy) Under what conditions can the shape fit in a square size DxD?
  2. For sizes larger than that, what is B? If your solution generates ambiguous solutions*, describe fully how to disambiguate. (Good enough to write a bit of software to do it.).

* E.g. if you take an inverse trig function, define how to select among quadrants.

  1. if L=D

  2. this feels inelegant and wrong, but:

  1. Clearly, some L>D can fit in the square:
  2. I defined the same \theta, but went a different (simpler) route. Going your route, the calculation of \theta at the right side of line 2 should be a difference rather than sum of angles. You should also further constrain cos^{-1} and tan^{-1}. (first quadrant?).

For those who want some numbers to play with, the part which inspired the problem has D=9", L= 11⅛", W=3¼".

Hint: ignore the circle other than specifying the height of the outer rectangle to be D. This can actually be solved without trigonometry, though a bit of first week trig will simplify the derivation.

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