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My work has a bunch of iMac boxes and because of their slightly wedged shape we are curious how many it would take to make a complete circle.

We already did some calculations and also laid enough out to make 1/4 of a circle so we know how many it would take, but I'm curious to see how others would approach this problem mathematically and see if you came up with the same answer.

Our stock consists of 12 of the 27" iMac boxes and 16 of the 21.5" iMac boxes.

The 21.5" box has the following dimensions

top: 5"

height: 21.25"

bottom: 8.75"

and the 27" box is

top: 5.75"

height: 23.75"

bottom: 9.5"

21.5" iMac box picture for reference

enter image description here

Community
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Justin Garrison
  • 291

2 Answers2

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Trying to avoid trigonometric functions: The outer circle must be longer than the inner circle by $2\pi$ times the height, so compute $$\frac{2\pi\text{height}}{(\text{bottom}-\text{top})} $$ as a good approximation.

Hagen von Eitzen
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Suppose we have $n$ wedge-shaped boxes wrapped around to make a circle, with an inner length (per box) of $C_1$, an outer length of $C_2$, and a 'diametric' length of $D$. Then putting the $n$ boxes together will yield an inner circle with circumference $n\times C_1$, and an outer circle with circumference $n\times C_2$. By the formula for the area of a circle, the diameter of the inner circle is then $\frac{nC_1}{\pi}$, and similarly for the outer circle, so the difference in diameters is $\frac{n}{\pi} (C_2-C_1)$ - but this has to be exactly equal to the twice the diametric length, $2\times D$ (since there we add an additional distance of $D$ on either side of the diameter). In other words, we have $\frac{n}{\pi}(C_2-C_1)=2\times D$, or solving for $n$, $n=\dfrac{2\pi D}{C_2-C_1}$.

For instance, for your first case, we have $C_1=5$ (inches), $C_2=8\frac34$, and $D=21\frac14$, so we get $n=\dfrac{2\cdot\frac{85}{4}\cdot\pi}{\frac{35}{4}-\frac{20}{4}} = \frac{34}{3}\pi\approx 35.6$ boxes. Unfortunately, since this isn't very close to an integer, any circle you make will be either a little tight or a little loose. I'll let you solve the second case yourself...

Steven Stadnicki
  • 51,746
  • @Rothgar Oops - I missed a factor of two in my answer, good catch! Let me correct that. – Steven Stadnicki Jul 17 '13 at 23:57
  • Thanks, that makes more sense now and the formula relflects the same answer given by @hagen-von-eitzen – Justin Garrison Jul 18 '13 at 00:01
  • Wait a sec, actually doing the math on this the number comes out to (170pi)/15 = 34pi/3 = 7*pi = 35 boxes. Let me double check my math. – Justin Garrison Jul 18 '13 at 00:06
  • @Rothgar Glah, I need to double-check mine - I divided 85 by 15 and got 7! (As opposed to $\frac{85}{17}=5$). Hang on, more corrections! – Steven Stadnicki Jul 18 '13 at 00:07
  • @Rothgar There, it should truly be fixed now; I didn't mean to live up to the cliche of mathematicians (or even enthusiastic amateurs) being bad at arithmetic! – Steven Stadnicki Jul 18 '13 at 00:10
  • I deleted my first comment because I remembered that when we actually laid out the boxes 11 per quarter was too much (44 total). We actually used 9 of the small boxes to roughly make 90 degrees (quarter circle) which lines up perfectly with your 35.6 boxes. Thanks – Justin Garrison Jul 18 '13 at 14:48