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Imagine that you have 4 spheres of equal size of the diameter 6cm. You then place two of them on opposites sides inside a cylinder with the diameter 14 cm and following that placement you put the remaining two with identical positioning, but rotated 90 degrees; meaning that they'll be placed slightly higher than the first two spheres as the two new ones essentially are "leaning" against the first two. How does one calculate the height, from the bottom of the cylinder, to the top of the two highest points of the spheres?

English is not my first language, sorry for any issues. I am also fairly new to this math forum, so pardon any mistakes. Thank you in advance!

1 Answers 1

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  1. Let $r=\frac{6}{2}=3$ denote the radius of spheres, $R=\frac{14}{2}=7$ denothe the radius of cylinder
  2. Notice, that the distance between middle points of two spheres on the same level is equal to $a=2R-2r=8$
  3. Notice, that distance between middle points of two spheres on different levels is equal to $b=2r=6$

  4. Notice, that if you look at the picture from above, the centres of the spheres are vertices of the sqare with diagonal $d=a=8$. On this 2D-picture the distance between two centres of spheres is then equal the length of this square edges, ie. $c=\frac{d}{\sqrt{2}}=4\sqrt{2}$

  5. Let $h$ be the difference between levels of centres of balls. After Pythagoras: $$h^2+c^2 = b^2$$ $$h^2=-c^2 + b^2 = 36-32=4$$ $$h=2$$

  6. The height of the whole cylinder is then equal to $H=h+2r = 8$
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    Lovely! Indeed, in step 4, you can say that from above, by symmetry, the four sphere-centers form a square (so the diagonals obviously meet at right angles) – 2017-01-19
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    @JohnHughes I've changed steps 4 and 5 to make this solution clearer. – 2017-01-19
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    Sorry, just gotta make sure I've understood correctly. How could you tell step 3 that the distance from their centers is 6? – 2017-01-19
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    @MiltonMiller - these are two tangent spheres. – 2017-01-19