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enter image description hereI take a sphere of diameter d and remove two ends to create two bowls each having a depth of d/4. If I bring these two bowls together it forms a 3D, flying saucer shaped, Vesica Piscis whereby the saucer's diameter around (x), divided by its height/depth (d/2) equals Root3. I drop three smaller spheres into one bowl & they bunch together around its central vertical axis...I sit another three into the first to form a six sphered octahedron...and then I place the second bowl over...the upper three spheres sit as perfectly in the top bowl as the lower sit in the bottom....a perfect fit of the six within the flying saucer.. How do I determine the diameter of each of the six smaller spheres please?

I say how but in all honesty I probably wouldn't understand any process offered...personally, with little knowledge of math, I just assume there to be some constant appertaining to the diameter of the original sphere and the diameter of the smaller?

Eg: Six regular 40mm ping-pong balls require a Vesica/Saucer formed from a sphere of diameter 40 x some Constant?

Thanks/Gill

Gill Simo
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 13 '21 at 22:44
  • It would be great if you could provide a diagram – TheBestMagician Sep 13 '21 at 22:48
  • You should prepare an image to illustrate the configuration you have. It is really hard to figure out that out from your description. – achille hui Sep 13 '21 at 22:49
  • Six spheres in a flying saucer and a bottle of rum :) I think one can carefully (and slowly) read and follow your description, and it makes sense, I believe I see what shape you are talking about. Of course a picture would help, but my point is that it is not THAT HARD to understand what you are asking (though it takes time). Re the answer, that's another matter, I may think about it. Welcome to MSE ! This may also help understanding: https://en.wikipedia.org/wiki/Vesica_piscis – Mirko Sep 13 '21 at 23:18
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    Restatement of problem. Let O,N,S be the center, north and south pole of a sphere of diameter x, let U be the midpoints of ON, and V the midpoint of OS. Cut the sphere with a horizontal plane through U and another horizontal plane through V. Throw out the piece between the planes and put the cap(at the top) onto the cup(at the bottom) to form the flying saucer. Take an octahedron with side y, with horizontal top and bottom triangular faces, center a sphere of diameter y at each vertex, and assume this all fits snugly into the flying saucer. What is the ratio x:y? It my interpretation correct? – Mirko Sep 14 '21 at 00:37
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    @achillehui While I have to agree that a picture would help, I also tend to believe that in this case the configuration may be a bit overloaded, and that making a good 3D picture may not be a simple task in itself. The result being that a verbal description may actually be a pretty good choice of a medium, after all, to convey the problem. I believe I understood the question (though I have yet to think of the answer), and I attempted an alternative verbal description in the preceding comment, which hopefully (...pretty sure) corresponds to the same question. – Mirko Sep 14 '21 at 01:14
  • @Mirko there is no need of a really good 3D picture, a quick sketch is enough. The point is just based on the description, it is not immediate clear to me the 6 spheres are inscribed inside the "flying saucer" as you suggest or the centers of the spheres lies on the surface of the saucer. One simply should not need to spend multiple minutes to decrypt a description to figure out the configuration. – achille hui Sep 14 '21 at 01:24
  • @achillehui "the upper three spheres sit as perfectly in the top bowl as the lower sit in the bottom." I guess it is a matter of taste, I enjoyed deciphering the meaning! One need not spend any time at all at this website to begin with. I was just trying to suggest that the description was not that bad, and to offer my interpretation, so as to try to help others understand the statement and perhaps contribute to the solution, as I think the problem is interesting. Thank you for your time and reply (didn't mean to involve you in an argument as to what is the proper way to state this problem)! – Mirko Sep 14 '21 at 01:51
  • Thank you all for the interest/effort thus far...I must apologise if my inability to state myself clearly leads to heated debate on the matter. – Gill Simo Sep 14 '21 at 14:14
  • I have attempted to add a simple diagram...hopefully I've succeeded & it helps to clarify what is in essence very simple. A baby toy as such...a clear perspex flying saucer or hang drum shape with a colourful six ball octahedron held firmly/centrally within....If the six colourful balls are each 40mm in diameter, for instance, then what diameter ball is the flying saucer cut from? – Gill Simo Sep 14 '21 at 14:22
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    +1 good illustration. – achille hui Sep 14 '21 at 14:35

1 Answers1

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Let $1$ be the radius of each small sphere (and thus $2$ the edge of the octahedron), and $r$ the radius of the large spheres forming the vesica. The center $O$ of the octahedron is also the center of the vesica, and is at a distance $r/2$ from the center $C$ of one of the large spheres.

Point $O$ is at a distance $\sqrt{2/3}$ from the center $G$ of the upper face of the octahedron, while $G$ is at a distance $2/\sqrt3$ from a vertex $A$ of that face. But $AGC$ is a right triangle and $AC$ must be equal to $r-1$ if the large sphere has to be tangent to the sphere centred at $A$. Hence we have the equation:

$$ (r-1)^2=(r/2+\sqrt{2/3})^2+(2/\sqrt3)^2, $$ which can be solved to: $$ r= \frac{2}{9} \left(6+\sqrt{6}+\sqrt{69+12 \sqrt{6}}\right) \approx4.08197. $$

enter image description here

Intelligenti pauca
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  • I had a question but realized I knew the answer: If $T$ is the point where the small and large spheres touch, you are using that $C,A$ and $T$ are on the same line. (Thanks for the picture, though I had just come up with the correct pic in my head before that.) I asked Computer Algebra Reduce to solve the above equation, it came with a different form of the answer, but obviously one ought to be able to get from one form to the other. Numerically it seems the same: $\frac29(\sqrt{3}\sqrt{4\sqrt{6}+23}+\sqrt{6}+6)\approx4.08196855972$ (coincidentally close to $\frac{10}{\sqrt6}\approx4.0824829$) – Mirko Sep 14 '21 at 20:40
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    @Mirko I checked with Mathematica and can confirm that both expressions are the same. I edited my answer because your form is simpler. – Intelligenti pauca Sep 14 '21 at 20:56
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    A very big thank you for taking the time/effort to answer this...as I imagined, I haven't the slightest notion re the math involved but hopefully others here will appreciate the artistry displayed. – Gill Simo Sep 15 '21 at 11:11
  • Please may I display my ignorance one more time.....am I correct then to assume that my six 40mm ping-pong balls will require a saucer cut from a sphere a tad under 164mm in diameter? – Gill Simo Sep 15 '21 at 11:14
  • @GillSimo Yes, that's correct: the value of $r$ given above is just the constant you asked for at the end of your question. – Intelligenti pauca Sep 15 '21 at 15:33
  • Brilliant....really can't thank you's enough! – Gill Simo Sep 15 '21 at 19:04