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I have drilled a hole.........

Well, I did not drill it, but about 50 years ago a colleague produced this question from his book:


"A cylindrical hole six inches long has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?"

Enjoy!

Clive
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  • Simon Barker:

    For an accurate result, there's a tricky complication.  The hole is being drilled into a curved convex surface.  If you drill a hole that's six inches long once it's drilled, then you have drilled away slightly more than six inches of material.


    From memory, the solution in my colleagues book was in part reliant on the knowledge that the questioner was always 100% right. Therefore you had all the information. I have seen a truly mathematical solution which I'm afraid is beyond my maths, and which I think surprised the person doing it when he found that the answer simply that of a solid sphere 6 inches in diameter. The simple answer is that if I have all the information I need, then knowing that the required diameter of the sphere is related to the diameter of the hole,  I can choose the diameter of the hole to suit my calculation.  (I think that makes sense?)

    Cheers!

    Clive


Reply
  • Simon Barker:

    For an accurate result, there's a tricky complication.  The hole is being drilled into a curved convex surface.  If you drill a hole that's six inches long once it's drilled, then you have drilled away slightly more than six inches of material.


    From memory, the solution in my colleagues book was in part reliant on the knowledge that the questioner was always 100% right. Therefore you had all the information. I have seen a truly mathematical solution which I'm afraid is beyond my maths, and which I think surprised the person doing it when he found that the answer simply that of a solid sphere 6 inches in diameter. The simple answer is that if I have all the information I need, then knowing that the required diameter of the sphere is related to the diameter of the hole,  I can choose the diameter of the hole to suit my calculation.  (I think that makes sense?)

    Cheers!

    Clive


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