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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
  • There's some ambiguities and missing information in that question.
    • What's the width (diameter) of the hole?

    • Does "straight through the centre" mean right through the sphere and out the other side, or does it only mean the hole was deep enough to get beyond the middle of the sphere?


    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 incles long once it's drilled, then you have drilled away slightly more than six inches of material.
  • Simon Barker:

    There's some ambiguities and missing information in that question.
    • What's the width (diameter) of the hole?

    • Does "straight through the centre" mean right through the sphere and out the other side, or does it only mean the hole was deep enough to get beyond the middle of the sphere?


    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 incles long once it's drilled, then you have drilled away slightly more than six inches of material.


    Sorry about this Simon, but there is no missing information! However the hole is all the way through.

    Clive


  • If the hole is truly cylindrical, as opposed to having domed ends, then it is infinitesimally thin. So the volume remaining is that of a sphere of a 3" radius i.e. 113 and a bit cubic inches.
  • Good answerComet
  • I'm not sure what it means for a hole to have domed ends.  What's the shape of something that isn't there?
  • This is I fear also the basis of a question from the Oxford Physics entrance exam from some time in the 1970s or 1980s. I recall  a metric version but otherwise  exactly the same from one of the past papers I looked at when I was looking at old papers to get a feel before sitting the entrance exam myself. ~(*)

    The key point is that once you know the length of the cylinder, you can express the radius of the sphere in terms of that length and the diameter of the cylinder bore. What you realise when you write it down, is that the remaining volume ends up  independant of the drill size, as to keep the hole l ngth constant, the sphere gets bigger as the drill does. I could  come back and show it on a bit of paper if it is not obvious.

    Mike

    (*) now I think they rely more on A level than they used to.
  • 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


  • OK, the internet connects people who would have never met.


    It has its own web page on wikipedia

    The "Napkin ring" problem 


    The maths is already there, and more neatly than I did it all those years ago.

    Mike
  • Back when I was working as a carpenter I worked for a guy renovating timber framed buildings, he had his own saw mill, when we needed wood we would go up the yard, choose a suitable tree trunk, bring it down the yard using a Rapier diesel electric crane and put it on the travelling table of the 46 inch Stenner band mill to cut it to size or if it was too big cut it down to size using a Stihl chainsaw mill with an aluminium ladder as a guide.


    When buying the timber as round sections we measured it in Hoppus feet, I have several Hoppus ready reckoners on my bookshelf. The gaffer had a Hoppus tape measure, but they are like hens teeth unlike the ready reckoners.


    You can easily measure without a tape, simply loop a length of cord around the round section of timber, hold it to with your fingers then fold the length that is the circumference into four then measure that and treat it as one side of a square, it’s close to the true measurement and averages out the segments fairly accurately, but not quite though that doesn’t actually matter if everyone knows what the means and units of measurement are.


    That was actually one of the best and most enjoyable jobs I ever had, but good things don’t last forever.


    There’s a Saturday night challenge, work out the area of a circle by using a quarter of the circumference squared rather than by A = π r² and compare the results.




  • Not a brilliant explanation, the Hoppus tape merely gives you a quarter of the circumference, hence its other name “quartering tape” a length of cord doubled up twice and an ordinary ruler is a simple alternative.