Sphere puzzle

Discussion in 'Math' started by zgozvrm, Jan 18, 2010.

  1. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    Here's a puzzle for you...


    Assume the Earth to be a perfect sphere (which it is not) having a diameter of exactly 7,926.34 miles.

    Given two coordinates, find the shortest distance between them (without digging any tunnels or trenches, or going under water).

    In other words, what is the shortest distance you must travel on the surface of the sphere to get from one point to the other?



    Starting point:
    43° 55' 0.80", -118° 29' 28.37"

    Ending point:
    32° 11' 52.54", -92° 17' 59.30"
     
  2. t_n_k

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    Mar 6, 2009
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  3. someonesdad

    Senior Member

    Jul 7, 2009
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    Yep, I agree with the great circle path.

    Here's a cute sphere puzzle I remember from decades ago. Given a sphere with a cylindrical hole 6 units long with the cylinder's longitudinal axis passing through the center of the sphere. What's the volume of the remaining material?
     
  4. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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  5. zgozvrm

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    Oct 24, 2009
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    There's not enough information here:

    The hole is 6 units long. Measured from where? The edge of the hole? The center of the cylinder (where there used to be dirt ... that would be a bit higher than the edges, due to the curvature of the sphere)?

    What is the diameter of the hole?

    What is the diameter of the sphere?
     
  6. t_n_k

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    Based on the Wikipedia method with a diameter of 7926.34 miles I get 1630.3858 miles.
     
  7. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    Using my own method, I got exactly the same!
    (I actually came up with 1630.385791 miles).
     
  8. someonesdad

    Senior Member

    Jul 7, 2009
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    That's part of the fun of the puzzle -- there is enough information to solve it. Put some effort into it and you'll learn an interesting tidbit. Oh, and it's a little bit like life -- most problems aren't specified exactly and you might have to make some assumptions.

    If others solve the problem, I suggest that the answer not be posted, as that removes the incentive for others to investigate.

    I saw this problem in the 80's or 70's and originally set out to solve it by integrating. But, as I recall, you can work it out with high school math and perhaps a peek into a handbook for some formulas.
     
  9. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    I can have a sphere of 30 units diameter with a cylindrical hole that is of 2 units diameter and 6 units deep (long) such that the hole's longitudinal axis passes through the center of the sphere. I can then widen the hole to 4 units diameter, there will be less material left in the sphere in the 2nd instance than in the 1st instance.

    I can put a similar hole in a sphere of 50 units diameter; it will have a significantly greater amount of material remaining. With the information given, there is no correlation between the radius of the hole and the radius of the sphere, so there is no way that those values can "cancel out."

    I still fail to believe that there is enough information to solve this problem. Unless, by solving, you are looking for a generalization such as:

    (4/3 * ∏ * R^3) - (6 * ∏ * r^2)

    where "R" is the radius of the sphere and "r" is the radius of the cylinder.

    But, that's hardly a solution.
     
  10. t_n_k

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    I get a remainder of 113.0973355 cubic units.
     
  11. Tesla23

    Active Member

    May 10, 2009
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    If you widen the hole without changing the radius of the sphere you change the length.

    It is a well posed problem (and relatively well known).
     
  12. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    I guess I don't understand what is meant by the "length" of the hole.

    A cylinder needs 2 measurements to define it: one to describe the size of the circle at the ends of the cylinder (radius, diameter, or circumference) and another describing the length of the cylinder.

    I took the length of the hole to mean the depth of the hole and, therefore the length of the cylinder.

    If, instead, the length of the hole is meant to be the diameter of the hole, we still have the issue of not knowing how deep the hole is (i.e., how long the cylinder is).

    Either something is missing from the question, or the question is poorly posed.
     
  13. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    Now, if you mean to say that the hole goes completely through the sphere (which cannot be implied by the question), then things are different. Then the definition of the length of the hole needs to be made:

    The shape of the missing material is not cylindrical since the ends of the "plug" are not 2-dimensional circles (they are parts of the sphere). In my view then, the length of such a hole would be the maximum distance, or the length of the longitudinal axis of the "plug" where it intersects with the surface of the sphere (therefore, the diameter of the sphere).

    I'm guessing that you are not measuring the length of the hole in this manner, rather that you are measuring from the outer edge of the hole.

    This, to me, is an inaccurate way of measuring the depth of the hole (or to say it another way, "the distance drilled"). If I were to drill a hole completely through a sphere of diameter 10" (passing through its center) with a drill bit having a diameter slightly less than this such that the walls of the hole (at the circumference) are only 1" tall, I would not call that a 1" long hole, since I had to drill 10" through the sphere to get the complete hole. If I had only drilled a distance of 1", I would not have passed completely through the sphere.

    But, judging from the replies, this is exactly what you mean when describing the length of the hole.

    If this is indeed the case, then 2 things need to be clarified in the original question:
    1) that the hole passes completely through the sphere
    2) that the walls of the hole measure 6 units long

    As I said before, something is missing from the question. Either dimensions, or clarity (or both).


    Edit: Also, you need to restrict the diameter of the sphere to more than 6 units.
     
    Last edited: Jan 20, 2010
  14. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    With these assumptions, I also get a remainder of 113.0973355 cubic units.

    And for any sphere having a round (cylindrical-like) hole completely through it such that the longitudinal axis of the hole passes through its center and the length of the side of the hole measures X units in length (assuming the sphere's diameter to be greater than X units), the volume of the remaining material is calculated by:

    (X^3) * ∏ / 6 cubic units

    So, for a hole whose sidewalls measure 3 units in length, the volume of the remaining material would be found by:

    (3^3) * ∏ / 6 = 9 * ∏ /6 = 3∏ / 2 ≈ 4.71238898 cubic units
     
    Last edited: Jan 20, 2010
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I'm sure you really meant 14.13716694 ..... or thereabouts

    The formula makes sense.
     
  16. t_n_k

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    Interesting also that the material remainder is equivalent to a spherical volume of radius half that of the cylinder wall length.
     
  17. zgozvrm

    Thread Starter Member

    Oct 24, 2009
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    Yes ... I got in a hurry and only squared the 3 when I should have cubed it.
     
  18. Tesla23

    Active Member

    May 10, 2009
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    That's the key to the easy solution - if the problem is well posed, then the volume stays constant as the hole radius goes to zero, hence the answer is the volume of a sphere of diameter equal to the hole length.
     
  19. thatoneguy

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    Feb 19, 2009
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    You also need to be careful to not hurt any of the Inner Earth People
     
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