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Circumference of a Ball

  1. Aug 23, 2009 #1
    Hi all,

    Hope you can help. How can I get the circumference of a specific point on a ball, more to the point in this example, the ball is in question is planet earth. Using the polar circumerence I have got the polar diameter and radius (r) (that's the easy bit). Now, what I figured is that from there I could use the degrees north / south figure to give me a percentile of how far up the planet you are - e.g. 45 degrees is 50%. So, using this example I have a a right angled triangle with r being the hypotenuse as this should be constant, (r*0.5) being another side of the triangle right? So from there it should surely be sqrt(c^2-a^2) to get b^2 and then sqrt this to get b? This being the new radius of the point up the planet where the new circumference is to be drawn so that I can use degrees/min/sec to get a distance east or west. So then I can use 2b*pi to get my circ and work from here (though what I actually did was to do b/r to get a percentage and factor my base figures, such miles per degree to this percentage).

    It seemed so easy in theory but then measuring out the distance I'd calculated in google maps comes out with something entirely different.

    BTW, this isn't homework - I'm 27 years old and mad / sad enough to be spending time making a spreadsheet that calcuates distance between two points based on GPS co-ordinates. Why? Just because :)
     
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  3. Aug 23, 2009 #2

    HallsofIvy

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    The "circumference of a specific point on a ball"? Do you mean the length of circle at a given longitude? That is the circumference of the circular cross section at that latitude but not a circumference of the sphere. Yes, if your circle is at, say, [itex]\theta[/itex] degrees (North) latitude, then drawing a line from the north pole to the center of the sphere and dropping a perpendicular from the point on the sphere to that line, you get a right triangle having R, the radius of the sphere, as hypotenuse and angle [itex]90- \theta[/itex]. Calling the radius of the circle r, it is the side "opposite" the angle and so we must have [itex]r/R= sin(90- \theta)= cos(\theta)[/itex] and so [itex]r= R cos(\theta)[/itex] and the circumference of that circle is [itex]2\pi r= 2\pi R cos(\theta)[/itex].

     
  4. Aug 26, 2009 #3
    Okay, I'm not following this entirely. If we have two of the sides of a right-angled triangle then why can the other not be worked out using pythagoras' theorum? Going into real basic maths here, but I was always taught that A2+B2=C2, so surely if you have A and C then C2-A2=B2?
    Then it'd just be a case of square rooting to get a radius at that point, doubling it for the diameter and then times'ing by [itex]\pi[/itex]?
    I'm not getting why trigonometry must be used? As your example above.
     
  5. Aug 26, 2009 #4

    HallsofIvy

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    What two sides are you talking about? Referring to the calculation I gave, we know one side of a right triangle: the hypotenuse which is the radius of the earth. The only other piece of information you have is the angle at the center of the earth- the "degrees north / south figure" you refer to.
     
  6. Aug 26, 2009 #5
    Thanks for that. I figured that I had just described my aim in a poor manner and have just drawn out a cross-section diagram in paintshop to graphically show what I mean to achieve.

    In doing so I have realised that I have made a fundamental and stupid error in thinking that 45 degrees north, as 50% of 90 degrees would have given me a point on the sphere that would cut across and be 50% of the radius, giving me two sides of the triangle and hence requiring only one other side to be discovered, using pythagoras theorum. Of course this is not true, if the we had been working on a straight line between the polar and equatorial positions, say for example at 0o0'0".00 North / 0o0'0" East and 90o0'0" North / 0o0'0" East then this would have been true, but I had not accounted for the curved surface of the sphere.

    So going back to your method of finding the triangle, knowing only the hypotenuse and the internal angle, could you explain to me again your method of gaining the other sides, circumference of the circle at position of X (say X is 45o0'0" North). From the point of view of poor little me, who has not had any need for trigonometry and hence not used it for circa 12 years. I understand sin / cos / tan etc but could you break your formula down a little please?

    Thanks in advance and sorry for being a total pain in the a**
     
  7. Aug 26, 2009 #6

    arildno

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    V8Maverick:
    I don't get your usage of the word "circumference".

    Do you mean the arc length as traced out on the surface of the ball between points A and B, given the internal angle as measured from the centre of the ball?
     
  8. Aug 27, 2009 #7
    My usage is hence:

    Take your ball, the obvious circumference would be it's widest point when held up, this could be the equator or from pole to pole, or any other point that runs straight through the centre of the ball. In the case of planet earth, this is defined as 40075.02km equatorially and 40007.86km meridianal (or so I'm led to believe). But the value or "circumference" I'm interested in getting is that if you dropped a ring onto planet earth that was the correct size to come to rest at 45o North, what would the cirumference of this ring be? Another way to think of it would be if I'd simply cut the top of the planet off at 45o North, what would the radius / circumference be of the plateau that I've just created be?
     
  9. Aug 31, 2009 #8
    Anyone?
     
  10. Aug 31, 2009 #9

    HallsofIvy

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    So you are asking about the circumference of a circle of fixed latitude. I answered that in my first response: if the radius of the earth is R then the circumference of the circle at latitude [itex]\theta[/itex] is [itex]2\pi R cos(\theta)[/itex].
     
  11. Aug 31, 2009 #10

    arildno

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    Well, let R be the radius of the ball, and let [itex]\theta[/itex] be the angle with which the radius vector makes with the equatorial plane.

    Clearly, the radius r of the circle located at a particular value of [itex]\theta[/tex] is [itex]r=R\cos\theta[/tex]
     
  12. Aug 31, 2009 #11

    DaveC426913

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    And if you do this kind of measurement enough, you'll instinctively know that, at 45 degrees, it is going to be .707 of the circumference.
     
  13. Aug 31, 2009 #12
    I would hate to drop the bomb and tell you that the earth isn't exactly spherical. Would this have an affect and how accurate do you need to be?
     
  14. Aug 31, 2009 #13

    DaveC426913

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    Waitaminnit.

    If this is your application then all this discussion is moot.

    The shortest distance between any two points on the Earth is going to follow a great circle i.e. a circle whose centre coincides with of the Earth's centre. This vastly simplifies your calculations. All this discussion about radii at a given latitude is completely unnecessary.
     
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