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3D Geometry problem.

  1. Jan 24, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    a. Suppose you do not know your location (on Earth) or the direction of north. Now suppose you track one particular star in the sky. You measure its exact position in the sky and record the exact time of the measurement. How many such measurements are necessary to deduce the length of the sidereal day? Your answer should not involve any formulas or calculations. It should just be a positive integer representing the number of measurements needed together with a brief explanation.

    b. Suppose you were also able to locate the North Star. Would that allow you to deduce the length of the day with fewer measurements?

    Note: For part b, instead of knowing the location of the North Star, it is also enough to either know your latitude or know which direction is north.

    3. The attempt at a solution

    Okay... Part a. I am not sure how to do, but I think I know part b.

    PART A:
    I am not sure. My guess was:

    With 3 equidistant points you could trace a smooth arc length in the sky. I could then use the ratio between this arc length and the time taken and the ratio between the circumference of the star's path and the length of the sidereal day to calculate the length of the sidereal day. The only problem is, I don't know if I can calculate the circumference of the star's path given only an arc length. Intuitively I feel there should be a way, but I have searched long and hard and have not come up with one. Point me in the right direction?

    PART B:
    The distance between the star being tracked and the north star would be the radius of the star's path. Another position would allow me to calculate the angle between the vectors from the north star to each of the two positions. The ratio between this angle and the arc length would equal the ration between the total circumference and the length of the sidereal day.

    Thanks in advance.
     
  2. jcsd
  3. Jan 24, 2012 #2

    Simon Bridge

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    With the pole-star known - you know where the center of rotation is.
    This makes the calc easy.

    If you don't have the center pinpointed for you, your first step will be to locate it.
    If you have three points on the circumference of a circle, that means you have two chords. Draw the picture... really, draw it: draw a circle with two chords. What can you do (say, with dividers and a pencil) that will get the chords to point at the center?

    I don't think you are expected to produce a formula
     
  4. Jan 24, 2012 #3

    ElijahRockers

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    Ok, so I drew a picture, and I believe it would take 4 measurements, taken at equal time intervals. Assuming the points are labeled sequentially, A, B, C, and D:

    Line AC and BD. draw a ray from B, bisecting AC. draw another ray, from C, bisecting BD. Where these two rays intersect, should be the north star...

    I couldn't figure out a way to do it with 3 points.
     
    Last edited: Jan 24, 2012
  5. Jan 25, 2012 #4

    Simon Bridge

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    well done:
    You know you can bisect a line segment?

    3 points A B C, not equal times.
    put point of compass at A, draw a wide arc radius |AB|
    repeat for B - give you two points.
    repeat for BC, confirm with AC.

    Where I live I get the south celestial pole ... thereabouts.
     
  6. Jan 25, 2012 #5

    ElijahRockers

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    Hm, I will have to try that when I have more time, but I am having a hard time visualizing it at a glance. Thanks for your help!
     
  7. Jan 25, 2012 #6

    Simon Bridge

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    Think like this: what you did was bisect an angle - you can visualize that ok.
    Now imagine the angle in question is 180 degrees - that's a line segment.

    Other way:
    the definition of a line is: mark to points P and Q - the line is the set of all points X such that |XP| = |XQ|. This line will be perpendicular to PQ. You can fins two points on the line with a compass provided it is set to any length greater than |PQ|/2.

    The circle about P is the set of all points equidistant from P, circle about Q is same for Q, where these two circles cross is the the two points where these distances are the same.
     
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