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Motion in Space calculations?

  1. Jun 29, 2010 #1
    Hello All,
    Noobie question. I am using Meeus' Astronomical Algorithims to creat a spreadsheet to transform coordinates. I had no problem with precession, nutation, aberration, etc. But, I don't have as good a math background for Motion in Space.
    I can follow down to the final formulas for x', y' z'.
    I researched further and found Meeus' source in the Sky Catalogue 2000.0 and that really didn't help me either.
    I just don't have the math background.
    Can anyone expound on what I do AFTER I solve for x', y' and z'?

    Any help greatly appreciated.
    Thanks, Scott

    P.S. I added a zip of a bmp of the relevent page in the Sky Catalogue.
    In any explaination, please keep my genetic deficit in mathematics in mind.
     

    Attached Files:

    Last edited: Jun 29, 2010
  2. jcsd
  3. Jun 29, 2010 #2

    Filip Larsen

    User Avatar
    Gold Member

    The text describes how you from the position of a star given by [itex]\alpha_0, \delta_0, r_0[/itex] for some time [itex]t_0[/itex] (like 2000.0) and its change per year given by [itex]\Delta \alpha, \Delta \delta, \Delta r[/itex] can find the position of the star [itex]t[/itex] years later for another time [itex]t_1 = t_0 + t [/itex], assuming that the star's proper motion can be approximated with a line in three-dimensional space over that time.

    Imagine that you have a globe (the surface of which are representing the spherical coordinates [itex]\alpha, \delta[/itex]) and you let a pencil represent the true motion of a star from the start to the end of the pencil, and you place it somewhere near the pole of your globe, then it is not hard to see that it is a complicated matter to describe the spherical coordinates of the end-point of the pencil given only the length of the pencil and its starting point in spherical coordinates. If the pencil was very short relative to the globe and close to equator you may just "add the spherical coordinates", but not when its near the pole.

    So, the text describes how you can find the spherical coordinates of a star after [itex]t[/itex] years by first transforming the spherical coordinate [itex]\alpha_0, \delta_0, r_0[/itex] and its rate of change to the Cartesian coordinates [itex]x_0, y_0, z_0[/itex] and rate of change [itex]\Delta x, \Delta y, \Delta z[/itex]. From this you can find the new Cartesian position [itex]x_1, y_1, z_1[/itex] of the star after [itex]t[/itex] years by assuming it follows a simple line (where I have taken the liberty to use subscript 1 instead of a accent).

    Now, finally, you want to change this new Cartesian position back into a spherical coordinate by using the first set of equation "in reverse", but now relating the new position, that is, starting with

    [tex]
    \begin{flalign*}
    x_1 & = & & r_1 \cos \delta_1 \cos \alpha_1 \\
    y_1 & = & & r_1 \cos \delta_1 \sin \alpha_1 \\
    z_1 & = & & r_1 \sin \delta_1
    \end{flalign*}
    [/tex]

    you can first find

    [tex]
    r_1 = \sqrt{x_1^2 + y_1^2 + z_1^2}
    [/tex]

    which you can insert in the equation for [itex]z_1[/itex] to get

    [tex]
    \delta_1 = \arcsin\left( z_1/r_1 \right)
    [/tex]

    where you use the inverse sine function. Next you can divide the equations for [itex]y_1[/itex] and [itex]x_1[/itex] getting

    [tex]
    \frac{y_1}{x_1} = \frac{r_1 \cos \delta_1 \sin \alpha_1}{r_1 \cos\delta_1 \cos\alpha_1} = \frac{\sin\alpha_1}{\cos\alpha_1} = \tan\alpha_1
    [/tex]

    which means you can find [itex]\alpha_1[/itex] by taking the inverse tangent:

    [tex]
    \alpha_1 = \arctan\left( y_1 / x_1 \right)
    [/tex]

    In a spreadsheet you should have the function ATAN2 available that takes the two arguments above, [itex]y_1[/itex] and [itex]x_1[/itex], and returns an angle in radians in the interval [itex]-\pi/2[/itex] to [itex]\pi/2[/itex] (or -180 to 180 if you convert the radians to degrees first). If you get a negative angle you can add [itex]2\pi[/itex] (or 360 degree) to it to get a positive angle.

    Also, in case you are not already aware of it, note, that when [itex]\alpha[/itex] is the equatorial rectascension coordinate it is often is given as an hour measure instead of degrees; in this case just multiply the value with 15 to get the angle in degrees.

    I hope this helped a bit, otherwise feel free to ask.
     
  4. Jun 29, 2010 #3
    Thank you Filip.

    Excellent explaination and thanks for deriving the formulae.

    I managed to get the declination perfect and have the ascension very close.
    A little more tinkering and hopefully I should get it.

    Thanks again, Scott
     
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