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anonymous

  • one year ago

Spherical Geometry Let A = (\(\theta_{A}, \psi_{A}\)) be a point on the earth at latitude \(\psi_{A}\) and longitude \(\theta_{A}\). Let B = (\(\theta_{B}, \psi_{B}\)) be another point on the earth. Let R be the radius of the earth. Prove that the distance |AB| between A and B is given by \(\cos(\frac{|AB|}{R}) = \sin\psi_{A}\sin\psi_{B} + \cos\psi_{A}\cos\psi_{B}\cos(\theta_{B}-\theta_{A})\)

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  1. Astrophysics
    • one year ago
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    @ganeshie8

  2. Jhannybean
    • one year ago
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    @ikram002p @zzr0ck3r

  3. ikram002p
    • one year ago
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    this need a long type -,- (if u wanna prove an exist theorem) and i have to go now, unless u wanna it directly

  4. anonymous
    • one year ago
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    If you don't have the time it's okay. It seems like that the problem is asking to prove the law of cosines but with a twist to it. Unless it's simpler than that.

  5. anonymous
    • one year ago
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    So, it looks really similar to: \(\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(C)\) But of course the cosines and sines are flipflopped and the above formula assumed unit sphere. So knowing how to adapt the formula and then see how things got switched up like they did.

  6. ganeshie8
    • one year ago
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    \[d^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\] change to spherical coordinates and simplify ?

  7. anonymous
    • one year ago
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    So represent all the points like \((\rho, \theta, \phi\)) and see if I can get something?

  8. ganeshie8
    • one year ago
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    Exactly! Also \(\rho \) = constant since we are "on" the sphere

  9. ganeshie8
    • one year ago
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    we may first work the formula in regular spherical coordinates changing them to latitude/longitude in the end shouldn't be hard

  10. ganeshie8
    • one year ago
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    No wait, that wont work. that formula still gives the direct distance between those two points

  11. anonymous
    • one year ago
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    Oh. Lol, okay.

  12. anonymous
    • one year ago
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    |dw:1438756659753:dw|

  13. ganeshie8
    • one year ago
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    |dw:1438756776294:dw|

  14. ganeshie8
    • one year ago
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    once we know the distance \(d\), its not hard to find the distance along that arc because we know that the shortest distance is always along the "circle whose center is same as the center of the sphere"

  15. anonymous
    • one year ago
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    So we're looking at a triangle with a vertex at the origin, right?

  16. ganeshie8
    • one year ago
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    Yep, isosceles triangle

  17. ganeshie8
    • one year ago
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    the required distance on the sphere is simply \(Rt\) we can find the angle \(t\) using law of cosines

  18. anonymous
    • one year ago
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    |dw:1438757025200:dw| Oh, so we do want to use \(s = r\theta\) But we want law of cosines to get the \(\theta\) first?

  19. ganeshie8
    • one year ago
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    Yes maybe call that angle between radii with some other name because you're already using \(\theta\) for latitude

  20. anonymous
    • one year ago
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    |dw:1438757317770:dw|

  21. anonymous
    • one year ago
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    \(d^{2} = R^{2} + R^{2} - 2R^{2}\cos(t)\) \(\frac{2R^{2}-d^{2}}{2R^{2}} = \cos(t)\)

  22. anonymous
    • one year ago
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    \(\vec{A}\cdot\vec{B} = ||\vec{A}||||\vec{B}|| \cos(t)\) Or maybe this idea. Not that it does anything better, lol.

  23. ganeshie8
    • one year ago
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    |dw:1438757456725:dw|

  24. anonymous
    • one year ago
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    But even then, I don't know how it works where you'd translate that to the actual angles on the triangle. I mean, that is just finding the angle between the vector representations of the points. Not that we need it for this problem, but I wouldn't know how to use that to get the angles of the triangle. Oh, okay, back to spherical coordinates.

  25. anonymous
    • one year ago
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    |dw:1438757852912:dw| So if we need \(\phi\) we want \(\psi = \frac{\pi}{2} - \phi\) I guess.

  26. ganeshie8
    • one year ago
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    Consider two points \(A, B\) in space whose coordinates in cartesian and spherical are given by : \(A = (x_1,y_1,z_1) = (R, \theta_1,\phi_1)\) \(B = (x_2,y_2,z_2) = (R, \theta_2,\phi_2)\)

  27. anonymous
    • one year ago
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    Oh, wait! So that's why the cosines and sines are flipped from the spherical law of cosines! \(z = \rho \cos(\phi)\) \( = \rho \cos(\frac{\pi}{2} - \psi) = \rho \sin(\psi)\)

  28. ganeshie8
    • one year ago
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    Lets compute the direct distance between them in terms of angles : \[\begin{align}d^2 &= (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 -R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 -R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2-R\cos\phi_1)^2\\~\\ &=\cdots \end{align}\] some simplification is possible here..

  29. ganeshie8
    • one year ago
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    that conversion from regular spherical to lat/long looks good to me

  30. anonymous
    • one year ago
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    A few things simplify if you expand that out, but its still nasty as a direct distance.

  31. ganeshie8
    • one year ago
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    \[\begin{align}d^2 &= (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 -R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 -R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2-R\cos\phi_1)^2\\~\\ &=2R^2[1-\sin\phi_2\sin\phi_1\cos(\theta_2-\theta_1)-\cos\phi_2\cos\phi_1] \end{align}\]

  32. anonymous
    • one year ago
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    Okay, Im with ya there. So that's definitely close to what we need.

  33. anonymous
    • one year ago
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    \(2R^{2}[1 - \sin(\frac{\pi}{2} - \psi_{2})\sin(\frac{\pi}{2} - \psi_{1})\cos(\theta_{2} - \theta_{1}) - \cos(\frac{\pi}{2} - \psi_{2})\cos(\frac{\pi}{2} -\psi_{1}) ]\) = \(2R^{2}[1 - \cos\phi_{2}\cos\phi_{1}\cos(\theta_{2}-\theta_{1}) - \sin\psi_{2}\sin\psi_{1} ]\)

  34. ganeshie8
    • one year ago
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    we may simply equate that to another \(d^2\) that we get using cosine law in the triangle

  35. anonymous
    • one year ago
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    Ah. With the angle t still or do we need \(\theta\) and/or \(\psi\)?

  36. ganeshie8
    • one year ago
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    |dw:1438759481827:dw|

  37. ganeshie8
    • one year ago
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    our goal is to get \(Rt\) in terms of angles

  38. anonymous
    • one year ago
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    \(2R^{2} - 2R^{2}\cos(t) = 2R^{2}[1-\cos\psi_{2}\cos\psi_{1}\cos(\theta_{2}-\theta_{1}) - \sin\psi_{2}\sin\psi_{1}]\) \(\implies \cos(t) = \cos\psi_{2}\cos\psi_{1}cos(\theta_{2} - \theta_{1}) + \sin\psi_{2}\sin\psi_{1}\)

  39. anonymous
    • one year ago
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    Oh, so then you have \(s = Rt \implies t = \frac{s}{R}\) . And the arc length would be the measure of the side, so can say s = \(|AB|\). Well, at least I hope I can just go ahead and make that claim, but seems legit :D

  40. ganeshie8
    • one year ago
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    looks good! \(s = \text{shortest distance between A,B over sphere}=|AB|\)

  41. anonymous
    • one year ago
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    Okay, awesome :D Do you think thatll be the approach to a lot of these spherical problems? Like, even though things are in terms of vectors or long/lat, just turn it all into spherical coordinates? Or are there different ways to do the algebra without leaving vectors and long/lat?

  42. ganeshie8
    • one year ago
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    Honestly, I don't know if there are other simpler ways... I used spherical coordinates because thats the only thing I can think of haha!

  43. ganeshie8
    • one year ago
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    Alternatively, we may directly optimize the arc length over sphere... there is some method in calculus of variations... but its not on top of my head at the moment..

  44. ganeshie8
    • one year ago
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    does this make sense ? http://mathworld.wolfram.com/SphericalDistance.html

  45. ganeshie8
    • one year ago
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    This shouldn't matter much but I think the relation between regular spherical coordinates and lat/long is \(\psi = \color{red}{|}\frac{\pi}{2} - \phi\color{red}{|}\) .

  46. anonymous
    • one year ago
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    Yeah, it makes sense in context with everything else. Getting used to the idea that the sides are basically angles now, lol. So yeah, thats angle between vectors formula, but the "angle" is the distance. So maybe it wouldve worked out that way as well. Now I'll have to look through notes and see if I can find out how you get the angle between the actual sides. The angles we're getting are actually sides in disguise. Now need ways to get the actual angles, lol

  47. ganeshie8
    • one year ago
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    https://en.wikipedia.org/wiki/Spherical_law_of_cosines

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