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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 \(\cos(\frac{AB}{R}) = \sin\psi_{A}\sin\psi_{B} + \cos\psi_{A}\cos\psi_{B}\cos(\theta_{B}\theta_{A})\)
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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Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0@ikram002p @zzr0ck3r

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0this need a long type , (if u wanna prove an exist theorem) and i have to go now, unless u wanna it directly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[d^2 = (x_2x_1)^2+(y_2y_1)^2+(z_2z_1)^2\] change to spherical coordinates and simplify ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So represent all the points like \((\rho, \theta, \phi\)) and see if I can get something?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Exactly! Also \(\rho \) = constant since we are "on" the sphere

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4we may first work the formula in regular spherical coordinates changing them to latitude/longitude in the end shouldn't be hard

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4No wait, that wont work. that formula still gives the direct distance between those two points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438756659753:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1438756776294:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4once 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"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we're looking at a triangle with a vertex at the origin, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yep, isosceles triangle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the required distance on the sphere is simply \(Rt\) we can find the angle \(t\) using law of cosines

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438757025200:dw Oh, so we do want to use \(s = r\theta\) But we want law of cosines to get the \(\theta\) first?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes maybe call that angle between radii with some other name because you're already using \(\theta\) for latitude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438757317770:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(d^{2} = R^{2} + R^{2}  2R^{2}\cos(t)\) \(\frac{2R^{2}d^{2}}{2R^{2}} = \cos(t)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\vec{A}\cdot\vec{B} = \vec{A}\vec{B} \cos(t)\) Or maybe this idea. Not that it does anything better, lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1438757456725:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438757852912:dw So if we need \(\phi\) we want \(\psi = \frac{\pi}{2}  \phi\) I guess.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Consider two points \(A, B\) in space whose coordinates in cartesian and spherical are given by : \(A = (x_1,y_1,z_1) = (R, \theta_1,\phi_1)\) \(B = (x_2,y_2,z_2) = (R, \theta_2,\phi_2)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, wait! So that's why the cosines and sines are flipped from the spherical law of cosines! \(z = \rho \cos(\phi)\) \( = \rho \cos(\frac{\pi}{2}  \psi) = \rho \sin(\psi)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Lets compute the direct distance between them in terms of angles : \[\begin{align}d^2 &= (x_2x_1)^2+(y_2y_1)^2+(z_2z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2R\cos\phi_1)^2\\~\\ &=\cdots \end{align}\] some simplification is possible here..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that conversion from regular spherical to lat/long looks good to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A few things simplify if you expand that out, but its still nasty as a direct distance.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\begin{align}d^2 &= (x_2x_1)^2+(y_2y_1)^2+(z_2z_1)^2\\~\\ &=(R\sin \phi_2\cos\theta_2 R\sin \phi_1\cos\theta_1 )^2 \\&+ (R\sin \phi_2\sin\theta_2 R\sin \phi_1\sin\theta_1 )^2\\&+( R\cos\phi_2R\cos\phi_1)^2\\~\\ &=2R^2[1\sin\phi_2\sin\phi_1\cos(\theta_2\theta_1)\cos\phi_2\cos\phi_1] \end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, Im with ya there. So that's definitely close to what we need.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(2R^{2}[1  \sin(\frac{\pi}{2}  \psi_{2})\sin(\frac{\pi}{2}  \psi_{1})\cos(\theta_{2}  \theta_{1})  \cos(\frac{\pi}{2}  \psi_{2})\cos(\frac{\pi}{2} \psi_{1}) ]\) = \(2R^{2}[1  \cos\phi_{2}\cos\phi_{1}\cos(\theta_{2}\theta_{1})  \sin\psi_{2}\sin\psi_{1} ]\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4we may simply equate that to another \(d^2\) that we get using cosine law in the triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah. With the angle t still or do we need \(\theta\) and/or \(\psi\)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1438759481827:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4our goal is to get \(Rt\) in terms of angles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(2R^{2}  2R^{2}\cos(t) = 2R^{2}[1\cos\psi_{2}\cos\psi_{1}\cos(\theta_{2}\theta_{1})  \sin\psi_{2}\sin\psi_{1}]\) \(\implies \cos(t) = \cos\psi_{2}\cos\psi_{1}cos(\theta_{2}  \theta_{1}) + \sin\psi_{2}\sin\psi_{1}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4looks good! \(s = \text{shortest distance between A,B over sphere}=AB\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Honestly, I don't know if there are other simpler ways... I used spherical coordinates because thats the only thing I can think of haha!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Alternatively, 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..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does this make sense ? http://mathworld.wolfram.com/SphericalDistance.html

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4This shouldn't matter much but I think the relation between regular spherical coordinates and lat/long is \(\psi = \color{red}{}\frac{\pi}{2}  \phi\color{red}{}\) .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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
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