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Does anyone have a quick method to do this?

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Does anyone have a quick method to do this?

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[tex]x = \rho \sin{\theta} \cos{\phi}[/tex]

[tex]y = \rho \sin{\theta} \sin{\phi}[/tex]

[tex]z = \rho \cos{\theta}[/tex]

Where [itex]\phi[/itex] is the longitude, [itex]\theta[/itex] is the latitude, and [itex]\rho[/itex] is the radius of the Earth.

cookiemonster

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I believe [tex]z = \rho \cos{\phi}[/tex] and [tex]x = \rho \cos{\theta} \sin{\phi}[/tex]cookiemonster said:

[tex]x = \rho \sin{\theta} \cos{\phi}[/tex]

[tex]y = \rho \sin{\theta} \sin{\phi}[/tex]

[tex]z = \rho \cos{\theta}[/tex]

Where [itex]\phi[/itex] is the longitude, [itex]\theta[/itex] is the latitude, and [itex]\rho[/itex] is the radius of the Earth.

cookiemonster

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HallsofIvy

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uart

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deltabourne said:I believe [tex]z = \rho \cos{\phi}[/tex] and [tex]x = \rho \cos{\theta} \sin{\phi}[/tex]

All you're doing there

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HallsofIvy

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uart

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Yes it looks like both conventions are in common use unfortunately. Here is what Mathworld has to say about it.HallsofIvy said:Maybe this is an "America against the rest of the world" thing but every text I've ever seen defines φ to be the angle the straight line from (0,0,0) to the point makes with the positive z axis while θ is the angle the projection of that line onto the xy-plane makes with the positive x-axis.

MathWorld said:A system of curvilinear coordinates which is natural for describing positions on a sphere or spheroid. Define [tex]\theta[/tex] to be the azimuthal angle in the xy-plane from the x-axis and [tex]\phi[/tex] to be the polar angle from the z-axis with ...

Unfortunately, the convention in which the symbols and are reversed is frequently used, especially in physics, leading to unnecessary confusion. The symbol [tex]\rho[/tex] is sometimes also used in place of r. Arfken (1985) uses [tex](r, \phi, \theta)[/tex], whereas Beyer (1987) uses [tex](\rho, \theta, \phi)[/tex]. Be very careful when consulting the literature.

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HallsofIvy

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Aha! So instead of "America against the world", it is "Physicists against Mathematicians"!

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The 'quick and dirty' method (assuming the Earth is a perfect sphere):kronchev said:Does anyone have a quick method to do this?

x = longitude*60*1852*cos(latitude)

y = latitude*60*1852

Latitude and longitude must be in decimal degrees, x and y are in meters.

The origin of the xy-grid is the intersection of the 0-degree meridian and the equator, where x is positive East and y is positive North.

So, why the 1852? I'm using the (original) definition of a nautical mile here: 1 nautical mile = the length of one arcminute on the equator (hence the 60*1852; I'm converting the lat/lon degrees to lat/lon minutes).

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X = (N+H) cos(phi) cos(lambda)Does anyone have a quick method to do this?

Y = (N+H) cos(phi) sin(lambda)

Z = [N(1-e^2)+H] sin(phi)

I have solved the inverse problem analytically as well as with other better methods.

This involves solving a complicated quartic eqation. See Vanicek & Krakiwsky,

There are other less efficient methods online. See Mathworks, e.g.

Ben Palmer

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