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Hi,
I'm trying to figure out a formula that will provide the angle formed by the intersection of the galactic plane and a specified line of constant RA. Equivalently stated, I need a formula that will provide the "bearing angle" of the galactic plane at a given RA. I know the galactic plane is inclined by roughly 62° to the celestial equator. But I would need to know the "reference" RA at which the galactic plane crosses that line of RA at 90°, and I would need to know a lot more spherical geometry than I currently know <g>, in order to figure this out. Can anyone help or point me to a reference? I've googled this and haven't found anything. Thanks, Jim S. |
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OTTOMH, the galactic plane crosses the equator somewhere in or near Orion. As far as doing the math, the only way I would know how to figure that is to plot the RA, Dec in Cartesian XYZ coordinates, rotate it to the galactic plane, and use an arctangent to get galactic latitude and longitude.
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| grant hutchison |
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This message has been deleted by grant hutchison.
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The north galactic pole is at α = 192.86º, δ = 27.13º, which implies that the galactic equator passes northwards across the celestial equator at right ascension α = 282.86º, making an angle of 62.87º.
So you can build a right spherical triangle using the galactic equator as the hypotenuse, with the celestial equator and your chosen line of RA as the other two sides. A Google search should turn up a toolbox of spherical trig formulae which will let you calculate the angle you're interested in within that triangle. Grant Hutchison |
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Thanks Grant! That was exactly what I needed. Because it's a "right" spherical triangle, the formula for the missing angle reduces to the following, simple form:
JS |
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Thanks for all the help folks.
One more question: Does anyone know of some freeware that can generate a "reference" image of stars (e.g., from sky2000 or gsc) at a specified resolution, center, projection (e.g., galactic cylindrical), etc.? JS |