# Calculating the orientation of the night sky

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How would one calculate the rotation/tilt of the earth to simulate the Night Sky in a self-written tool or app. I am trying to built an app for my telescope to show me on my phone what I am looking at. I can already simulate the exact positions of stars at the 1.1.2000 at 00:00:00 but I dont know how the "earth" should be rotated so I can simulate the sky from my current position and accurate time.

There are App and tools like stellarium that can achive this but I just cant get the angles right:

If anyone could point me in the right direction and tell me what I need to change/calculate it would help me a lot.

If you do not require arcminute precision, you can approximate Greenwich sidereal time as the Earth rotation angle $$heta(t)$$. IERS Technical Note 32 §5.4.4 gives egin{align} heta(t) &= 2 pi (0.77905~72732~640 + 1.00273~78119~11354~48~ t) &approx 280.46^circ + 360.985612^circ~ t end{align} where $$t$$ is a real number of days since JD 2451545.0 (2000-01-01 12:00 TT ≈ 11:59 UTC).

For an observer at north latitude $$phi$$ and east longitude $$lambda$$, things should line up like this:

Local sidereal time $$= mathrm{LST} approx heta(t) + lambda$$

Zenith (RA, Dec) $$= (alpha, delta) = (mathrm{LST},~ phi)$$

North horizon $$(alpha, delta) = egin{cases} (mathrm{LST + 12h},~ 90^circ - phi) & mathrm{if}~phi >= 0 (mathrm{LST},~ 90^circ + phi) & mathrm{if}~phi < 0 end{cases}$$

East horizon $$(alpha, delta) = (mathrm{LST + 6h},~ 0^circ)$$

The transformation between equatorial and horizontal coordinates can be composed of two rotations, similar but not necessarily identical to those in Wikipedia: Celestial coordinate system. For example, you could:

1. Start with the north horizon vector pointing at the north celestial pole, and the zenith vector pointing at (0h, 0°).
2. Rotate the ground counterclockwise as seen from the north horizon vector by LST.
3. Rotate the ground clockwise as seen from the east horizon vector by $$phi$$.

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