# How to find the rising time of constellations at a specific longitude and latitude on earth?

I wish to find the rising time of constellations I am writing a computer program. I have successfully determined the rising time of planets and ascendant (zodiac sign on the eastern horizon) by the formulae described here https://en.wikipedia.org/wiki/Ascendant. It is also given that also different zodiac signs ascend at a different rate. I want to create a table of ascendant for a given day. I know that the position of a star is fixed but I am unable to determine the change with respect to the earth rotations. For rising and setting time calculation I am referring "Astronomical Algorithms Jean Meeus".

Please give me any formulae or something that I can use to determine the table of the ascendant. I don't want to refer the existing programs that are available for free as they are too difficult to understand.

Thanx in advance

• "I don't want to refer the existing programs that are available for free as they are too difficult to understand.": I think you are taking the right approach writing your own software, then. You may well find that after you have learnt what you need to know to make your software work and agree with other softwares already out there, you will understand the existing programs. Good luck. Anyhow, @PM2Ring 's formulas will get you a long way. Dec 12, 2016 at 9:05

## 1 Answer

Given

$$\theta = \tan^{-1}\left(\frac{\cos t}{-(c\sin t + k)}\right)$$

We want to find the derivative, $\frac{d\theta}{dt}$

Let

\begin{align} u & = \cos t\\ v & = -(c\sin t + k)\\ y & = u / v\\ \theta & = \tan^{-1} y\\ \end{align}

Taking derivatives,

\begin{align} du & = -\sin t \,dt\\ dv & = -c\cos t \,dt\\ dy & = (v \, du - u\, dv) / v^2\\ d\theta & = \frac{dy}{1 + y^2}\\ \end{align}

Substituting, \begin{align} dy & = (v \, du - u\, dv) / v^2\\ & = ((c\sin t + k)\sin t \,dt + c\cos^2 t\,dt) / (c\sin t + k)^2\\ & = (c\sin^2 t + k \sin t + c\cos^2 t) dt/ (c\sin t + k)^2\\ & = \frac{(c + k\sin t) dt}{(c\sin t + k)^2}\\ \\ 1 + y^2 & = 1 + \cos^2 t / (c\sin t + k)^2\\ & = \frac{(c\sin t + k)^2 + \cos^2 t}{(c\sin t + k)^2}\\ \end{align}

Therefore, $$\frac{d\theta}{dt} = \frac{c + k\sin t}{(c\sin t + k)^2 + \cos^2 t}$$