# Sun's height above the horizon

1. Dec 29, 2007

### Irid

Hi,
I'm working on a lengthy problem, and one part asks to find the height of the Sun above the horizon as a function of time. I came up to this solution:

$$\tan (\theta+h) = \frac{\cos (\omega t)}{\tan \phi}$$

where $$\theta$$ is Sun's height below (or above) the celestial equator (i.e. -23.4 deg in winter solstice), while $$\phi$$ is the latitude of observation place. $$\omega t$$ is the phase of Sun's revolution if noon is the point of reference. I've checked many times that this formula is correct, however I have problems and I suspect that this might not be correct after all, because I have a factor of about 2 missing. Can you verify that this is a correct result?
I could post my solution to obtain this formula, if necessary.

2. Dec 29, 2007

### Shooting Star

What happens to your formula at the equator, where phi=0?

Could you show us some of the calculatons? What is h?

Last edited: Dec 29, 2007
3. Dec 30, 2007

### Irid

h is height of Sun. You're right, I forgot to check the condition for phi=0. So this formula is incorrect and I've already found a fault in my solution. I'm working on a correct version, but this seems rather complicated... Thanks for you help!

4. Dec 31, 2007

### Irid

OK, so now I've obtained a new result for the height of Sun h above the horizon:

$$\sin h = \cos \theta \cos \phi \cos (\omega t) + \sin \theta \sin \phi$$

where $$\theta$$ is Sun's height below (or above) the celestial equator (i.e. -23.4 deg in winter solstice), while $$\phi$$ is the latitude of observation place. $$\omega t$$ is the phase of Sun's revolution if noon is the point of reference. Can somebody verify if this is correct?

5. Jan 4, 2008

### Shooting Star

At wt = pi/2, that is at sunset, the sun is at the horizon, and its elevation should be zero. But your formula does not give this for a non-zero phi.

6. Jan 5, 2008

### kamerling

The sun doesn't always set at the same time if you're not at the equator

7. Jan 5, 2008

### Shooting Star

On the equinoxes, day and night are equal at any latitude, and wt = pi/2 at sunset. So sin h should be zero. But the formula gives otherwise.

I feel sorry for the OP, but it looks as if he doesn't give up easily. I'm sure he'll derive the right formula eventually.

8. Jan 6, 2008

### Irid

On an equinox day Sun's declination is zero, so theta=0. Then the formula becomes

$$\sin h = \cos 0 \cos \phi \cos \omega t + \sin 0 \sin \phi = \cos \phi \cos \omega t$$

At sunset wt=pi/2, and it gives h=0, no matter what phi. Maybe I misunderstood your point, but it seems incorrect.

9. Jan 6, 2008

### Shooting Star

You're correct. I mistakenly took theta to be 23.5 deg on the equinox. Sorry.

I don't know how you have derived the formula. That's why I wanted to check it by putting special values.

What results does it give when the latitude is beyond the Arctic circle?

10. Jan 6, 2008

### Irid

This formula gives polar day or night (h>0 or h<0) values for appropriate Sun's declination and observation latitude figures.