Sun's height above the horizon

[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.

 

[Shooting Star]

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

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

 

[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!

 

[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?

 

[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.

 

[kamerling]

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.

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

 

[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.

 

[Irid]

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.

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.

 

[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?

 

[Irid]

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