<p>The problem is not with the spherical triangle. A triangle can always be formed between three points on a sphere as long as the three points are not on a great circle. (The three points are the celestial pole P, the zenith Z, and the object X in the figure.)</p>
<p>The problem is the triangle cannot be drawn <strong>with an altitude of 0</strong> for X if:</p>
<ul>
<li>the object X never rises.</li>
<li>the object X never sets.</li>
</ul>
<p>Since the spherical triangle cannot be created in the two cases, the value of the hour angle H cannot be found.</p>
<p>Or another way to say it, the time 2*H is the time from rising to setting. When the object never rises, there is no value of H that satisfies the rising and setting formula. If the object never sets, there is no value of H that satisfies the rising and setting formula. In both cases, the right side of the equation is greater than 1 or less than -1, but the value of cos H is limited to ranges between -1 and 1.</p>
<p><a href="https://i.sstatic.net/W8bhj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/W8bhj.png" alt="modified figure from post https://astronomy.stackexchange.com/questions/44796/straightforward-derivation-of-the-sunrise-equation" /></a></p>

CC BY-SA 4.0
2022-09-28T15:38:59.863Z
Are you saying that if the object never sets it is on the meridian PZ? Sorry, I can't quite see why that is.
Peter
CC BY-SA 4.0
2022-09-28T15:59:22.787Z
John, your description of the hour angle is incorrect, the hour angle is based only on the angular distance from the meridian.  What you said applies more to the right side of the equation in the OP.
Greg Miller
CC BY-SA 4.0
2022-09-28T17:30:00.287Z
You're right @Peter. It was confusing, and I am not sure what I was trying to say!
JohnHoltz
CC BY-SA 4.0
2022-09-28T17:31:11.383Z
I made some changes @GregMiller to try to clarify that H in this discussion is the hour angle when rising or setting, not a general hour angle.
JohnHoltz
CC BY-SA 4.0
2022-09-28T15:00:12.553Z
2022-09-28T17:29:33.163Z
2022-09-28T17:29:33.163Z
JohnHoltz
<p>In the rising and setting formula<span class="math-container">$$\cos H=-\frac{\left(\sin\nu+\sin\phi\sin\delta\right)}{\cos\phi\cos\delta},$$</span> <span class="math-container">$H$</span> is hour angle, <span class="math-container">$\nu$</span> is vertical shift, <span class="math-container">$\delta$</span> is declination and <span class="math-container">$\phi$</span> is the observer's latitude. If <span class="math-container">$\cos H&gt;1$</span>,the object never rises; if <span class="math-container">$\cos H&lt;-1$</span>, the object never sets. My question is why does <span class="math-container">$H$</span> not having a solution in the above formula imply an object either never rises or never sets?</p>
<p>A little background. I understand the derivation of the above formula. It's a variation of <span class="math-container">$$\cos H=\frac{\sin a-\sin\delta\sin\phi}{\cos\delta\cos\phi},$$</span> where <span class="math-container">$a$</span> is altitude, and <span class="math-container">$\nu=-a$</span>. The derivation was explained nicely by @HDE 226868 in the answer to <a href="https://astronomy.stackexchange.com/questions/44796/straightforward-derivation-of-the-sunrise-equation">this</a> question. According to the diagram in that answer, no solution for <span class="math-container">$H$</span> implies some sort of failure of the spherical triangle PZX. But I can't visualise what such a failure actually means. Can anyone help? Thanks</p>

<positional-astronomy><declination><sunrise><sunset>
Meaning of $\cos H$ in the rising and setting formula
Peter
# Meaning of $\cos H$ in the rising and setting formula

(Asked by: username_0 on 2022-09-28)

In the rising and setting formula$$\cos H=-\frac{\left(\sin\nu+\sin\phi\sin\delta\right)}{\cos\phi\cos\delta},$$ $H$ is hour angle, $\nu$ is vertical shift, $\delta$ is declination and $\phi$ is the observer's latitude. If $\cos H>1$,the object never rises; if $\cos H<-1$, the object never sets. My question is why does $H$ not having a solution in the above formula imply an object either never rises or never sets?

A little background. I understand the derivation of the above formula. It's a variation of $$\cos H=\frac{\sin a-\sin\delta\sin\phi}{\cos\delta\cos\phi},$$ where $a$ is altitude, and $\nu=-a$. The derivation was explained nicely by @HDE 226868 in the answer to [this](https://astronomy.stackexchange.com/questions/44796/straightforward-derivation-of-the-sunrise-equation) question. According to the diagram in that answer, no solution for $H$ implies some sort of failure of the spherical triangle PZX. But I can't visualise what such a failure actually means. Can anyone help? Thanks

**Tags:** positional-astronomy, declination, sunrise, sunset


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## 1 answers

### Answer by username_1 on 2022-09-28. Score: 3

The problem is not with the spherical triangle. A triangle can always be formed between three points on a sphere as long as the three points are not on a great circle. (The three points are the celestial pole P, the zenith Z, and the object X in the figure.)

The problem is the triangle cannot be drawn **with an altitude of 0** for X if:

* the object X never rises.
* the object X never sets.

Since the spherical triangle cannot be created in the two cases, the value of the hour angle H cannot be found.

Or another way to say it, the time 2\*H is the time from rising to setting. When the object never rises, there is no value of H that satisfies the rising and setting formula. If the object never sets, there is no value of H that satisfies the rising and setting formula. In both cases, the right side of the equation is greater than 1 or less than -1, but the value of cos H is limited to ranges between -1 and 1.

[![modified figure from post https://astronomy.stackexchange.com/questions/44796/straightforward-derivation-of-the-sunrise-equation](https://i.sstatic.net/W8bhj.png)](https://i.sstatic.net/W8bhj.png)

> **Comments:**
> **username_0**: Are you saying that if the object never sets it is on the meridian PZ? Sorry, I can't quite see why that is.  
> **username_2**: John, your description of the hour angle is incorrect, the hour angle is based only on the angular distance from the meridian.  What you said applies more to the right side of the equation in the OP.  
> **username_1**: You're right @username_0. It was confusing, and I am not sure what I was trying to say!  
> **username_1**: I made some changes @GregMiller to try to clarify that H in this discussion is the hour angle when rising or setting, not a general hour angle.  

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