<p>I found the text you quoted confusing - obviously the reason for your question. In particular it wasn't clear whether arcs were measured as distance over the ground or radians/angles. HTH.<a href="https://i.sstatic.net/LjKIa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/LjKIa.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-03-11T23:07:24.297Z
2024-03-11T23:07:24.297Z
stretch
<p>So I just got reading- &quot;<em>Astronomy: Principles and Practice, Fourth Edition</em> by A. Roy and D. Clarke&quot; and I couldn't understand some lines (a possible derivation for some geometrical formula, which I have no clue about)</p>
<p>It stated the following:</p>
<blockquote>
<p>The length of a small circle arc, such as <span class="math-container">$FG$</span> is related simply to the length of an arc of the great circle whose plane is parallel to that of the small circle In figure 7.1, let <span class="math-container">$r$</span> be the radius of the small circle <span class="math-container">$EFGHE$</span>. Then
<span class="math-container">$$FG = r × ∠FKG$$</span></p>
<p>Also
<span class="math-container">$$BC= R×∠BOC$$</span></p>
<p>Both <span class="math-container">$OB$</span> and <span class="math-container">$KF$</span> lie on plane <span class="math-container">$PFBQ$</span>; <span class="math-container">$KF$</span> also lies on plane <span class="math-container">$EFGH$</span> while <span class="math-container">$OB$</span> lies on plane <span class="math-container">$ABCD$</span>. Therefore, <span class="math-container">$KF$</span> must be parallel to <span class="math-container">$OB$</span>, since plane <span class="math-container">$EFGH$</span> is parallel to plane <span class="math-container">$ABCD$</span>. Similarly, <span class="math-container">$KG$</span> is parallel to <span class="math-container">$OC$</span> Then
<span class="math-container">$$∠FKG = ∠BOC$$</span>
Hence,
<span class="math-container">$$FG = BC × r/R$$</span></p>
<p>In the plane triangle <span class="math-container">$KOF$</span>, right-angled at <span class="math-container">$K$</span>, <span class="math-container">$KF=r$</span>; <span class="math-container">$OF = R$</span>. Hence,
<span class="math-container">$$FG = BC × \sin KOF$$</span>
But <span class="math-container">$∠POB = 90^\circ$</span> so that we may write alternatively
<span class="math-container">$$FG = BC\cos FB$$</span>
If the radius of the sphere is unity,
<span class="math-container">$$PF = ∠POF = ∠KOF$$</span>
and
<span class="math-container">$$FB = ∠FOB$$</span>,
so that we have
<span class="math-container">$$FG = BC\sin PF$$</span>
and
<span class="math-container">$$FG = BC\cos FB$$</span>.</p>
</blockquote>
<p>Here is the figure 7.1 it was talking about:
<a href="https://i.sstatic.net/M7KHp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/M7KHp.png" alt="figure 7.1" /></a></p>
<p>I couldn't understand what it wanted to derive. Also how can <span class="math-container">$\sin PF$</span> and <span class="math-container">$\cos FB$</span> be possible? aren't <span class="math-container">$PF$</span> and <span class="math-container">$FB$</span> arcs?</p>

CC BY-SA 4.0
2024-03-10T14:39:40.743Z
@ScienceSnake this addresses the issue, thanks, you can answer this question if you want to, instead
Celestia Novalis
CC BY-SA 4.0
2024-03-11T03:19:13.237Z
I finally got one of the possible reasons why- the author wanted to derive the formula for *departure*
Celestia Novalis
<positional-astronomy><spherical-astronomy>
Relation of length of small circle with that of a great circle
Celestia Novalis
# Relation of length of small circle with that of a great circle

(Asked by: username_0 on 2024-03-10)

So I just got reading- "*Astronomy: Principles and Practice, Fourth Edition* by A. Roy and D. Clarke" and I couldn't understand some lines (a possible derivation for some geometrical formula, which I have no clue about)

It stated the following:

> The length of a small circle arc, such as $FG$ is related simply to the length of an arc of the great circle whose plane is parallel to that of the small circle In figure 7.1, let $r$ be the radius of the small circle $EFGHE$. Then
> $$FG = r × ∠FKG$$
>
> Also
> $$BC= R×∠BOC$$
>
> Both $OB$ and $KF$ lie on plane $PFBQ$; $KF$ also lies on plane $EFGH$ while $OB$ lies on plane $ABCD$. Therefore, $KF$ must be parallel to $OB$, since plane $EFGH$ is parallel to plane $ABCD$. Similarly, $KG$ is parallel to $OC$ Then
> $$∠FKG = ∠BOC$$
> Hence,
> $$FG = BC × r/R$$
>
> In the plane triangle $KOF$, right-angled at $K$, $KF=r$; $OF = R$. Hence,
> $$FG = BC × \sin KOF$$
> But $∠POB = 90^\circ$ so that we may write alternatively
> $$FG = BC\cos FB$$
> If the radius of the sphere is unity,
> $$PF = ∠POF = ∠KOF$$
> and
> $$FB = ∠FOB$$,
> so that we have
> $$FG = BC\sin PF$$
> and
> $$FG = BC\cos FB$$.

Here is the figure 7.1 it was talking about:
[![figure 7.1](https://i.sstatic.net/M7KHp.png)](https://i.sstatic.net/M7KHp.png)

I couldn't understand what it wanted to derive. Also how can $\sin PF$ and $\cos FB$ be possible? aren't $PF$ and $FB$ arcs?

**Tags:** positional-astronomy, spherical-astronomy
> **Comments:**
> **username_0**: @ScienceSnake this addresses the issue, thanks, you can answer this question if you want to, instead  
> **username_0**: I finally got one of the possible reasons why- the author wanted to derive the formula for *departure*  

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## 1 answers

### Answer by username_1 on 2024-03-11. Score: 0 | Accepted answer

I found the text you quoted confusing - obviously the reason for your question. In particular it wasn't clear whether arcs were measured as distance over the ground or radians/angles. HTH.[![enter image description here](https://i.sstatic.net/LjKIa.png)](https://i.sstatic.net/LjKIa.png)


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