This is not the best place to ask such doubts, so please recommend where I can. I currently don't have the facility to ask a teacher specialized for astronomy.

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

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? Could anyone please help, I will be grateful!

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    $\begingroup$ The author says FB=∠FOB, because for a unit sphere or circle (and angles measured in radians) the length of an arc is identical to the measure of angle subtending it. So cos BF is cos∠FOB. As for what the author is trying to derive, he says it right at the top: it's to relate the length FG to BC. Which he does in the last two lines. $\endgroup$ Commented Mar 10 at 14:28
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    $\begingroup$ @ScienceSnake this addresses the issue, thanks, you can answer this question if you want to, instead $\endgroup$ Commented Mar 10 at 14:39
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    $\begingroup$ I finally got one of the possible reasons why- the author wanted to derive the formula for departure $\endgroup$ Commented Mar 11 at 3:19

1 Answer 1


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


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