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An aircraft is flying horizontally in a circle of radius $b$ with constant speed $u$ at an altitude $h$. A radar tracking unit is located at $C$. Write expressions for the components of the velocity of the aircraft in the spherical coordinates of the radar station for a given position $\beta.$

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At some point in the solution to this problem they state that $\theta=\beta/2$. I don't see how that is evident from the figure. Can anyone show me how this is deduced?

I also wonder why

$$\sin\phi=\frac{h}{R}\implies\dot{\phi}=-\frac{h\dot{R}}{rR}.$$

If I differentiate both sides I get

$$\dot{\phi}\cos{\phi}=-\frac{h\dot{R}}{R^2}\implies\dot{\phi}=-\frac{h\dot{R}}{R^2\cos{\phi}}.$$

What am I missing?

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    $\begingroup$ To address why $\theta = \frac {\beta}2$, see the diagram from top such that O is the centre of a circle, and C and the aircraft are points on it. You can then deduce the relation by marking the angles there and using angle sum property of a triangle. $\endgroup$
    – Eagle
    Jul 18, 2019 at 16:24

2 Answers 2

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Along the $x-$direction, the airplane moves a distance of $b \sin \beta = 2 b \sin \frac{\beta}{2} \cos \frac{\beta}{2}$. Along the $y-$ direction, the distance moved is $b(1-\cos \beta) = 2 b \sin^2 \frac{\beta}{2}$. Now $\tan \theta = \frac{2 b \sin^2 \frac{\beta}{2}}{2 b \sin \frac{\beta}{2} \cos \frac{\beta}{2}}$ (ratio of distance moved in $y-$ direction to distance moved in $x-$direction) which is equal to $\tan \frac{\beta}{2}$. Thus, $\theta = \frac{\beta}{2}$.

As for the second part of your question, you need to use the fact that

$\cos \phi = \frac{r}{R} $ to get rid of $\phi$.

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  • $\begingroup$ Carrying O-plane-z=h triangle to the z=0 plane would be easier. Then, by the isosceles, $2\times(90-\theta)+\beta=180$. $\endgroup$
    – Xfce4
    Sep 24, 2021 at 22:06
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Looking down the $z$ axis toward $C$, you will see that $r$ is the base of an isosceles triangle whose other two sides have length $b$. Of the two equal angles in this triangle, one is $90-\theta$ and the other is the sum of $\theta$ and $90 - \beta$. Equating the two angles: $90 - \theta = \theta + 90 - \beta$, which leads to $\beta=2\theta$.

To see how one angle is the sum of $\theta$ and $90-\beta$, it may help to draw a horizontal (parallel to the $x$ axis) line through the plane's position.

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