The calculation formula of disparity angle and its derivation

(Image source: "Astronomical Algorithm")

What is the "parallax angle"? The parallax angle is the angle between the equatorial meridian of the celestial body and the plane meridian of the celestial body, before the Zhongtian is negative and the transit is positive. For first quarter, the parallax angle is the degree to which it "falls":

(Image source: "Astronomical Algorithm")

So how do we calculate the angle? The Book of astronomical algorithms gives a formula:

\ (\tan{q} = \frac{\sin{t}}{\tan{\varphi}\cos{\delta}-\cos{t}\sin{\delta}} \)

(where \ (q\) is the celestial parallax angle, \ (t\) is the celestial time angle, \ (\delta \) is the celestial body's red latitude, \ (\varphi \) for the local latitude (north latitude is positive))

But there is no derivation process in the book. Here we complement the derivation process (for example in the Northern Hemisphere):

, draw the celestial sphere, the earth plane, the sky equator, the parallel circle. Set the celestial Center for \ (o\), North Day Extreme \ (p\), Zenith for \ (z\), celestial body for \ (b\).

As spherical triangle \ (bpz\), within the spherical triangle, the angle of the (q\) is (\ANGLE{ZBP} \). The definition of the time angle is easy to know \ (\ANGLE{ZPB} = t \).

Over \ (b\), \ (p\) for large arc \ (bp\) cross-day equator in \ (B ' \). Easy to know arc \ (BP = {90}^{\circ}-\delta\).

by local latitude for \ (\varphi\) Easy to know, arc \ (ZP = {90}^{\circ}-\varphi \).

Now that we know the sides of spherical triangle \ (bpz\) and their angles, the question is how to ask for another corner. This is more troublesome. You can first use the cosine theorem to find the third side, and then use the sine theorem to find the angle, but this will certainly be a collapse. Or take a three-side corner and push it one step at a pace:

For example, z\ \ (ZZ \) is perpendicular to plane \ (pob\), perpendicular is \ (Z ' \). Over \ (z ' \) make \ (z ' t_1\) (bp\) in \ (t_1\), over \ (z ' \) make (z ' t_2\) cross \ (op\) to \ (t_2\). Connection \ (zt_1\), \ (zt_2\).

According to the definition of spherical triangle internal angle (\angle{zt_2z '} = \angle{p} = t\), \ (\angle{zt_1z '} = \angle{b} = q \).

So all known and unknown quantities are shown on the graph. Then from \ (oz=1\) Start step by step to the length of the line (ZZ ' \) and \ (Z ' t_1\), and then calculate the \ (\tan{q} \), you can get the formula above. This process has been written in the picture, no longer repeat it.

When the celestial body rises/falls, (great circle) arc \ (ZB = {90}^{\CIRC} \), spherical triangle \ (zbp\) Three sides are known, directly apply the cosine theorem:

\ (\cos{zp} = \cos{bz}\cos{bp} + \sin{bz}\sin{bp}\cos{\angle{zbp}} \)

\ (\cos{({90}^{\circ}-\varphi)} = \cos{{90}^{\circ}}\cos{({90}^{\circ}-\delta)} + \sin{{90}^{\circ}}\sin{({90}^{\ Circ}-\delta)}\cos{q} \)

\ (\sin{\varphi} = \cos{\delta}\cos{q} \)

\ (\cos{q} = \frac{\sin{\varphi}}{\cos{\delta}} \)

In this case, you do not need to know the time angle \ (t\). (In other words, the time angle of the rise/fall can be calculated.) ）

The calculation formula of disparity angle and its derivation