Spherical Trigonometry

and

Celestial Coordinates

When we look at the sky and seek to map the things that we see on it, we imagine ourselves sitting at the center of a spherical shell upon which some god has inscribed invisible lines that we can only see with our Platonic vision. We use those lines to give numerical expression to the location of various skymarks relative to some well-defined pair of fiducial points. Astronomers use the First Point in Aries as one such point, common to both the Equatorial and Ecliptic systems. For astronomers the First Point in Aries denotes the point on the sky that the sun occupies at the beginning of spring. Representing one of the equinoxes, that point lies where the Ecliptic crosses the Celestial Equator, where the sun goes from south of the Equator to north of it. Because of the precession of Earth’s axis, that point no longer lies in Aries: it lies now in Pisces, just a little southeast of the pattern that astronomers call the Circlet.

If we use the Equatorial System of coordinates, based upon projecting Earth’s Equator onto the sky, then we refer the locations of skymarks to the north celestial pole and the First Point in Aries in their declinations and right ascensions. Imagine standing at Earth’s center with the North Pole directly overhead and the First Point in Aries directly ahead of you. We describe the position of an object on the sky by Right Ascension measured in hours, minutes, and seconds east (to your left) along the Celestial Equator from the First Point in Aries and by Declination measured in degrees, minutes, and seconds north (positive) or south (negative) of the Celestial Equator. In that system of coordinates the north pole of the Ecliptic lies 23.43928 degrees from directly overhead to your right, toward the point of winter solstice, and ahead of you the Ecliptic crosses the Celestial Equator, from lower right to upper left, at a 23.43928-degree angle.

If we use the Ecliptic System of coordinates, based upon projecting the plane of Earth’s orbit outward, then we refer the locations of skymarks to the north Ecliptic pole and the First Point in Aries in their ecliptic latitudes and ecliptic longitudes. Again imagine standing at Earth’s center and facing the First Point in Aries, only this time with the north pole of the Ecliptic plane directly overhead. In this case Earth’s north celestial pole lies 23.43928 degrees to your left from directly overhead. We describe the position of an object on the sky by Ecliptic longitude, measured in degrees, minutes, and seconds east (to your left) along the Ecliptic from the First Point in Aries and by Ecliptic latitude, measured in a direction perpendicular to the Ecliptic in positive degrees, minutes, and seconds of arc northerly or in negative degrees, minutes, and seconds of arc southerly.

Because we can conceive our idea of the sky as a spherical shell, we can take any trio of points on that shell as marking the vertices of a spherical triangle that we can solve by using the techniques of spherical trigonometry. We label those points A, B, and C and note that they define two trios of angles – those measured at the vertices and those referred to the center of the sphere.

We note that the sides of the triangle are simply arcs of Great Circles drawn on the sphere, circles whose centers coincide exactly with the center of the sphere. If we make the radius of the sphere equal to one unit of whatever units we want to use, then the lengths of those arcs equal the angles (measured in radians) that those arcs subtend as seen from the center of the sphere. For convenience we label those arcs with the lower-case versions of the letters with which we label the opposing vertices; that is, for example, the arc opposite A, passing between B and C, we label a, the arc opposite B we label b, and the arc opposite C we label c. Even though we use angular measure to describe them, we nonetheless refer to a, b, and c as the sides of the spherical triangle.

The angles of the spherical triangle measure the separation between two arcs as they meet at one of the triangle’s vertices. Imagine that we have drawn across each vertex a straight line tangent to the surface of the sphere and lying in the plane of each of the Great Circles whose arcs meet at that vertex. We identify the angle at which those straight lines meet as the angle of the vertex (think of it as a kind of limit), taking the angle that lies inside the triangle as the vertex’s angle. At the vertex A we label the angle a, at B we label the angle b, and at C we label the angle g.

In astrogation we often have a description of something that involves a spherical triangle, but we usually do not have all of that triangle’s sides and angles, so we must solve the triangle for the missing sides and angles. Spherical trigonometry gives us two tools for solving spherical triangles – the cosine rule and the sine rule. We can deduce those rules readily if we label the center of the spherical shell M and re-conceive the lines extending from that point to the vertices (which we have labeled A, B, and C) as vectors of unit length.

The Cosine Rule

We want to create two plane right triangles with the line MA as their common side. Imagine the line MA as the body of a butterfly whose head lies at the point M and whose wings coincide with the two triangles we want to examine. Each of those wings is bounded by two straight lines emanating from the point M and the arc that connects them. We will deduce the cosine rule from the relationship between those two triangles via the angle between them.

First, extend the line MB beyond the spherical shell. Then extend the line tangent to the sphere at A and parallel to the arc c (that is, lying in the plane defined by the arc) until it meets the extended line MB at a point that we label L. We now have a right plane triangle with hypotenuse ML, first side MA, and second side AL. Because the triangle has the angle c between lines MA and ML, we can express the lengths of the sides in terms of the unit length MA: we have

(Eq’n 1)

and

(Eq’n 2)

Second, extend the line MC beyond the spherical shell. Then extend the line tangent to the sphere at A and parallel to the arc b until it meets the extended line MC at a point that we label K. We now have a right triangle with hypotenuse MK, first side MA, and second side AK. Because the triangle has the angle b between lines MA and MK, we can express the lengths of the sides in terms of the unit length MA: we have

(Eq’n 3)

and

(Eq’n 4)

Now calculate the vector dot (or inner) product between the vectors coinciding with the lines AL and AK; that is, the vectors that have the same length as those lines do and point in the directions from A to L and from A to K. That product equals the lengths of the vectors multiplied together and then multiplied by the cosine of the angle between them. We have the vectors defined by subtraction of the radial vectors that they join;

(Eq’n 5)

and

(Eq’n 6)

Those vectors give us the dot product as

(Eq’n 7)

Making the appropriate substitutions from Equations 1 through 4 and dividing out the square of MA transforms that equation into

(Eq’n 8)

Multiplying that equation through by cosbcosc and rearranging it gives us the cosine rule,

(Eq’n 9)

If we know one vertex angle and the two adjacent sides, we can solve that equation for the remaining side. And if we know all three sides of the spherical triangle, we can solve for one of the vertex angles by using Equation 9 in the form

(Eq’n 10)

If we repeat that analysis using the lines MB and MC as the common side of our two plane right triangles, we get the other two versions of the cosine rule;

(Eq’n 11)

and

(Eq’n 12)

We also have the appropriate analogues of Equation 10. With those equations we can solve almost any spherical triangle that we are likely to encounter in astronomy. Although we won’t often use it, I now include the sine rule for completeness.

The Sine Rule

Go back to the triangles MAL and MAK and calculate the vector cross (or
outer) product of the vectors **AL** and **AK**. The resulting vector
points in the direction of MA, which has unit length, so we can calculate the
magnitude of the new vector by taking the dot product of it with MA. We get

(Eq’n 13)

in which equation I have exploited the fact that the dot product of a vector with a cross product in which that vector is a factor equals zero. Making the appropriate substitutions, including

(Eq’n 14)

and

(Eq’n 15)

we get

(Eq’n 16)

Multiplying that by cosbcosc gives us

(Eq’n 17)

If we apply that analysis to our other two pairs of right plane triangles, we get

(Eq’n 18)

and

(Eq’n 19)

But vector analysis tells us that, for any
vectors **U**, **V**, and **W**, we have as true to mathematics

(Eq’n 20)

so we can combine Equations 17, 18, and 19 into

sinbsincsinA=sincsinasinB=sinasinbsinC

(Eq’n 21)

If, at last, we divide that equation by sinasinbsinc, we get the sine rule

(Eq’n 22)

which tells us that if we know one vertex angle and the opposite side in a spherical triangle and one other side or vertex angle, then we can calculate the other opposite vertex angle or side. I normally don’t use this rule, because I have found that it can give wrong answers in certain circumstances and that it can be difficult to find the errors.

Converting Celestial Coordinates

When we want to convert the celestial coordinates of some skymark, such as a star, from one system of coordinates, say the Equatorial system, to another, say the Ecliptic system, the problem boils down readily to that of solving a spherical triangle. If we consider stars and other far distant objects, then the difference between referring the origins of our coordinate systems to Earth’s center or the sun’s center makes a difference so small in our coordinate conversions that we may disregard it without creating a noticeable astrogational error. If we consider bodies within the solar system, then we have descriptions of their orbits that lend themselves readily to expression in Ecliptic coordinates, making conversion unnecessary. Now to illustrate the solution of a spherical triangle, I will use the specific example of Alpha Centauri.

Again imagine that we are standing at Earth’s center looking toward the First Point in Aries with the north celestial pole directly overhead. We label that pole point A on our triangle. We find the star whose coordinates we want to convert and label its location on the sky point B. And we label the north Ecliptic pole, which lies 23 degrees, 26 minutes, 21.412 seconds of arc directly to our right from the north celestial pole, point C. On our spherical triangle those identifications make the arc from A to C equal to the side b=23.43928 degrees. The arc AB, extending directly from the north celestial pole to our star’s location subtends an angle equal to ninety degrees minus the star’s declination (d), so we have the side

(Eq’n 23)

The angle A, lying between the arcs AB and AC (which, when extended, passes through the point of the winter solstice), equals our star’s right ascension (r) plus ninety degrees, so we have

(Eq’n 24)

in which we convert the hours, minutes, and seconds of right ascension into the equivalent degrees at the rate of fifteen degrees per hour. The line extending from the north Ecliptic pole to the location of our star, the arc BC, equals ninety degrees minus the star’s Ecliptic latitude (j), so we have

(Eq’n 25)

And the angle C between the arcs AC and BC represents the complement of [the Ecliptic longitude of our star (q=its angular distance along the Ecliptic east of the First Point in Aries) plus ninety degrees], so we have

(Eq’n 26)

We now want to solve our spherical triangle for a and C so that we can calculate the Ecliptic coordinates from them. As of January 2000 we have Alpha Centauri’s right ascension and declination as

(Eq’n 27)

so we have our initial input as

(Eq’n 28)

We can see immediately that we have a special problem here. The angle A goes beyond 180 degrees, which means that we have our spherical triangle turned inside out. We need to have A, B, and C as internal angles, so we must use A=50.10 degrees. Now we can use Equation 9 to solve the triangle for cosa=-0.676762, whose arccosine gives us a=132.587 degrees. Next we use Equation 12 to solve for cosC=-0.861524227, which gives us C=149.488 degrees.

Equation 25 gives us the correct Ecliptic latitude easily enough, but Equation 26 does not yield the correct Ecliptic longitude. We see why that happens when we recall that we originally defined the angle C as measured westward from the line passing through the point of the summer solstice, which lies 90 degrees east of the First Point in Aries. In our inside-out spherical triangle we must now measure the angle C eastward from that line, so we calculate the correct Ecliptic longitude by simply adding 90 degrees to C. Thus we get for Alpha Centauri;

(Eq’n 29)

But we also want to add the proper motion of Alpha Centauri to those figures. In Equatorial coordinates Alpha Centauri’s position on the sky shifts by 0.113333... degree per century in right ascension and by -0.416666... degree per century in declination. We can’t convert those shifts directly, so let’s add ten centuries’ worth of drift to today’s Equatorial coordinates, convert those into the equivalent Ecliptic coordinates, subtract today’s Ecliptic coordinates from our results and then divide that difference by ten to obtain the per century drift of Alpha Centauri’s Ecliptic coordinates.

One thousand years from now Alpha Centauri will occupy the point on the sky with Equatorial coordinates of declination -65 degrees and right ascension 221.33 degrees. Thus we have to start

(Eq’n 30)

If we apply the cosine rule of Equation 9 to those figures, recalling that we must use A=38.67 degrees, we get cosa = -0.70027, whose arccosine gives us a=134.449 degrees. Applying Equation 12 then gives us cosC=-.929074847, which gives us C=158.291 degrees. We thus obtain Alpha Centauri’s Ecliptic coordinates for AD 3000: j’ = -44.449 degrees and q’ = 248,291 degrees. Carrying out the subtraction then tells us that over the next millennium Alpha Centauri will drift across the sky by -1.862 degree in Ecliptic latitude and by 8.803 degrees in Ecliptic longitude.

We thus calculate the Ecliptic coordinates of Alpha Centauri as

(Eq’n 31)

in which t represents time elapsed, in centuries, since January 2000.

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