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The intense mathematics...

Well, not really. Since I don't know any intense mathematics, you will find only the basics here.

It is easiest to begin on familiar ground and then proceed into the Unknown, so let us begin by thinking of right triangles in the plane. These were the familiar kind from high school and traditional favorite of civil engineers. When we want to know "everything there is to know" about any triangle, we conclude that there are six bits of information to tell the whole story: the lengths of the 3 sides and the measurements of the 3 angles. Of course, in a right triangle, 1 of the angles is spoken for--it is a right angle of exactly 90 degrees. This leaves 5 things yet to discover about any particular right triangle. We all remember those days of geometry and trigonometry when we first learned that the remaining 5 parameters may not be just any values; rather, they are constrained by special relationships that help us turn one choice of one piece of information into a definite conclusion about some other piece of information. A summary of some useful relationships may be found in this diagram:

A right triangle in the plane has many useful relationships between its 3 sides and 2 base angles. I prefer to use capital Latin letters for sides, and lower case Latin letters for angles.

So much for the familiar...now let us consider something new.

Specifically, what happens when we look in the sky at a familiar formation of stars such as the Summer Triangle, which has the bright stars Deneb, Vega, and Altair as its vertices? Well, in posing the question I answered it: we see something that looks like it "ought to be" a triangle. Of course it isn't really, because it is not in one flat plane. Instead, it is a curved region of the Celestial Sphere which happens to have 3 corners.

If you want to see a down-to-earth example of this phenomenon, go peel an orange. Cut a large piece of peel from it, and give that piece 3 corners. You will see that you can't really pound it flat on a table top (not a big piece, anyway) without it tearing. You can't take such a piece that is a 3-dimensional curve and flatten it to 2 dimensions without ruining it. Yet from its very "3-corneredness" we feel like this ought to behave as a triangle.

As things turn out of course, it is a special sort of triangle which we refer to as a spherical triangle. Spherical triangles also have 6 parameters: 3 angles, which are the corners of our orange peel, and 3 sides, which are really segments of 3 circles. This last point is really crucial to seeing the difference between plane triangles and spherical triangles. When we are solving spherical triangles for Navigation, we have no means of measuring the lengths of their sides. Since this is the case, when we refer to the "sides" of a spherical triangle, we are actually speaking of the central angles that subtend the sides of our spherical triangle. This is something we can measure since we are standing at the center of the Celestial Sphere, the Earth, and looking outward to those triangles. We call a spherical triangle a right spherical triangle when one of its 3 vertices is a right angle. An example of a right angle in a sphere occurs where any meridian cuts the Equator-- they meet at 90 degrees, a right angle.

After all this talk, it must be time for a diagram:

Here is spherical right triangle with a "Circle of 5 Parts" to help find relationships. We don't have relatively simple formulas to get relationships, as we do with plane right triangles. In fact, spherical triangles have an elaborate set of relationships that would be almost impossible to memorize, so the great mathematician Napier devised this clever and handy way to find them instead. We draw a circle and divide it into 5 wedges. We pick one of the dividing lines to represent the right angle, which I have done by making it bold. Then we "work our way around" the triangle, dropping each bit of information in turn into each wedge. In the 2 wedges next to the right angle, parameters are taken "as is", but for the remaining 3 wedges, we take the complement of the parameters. To find relationships, pick a wedge with some parameter of interest in it, and take the sine of that wedge. It will be equal to the product of the tangents of the "adjacent" 2 wedges, as well as the product of the cosines of the "opposite" 2 wedges. Then applying trig Identities simplifies the answer. In Celestial Navigation we are not interested in the "adjacent" wedges, so I omitted the tangent-cotangent relationship from the final answer. You will see the reason for this in a few moments...

So at this point it is natural to wonder, why is this of advantage to the Navigator?

Frankly, in his everyday work it isn't--That is how it ended up in an "Appendix for further information" instead of being in the Main Pages with the essentials. However, it is important in a broader sense because it is the theoretical groundwork which allows for us to compute altitudes for any celestial body, regardless of whether it is on our meridian or not. When we say the object is not on our meridian, that means that its vertical circle is not the same as our principal vertical circle; that is, its azimuth will be some value besides due North or due South. It also implies that its hour circle is not on our meridian either.

Since its hour circle is not on our meridian, obviously, there must be some angle between our meridian and its hour circle. (Remember the Diagram on the Plane of the Equator?) If we measure this angle only westward from our meridian, we call it LHA. But it is sometimes more convenient to measure eastward for bodies that are still rising. When we do this, we call the angle the meridian angle and denote it t, because it is intimately related to time. Methods of sight reduction that involve direct computation, rather than a table look-up, tend to invoke meridian angle more than LHA, because trig tables are written for angles less than 90, but LHA is between 270 and 360 for bodies which are rising. I have introduced t because it makes the diagram simpler, but keep in mind that when I write t, I could as easily say it is LHA--they refer to the same thing.

There is no easy way to do this, so let us just plunge right in...

What a busy diagram! Try not to be overwhelmed, and we will go through it all bit-by-bit. Our "god's-eye" viewpoint is from directly above our meridian, which appears as a vertical line. At present we are interested in only that quarter of it which runs from the Pole, P, to the Equator, which we already know can be used synonymously with Celestial Equator. We know from studying our Diagrams on the Plane of the Meridian that the "distance" along the meridian from zenith, Z, to Equator is exactly our Latitude, L. In this case, we are observing a star which is in Declination of the SAME name as our Latitude (Think of them as both North, or both South, as you prefer.) Its hour circle is shown as the left-hand edge of the sketch, and we know it has a declination above the Equator, Dec., so the distance remaining to the Pole along its hour circle must be (90 - Dec).

In this diagram, I have made no attempt to depict the horizon, but we know that the star appears somewhere above it just the same. We will see the star in some vertical circle at altitude H. My zenith is always 90 degrees from the horizon, so the "distance" along the vertical circle from the body to my zenith must certainly be (90 - H), as sketched in. Then we come to an interesting complication.

Let us return to those familiar plane triangles from high school for a moment. You of course remember that a right triangle is only one special case of triangles in general. We have a set of rules for solving general triangles, but they are not as neat as the rules for right triangles. Therefore, when we came to a general triangle that looked too nasty to deal with, our teachers told us time and again: "Drop a perpendicular" to split the ugly general triangle into 2 neat right triangles.

So it is with spherical triangles. We typically get general spherical triangles in celestial navigation problems. Although there are rules for solving such general spherical triangles, I don't know any; so we are going to use the old procedure of "drop a perpendicular" to split our general spherical triangle into a pair of right spherical triangles. This is the meaning of the arc marked R. It is drawn from the body to our meridian, and intersects our meridian at right angles in declination K. In actuality, this intersection can be either "above" or "below" the zenith depending on the circumstances of the observation: I just drew it "above" for purposes of illustrating only one case. The possibility of other cases gives rise to that really awkward notation: (K ~ L), which means "take either the sum of K and L, or the difference between K and L, as appropriate for the case at hand." Of course, anyone actually using the formulas to solve computed altitudes will need to have a couple of rules to determine which case is which. I won't go into that on this Web Page...

The last noteworthy item on the busy diagram is the 2 small angles which I have named e and f. Taken together, they are the parallactic angle, which is actually of no consequence to the Navigator, mainly because there is no way for the Navigator to conveniently measure it, and it is not needed for the solution of altitude and azimuth, which is all we desire for the altitude-intercept method. These two small angles are important now, though, because they represent pieces of spherical triangles which we must avoid in our calculations, if possible, since we cannot measure them and we are not interested in making work for ourselves by solving for information without an immediate practical value. In fact, we would prefer not to solve for R and K since they don't really help our navigation any. Unfortunately, we cannot avoid them.

Now that we understand the diagram, the rest is fairly easy. We just draw a Circle of 5 Parts for each of the 2 right spherical triangles we have created and write down the correct trigonometric relationships.

This Circle of 5 Parts is for the upper right spherical triangle. We construct it by simply following the same rules we used for the general right spherical triangle. I have shaded the wedge containing the unknown angle f as a reminder that we want to avoid it. Notice that a convenient thing happens. We have information that is known, namely Declination and meridian angle (or LHA, as you prefer) which allows us to determine the perpendicular, R. With that in hand, we also get the declination of the perpendicular, K Notice that these two equations invoke only "opposite" wedges in order to avoid the shaded wedge; thus, we do not need tangent-cotangent relationships.

This Circle of 5 Parts is for the lower right spherical triangle. I have shaded the wedge containing the unknown angle e as a reminder that we want to avoid it. Latitude and K work together with R to reveal for us the goal of this whole exercise: H, which in the altitude-intercept problem we call Hc This explains why we could not avoid R and K, even though they aren't "the answer" we seek-- they are intermediates necessary to bridge from the upper triangle of data we know, to the lower triangle of data we desire to know. Strangely enough, we see altitude is used to compute azimuth in our 4th and final equation. Now we have both altitude and azimuth, which is all we needed for the altitude-intercept method, so we are done! Please note that the azimuth number computed this way will not really be Zn, which means "azimuth from True North", in all cases. I simply used the notation to avoid confusion with the Z for zenith. Anyone using the formulas would also need rules for converting the computed azimuth to True Azimuth in cases when it is necessary. That is another set of rules I will skip on my Web Pages.

"Philosophy and Farewell"

If you have come with me this far, I admire your tenacity and depth of understanding in this subtle and beautiful world we call Celestial Navigation. You are now among the Keepers of the Lore, those who understand an ancient mathematical discipline which states for us an unexpected humanistic Truth:

You do have a place in the World, and indeed; objects in the farthest reaches of the Universe have some very personal relationship to your exact place.

"I wish you a pleasant voyage."

Go to the Celestial Navigation Appendices.

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