Figure 1
Figure 2
Sometimes the various angles on the celestial sphere are interrelated and can be easier to understand by drawing a time diagram. Time diagrams are used to understand the relationships between the values of the various hour angles.
In a time diagram you are viewing the celestial sphere from the celestial south pole. The center represents the celestial south pole, "Ps", the outer circumference represents the celestial equator, and a line drawn from the center to the edge represents a celestial meridian or an hour circle containing a body. Most of the time, the line drawn from the center to the 12 o'clock position represent either the Greenwich celestial meridian, labeled "G", or the local celestial meridian, labeled "M". Small "g" and "m" represent the lower branches of their meridians.
In figure 1 this is a time diagram where GHA of the sun is. The upper branch of the sun’s hour circle is shown as a solid line. The angle or arc of the celestial equator between the Greenwich meridian and the sun’s hour circle is 90°. The GHA of the sun at this instant is 90°. GHA is always measured westward from G. Local hour angle (LHA) is the name given to the angle of arc (expressed in degrees, minutes, and tenths of minutes) of the celestial equator between the celestial meridian of a point on the celestial sphere and the hour circle of a heavenly body. It is always measured westward from the local meridian through 360°.
Let’s try an example problem of LHA on a time diagram. Say you are at longitude 135° from M toward Greenwich which means that Greenwich will be shown east of M. Think about this for a minute, you are to the west of Greenwich, which means Greenwich is to the east of you. Now that you know where Greenwich is and where you are, let’s figure the LHA of the sun as it is shown above in figure 1.
Figure 2 shows you that the sun is 90° west of Greenwich. We know that the LHA is always measured westward from your location meridian (M) to the hour circle of the body (in this example, the sun). The LHA here is the whole 360° around minus the 45° between the sun’s hour circle and M. This 45° can be found by looking at figure 2 or by subtracting 90° from 135°. Let’s think this over, you are 135°W of Greenwich, G is135° clockwise from you. The sun is 90° W or counterclockwise from G, the difference is 45°. Subtract this 45° from 360° and you will get 315°, the LHA. Look at figure 2. You can see, the sun is east (clockwise on the diagram) of your local meridian (M). Now suppose that you are at the same longitude (135° W), but the GHA of the sun is 225° instead of 90°.
The time diagram will appear as shown in figure 3. The sun is now west of your meridian (M). The LHA is always measured westward from the local celestial meridian to the hour circle of the body. Now the LHA is the 90° from M to the sun’s hour circle. Here are two general rules that will help you in finding the LHA when the GHA and longitude are known:
1. LHA = GHA - W (used when longitude is west)
2. LHA = GHA + E (used when longitude is east)
When in west longitude you might have to add 360° to the GHA before the subtraction can be made. In east longitude, 360° is subtracted from the LHA if it exceeds this amount. You can check your work by referring to a time diagram. It gives you a graphic means of obtaining the data you need.
The GHA of a star is measured in the same direction from Greenwich to the star but because the SHA enters the picture here, your method of locating a star on the time diagram is a little bit different. First, you have to locate the vernal equinox by its tabulated GHA. Let’s say the GHA of the vernal equinox for the time of your observation is 45°. You locate the vernal equinox 45° W from Greenwich. From the Nautical Almanac you find the SHA of the star in your looking for. You already know that the SHA is measured to the west from the vernal equinox (first point of Aries). All you have to do here is find the SHA of this star and measure the SHA westward from the vernal equinox, you then have the star located on your time diagram.
Note: Meridian angle "t", this value is measured either eastward or westward from the local meridian to the hour circle of the body. Its value cannot exceed 180 degrees. If LHA exceeds 180 degrees "t" will be equal to 360 degrees minus LHA.
Try this example problem:
You are in longitude 104° east. The Greenwich hour angle of a star is 211°. The meridian angle "t" of the star is?
A. 45 degrees east
B. 45 degrees west
C. 315 degrees east
D. 315 degrees west
Step 1: Draw a basic time diagram as with the local celestial meridian "M" up. The solid part of the line is the upper branch of your meridian (the part you are on), and the dotted line represents the lower branch. Next put in the known values. Place "G," the meridian of Greenwich, 104 degrees west of "M" because you are east of Greenwich.
Step 2: Next measure westward from Greenwich (counterclockwise) the 211 degrees of GHA and draw in the hour circle of the star.
Note: You do not have to use exact measuring methods, a close approximation will be good enough.
Step 3: LHA is the measurement westward between the local celestial meridian and the hour circle of the body, from 0 degrees to 360 degrees. Draw the LHA on your time diagram. Look at the angles you have drawn, you should see that the LHA of the star is equal to the longitude plus the GHA of the star (104 + 211 = 315).
Step 4: Remember the note above about the definition of meridian angle, "t". The meridian angle is the measurement between the local celestial meridian and the hour circle of the body, its value cannot exceed 180 degrees, and it is labeled east or west. In short, "t" is the shortest angular distance between "M" and the body's hour circle. In this problem you can see that meridian angle, "t", is equal to 360 degrees minus the LHA determined above and is east (360 - 315 = 45° East).
