Saturday, February 27, 2010

Celestial Navigation

I stumbled upon Celestial Navigation, or rather the technique and the equations, in the same way that, I think that many inventions were made; where someone was thinking about the problem in their head for days and days, without a solution, and then an ordinary event puts the pieces in place and you have a moment where you say “Eureka!” Sometime in the fall I looked up on Wikipedia briefly a the page on Celestial Navigation, and it gave me a picture that had two overlapping cirlces. I don’t remember anything else about it, but those two circles stuck with me, though they didn’t explain the actual equations I would use.

The idea is that you take a sight with your sextant from the sun, and you plot a circle on the earth. This works, because if you accept that the earth is a sphere (a modern invention) then at any time, the sun is shining directly down at some point on that sphere. Therefore, the angle that you would measure with your sextant at that spot is 90 degrees (with respect to the horizon). So its straight overhead. Now, if you are 1000 miles away from that spot, the sun will not be directly overhead. Suppose you are 1000 miles north. Now, the sun is a little south of you, and you cast a small shadow at your feet. You take your sextant out and measure the angle between the sun and the horizon and you find its not 90 degrees anymore (because its not directly overhead). Say its 14 degrees from overhead, so 76 degrees from the horizon.
Suppose you were at that same moment to the south 1000 miles, you could also find 76 degrees as the angle between the sun and horizon, except that the sun is to the north of you. Now, since the Earth turns, (and the sun appears to rise, go west and set), there are times that the sun passes directly overhead at noon, but is not directly overhead before that time (like at sunrise). The earth is also divided up into time zones, so suppose that you were directly west of that same spot we were using before (where the sun was directly overhead). If you are looking at the exact moment that the two other sights were being taken, the sun will not be directly overhead for you, instead its slightly to the east. In fact, it if you’re 1000 miles to the west, the sun should be at 76 degrees from the horizon. Same in the other direction, and all directions in between. So, the sun sight that you take tells you how far away you are from where the sun is shining directly down. And you can draw a circle with a radius of that distance, if you only knew the location that the sun is shining directly down upon.
It just so happens that you can figure this out. Every day, the sun crosses the Prime Meridian (the line due north-south where Greenwich, England sits). The time that it crosses should be Noon exactly, but in fact it varies throughout the year, according to the orbit of the earth. There is an offset diagram on some older maps that looks like a little figure 8, which tells of the offset in minutes. So suppose that the sun crosses the prime meridian at 12:00, and at that moment we take a sight of the sun from somewhere on the earth. If we have a clock onboard that is set to Greenwich Mean Time (GMT, also known as Z) then we can say with certianty that at 12:00 GMT, the sun is shining over the Prime Meridian somewhere (remember the prime meridian is a line, not a location). So that tells us the Sun’s East-West location (Longitude), but we still need to find the North-South position (Latitude). But that’s easy, right? The sun is always over the equator…? Actually, no. On March 21 or on September 21 (the Equinox), the sun shines directly over the Equator, but on all other days, the sun is either to the North or to the South, which is why we have seasons. If the sun is to the North, then its higher in the sky in the Northern Hemisphere, and warmer in Washington. Conversely, in the winter, the sun is to the south, and lower in the sky. The sun will travel 21.5 degrees north from the Equator to the Tropic of Cancer over the period of March 21 to June 21, and then back to the Equator from June 21 until September 21. Then it starts to head south until December 21 (the Winter Solstace), where it is directly over the Tropic of Capricorn (at 21.5 degrees South), and then it heads back north again. If you were to model the path as a graph, where the north-south location is shown as the Y axis, and the year was spread out on the X axis, you would get a sine wave:






So, if you know what day out of the year it is, you should be able to find the sun’s Latitude position, using this graph.
Now we have the Latitude and Longitude of the sun (or of the position directly underneath) if we know the time and the day that we are looking. Then we take a sight of the angle between the sun and the horizon and we find our distance (as a radius) from that point. The way that the radius works is this: Suppose you were directly under the sun, you measure 90 degrees, and the radius from that spot is zero. If you move away 1000 miles, you measure 76 degrees on your sextant, (because you are 14 degrees away from 90). And, since the earth is divided up ever so conviently into degrees as distance, 14 degrees on the sextant is 14 degrees of Earth. So 1000 miles is 14 degrees. If we were to go until the sun dropped down to the horizon, we would measure the sun to be 0 degrees from the horizon, and our distance from the point where it was directly overhead would be 90 degrees of the earth. (We could convert that into miles, if you like, where 60 nautical miles is 1 degree, and about 70 regular miles is one degree, but since the position we want to find is going to be in degrees anyway we don’t have to.) The only problem is that we don’t know the direction we are traveling. This really bothered me for a long time, so I gave up on the two circle method that I had seen in the Wikipedia page and so I instead made a circle and a line.




If I know that I am some radius from the sun (on the circle) and I use my compass to tell me which way the sun is from myself (the line), I can calculate the intersection pretty easily. So to make it into equation form, so I don’t have to draw out the circle on a piece of map every time, I need a couple things: the sight, the bearing (from the compass) that the sun is away from me, the latitude of the sun, and the longitude of the sun (from the almanac). The location of the sun is the center of the circle. I use them and then add my extra latitude and longitude (or subtract it, depending on which way I am on the circle) to get my position. If I measure a 0 degree bearing, it means that I am due south of the sun’s location, so I should subtract the radius of the circle from the sun’s location. (Note: I am making South latitude be negative, as well as East longitude be Negative, so that the numbers work out.)

Latitude = Sun’s Latitude - Radius

But the radius is in degrees, and is 90 degrees minus my sun sight with the sextant

Latitude = Sun’s latitude – (90 - Sight angle)

But this is only correct for when the bearing is 0 degrees, and I am due south of the sun. What about a general equation that works in all cases? If I move to a bearing of 90 degrees, it means I am due west of the sun, and my latitude is the same as the sun’s, so I am subtracting 0 degrees from the sun’s latitude.
A Cosine wave fits here, because at 0 degrees, it is 1, and at 90 degrees it is 0, and then negative 1 at 180 degrees.

Latitude = Sun’s Latitude – (Cos (bearing) x (90 – Sight angle))

There we have it! And for longitude, I found the same equation, except that at 90 degrees bearing, we are the radius of the circle west (and therefore more longitude) of the sun. So this is fit with a sine wave.

Longitude = Sun’s Longitude + (Sin (Bearing) x (90 – Sight angle))

So that was my bastardized celestial, with a compass and a sextant. Unfortunately, I quickly realized that the accuracy of a compass is really bad, so you’ll find that the accuracy is plus or minus about 10 degrees, which is 700 miles on the earth. That just doesn’t cut it. So I needed to go back to the double circles and make it all work out.
At that time, I was tutoring algebra, specifically different graphs of different equations, like parabolas, lines, and circles! This was my Eureka moment, when I saw that the equation of a circle is so simple.

x^2 + y^2 = radius^2 (Where x^2 means: x squared)

This is the equation that I would use, I just needed to modify it a little bit. If you have a circle that is not centered at 0,0 on the graph, it has offset numbers, and they are in the equation.
So if you had a circle that was centered on 2,3, the equation would be

(x-2)^2 + (y-3)^2 = radius^2

Now, remember, the center of the circle is the location where the sun is shining directly down on the earth, and we can calculate that position knowing the time and date (or look It up in an almanac). The radius is again 90 degrees – sight angle. What we do is substitute in the sun’s position into the offsets for the circle (so that it is centered where the sun is, and plug in our radius, and we have one equation and two unknowns (x and y). Then we take another sight at a later time (hence drawing another circle) and we have a new sun position, and a new radius, and we then have two equations and two unknowns, which is a simple algebra problem.
So it would look like this: Suppose you had a sight of 50 degrees at a time when the sun’s position was 105 degrees west, 5 degrees north.

(x-105)^2 + (y-5)^2 = (90 –50)^2

Then we maybe take another sight a couple hours later, and find that the sight was 40 degrees, and the sun’s position was 150 degrees west, 5 degrees north.

(x-150)^2 + (y-5)^2 = (90 –40)^2

If we combine the two equations and solve for x and y, we’ll find our position. This next part is just the math. So first we will expand each equation in terms of x…

A: (x-150)^2 + (y-5)^2 = (90 –40)^2

(x-150)^2 + (y-5)^2 = (50)^2

(x-150)^2 + (y-5)^2 = 2500

x^2 – 300x + 22500 + (y-5)^2 = 2500

B: (x-105)^2 + (y-5)^2 = (90 –50)^2

(x-105)^2 + (y-5)^2 = (40)^2

(x-105)^2 + (y-5)^2 = 1600

x^2 – 210x + 11025 + (y-5)^2 = 1600

Then Subtract B from A

x^2 – 300x + 22500 + (y-5)^2 = 2500
(x^2 – 210x + 11025 + (y-5)^2 = 1600)

x^2 x^2300x + 210x + 22500 - 11025 + (y-5)^2(y-5)^2 = 25001600

And we get:

– 90x + 11475 = 900

So if we solve for X, we get:

x = 117.5

Since it is positive, that means it is 117.5 Degrees West Longitude. (you might expect X to be Latitude, but on an X, Y coordinate system, X measures the side to side portion, the same that Longitude does. Then we can take that and plug it in to one of the initial equations and solve for Y. Note: you will get two Y values, one positive, and one negative. This is because there are two points at which the circles cross. You have to use your brain to figure out which one you are at.

(x-105)^2 + (y-5)^2 = 1600

(117.5 – 105)^2 + (y-5)^2 = 1600

12.5^2 + (y-5)^2 = 1600

156.25 + (y-5)^2 = 1600

(y-5)^2 = 1600 – 156.25

(y-5)^2 = 1443.75

y – 5 = ± √1443.75

y – 5 = ± 37.996

y = ± 42.996

So that is 42.996 degrees North Latitude (if positive) or 42.996 degrees South Latitude (if negative). If I am taking the sight and I know that I’m in the Northern Hemisphere, suppose I have just left Seattle a couple days ago, I can assume that the 42.996 degrees is North Latitude, and that I haven’t jumped to the south pacific near Chile. So there you have it! The position, after two sights, and a lot of math is as follows:

117.5 degrees West Longitude, 42.996 degrees North Latitude.

2 comments:

mlloyd said...

So have you asked someone who already knew how to navigate celestially, if the way you have figured is the same as the way they figure? I was wondering, when you talked about the inaccuracy of a compass, were you using a nautical compass with the sighting mirror? If so, you are saying that these are not very accurate, because I remember learning how accurate Cap'n Lewis was in calculating his distance traveled using dead reckoning, of course he used a compass.

hank said...

I've given your figures and your logic a quick look-over, and it seems to work out. But is it really worth it? The usual intercept method will give the same result, the major inconvenience being the use of sight reduction tables--an annoying and cumbersome error prone procedure at best! And of course, you need not rely on bearings either. I recommend Pepperday's S-Tables if space is at a premium (only 10 pages!)

My compromise is to use a pocket calculator to reduce the sight and derive an LOP instead of the tables. The math (pg 279 of the Nautical Almanac) is straightforward, and any $10 calculator that can handle trig will do the job. You can carry a few spares if you are concerned about being overly dependent on high tech. They don't take up much room, and you can get them solar powered so batteries for them will not be an issue.

You can go full Luddite with just an almanac (even an out of date one, stellar Declination changes little over the years) and a sextant to do latitude sailing by the stars, like the old timers did. No tables, calculator or timepiece is required. Meridians are for sissies, real sailors just need Parallels to find their way home.