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    Re: A simple three-body fix puzzle
    From: Tom Sult
    Date: 2010 Dec 11, 13:27 -0600
    Yes. This is the view I have always held. I was simple confused by the groups use of big words! ;)

    Thomas A. Sult, MD
    Sent from iPhone

    On Dec 11, 2010, at 1:03, Frank Reed <FrankReed@HistoricalAtlas.com> wrote:

    Tom, you wrote:
    "How can that be true, If you are saying that the distribution about the LOP is normal then you have forever lower probability of being at the extreme edges of the distribution."

    But those bands of lower probability around the LOPs overlap. That means that the location with the highest probability is still inside the triangle, though it's a flat, low hill of probability so many points have nearly identical probability. The exact choice of a point, within reason, has no huge navigational significance (unless you get it REALLY wrong, for example by placing it always at the center of the triangle --see below). The reason to "get it right" is primarily that this rather academic problem is a strong foundation upon which other cases are built. Get this right, and you can understand a lot of other issues.

    And you wrote:
    "Only if the normal distributions overlap to be summed would you have a high probability of being at the center."

    Yeah, there you go. They do overlap. So you do see it.

    And you wrote:
    "And at that we have discovered by several means that the probability of being in the triangle is lower than being outside of it."

    On average, yes. It is after all, bigger out there! So even though the probability of being within some small distance of any specific point outside the triangle is lower than the probability of being within some small distance of any specific point inside the triangle, the total probability of being ANYWHERE outside the triangle ends up being bigger.

    Imagine this:
    Suppose we take an area on the ocean one degree on a side and we divide up it into little patches one tenth of a minute of arc on a side (one tenth of a nautical mile). That gives us a grid of 600 by 600 small patches, for a total of 360,000 squares in all. Now suppose I do some sights near the center of this one degree wide zone and I find three LOPs that cross in a nice equilateral triangle where the side length is one nautical mile --a fairly typical celestial navigation triangle. Note that there are ten of those little patches along one side of the triangle so the total area of the triangle is 43 patches. Now we can ask, what is the probability of being in that square patch right at the center of the triangle where I mark my fix? It's pretty low. I'm not gonna work out exact numbers, mostly because I don't want to waste more time on it, but it's a low number. Let's say 0.65%. So if I did this set of sights a hundred times, the probability of being inside that SPECIFIC patch right at the center where the fix is marked is less than even money.

    Now let's examine other square patches in the triangle. As we move away from the center, the probability of being in any small square patch falls off. It's lower and lower as we move away from the center. But the probability of being somewhere in the triangle must total to 25% so, on average, it's about 0.58% in each square (0.58% multiplied by 43 patches adds up to 25%). When we reach the patches right at the corners of the triangle, it would be lower than that, perhaps 0.53% in each square. You can see that the probability DENSITY is falling away rather gradually, so there's not much difference. We're less likely to be in any one of the corners than in the center, but not by a whole lot. So we keep on going... The probability density actually falls off radially with distance away from the center patch so that some points just outside the triangle actually have higher probabilities than the corners of the triangle (they're closer to the triangle's center), but eventually we get to patches where the probability of being inside a particular patch is 0.10%, but there are lot of those little squares all at the same distance from the triangle's center. They add up to a greater total NET probability. Even further out, there are square patches where the probability is only 0.01% --the odds are 10,000-to-one that you are in one of those squares-- but there are lots and lots of those low probability patches. When you add up ALL of those squares outside the triangle, you find that the odds of being outside are THREE TIMES greater than being inside the triangle. But the patches with the highest probability within them are all closest to the center of the triangle and the highest one, by just a bit, is right at the center.

    You wrote:
    "It gives me a headache. The position should be at the center of the cocked hat for no there reason than it looks like it should be."

    The headache, I can understand! But that's why God invented aspirin :). The square with the highest probability is at the center of the triangle ONLY when the triangle is equilateral. If you have a long skinny triangle, the correct fix is close to the short side. That's the "square one" lesson in this. Putting the fix at the center of the triangle can be LITERALLY wrong. Any of the various methods that are out there for finding the fix in the triangle will get you close, and, after you have seen a number of examples for different shapes of triangles, the "eyeball" method will work quite well, too, for any practical navigation purposes.

    And you concluded:
    "We must simply place a dot on the chart then look out of the wheel house and make sure we are not headed for the rocks."

    When you plot your LOPs and place your dot, you need to be able to say how much confidence you have in that point. What this is all about is placing rational, statistical ranges and limits around it. Far too many navigators believe that the center of the triangle is the point with the highest probability (it is so only in the equilateral case) and also think that the bounds of the triangle itself are the outer limits of any significant probability. BOTH of these beliefs are wrong and might well put a navigator in danger. Instead, we need to know where that center goes and how to at least visualize an error ellipse around that fix. Imagine drawing that triangle with a nice "number 2" pencil and then smudging it into a a roughly oval shape. This rough picture can be codified and quantified. That's what the mathematical analysis is for. Of course, the actual practitioner doesn't need to learn or study the mathematical underpinnings any more than a medical doctor needs to become an expert in molecular biology. But in both of these cases, every little bit of such knowledge helps.

    Incidentally, Tom, I saw on a blog not long ago that you bought the Starpilot app for your iPhone (this past Spring?). Luis Soltero, the developer, was one of our speakers for the Navigation Weekend at Mystic a few years ago although at that time he had not yet ported it to the iPhone (come to think of it, it wasn't out yet). You may have noticed in that software an option that says "Fix by Computation". It might interest you to know that this fix is generated by the very equations and methods that we have been talking about here. So if you want to experiment with the results, instead of puzzling over the math itself, you can enter some three-body sight data and find out where the fix ends up.


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