NavList:

We have discussed the cocked hat-three body fix a number of times over the years and this has been the most detailed and informative by far. But after the academic discussion I have to ask "what is the point we call a fix used for?" to get it back into the realm of practical navigation. Celestial navigation is an enroute navigation system and is not for threading your way through a narrow channel. Its purpose is to get you close enough, at the time of landfall, so that you can transition to the more accurate approach systems such as lighthouses and then even more precise, channel markers. All this quest for precision, I think, comes from GPS as now those interested in CELNAV can compare their work to the much more accurate GPS and now feel that their navigation is not good enough. Only since GPS have navigators been able to make this comparison while offshore. Prior to that they were satisfied with the CELNAV as there was only one time in a voyage when you could judge your own navigation, at landfall. If the lighthouse came into view in front of you at about the time you expected it then you were satisfied. When I crossed the Atlantic last year on the Royal Clipper I was the only passenger on deck at two in the morning looking for the lights of Barbados to come up over the horizon and I was quite satisfied to see the lights ahead at about the time I had expected them based on my prior evening's round of sights. Many of us, myself included, take sights from known positions ashore and want to compare our celestial positions so that we can judge our own workmanship, since we all take pride in this skill, but this is not navigation, which is directing a vessel (or plane) over the surface of the earth. I have brought this point up before. So what is the fix used for when navigating? It is used for planning the next course to the destination and to ensure that you are staying on course closely enough to avoid dangers on the way. So, for example, say you have sailed for a day since your last fix, covering about 120 NM and you take a three star fix and the triangle is 1 NM in breadth. You take as the fix a point somewhere inside the triangle. What do you do with this information? You lay off your new course to the destination. If the destination is a thousand miles ahead of you the difference in the new course from a point anywhere in the triangle is only three minutes. "Captain, if I use the north point of my cocked hat the course to New York is two seventy degrees but if you want me to use the southern point then your new course is 270 degrees, three minutes." You better have a damned good helmsman to steer that course. You might like to see if there is a current setting you off course since the last fix. After sailing 120 NM a fix taken at either extremity of the triangle would only amount to a half a degree difference, not enough to separate current from steering errors and sea state. (In flight navigation the fixes are used to determine the wind encountered since the last fix so that the wind information can be used in computing a heading to stay on the next course leg. This is more significant for planes as the wind speed can be a significant percentage of the air speed, much greater than the percentage of drift in mid ocean.) Then how closely do we have to stay on our preplanned trans-ocean course? On a two thousand mile ocean crossing you can be a hundred NM off course at the mid-point and only increase the total distance by just 0.5%, 10 miles, five minutes in a slow plane, a half hour in a fast ship, or two hours in a sailboat. So my advice is to avoid the whole cocked hat conundrum by only taking two star fixes then there is no triangle that you will be outside of 3/4ths of the time. Think about this. If you had looked up in the sky at twilight and, due to clouds, you were only able to take two well spaced sights you would still be satisfied with your fix and the fix would be accurate enough for the purposes of guiding the vessel to the destination. Or, if you do get three stars, then just accept that the third line is mainly useful as a check of the two body fix to find gross errors so that you can have more confidence in your fix. Aircraft GPS units are required to have RAIM, Receiver Autonomous Integrity Monitoring, so that a pilot can be sure that the accuracy of the GPS position is within strict narrow limits to allow an instrument approach to a runway hidden in the clouds. It has to take you to the center line of the runway and you have to know that it is accurate enough to do that so that you don't go flying into a school. To do this you must have at least 5 satellites in view at good azimuths. This is one more than required for a three dimensional fix and the GPS checks the pseudorange expected from the extra satellite from the computed fix and if the pseudorange differs too much the unit flags the fault and you have to use a different navigation system to complete your flight.  The third star LOP serves the same function as the fifth pseudorange confirming the accuracy of the two body fix. If you are sailing near to danger then don't use the center of the triangle as the fix, use the point closest to the danger and plan your course based on that while keeping in mind that you might be outside the triangle, and closer to the danger, by three sigmas. It is perfectly valid to use any of the corners of the triangle as your fix since each corner is a valid two body fix which you would be perfectly happy with if you had only been able to get two stars. I know that adding more LOPs to the solution will also mathematically improve the accuracy of the fix a bit but since a two body fix is accurate enough for practical navigation there is no compelling need to use the third (or fourth or fifth) LOP for this purpose, just use the extra LOP for confidence in the two body fix. There is no good reason to polish the cannonball. gl --- On Fri, 12/10/10, Peter Fogg <piterr11@gmail.com> wrote: From: Peter Fogg <piterr11@gmail.com> Subject: [NavList] Re: A simple three-body fix puzzle To: NavList@fer3.com Date: Friday, December 10, 2010, 11:19 PM  Frank Reed wrote: 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. This is a very good explanation, Frank.  You have written an eloquent word-picture that is possibly the clearest post yet on this subject, which has been bounced about here for years.