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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Finding position ab initio
From: Bill Noyce
Date: 2002 Mar 22, 10:22 -0500
From: Bill Noyce
Date: 2002 Mar 22, 10:22 -0500
George Huxtable suggests using Chuck's Venus and Moon altitudes to locate his position, and shows how we could get an AP to start with by drawing circles on a globe: > For a first shot (as we have no idea yet where Chuck was observing from) > the geographical positions of Moon and Venus at that time could be marked > on a globe. Chuck's measured altitudes for these bodies should be corrected > in the ordinary way (ie corrections as for altitudes, not corrections as > for lunar distances), and then the corrected altitudes converted to zenith > distance by taking the 90-degree complement. Now draw circles of that > radius about the two GPs on the globe. These intersect in two places, so > you have to guess which one is most likely. This paragraph could be > bypassed if we had a rough idea of whereabouts Chuck lived, in advance. Instead of using this procedure, we could use the method of "double altitudes" to find the latitude directly. It involves three spherical triangles, and repeated use of the formula cos a = cos b * cos c + cos A * sin b * sin c where lower-case letters represent the length of a side, and upper-case letters represent the angle opposite that side. This is the same formula used in sight reduction, except that altitude, latitude, and declination are each the complement of a side in the navigational triangle, so sines and cosines of the sides get interchanged. The formula as I've presented it directly gives the third side if we know two sides and their included angle; a little algebra rearranges it to give an angle if we know three sides: cos A = (cos a - cos b * cos c) / (sin b * sin c) The first spherical triangle has vertices of the pole and the two bodies (Venus and the Moon, in this case). We know the two declinations, which gives us two of the sides, and normally know the difference in GHA, so we can use the formula to compute the distance between the two bodies. In our example we already know the distance between the bodies, too, so we have all three sides of this triangle. We need to use the formula once more to compute the azimuth angle from one body (say Venus) to the other -- we'll use this for solving the third triangle later on. The second spherical triangle has vertices of the two bodies and the observer's zenith. Two of its sides are the zenith distances of the bodies (complements of the measured altitudes, after correction for IE, dip, refraction, semi-diameter, and parallax). The third side is the distance between the bodies, either computed from the first triangle, or using the cleared distance from our lunar work. We need to use the formula to compute the angle of this triangle at the same body as we did above. So in our example, this is the angle Zenith-Venus-Moon. The third spherical triangle has vertices of the pole, the observer's zenith, and the chosen body (Venus). Its sides are the body's co-declination, the measured zenith distance (corrected), and the observer's unknown latitude. Further, we know something about the angle of this triangle at the chosen body. Because we already know the angles Pole-Venus-Moon and Zenith-Venus-Moon, a simple addition or subtraction will give the angle Pole-Venus-Zenith. (A diagram is probably needed here, to decide whether to add or subtract.) From this angle and the other two sides, the formula gives us the third side - our co-latitude. This method of finding latitude by two observations can be used without a chronometer, as long as we can get the declinations of the bodies and their difference in GHA. For two fixed stars, you hardly even need a calendar for that; for the sun it helps to know the approximate hour and the date. If the two bodies (or even the same one) are observed at different times from the same location, we need a watch to determine the difference in GHA, but it need not be set to any particular time, and need not have particularly good long-term accuracy. Once we have latitude, we can use either (or both) of the observations as a "time sight". This uses the same formula to give us the LHA of the observed body, given latitude, declination, and altitude. Since in our case we know the time, and thus GHA, our longitude follows directly. Perhaps this is only of academic interest -- certainly the method that finds an approximate position first lets us use a more familiar method (and tables). But I was fascinated when I first read about it, with pictures, here: http://www.northwestjournal.ca/dtnav.html and go to Article VI.