Navigation FormulaDate: 10/21/96 at 9:9:50 From: Jerry Jost Subject: Re: navigation formula Hi, I'm writing a program that deals with aircraft navigation. The only part of the equations that I'm still having trouble with is finding the formula that gives the final longitude and latitude if the starting longitude and latitude, course, and direction of a plane are known. Let me put it in terms of what the program requires. I'm taking off in an aircraft from an airport whose latitude and logitude are known (LatOrigin and LonOrigin). I'm flying a course of 274 degress true (not magnetic). I fly for a distance of 73.9 nautical (or statute) miles. Given these variables, where would I end up (LatDest and LonDest)? I guess in mathematical terms, the distance would be referred to as a great circle distance. Stating the question in another way; I left Minneapolis airport and flew X nautical miles at a course of Y degress (true). How do I determine my new current position (in Latitude and Longitude)? Thank you very much for your help, Jerry Date: 10/24/96 at 16:46:31 From: Doctor Jerry Subject: Re: navigation formula Jerry, Since you refer to a course, I think you must mean a constant compass heading, that is, all meridians are crossed at the same angle. If so, then you are not talking about great circle navigation (except in a few special cases); rather, you are talking about rhumb line navigation, which is based on Mercator's projection. The answer is relatively complicated, depending on your background. I feel certain that you need some understanding of calculus to do this problem. On the chance that you may not have had a calculus course, I won't go into extreme detail and, moreover, I'll assume a few things to make the discussion easier for me. If you indeed want to see the details, complete with graphs, I guess we'd have to talk about that since it involves some effort. I expect that this is written up somewhere, but I'm not an expert on practical navigation. What I'll assume is such things as the radius of the earth is one unit, longitude is measured from the (x,z)-plane, and I'll use co-latitude instead of latitude. You can see that I'll be working on the unit sphere, using spherical coordinates phi and theta, which I'll abbreviate to p and t. These will be in radians. I'll give some details, just to give an idea of what's involved. Under Mercator's projection, a point (p,t) on the sphere is mapped onto the point (X,Y) = (2t/pi,-(2/pi)ln(tan(p/2))) in an (X,Y)-plane. This is correct for certain map conventions. Minor changes for other conventions. Let Y(p) = -(2/pi)ln(tan(p/2)). If one flies or sails on a rhumb line, making an angle of b with the meridians, then p and t are related by the equation t = (pi/2)[(tan b)Y(p) - k], where k is a constant and can be determined by knowing a single point (p1,t1) through which the rhumb line passes. This could be, for example, your starting point. Let H(p) = (pi/2)[(tan b)Y(p) - k]. We can now write a parametric vector equation for the rhumb line on the sphere. It would be: r(p) = {cos(H(p))*sin(p),sin(H(p))*sin(p),cos(p)}, p >= p1. There are formulas for calculating arc length of such a curve. One calculates r'(p) and then its length |r'(p)|. The arc length from p = p1 to p = p2 is integral from p1 to p2 of |r'(p)|dp. I know this is too much, too fast. But you may decide this stuff isn't for you. Anyway, we don't know p2; what we are given is (p1,t1) and the arc length s. We must solve the equation for p: integral from p1 to p of |r'(q)|dq = s I used q as a "dummy" variable, to avoid confusion with the variable p. Newton's method can be used. Each iteration will require a numerical integration. I tested this whole scheme by taking a beginning point (p1,t1), and ending point (p2,t2), and calculating the arc length s. I then did Newton's method starting with (p1,t1) and s, and, to my delight, got p2. The number t2 then comes from H(p2). Are you interested in continuing this discussion? I wrote up a student project, for a calculus course, on a Mercator's project last year. I have been asked to write up a key answer sheet and your question pushed me into doing this a little earlier than I planned. Thanks! This was good for me. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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