Calculating the Shortest Flight PathDate: 10/30/97 at 20:48:26 From: Phil Subject: Calculate angle between two places Using flight simulator, flying long distance between two points with given latitude/longitude, calculate the flight path (angle) which is the shortest route. Is this spherical trigonometry or can it be solved using simpler geometric equations? Date: 10/31/97 at 08:27:38 From: Doctor Jerry Subject: Re: Calculate angle between two places The bearing of the great circle path (shortest) at the initial point can be calculated, as you suspect, by spherical trigonometry. The word "simpler" is tricky, however. For me, knowing almost no spherical trigonometry, it's simpler to use spherical coordinates, vectors, and dot products. This gives a nice method for calculating the bearing. If you are interested, please write back. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 10/31/97 at 08:52:34 From: Phil Subject: Re: Calculate angle between two places Thanks for replying. Yes, I would like to know how to calculate the shortest path (angle) between two sets of co-ordinates (lat/long) using the method you mentioned. Thanks in anticipation. Date: 10/31/97 at 15:26:02 From: Doctor Jerry Subject: Re: Calculate angle between two places Hi Phil, It's easiest for me to explain in terms of an example. I'll use spherical coordinates, which are closely related to the geographic coordinates of latitude and longitude, and I'll work on a sphere with radius 1. I think the solution procedure will be clear for the general case. I don't know that the solution I'll explain is the simplest. If one were to read a book written for navigators, I presume that relatively simple ways to calculate the bearing of a great circle route would be explained. However, from my experience in looking at such books, to understand the procedures I would have to read more of the book than I'd want. I'll consider points S (starting point) and E (ending point); specifically, S has spherical coordinates (1,pi/4,pi/6) and E has spherical coordinates (1,pi/6,pi/3). These correspond to (on the unit sphere) points with latitudes 45 deg N and 60 deg N and longitudes differing by 30 deg. I'll use the spherical to rectangular transformation r(phi,theta)={r*sin(phi)*cos(theta),r*sin(phi)*sin(theta),r*cos(phi)}. Start by calculating the rectangular coordinates of S and E; call these P1 = r(pi/4,pi/6) and P2 = r(pi/6,pi/3). We may think about P1 and P2 as vectors from the origin. From these two vectors we form mutually perpendicular vectors u and v lying in the plane containing P1 and P2. The intersection of this plane with the sphere is the great circle of the shortest route. I'll explain in terms of the Mathematica code I used to check all of this. The code is a little peculiar, but I think you can follow it with no difficulty. r[phi_,theta_]:={Sin[phi]*Cos[theta],Sin[phi]*Sin[theta],Cos[phi]} The curly brackets make r[phi,theta] a vector. Let P1 = r[Pi/4,Pi/6] and P2 = r[Pi/6,Pi/3]. Let u = P1. The vector w w = P2-((P1.P2)/(P1.P1))P1 is a vector perpendicular to u = P1. The expression P1.P1 is dot product. Lgth is the length function. Now define v = w/Lgth[w]. The great circle route through S and E is the parametric curve r1[t_]: = Cos[t]*u+Sin[t]*v for 0 <= t <= 2*pi. Note that r1[0] is the point S. To calculate the bearing I'll calculate the tangent vector r1'[0] and dot it with the north-pointing unit vector m through S. Differentiating r1, r1'(t) = -sin(t)*u+ cos(t)*v. Hence, r1'(0) = v, which was obvious anyhow. Also, m = {-sqrt(3)/(2*sqrt(2)), -1/(2*sqrt(2)), 1/sqrt(2)} This vector can be calculated by forming the negative of the tangent vector to the curve with parametric equation r(phi,pi/6), and evaluating at phi = pi/4. The bearing is then b1 = ArcCos[(m.v)/(Lgth[m]*Lgth[v])] In particular, b1 = ArcCos[3*sqrt(3)/(8*sqrt(5))+sqrt(15)/8]=0.684719 or 39.23 degrees. Although this procedure probably will require some special cases for points in different hemispheres and other quirks of spherical or geographic coordinates, it is fairly straightforward. I checked all of this by doing it a different way. I assumed the equation of the plane through the center of the earth, S, and E had the form A*x+B*y+z = 0 and, with the rectangular coordinates of S and E, calculated A and B. This gave the plane -2y+z = 0. I solved this simultaneously with x^2+y^2+z^2 = 1, thinking of z as fixed. This gave me parametric equation (parameter z) for the great circle. I found r2[z_]: = {Sqrt[4-5z^2]/2,z/2,z} I calculated r2'[0] and dotted it with the meridian vector as before and got the same bearing. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 06/14/2005 at 06:28:49 From: Nick Subject: Distance Between Points on the Earth I have read your answer, and feel that a solution can be derived quite simply without using dot products, scalar products or coordinate transformations: Calculating the "Great Circle" distance from P1 (Longitude X1 West, Latitude Y1 North), to P2 (X2,Y2). (i) - The length of a "vertical" drop (vertical means parallel to the north/south axis) from P1 to the equatorial plane is given by RsinY1 where R is the Earth's radius. This vertical will intersect the equatorial plane at a point a distance RcosY1 from the Earth's centre. Call this point P3. (ii) - The length of a vertical drop from Y2 to the equatorial radius is given by RsinY2. To a point RcosY2 from the earth's centre (P4). (iii) - Thus we have a trapezium with three orthogonal sides: 1) - The "base" is P3/P4, and its length can be calculated since the lengths of two sides and the angle between them are known, using the Cosine Rule: (a^2 = b^2 + c^2 - 2bc*cosA) where: "a" is base P3/P4; b and c are the two distances from the Earth's centre, RcosY1 and RcosY2; angle A is the angle between them which is the difference between the two longitudes: X2-X1. So: (P3/P4)^2 = (RcosY1)^2 + (RcosY2)^2 - 2(RcosY1)(RcosY2)cos(X2-X1) 2,3) - The two parallel sides of the trapezium are orthogonal to the base (P3/P4) and are the "verticals" given in (i) and (ii) with lengths of RsinY1 and RsinY2. (iv) - The length of the fourth side is the straight line distance between P1 and P2. The trapezium is effectively a rectangle joined with a right-angled triangle. The "fourth side" of the trapezium is the hypotenuse of the triangle; the base of the triangle is the same as P3/P4 (see (iii)-1); the third side of the triangle is the difference between the lengths of the two parallel sides of the trapezium: RsinY1 - RsinY2. Pythagoras gives us: (P1/P2)^2 = (RsinY1 - RsinY2)^2 + (P3/P4)^2 (v) - Knowing the straight line length of P1/P2 enables the calculation of the angle subtended between P1, P2 and the centre of the Earth, using the Cosine Rule A = arccos(b^2 + c^2 - a^2)/2bc, where b and c both = R and a = P1/P2. So: A = arccos({R^2+R^2-[(RsinY1-RsinY2)^2+(RcosY1)^2+(RcosY2)^2-2 (RcosY1)(RcosY2)cos(X2-X1)]}/2R^2) Removing the R^2 terms gives: A = arccos({2-(sinY1-sinY2)^2-(cosY1)^2-(cosY2)^2+2(cosY1)(cosY2)cos (X2-X1)}/2) Expanding (sinY1-sinY2)^2 gives: A = arccos({2-(sin^2Y1+sin^2Y2-2sinY1sinY2)-cos^2Y1-cos^2Y2+2(cosY1) (cosY2)cos(X2-X1)}/2) Then: A = arccos({2-(sin^2Y1+sin^2Y2-2sinY1sinY2+cos^2Y1+cos^2Y2)+2(cosY1) (cosY2)cos(X2-X1)}/2) Since sin^2 + cos^2 = 1, this simplifies to: A = arccos({2-(1+1-2sinY1sinY2)+2(cosY1)(cosY2)cos(X2-X1)}/2) And then to: A = arccos({2sinY1sinY2+2(cosY1)(cosY2)cos(X2-X1)}/2) And finally to: A = arccos{sinY1sinY2+cosY1cosY2cos(X2-X1)} (vi) - The "great circle" distance along the Earth's surface between the two points is trivial from here, since one nautical mile is defined as the distance travelled when moving through one minute of arc along the Earth's circumference. So just multiply the angle A, in degrees, by 60. From nautical miles, the distance can easily be converted to miles, km or another linear measure! Date: 06/14/2005 at 15:04:18 From: Doctor Rick Subject: Re: Distance Between Points on the Earth Hi, Nick. Doctor Jerry offered a method without spherical TRIGONOMETRY, which is a separate subject, not taught much anymore because spherical geometry and vector algebra can do the same things and are more general in application. Simplicity may be in the eye of the beholder. For one who is familiar with cartesian-spherical coordinate transformations and vector algebra, the method shown in the archives is much simpler. However, it is surely comforting for those who do not understand these methods to see that they are not required. In fact, much of your work is a re- derivation and confirmation of the formulas for the coordinate transformations and vector algebra. - Doctor Rick, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/