Bearing CalculationDate: 09/01/97 at 00:39:54 From: Hrvoje Vojnic Subject: Bearing calculation Hello! I have a question regarding the calculation of bearings. Given two cities of geographic coordinates (xA,yA) and (xB,yB) is there a formula which calculates the bearing from city A to city B? If so, would you please write it, along with any specifics which it may include (angle in radians or degrees, distance in miles or kilometers, etc.) Thank you in advance! Best regards, Hrvoje Date: 09/01/97 at 14:56:04 From: Doctor Jerry Subject: Re: Bearing calculation Dear Hrvoje, Do you mean the angle made by the great circle route with the meridian through city A? Please let me know and I'll try to answer your question. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/04/97 at 01:49:03 From: Hrvoje Vojnic Subject: Re: Bearing calculation Hello again, Dr. Jerry! Regarding my question, (bearing from city A to city B, given their geographic coordinates) I apologize for not being precise. It is exactly like you understood my question, I'm interested how to calculate the angle between the great circle route from city A to city B and the meridian of city A. That is, the bearing at city B from city A. I know how to calculate the great circle distance, but bearing is a puzzle. Thank you! Hrvoje Date: 09/04/97 at 12:17:30 From: Doctor Jerry Subject: Re: Bearing calculation Hi Hrvoje, 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 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). Think about the unit vector u joining the origin to the point (x1,y1,z1). The way I wrote the r function, using curly brackets { }, already puts us in the vector format. Form the unit vector v lying in the plane through the origin, P1, and P2 (the intersection of this plane with the sphere is the great circle we're working with), as follows. I'll give part of the Mathematica code I used to check all of this. The code is a litte peculiar, but I think you can follow it with no difficulty. r[phi_,theta_]:={Sin[phi]*Cos[theta],Sin[phi]*Sin[theta],Cos[phi]} P1=r[Pi/4,Pi/6]; P2=r[Pi/6,Pi/3]; u=P1; The vector w (below) is the vector along u onto which the vector P2 projects. The expression P1.P1 is dot product. Lgth is the length function. w=P2-((P1.P2)/(P1.P1))P1 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) + cos(t) v. Hence, r1'(0)=v. 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/