Bearing Between Two Points

Date: 12/19/2001 at 20:32:39
From: Doug
Subject: Latitude/Longitude calculations

I have found numerous solutions for finding the distance between two 
Lat/Long points on the earth (including the Haversine Formula), but 
I can't seem to find a reference that shows how to also calculate the 
direction between those points. At first blush it seemed easy but 
there appears to be something I'm missing. Is there an easy way to 
calculate the "heading" (relative to North = 0)between two 
coordinates?

Thanks.

Doug

Date: 12/20/2001 at 16:37:08
From: Doctor Rick
Subject: Re: Latitude/Longitude calculations

Hi, Doug.

My favorite web reference for navigational formulas is this:

  Aviation Formulary V1.33, by Ed Williams
  http://williams.best.vwh.net/avform.htm   

The algorithm it gives for bearing (or course) between two points is 
this:

   tc1=mod(atan2(sin(lon2-lon1)*cos(lat2),
           cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1)),
           2*pi)

The formula gives the *initial* heading for a great-circle route from 
point A to point B. The heading will change in the course of the 
trip. The quantities in the formula have these meanings:

  lon1 = longitude of point A
  lat1 = latitude of point A
  lon2 = longitude of point B
  lat2 = latitude of point B
  tc1 = direction of point B from point A (angle east of north)
  pi = 3.141596...

NOTE: The introduction to the Aviation Formulary page says this:

"For the convenience of North Americans I will take North latitudes 
and West longitudes as positive and South and East negative. The 
longitude is the opposite of the usual mathematical convention."

The algorithm as I show it above has been modified to use the usual 
convention that east longitudes are positive.

In the formula, atan2(y,x) is the arctangent, or inverse tangent, of 
y/x, with the additional feature of deciding which quadrant the angle 
belongs in, based on the signs of x and y. This is common in computer 
programming languages. 

Also, mod(a,2*pi) is the remainder you get when you divide a by 2*pi; 
that is, subtract the largest multiple of 2*pi less than a from a, 
and that's the answer.

I'll restate the algorithm in a form that makes the quadrant decision 
explicit (and also covers the cases for dlat and dlon equal to zero), 
in case you don't have access to an atan2 function.

  dlat = lat2 - lat1
  dlon = lon2 - lon1
  y = sin(lon2-lon1)*cos(lat2)
  x = cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1)
  if y > 0 then
    if x > 0 then tc1 = arctan(y/x)
    if x < 0 then tc1 = 180 - arctan(-y/x)
    if x = 0 then tc1 = 90
  if y < 0 then
    if x > 0 then tc1 = -arctan(-y/x)
    if x < 0 then tc1 = arctan(y/x)-180
    if x = 0 then tc1 = 270
  if y = 0 then
    if x > 0 then tc1 = 0
    if x < 0 then tc1 = 180
    if x = 0 then [the 2 points are the same]

Thanks to Kalpana who wrote in and helped me correct the original 
version of this algorithm.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 12/20/2001 at 22:47:05
From: Doug
Subject: Latitude/Longitude calculations

Thanks very much Dr. Rick. I will try the code you've suggested and 
let you know the result. For your information, it's to be used to 
calculate current speed and direction from an Acoustic Doppler 
Current Profiler. Normally the instrument utilizes the ship's gyro to 
determine direction, but it failed, and I need to use GPS positions 
at 2-minute intervals to determine the ship's direction. The current 
speed and direction relative to the ship can then be rotated to real-
world coordinates...

Doug 
    

Date: 06/15/2002 at 12:18:04
From: Hurle
Subject: Your formula for bearing

I know how to derive the formula 

                  sin(lat2) - sin(lat1)*cos(d)
   Bearing = acos ----------------------------
                     sin(d) * cos(lat1)

But how did you derive the alternative formula (which does not require
the pre-computation of the distance, d) that appears in your archive? 

Thank you. 

Hurle of Phoenix (actually "Surprise") Arizona

Date: 06/19/2002 at 08:45:23
From: Doctor Rick
Subject: Re: Your formula for bearing

Hi, Hurle.

I've wanted to get a derivation of the bearing formula into the 
Archives, but until now our correspondents have been content to get 
the formula without knowing why it works. Thanks for asking, and
motivating me to do the work!

Let's define three unit vectors, each in the direction of the line 
from the center of the earth to a point on the surface: N in the 
direction of the north pole, A in the direction of the initial point, 
and B in the direction of the final point on the course. Then the 
bearing we seek is the angle between the plane containing N and A, 
and the plane containing A and B. Thus it equals the angle between 
vectors perpendicular to these planes, namely, NxA and BxA.

Let point A have latitude lat1 and longitude 0 (we can rotate our 
coordinate system so this is true), and let point B have latitude 
lat2 and longitude dlon (the difference between the actual longitudes 
of A and B). Then we can calculate NxA and BxA:

  N = (0, 0, 1)
  A = (cos(lat1), 0, sin(lat1))
  B = (cos(lat2)*cos(dlon), cos(lat2)*sin(dlon), sin(lat2))

  NxA = (0, cos(lat1), 0)
  BxA = (sin(lat1)*cos(lat2)*sin(dlon),
         cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(dlon),
         -cos(lat1)*cos(lat2)*sin(dlon))

The usual way to find the angle would be to take the dot product of 
these vectors, divide by the product of the magnitudes of the vectors,
and take the arccosine. But that would be quite a mess. Instead we 
can take advantage of the fact that NxA is parallel to the y axis. 
The tangent of the angle between BxA and the y axis is the component 
of BxA in the x-z plane (the square root of the sum of the squares of 
the x and z components) divided by the y component:

  tan(theta) = sqrt((sin(lat1)*cos(lat2)*sin(dlon))^2 + 
                    (-cos(lat1)*cos(lat2)*sin(dlon))^2) / 
               (cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(dlon))

The numerator simplifies to

  sqrt(cos(lat2)^2 * sin(dlon)^2) * sqrt(sin(lat1)^2 + cos(lat1)^2)
  = cos(lat2)*sin(dlon)

Thus

  tan(theta) = cos(lat2)*sin(dlon) / 
               (cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(dlon))

That's the formula. If we take the arctan of both sides, we get a 
value in the range from -pi/2 to pi/2 radians. The problem is, this 
doesn't distinguish between opposite directions, NE vs. SW for 
instance. But the function atan2(y,x) returns the arctan of y/x with 
adjustments for the signs of x and y so that the angle returned is 
the angle of the cartesian point (x,y) in polar coordinates -- just 
what we need here. 

The only problem is that it returns a value in the range (-pi, pi]. 
The function mod(..., 2*pi) moves negative angles up to the range 
(pi, 2*pi). To get a final answer in degrees (0 to 360), you must 
multiply by 180/pi.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/19/2002 at 19:32:37
From: Hurle
Subject: Your formula for bearing

Thank you very much. I have not found this derivation anywhere 
else -- you are a wizard, I don't care what the other Math 
Doctors say about you!

Why is this used over the simpler version?

Date: 06/19/2002 at 20:44:33
From: Doctor Rick
Subject: Re: Your formula for bearing

Hi, Hurle.

I believe you're referring to the first version given here:

  Aviation Formulary V1.33, by Ed Williams
  http://williams.best.vwh.net/avform.htm#Crs

That version is only simpler if you don't count the calculations you 
had to do to find the distance between the points, which that version 
uses. Thus that version is better if you need to find the distance as 
well as the bearing, but if you only needed the bearing, then the 
formula I derived would be simpler.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/20/2002 at 09:25:01
From: Hurle
Subject: Thank you (Your formula for bearing)

You're right. Thanks once again. 

Hurle

Date: 01/31/2006 at 13:52:18
From: Bruce
Subject: spherical trigonometrical bearings for a dumb pilot!

Hello!  I am a rather unlearned (when compared to others on this 
site) pilot from South Africa.  I am trying to write an Excel sheet
that calculates great circle distances and bearings in the context
of a logcard.  The distance part I got right (as backed up by my 
helicopter's GPS system), while bearings I just cannot hack.  As I 
am in the Southern and Eastern Hemisphere, that might be causing 
the problem.

Also, I don't know what distance to insert in the formulas (i.e. nm, 
degrees, etc) and whether to use degrees or radians.  I keep getting 
answers like 708 or 1.257777 which don't equate to bearings.

Please help me with an idiot-proof formula, if possible relating 
to Excel, that I can try out.  Thank you in advance!!!

Date: 01/31/2006 at 20:04:12
From: Doctor Rick
Subject: Re: spherical trigonometrical bearings for a dumb pilot!

Hi, Bruce.

The main formula discussed on this page does not require input of the 
distance between points.  (There is a link to a different formula, 
that does require that input, near the bottom of the page.)

  tc1 = mod(atan2(sin(lon2-lon1)*cos(lat2),
            cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1)),
            2*pi)

A spreadsheet version of this formula is:

  BearingRad: =MOD(ATAN2(COS(LatA)*SIN(LatB)-SIN(LatA)*COS(LatB)
               *COS(LongB-LongA), SIN(LongB-LongA)*COS(LatB)),
               2*PI())

where I have defined cells containing formulas as follows:

  LatA:  =lat1*pi()/180
  LongA: =lon1*pi()/180
  LatB:  =lat2*pi()/180
  LongB: =lon2*pi()/180

and lat1, lon2, lat2, and lon2 are cells containing the coordinates 
in decimal degrees.  It appears that your coordinates are in degrees 
and decimal minutes, so you'll need to convert them to decimal 
degrees using formulas like this one:

  lat1: =lat1deg+lat1min/60*SIGN(lat1deg)

The BearingRad formula above gives the bearing in radians.  You'll 
need to convert that to degrees east of north with a formula like 
this:

  Bearing: =BearingRad*180/PI()

I took these formulas from my own spreadsheet for calculating 
bearings.  If this isn't enough to straighten out your problems, let 
me know what you do and what you get, so I can check it out.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/

Date: 02/05/2006 at 10:38:17
From: Bruce
Subject: Thank you (spherical trigonometrical bearings for a dumb pilot!)

Thank you so much Doctor Rick!  You are a legend - you really put it
so well and in terms that I, the maths wannabe, could understand.  
Keep up the good work and know you have a loyal fan from the 
southern hemisphere!!!

Bruce