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Location of Plane Flying on Great Circle

Date: 11/20/2003 at 06:11:38
From: Matteo
Subject: Great circle problem.

An aircraft is flying with the aid of an inertial navigation system 
(INS) connected to the autopilot.  The following two points have been 
entered in the INS computer; WPT 1: 60N 030W, WPT 2: 60N 030W.
When 025W is passed the latitude shown on the display unit of the 
inertial navigation system will be:

6011.0'N
5949.0'N
6005.7'N <<<<<< correct answer, but WHY??
6000.0'N

The INS navigates via a great circle.



Date: 11/20/2003 at 11:10:39
From: Doctor Rick
Subject: Re: Great circle problem.

Hi, Matteo.

Two of the options can be eliminated without calculation.  The two 
waypoints are at the same latitude.  A line of latitude is not a great 
circle; the great circle between the two points will go farther away 
from the equator than the line of latitude.  Therefore, at the 
midpoint of the trip, the latitude will be greater than 60 degrees N. 
This eliminates 59 deg 49.0'N and 60 deg N.

To distinguish between the remaining two options requires calculation.
I do not know what formulas you may have learned for this sort of
situation.  Formulas can be found here:

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

The formula "Latitude of point on GC" can be used to calculate the 
correct latitude in your problem:

"Intermediate points {lat,lon} lie on the great circle connecting 
points 1 and 2 when: 

  lat = atan((sin(lat1)*cos(lat2)*sin(lon-lon2)
        - sin(lat2)*cos(lat1)*sin(lon-lon1))/
          (cos(lat1)*cos(lat2)*sin(lon1-lon2)))

  (not applicable for meridians. i.e if sin(lon1-lon2)=0)"

In your problem, both waypoints are at the same latitude, and the 
point of interest is the midpoint of the route, therefore the highest 
latitude reached by the great circle.  Since the problem is simpler 
than the general case, I can find the answer from first principles; 
in the process I will come up with a formula that is simpler than the 
formula for the general case, found in the Formulary. 

Looking at the earth from above the north pole, we see the two 
waypoints as points on a circle (the line of latitude) of radius 
R*cos(lat1) where R is the radius of the earth.

                            ***********
                        ****           ****
                    ****                   ****
                 ***                           ***
               **                                 **
             **                                     **
            *                                         * 
           *                                           *
          *                                             *
         *                                               B
         *                                           -   |
        *                            R*cos(lat1) -       |*
       *                                     -           | *
       *                                 -               | *
      *                              -  dlon/2           |  *
      *--------------------------O-----------------------C--*
      *                              -                   |  *
       *                                 -               | *
       *                                     -           | *
        *                                        -       |*
         *                                           -   |
         *                                               A
          *                                             * 
           *                                           *
            **                                       **
              **                                   **
                ***                             ***
                   ****                     ****
                       ****             ****
                           *************

The angle AOB is the difference in longitudes, dlon (5 degrees in 
your problem).  Thus the distance OC is

  OC = R*cos(lat1)*cos(dlon/2)

Now we look at the earth from the side, edge-on to the great circle. 
We see a circle of radius R:

                            ***********
                        ****           ****
                    ****                   ***D
                 ***                         / ***
               **                           /     **
             **                            C        **
            *                             /|          * 
           *                             / |           *
          *                             /  |            *
         *                             /   |             *
         *                            /    |R*sin(lat1)   *
        *                            /     |              *
       *                            /      |               *
       *                           /       |               *
      *            R              /lat     |                *
      *--------------------------O---------E----------------* 
      *                          R*cos(lat1)                *
       *                         *cos(dlon/2)              *
       *                                                   *
        *                                                 *
         *                                               *
         *                                               *
          *                                             * 
           *                                           *
            **                                       **
              **                                   **
                ***                             ***
                   ****                     ****
                       ****             ****
                           *************

Now OE is the length OC from the previous figure.  We want to find the 
latitude of point D, which is lat in the figure.  We also know EC, 
because it is the perpendicular distance of point A or B from the 
equator:

  EC = R*sin(lat1)

Thus we can solve triangle OEC:

  tan(lat) = EC/OE
           = (R*sin(lat1))/(R*cos(lat1)*cos(dlon/2))
           = tan(lat1)/cos(dlon/2)

Therefore we can find the latitude of D as follows:

  lat = arctan(tan(lat1)/cos(dlon/2))
      = arctan(tan(60)/cos(5))
      = arctan(1.738666)
      = 60.0945 degrees

Convert from decimal degrees to degrees and minutes:

  60.0945 deg = 60 deg + 0.0945*60 min
              = 60 deg 5.670 min

Well, we've got our answer!  Do you think you'd be expected to go 
through all this 3-dimensional trigonometry, or to know the general 
formula, or have you been taught this specific formula for the 
maximum latitude reached by a great circle when two points at the 
SAME latitude are known?

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Geometry
College Higher-Dimensional Geometry
College Trigonometry
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Trigonometry

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