Location of Plane Flying on Great CircleDate: 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: 60°N 030°W, WPT 2: 60°N 030°W. When 025°W is passed the latitude shown on the display unit of the inertial navigation system will be: 60°11.0'N 59°49.0'N 60°05.7'N <<<<<< correct answer, but WHY?? 60°00.0'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/ |
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