View Full Version : Track Changes

25th Feb 2008, 00:07
Bored out of my mind at 40 West I found myself looking thru the Track Change Performa Chart.:eek:.(yes, all the newspapers were read)...

QUESTION: Is there a similar chart OR formula
which can produce a True/Mag Track from one lat/long position to another lat/long position? ie VOR to VOR :confused:


Spooky 2
25th Feb 2008, 08:12
Find a copy of the the Universal 10 degree tables. I have some for the NA, but don't ask me where they at this hour. I'll take a look and see if I can find a copy in a couple of weeks or so if no else pipes up with a better suggestion. This is Flight Navigator stuff!

27th Feb 2008, 00:29
Cheers! I'll see what I can find...
I thought I was going crazy with that question..must be the P3 in me :O

Fly Safe

Tim Zukas
28th Feb 2008, 20:37
If you're asking for formulas that give the distance and direction between two given lat-lons--

There's one set of formulas that gives accuracy to a millimeter, but programming them into your calculator will take a while

There's another much simpler set, about as accurate but only up to 100 km or so

And another even simpler set that gives distance correct to a few meters over lines several thousand km long, but dunno about the directions it gives

Or would you settle for a website that calculates it for you?

28th Feb 2008, 22:20
Distance was not too important, What's a few miles once you have the gas..
How to determine the "Track" was...

Where can I find it? What are the formulas??
As for programing it into my PDA...Let's just say Im better with a sharp pencil than a keyboard

Standing By....


Tim Zukas
28th Feb 2008, 22:30
(Edit: don't bother with the formulas below-- I forgot you probably want normal-section azimuth which is much simpler. I'll look that one up and get back to you.)

See if this one is any use to you


That's the up-to-100+ km formula. If you want to work on it let me know-- it might need a bit of clarification.

The long-distance formulas are at


Those may not be the best any more, but the improved formulas aren't on line AFAIK. In any case it does the job.

I was assuming you wanted something more accurate than you'd get by assuming the earth was a sphere, but of course those formulas are much simpler still.

29th Feb 2008, 16:14
Cheers!! :ok:
I'll give it a try.....


29th Feb 2008, 16:58
Hi Yflex.
I have developed a program for Visual Basic that take data from GPS and calculates the distance and the ETA. Use this code if that helps you. You must only write the correct coordinates in the same format they are written now and will give you distances on WPTDIS value.

LatWpt1 = (Val(Left(WPT1, 2)) * 3600) + (Val(Mid(WPT1, 3, 2)) * 60)
LatWpt1 = LatWpt1 + Val(Mid(WPT1, 6, 2))
LatWpt1 = LatWpt1 / 206265
If Mid(WPT1, 7, 1) = "S" Then LatWpt1 = 0 LatWpt1
LonWpt1 = (Val(Mid(WPT1,8,3)) * 3600) + (Val(Mid(WPT1, 11, 2)) * 60)
LonWpt1 = LonWpt1 + Val(Mid(WPT1, 13, 2))
LonWpt1 = LonWpt1 / 206265
If Right(WPT1, 1) = "E" Then LonWpt1 = 0 - LonWpt1

LatWpt2 = (Val(Left(WPT2, 2)) * 3600) + (Val(Mid(WPT2, 3, 2)) * 60)
LatWpt2 = LatWpt2 + Val(Mid(WPT2, 6, 2))
LatWpt2 = LatWpt2 / 206265
If Mid(WPT2, 7, 1) = "S" Then LatWpt2 = 0 LatWpt2
LonWpt2 = (Val(Mid(WPT2,8,3)) * 3600) + (Val(Mid(WPT2, 11, 2)) * 60)
LonWpt2 = LonWpt2 + Val(Mid(WPT2, 13, 2))
LonWpt2 = LonWpt2 / 206265
If Right(WPT2, 1) = "E" Then LonWpt2 = 0 LonWpt2

WptDis = Sqr((Sin((LatWpt1 - LatWpt2) / 2)) ^ 2 + Cos(LatWpt1) * Cos(LatWpt2) * (Sin((LonWpt1 LonWpt2) / 2)) ^ 2)
WptDis = 2 * (Atn(WptDis / Sqr(-WptDis * WptDis + 1)))
WptDis = WptDis * 3437.7475
WptDis = Round(WptDis, 2)

Tim Zukas
29th Feb 2008, 18:18
That all assumes a spherical earth, doesn't it?

But if we want more accurate results we assume the sphere is slightly squashed-- likely we'll use the WGS84 spheroid, where the distance between the poles divided by the equatorial diameter is 297.257223563 divided by 298.257223563.

About azimuth: those PDFs I linked give the azimuth of the geodesic-- the shortest path on the surface of the spheroid between Point 1 and Point 2. It eventually occurred to me that you would do as well or better with the normal-section azimuth, which is the azimuth (at Point 1) of the plane that passes through Points 1 and 2 and is vertical at Point 1. In other words, you're standing vertically at Point 1, facing Point 2-- which way are you facing?

That's much simpler. In the following formulas Z1, Z2 etc are just the intermediate results in the calculation; A and B are constants that depend on the spheroid you choose. For WGS84, A rounds to 1.0067394967 and B rounds to 0.0066943800. Point 1 is at latitude Lat1; Point 2 is at latitude Lat2, DLon degrees to the west of Point 1.

Z1 = A + tangent-squared of Lat2

Z2 = A + tangent-squared of Lat1

Z3 = B x square root of (Z1/Z2)

Z4 = (tangent Lat2) divided by (tangent Lat1)

Z5 = Z3 + (Z4/A)

Z6 = [(cosine DLon) - Z5] x sine Lat1

tangent of azimuth = (sine DLon)/Z6

This doesn't work if Point 1 is on the equator (zero in the denominator); in that case, use

tangent of azimuth = (-sine DLon) x A divided by (tangent Lat2)