What's 'Arccos'?Date: 06/17/2001 at 22:20:10 From: Daniel Subject: What's 'Arccos'? I'm trying to write a program in Visual Basic 6 about the distance calculation between two cities. I found a equation that you wrote in 1995, but I can't understand some parts. You said that with the latitudes and longitudes of two places (cities), when substituted into the equation: Arccos = {[cos(a1) cos(b1) coa(a2) cos(b2)] + [cos(a1) sin(b1) cos(a2) sin(b2)] + [ sin(a1) sin(a2)]}/360*2Pi*r I'd like to ask you what you mean by Arccos and Pi in this equation. Any help given by 'The Swat Team' will be greatly appreciated. Date: 06/18/2001 at 08:14:46 From: Doctor Jerry Subject: Re: What's 'Arccos'? Hi Daniel, pi means the constant 3.141592654... It is the ratio of the circumference of a circle to its diameter, which turns out to be the same for every circle. You can read about it in the Dr. Math FAQ on Pi: http://mathforum.org/dr.math/faq/faq.pi.html arccos is the name of the inverse cosine function, which is sometimes written as cos with an exponent of -1. You have written, in part, Arccos = {[cos(a1) cos(b1) ... This can't be correct, because Arccos alone doesn't have a meaning in this equation. You must mean the arccos(s) of something. Look under the worked examples on the Aviation Formula VI.30 page by Ed Williams: http://www.best.com/~williams/avform.htm#LL - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 06/18/2001 at 12:03:22 From: Doctor Peterson Subject: Re: What's 'Arccos'? Hi, Daniel. I want to add a couple of things. The 'arccos' function is the inverse of the cos function. That is, if it is true that cos(a) = x then it is also true that arccos(x) = a For example, cos(pi/4 radians) = cos(45 degrees) = 0.707 so arccos(0.707) = 45 degrees = pi/4 radians Second, in Visual Basic, the arccos function has to be implemented using the arctangent function Atn. Microsoft provides equivalents of the arccos and other functions here: How to Derive Inverse (ARC) and Hyperbolic Trig Functions - Microsoft Corporation http://support.microsoft.com/support/kb/articles/Q28/2/49.asp They give ARCCOS(Y) = -ATN(Y/SQR(1-Y*Y)) + Pi/2 If y is 1 or -1, this will fail, so you should protect against that case. Visual Basic help suggests this definition of Pi: Dim pi pi = 4 * Atn(1) Third, there are several formulas for geographical distance in our archives. Here are three, which I found by searching the archives for "distance latitude longitude": Using Longitude and Latitude to Determine Distance http://mathforum.org/dr.math/problems/longandlat.html Distance using Latitude and Longitude http://mathforum.org/dr.math/problems/reed12.31.97.html Deriving the Haversine Formula http://mathforum.org/dr.math/problems/neff.04.21.99.html The first is probably what you found; don't miss the comments at the bottom about radians and degrees; this will be true of Basic, which uses radians. The last uses "atan2(y,x)", which in Visual Basic will be replaced with "atn(y/x)" - again with some tricks needed for negative values. This is probably the best for you to use, for several reasons described there. Looking for a site that might help you use the limited set of functions available in Visual Basic, I ran across this page http://vbgraphic.altervista.org/geoalgo.htm#arctangent which gives a short version of atan2 you can use: Public Function Atn2(X As Double, Y As Double) As Single Const NearZero = 0.000000001 If Y = 0 Then Y = NearZero Atn2 = (Atn(Abs(X) / Abs(Y)) * Sgn(X) - 3.141592653 / 2) * Sgn(Y) End Function Note that this uses a different order of X and Y; it also gives the wrong sign for the answer. I suggest this instead, for your purposes: Public Function Atn2(Y As Double, X As Double) As Double Const NearZero = 0.000000001 Const Pi = 4 * Atn(1) If X = 0 Then X = NearZero Atn2 = (Pi / 2 - Atn(Abs(X) / Abs(Y)) * Sgn(X)) * Sgn(Y) End Function There are many other ways to accomplish the same thing. Finally, if you are going to be writing programs of this sort much, it would be an excellent idea to get a book on basic trigonometry so you will have the background needed to properly understand what you are doing. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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