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Haversine formula C++ help!

  1. Jan 25, 2014 #1

    perplexabot

    User Avatar
    Gold Member

    Hello all. So i am trying to calculate the distance between two points on the earth's surface. Here is what I have:

    Code (Text):

                    //calculate distance here
            dlat = abs(startLoc.lat - tarList[i].lat);
            dlon = abs(startLoc.lon - tarList[i].lon);
            a = pow(sin(dlat/2),2) + ( cos(startLoc.lat)*cos(tarList[i].lat)*pow(sin(dlon/2),2) );
            c = 2*atan2(sqrt(a),sqrt(1-a));
            tarList[i].distance = R * c;
     
    Where tarList and startLoc are the two locations.

    Also, the program takes user input in the form of:
    LATITUDEVALUE/(NorS) LONGITUDEVALUE(WorE) NAME
    e.g: 2341/N 41355/W Random City

    I have a part of the code that changes the sign of the latitude or longitude according to (N or S and W or E), here it is:

    Code (Text):

            //adjust sign of latitude according to N/S
            if(tarList[i].noSo == 'N'){
                tarList[i].lat = getLat(aString);
            }else{
                tarList[i].lat = -getLat(aString);
                cout<<"MINUS USED!"<<endl;
            }

            //adjust sign of longitude according to W/E
            if(tarList[i].weEa == 'E'){
                tarList[i].lon = getLon(aString);
            }else{
                tarList[i].lon = -getLon(aString);
                cout<<"MINUS USED!"<<endl;
            }
     
    I am not getting errors but I am not achieving correct answers. Does anyone know what the problem may be? Thank you physics forums as usual!
     
  2. jcsd
  3. Jan 25, 2014 #2

    jedishrfu

    Staff: Mentor

    Okay so here's the haversine formula:

    http://en.wikipedia.org/wiki/Haversine_formula

    are you using radian measure or degrees? It looks like you're using degrees so the question really is what are the args to the sin and cos in C++?

    Next, the wiki article doesn't mention inverting the lat / lon values.

    What I would do to determine whats going on is to print all intermediate results and compare them to what you calculate manually to see that they make sense. As an exmaple does the sin() give you what you expect...

    http://www.cplusplus.com/reference/cmath/sin/

    and here's a website that you can compare your results to:

    http://www.movable-type.co.uk/scripts/latlong.html
     
  4. Jan 25, 2014 #3

    perplexabot

    User Avatar
    Gold Member


    Hey, thank you for your reply. I will compare to my own calculations now and see what is happening. Thanks for the suggestion. I will update you on what happens.
    I don't understand what you mean by:
    "Next, the wiki article doesn't mention inverting the lat / lon values."
     
  5. Jan 25, 2014 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    The arguments to the C++ trigonometric functions are angles in radians, not degrees. If you aren't converting your latitude and longitude to radians that explains your incorrect results.
     
  6. Jan 25, 2014 #5

    perplexabot

    User Avatar
    Gold Member

    OK that definitely fixed the problem. Thank you.

    I am wondering if jedishrfu knew this but was trying to have me find out. Either way thank you too.

    The answers are very close but not quite. I will try to figure out why.
     
  7. Jan 25, 2014 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    What value are you using for the radius of the Earth? The equatorial radius is 6378.137 km (the distance from the Earth's center to the equator at sea level), while the polar radius is 6,356.752 km. For the purposes of this calculation, it's best to use a mean value of 6371 km.

    The haversine formula is an approximation that assumes a spherical Earth. From the above, this is a good but not great model. There's a 21 km difference between the equatorial and polar radii. A much better model is an oblate ellipsoid. This model is correct to within 100 meters or so. This means the haversine formula isn't quite correct. There is no simple closed formula for the length of the shortest path (i.e., "distance") between two points on an oblate spheroid. You'll need to use elliptical integrals.

    Or you could just use the haversine formula. The error isn't all that much except for extreme cases.
     
  8. Jan 25, 2014 #7

    perplexabot

    User Avatar
    Gold Member

    Interesting. I will be sticking to the haversine formula. By "not accurate," I mean I am comparing my answer to an "ultimate output," and they are close but not the same. By ultimate output, I mean an output of a program that is known to output the "correct results."
     
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