Help me write this formula in c++

nekavnick · #1 23-Aug-2008, 14:34

Hello!

I'm creating a program that calculates the distance between two cities. I have problems with converting the formula to c++ code.
Here is the formula

Code:

d=2*R*sin^-1 (sin^2 (a1-a2)/2 + cos*a1*cos*a2*sin^2(b1-b2)/2)^1/2

Here, d is the distance between the towns, a1 and a2 are their latitudes,b1 and b2 are their longitudes, and R is the radius of the earth.

The latitudes and longitudes are given in degrees but sin and cos functions work with radians.

I've been trying to make this two days and it doesn't work

Here is what I've done

CPP / C++ / C Code:

double d;
d=sin((Pi*((a1-a2)/2.0))/180.0);
d*=d;
double dd;
dd=sin((Pi*((b1-b2)/2.0))/180.0);
dd*=dd;
dd*=cos((Pi*a1)/180.0);
dd*=cos((Pi*a2)/180.0);
d+=dd;
d=pow(d, 0.5);
d=2*R/d;
cout << d << endl;

and also

CPP / C++ / C Code:

d=2*R/sin(Pi*pow( sin((Pi*(a1-a2)/2.0)/180.0)*sin((Pi*(a1-a2)/2.0)/180.0)+cos(Pi*a1/180.0)*cos(Pi*a2/180.0)*sin((Pi*(b1-b2)/2.0)/180.0)*sin((Pi*(b1-b2)/2.0)/180.0), 0.5)/180.0);

but both methods did not work.

Can you please tell me what I'm doing wrong?
Any help will be appreciated.

p.s. Sorry for my English. I'm still learning

ocicat · #2 23-Aug-2008, 15:57

Quote:

Originally Posted by nekavnick

I have problems with converting the formula to c++ code.
Here is the formula

Code:

d=2*R*sin^-1 (sin^2 (a1-a2)/2 + cos*a1*cos*a2*sin^2(b1-b2)/2)^1/2

What does cos*a1*cos*a2 mean? Is this cos(a1) * cos(a2)?

nekavnick · #3 23-Aug-2008, 16:12

Quote:

Originally Posted by ocicat

What does cos*a1*cos*a2 mean? Is this cos(a1) * cos(a2)?

Yes it is cos(a1)*cos(a2)
Sorry for the mistake

davekw7x · #4 23-Aug-2008, 16:16

Quote:

Originally Posted by nekavnick

Can you please tell me...

Without validating your formula and without looking very closely at your "simplifications," I would say that, for one thing, I am almost sure that you have not interpreted the following notation correctly:

Code:

sin^-1(xxx)

I believe that this should be taken to designate arcsin(xxx), not 1.0/sin(xxx). The standard library function asin() calculates the arcsin in radians.

Regards,

Dave

Footnote: I have always used the "haversine" formula (look it up on wikipedia or wherever...)

Code:

        d = R * acos(sin(lat1)*sin(lat2)+ cos(lat1)*cos(lat2)*cos(long2-long1));

Where R is the radius of the earth (something like 6371 km). I won't try to reconcile it with yours (there are many ways to express the same operations using various trig identities); but you can check it something like the following

Code:

Enter latitude and longitude of first  point in degrees:  58.2628 -122.1092
Enter latitude and longitude of second point in degrees:  57.4744 -8.1620
Distance = 5889.658973 km.

If you convert your latitude and longitude into radians before plugging into the formula, it might look a little cleaner:

CPP / C++ / C Code:

    const double M_PI = 3.14159265358979323846

    const double radians_per_degree = M_PI/180.0;
    double d;

    /* latitude and longitude in radians */
    double lat1, long1, lat2, long2;

    /* latitude and longitude in degrees */
    double lat1_degrees, long1_degrees, lat2_degrees, long2_degrees;
.
.
.
    /* get latitude and longitude in degrees */
        printf("Enter latitude and longitude of first  point in degrees: ");
        if (scanf("%lf %lf", &lat1_degrees, &long1_degrees) != 2) {
            /* bail out of the program or try to recover or whatever... */
        }
        printf("Enter latitude and longitude of second point in degrees: ");
        if (scanf("%lf %lf", &lat2_degrees, &long2_degrees) != 2) {
            /* bail out of the program or try to recover or whatever... */
        }


    /* then */
        lat1  = lat1_degrees  * radians_per_degree; /* now it's in radians */
        long1 = long1_degrees * radians_per_degree; /* now it's in radians */
        lat2  = lat2_degrees  * radians_per_degree; /* now it's in radians */
        long2 = long2_degrees * radians_per_degree; /* now it's in radians */
..
.
.
       d = /*function of lat and long in radians */

ocicat · #5 23-Aug-2008, 16:26

Quote:

Originally Posted by davekw7x

I would say that, for one thing, I am almost sure that you have not interpreted the following notation correctly:

Code:

sin^-1(xxx)

sin^-1(x) is frequently the way arcsin(x) is typeset in textbooks. I don't suspect that the OP intended it to be correct C syntax. In C, arcsine is defined as the function asin(double) prototyped in math.h as I am sure you know.

ocicat · #6 23-Aug-2008, 17:05

Quote:

Originally Posted by nekavnick

Here is the formula

Code:

d=2*R*sin^-1 (sin^2 (a1-a2)/2 + cos*a1*cos*a2*sin^2(b1-b2)/2)^1/2

The first thing is to build the solution incrementally to ensure that the solution is correct. After the implementation is validated, then you can worry about whether it is coded in an optimal fashion.

So, beginning with the following assumptions:

sin^-1(x) == arcsin(x)
sin^2(x) == (sin(x))^2

Decompose the argument to asin() as two separate terms:

Code:

sin^2 (a1-a2)/2

...which becomes:

CPP / C++ / C Code:

double latitude, term1;
latitude = (a1 - a2) * M_PI / 180.0;
term1 = pow(sin(latitude), 2.0) / 2.0;

Code:

cos*a1*cos*a2*sin^2(b1-b2)/2

...becomes:

CPP / C++ / C Code:

double longitude, term2;
longitude = (b1 - b2) * M_PI / 180.0;
term2 = cos(a1) * cos(a2) * pow(sin(longitude), 2.0) / 2.0;

So, putting this together for:

Code:

d=2*R*sin^-1 (sin^2 (a1-a2)/2 + cos*a1*cos*a2*sin^2(b1-b2)/2)^1/2

CPP / C++ / C Code:

double d;
d = 2 * R * pow(asin(term1 + term2), 0.5);

...assuming I have read your notation correctly.

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Help me write this formula in c++

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