Thread: Sqrt REAL problem

This is a funtion that detects when two circles colide, for that I calculate the distance between the centers of both circles and then compare it with the sum of the radius of both circles. I have no problem with that.

The PROBLEM is that sqrt seems to have a limit range and so if d is bigger than that range when you run the program it ends and says"sqrtOMAIN error,floating point errorOMAIN". So I have to make d smaller, I tried dividing by 1000 but then the formula loses its utility because the other member of the formula hasnt changed.

What can I do to make d smaller whithout altering the formula integrity??

Are there any other solution to my problem?

Code:

void colision(void)
{
  double d=(x-x_c)*(x-x_c)+(y-y_c)*(y-y_c);
  
  d=sqrt(d); 
  
  if(d<=RADIO_O+RADIO_J){  
    borrar_j();
    x=X_I;
    y=Y_I;
    hecho=1;
    vidas-=1;
    }
}

Honestly, I don't see why you need to use square root at all.

Code:

bool checkCollision( Circle a, Circle b )
{
    int distance;

    distance = calcDistance( a, b );

    if( distance >=  a.radius + b.radius )
        return 1;
    return 0;
}

Now, why do you need to use square root? I believe you can just use something simple to calculate the length of a line between a given set of points on a graph. (The math escapes me at the moment, but it has to do with something simple like: A^2 + B^2 = C^2. In other words, you can make it into a right angle triangle, and calculate the remainder's line length.)

Actually I guess that does require the square root... Hm.. I'm pretty sure there is a way to find the length of a line between two given poitns without using the square root.

[edit]
Just how far apart are we talking here? I can get the square root of a billion with no problems.

I do get odd behaviour if I pass it negative a billion though. Assuming your math is correct, you should never be passing it a value less than or equal to zero. Simple checks should solve that for you.

IE: If a.x_coord == b.x_coord, just simple subtraction works. If a.y_coord == b.y_coord, the same applies.
[/edit]

Quzah.