Inner angle between two lines

Question

Hi folks,

I have two lines: Line1 and Line2. Each line is defined by two points (P1L1(x1, y1), P2L1(x2, y2) and P1L1(x1, y1), P2L3(x2, y3)). I want to know the inner angle defined by these two lines.

For do it I calculate the angle of each line with the abscissa:

double theta1 = atan(m1) * (180.0 / PI);
double theta2 = atan(m2) * (180.0 / PI);

After to know the angle I calculate the following:

double angle = abs(theta2 - theta1);

The problem or doubt that I have is: sometimes I get the correct angle but sometimes I get the complementary angle (for me outer). How can I know when subtract 180º to know the inner angle? There is any algorithm better to do that? Because I tried some methods: dot product, following formula:

result = (m1 - m2) / (1.0 + (m1 * m2));

But always I have the same problem; I never known when I have the outer angle or the inner angle!

Thanks in advance for reading my trouble and for your time!

Oscar.

Answer 1

if (result > 180)
{
     result = 360 - result;
}

That way it will always be the inner angle. Just add it after you get result.

Answer 2

Getting the outer angle vs the inner angle is determined entirely by the order of your subtractions (think about it). You need to subtract the smaller theta from the larger in order to reliably always get the inner angle. You also probably want to use the atan2 function because of the type of data you're expecting.

Answer 3

Inner angle between 2 vectors (v1, v2) = arc cos ( inner product(v1,v2) / (module(v1) * module(v2)) ).

Where inner product(v1,v2) = xv1*xv2 + yv1*yv2

module(v) = sqrt(pow(xv,2) + pow(yv,2))

So, the answer of your question is implemented on the following example:

#define PI   3.14159258

int main()
{
    double x1,y1,x2,y2,y3;
    double m1, m2;
    double mod1, mod2, innerp, angle;

    cout << "x1 :";
    cin >> x1;
    cout << "y1 :";
    cin >> y1;
    cout << "x2 :";
    cin >> x2;
    cout << "y2 :";
    cin >> y2;
    cout << "y3 :";
    cin >> y3;

    m1 = atan((y2-y1)/(x2-x1)) * 180 / PI;
    m2 = atan((y3-y1)/(x2-x1)) * 180 / PI;

    mod1   = sqrt(pow(y2-y1,2)+pow(x2-x1,2));
    mod2   = sqrt(pow(y3-y1,2)+pow(x2-x1,2));
    innerp = (x2-x1)*(x2-x1) + (y2-y1)*(y3-y1);
    angle  = acos(innerp / (mod1 * mod2)) * 180 / PI;

    cout << "m1 : " << m1 << endl;
    cout << "m2 : " << m2 << endl;
    cout << "angle : " << angle << endl;
}

Answer 4

The whole point is much easier than the given answers:

When you use atan(slope) you lose (literally) one bit of information, that is there are exactly two angles (theta) and (theta+PI) in the range (0..2*PI), which give the same value for the function tan().

Just use atan2(deltax, deltay) and you get the right angle. For instance

atan2(1,1) == PI/4
atan2(-1,-1) == 5*PI/4

Then subtract, take absolute value, and if greater than PI subtract from 2*PI.