Jacopo Grassi Jacopo Grassi - 7 months ago 32
Java Question

Calculate angle between 3 points with vector, more precision needed (java)

I need to calculate the angle between 3 points. I've done that using vectors, it looks like it's working, but sometimes I get NaN as a result. To calculate the angle I used the arcos(dot(v1,v2)/(length(v1)*length(v2))) formula. Here's the code:

private static double angleBetween(Point previous, Point center, Point next) {
Vector2D vCenter = new Vector2D(center.x, center.y );
Vector2D vPrev = new Vector2D(previous.x, previous.y );
Vector2D vNext = new Vector2D(next.x, next.y );

Vector2D a = vCenter.minus(vPrev);
Vector2D b = vCenter.minus(vNext);

double dot = Vector2D.dot(a, b);

double la = a.length();

double lb = b.length();

double l = la*lb;

double acos = Math.acos(dot/l);

double deg = Math.toDegrees(acos);

return deg;

}


Vector2D class:

public double length() {
return Math.sqrt(x * x + y * y);
};

public static double dot(Vector2D v1, Vector2D v2) {
return v1.x * v2.x + v1.y * v2.y;
};

public Vector2D minus(Vector2D v) {
return new Vector2D(x - v.x, y - v.y);
};


Debugging the program I've discovered why this happens. for example let be:

center = (127,356)
previous = (117,358)
next = (137,354)

//calculating the vectors
a = (-10,2) //center - prev
b = (10,-2) //center - next

dot = -10*10 + 2*-2 = 104

la = sqrt(-10*-10 + 2*2) = sqrt(104) = see attachment
lb = sqrt(10*10 + -2*-2) = sqrt(104) = see attachment

l = la * lb = see attachment


acos = NaN
because
dot/l>1
because I lost precision because of
sqrt()
which didn't give me the exact value therefore
la*lb
isn't 104.

now as far as I know double is the most precise number type in java. How can I solve this problem?

attchment: la,lb,l values

PS It may looks like a very rare situation, but I'm experiencing quite a lot of them, so I can't just ignore it.

Answer

The best way to solve this problem is to use an appropriate data type like java.math.BigDecimal and define a precision for you computation using an instance of type java.math.MathContext. For instance:

double l = la*lb;
BigDecimal lWrapper = new BigDecimal(l, new MathContext(5));
l = lWrapper.doubleValue();

There is an other way to work around this problem. Use the following formula:

angle = atan(length(crossProduct(a, b)) / dotProduct(a, b)) // Because the domain of definition of the tan function is R

Derivation of the formula:

  cos(angle) = dotProduct(a, b)   / (length(a) * length(b)) and
  sin(angle) = length(crossProduct(a, b)) / (length(a) * length(b))

One has

 tan(angle) = sin(angle) / cos(angle)

so

tan(angle) = length(crossProduct(a, b)) / (length(a) * length(b)) / dotProduct(a, b)   / (length(a) * length(b))
tan(angle) = length(crossProduct(a, b)) / dotProduct(a, b)

Applying the invert function of tan:

angle = atan(length(crossProduct(a, b)) / dotProduct(a, b))

The cross product of A, B ∈ ℜ2:

|| A x B || = det(A,B) = ((A.x * B.y) - (A.y * B.x))

Remarks:

  1. ||x|| is the length of the vector x ⇔ length(a)
  2. ∀ A, B ∈ ℜ2: || A x B || equal the determinant of A, B
  3. You can use the sign(|| A x B ||) to find out the orientation