Jacopo Grassi - 1 year ago 68

Java Question

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`

`dot/l>1`

`sqrt()`

`la*lb`

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

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 ∈ ℜ^{2}:

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

Remarks:

- ||x|| is the length of the vector x ⇔ length(a)
- ∀ A, B ∈ ℜ
^{2}: || A x B || equal the determinant of A, B - You can use the sign(|| A x B ||) to find out the orientation

Source (Stackoverflow)