Removed..

Performance Tuning
#1

Removed…

#2

“The angle between two points” doesn’t mean anything.
For the angle between two vectors (dx1,dy1) and (dx2,dy2), just use :
Math.toDegrees(Math.atan2(dy1,dx1)-Math.atan2(dy2,dx2))

No ?

#3

this is ‘calculate angle of one line’ in effect.

#4

I think this is what you want:


  public static final float calcAngle(Point p0, Point p1)
  {
    float x0 = (float)p0.getX();
    float y0 = (float)p0.getY();
    float x1 = (float)p1.getX();
    float y1 = (float)p1.getY();
    
    // op = m sin(theta)
    // ip = m cos(theta), where m = |P0| * |P1|
    float op = x1*y0-x0*y1;
    float ip = x0*x1+y0*y1;   

    return (float)Math.toDegrees(Math.atan2(op,ip));
  }

But I’m not sure. Your code returns 45 for: (0,1) & (1,0) and something near 90 for (0,1) & (10000,0). So I might be not understanding the question.

#5

If you’re after the angle of the line that passes through points p1 and p2, then

Math.toDegrees( Math.atan2( p2.getY()-p1.getY(), p2.getX()-p1.getX() ) )

should do it.

The result will be in the range -180 to +180 degrees, measured anti-clockwise from East (see the comments here on angles in polar coordinates).

Simon

#6

Yup, that’s exactly what atan2 is for. You don’t have to bother with checking quadrants like you’ve been doing.

#7

Removed…

#8

The result above makes East = 0.
If you want North to be 0, East effectively becomes -90.
That means, add 90 to the value returned.

The result above gives you a value from -180 to 180 (where 0 is east).
If you want to make it 0 to 360, effectively everything is 180 more.
That means, add 180 to the value returned.

We want to do both.

Let’s convert it into an equation.


public double convertToNorth(double angle)
{
    return angle + 90;
}
public double convertTo360(double angle)
{
    return angle + 180;
}

Doing both means you’ve got to convert it to 360 and then make sure 0 is considered north. That means:


public double doBoth(double angle)
{
    //At this point, -180 to 180 and East is 0.

    angle = convertTo360(angle);

    //At this point, 0 to 360 and West is 0.
    
    angle = convertToNorth(angle);

    //Now, 0 to 360 and North is 0.

    return angle;
}

Make sense yet? What’s the final result? That’s right. Add 270.

::slight_smile:

#9

Removed…

#10 

public float calcAngle (Point p1, Point p2 )
{
    float f1 = (float)Math.toDegrees( Math.atan2( p1.getY()-p2.getY(), p1.getX()-p2.getX() ) );
    return (f1+270)%360;
}
#11

Do you need to know the angle?

I often find that it’s both faster and easier to work with vectors and dot products.

#12

One time I needed the angle specifically for where I was applying a rotation to Sprite (a goal compass). Otherwise I’ve always used vectors.

#13

I know I’m replying to an old post but this answer is just wrong I think.

The original results would give if:
north=90
west=180 or -180
south= -90
east= 0

adding 270 gives:

north=360
west=450 or 90
south=180
east=270

the only result I can think of that would be usefull is:

north=0 or 360
east or west=270
south=180
east or west=90

edit: This gives the right answer if the original answer was “polar”:
int nonPolar= 90-((polar+180)%360)
if(nonPolar<0)
{
nonPolar=nonPolar+360;
}