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[Release] Getting vector components by angle and length

 
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Odecey
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PostPosted: Mon Aug 24, 2009 6:37 am    Post subject: [Release] Getting vector components by angle and length Reply with quote
What I'm presenting here is a function which, as the title states, returns a vector with the properties specified. This is useful for when you want the player to be able to move in more than the four regular directions(N/S/W/E).
What you can use it for:
Say you have a game where you want to give an object complete freedom in which direction it should move. This function could then be used to determine the new position of the object when it moves, given the length it should move, and it's current direction.

N.B:
- The Y- axis is reversed.
- Angle is in degrees.
- Angle 0 goes along the X-axis.

Code:

public Point GetVectorByLengthAndAngle(double length, double angle)
{
    if (length == 0)
        return new Point(0, 0);
    else if (length < 0)
        throw new ArgumentException("Length cannot be negative.");
    if (360< angle || angle < 0)
           throw new ArgumentException("Angle cannot be negative or bigger than 360.");
    double angleBuf = (((90 < angle && angle <=180)||(270 < angle && angle<=360)) ? 90 - (angle % 90) :(angle % 90))/(180/Math.PI) ;
    return new Point((int)(length * Math.Cos(angleBuf) * ((90 < angle && angle <= 270) ? -1 : 1)),(int)(length * Math.Sin(angleBuf) * ((180 <= angle && angle <= 360) ? 1 : -1)));

        }

As always, all feedback is appreciated.
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Polynomial
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Joined: 17 Feb 2008
Posts: 524
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PostPosted: Mon Aug 24, 2009 7:20 am    Post subject: Reply with quote
Good work. The math is solid Smile

Instead of throwing an exception when length is negative, why not simply take its absolute value and flip the angle by 180 degrees? Also, it is reasonably simple to correct for negative angles and angles over 360.

Code:
public Point PolarToVector(double length, double angle)
{
    if (angle >= 360) {
        angle = angle % 360;
    } else if (angle < 0) {
        angle = 360 - (angle % 360);
    }
    if (length == 0) {
        return new Point(0, 0);
    } else if (length < 0) {
        length = -length;
        angle = (angle + 180) % 360;
    }
    double angleBuf = (((90 < angle && angle <=180)||(270 < angle && angle<=360)) ? 90 - (angle % 90) :(angle % 90))/(180/Math.PI) ;
    return new Point((int)(length * Math.Cos(angleBuf) * ((90 < angle && angle <= 270) ? -1 : 1)),(int)(length * Math.Sin(angleBuf) * ((180 <= angle && angle <= 360) ? 1 : -1)));
}


Just a side note: a length and angle from a known line is called a polar coordinate.
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Jani
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Joined: 29 Dec 2006
Posts: 804
PostPosted: Mon Aug 24, 2009 7:31 am    Post subject: Reply with quote
Just a question.. I'm not familiar with C#/Java at all.. Is this C# or Java? Seeing that it isn't C/C++..

Also, this could be way more simple.. :) Something like this (C), just add your twists, like inverted Y or sth.:
Code:
Point p;
p.x = (int)( r*cos(a*PI/180) );
p.y = (int)( r*sin(a*PI/180) );
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