Calculate Bounding box coordinates from a rotated rectangle, Picture inside.

Answer 1

I am not sure I understand, but a compound transformation matrix will give you the new co-ordinates for all points concerned. If you think the rectangle may spill over the imagable area post transformation apply a clipping path.

In case you are unfamiliar with the exact definition of the matrices take a look here.

Answer 2

Apply the rotation matrix to your corner points. Then use the minimum/maximum respectively of the obtained x,y coordinates to define your new bounding box.

Answer 3

• Transform the coordinates of all four corners
• Find the smallest of all four x's as min_x
• Find the largest of all four x's and call it max_x
• Ditto with the y's
• Your bounding box is (min_x,min_y), (min_x,max_y), (max_x,max_y), (max_x,min_y)

AFAIK, there isn't any royal road that will get you there much faster.

If you are wondering how to transform the coordinates, try:

x2 = x0+(x-x0)*cos(theta)+(y-y0)*sin(theta)
y2 = y0-(x-x0)*sin(theta)+(y-y0)*cos(theta)

where (x0,y0) is the center around which you are rotating. You may need to tinker with this depending on your trig functions (do they expect degrees or radians) the sense / sign of your coordinate system vs. how you are specifying angles, etc.

Answer 4

The method outlined by MarkusQ works perfectly but bear in mind that you don't need to transform the other three corners if you have point A already.

An alternative method, which is more efficient, is to test which quadrant your rotation angle is in and then simply compute the answer directly. This is more efficient as you only have a worst case of two if statements (checking the angle) whereas the other approach has a worst case of twelve (6 for each component when checking the other three corners to see if they are greater than the current max or less than the current min) I think.

The basic algorithm, which uses nothing more than a series of applications of Pythagoras' theorem, is shown below. I have denoted the rotation angle by theta and expressed the check there in degrees as it's pseudo-code.

ct = cos( theta );
st = sin( theta );

hct = h * ct;
wct = w * ct;
hst = h * st;
wst = w * st;

if ( theta > 0 )
{
    if ( theta > 90 degrees )
    {
        // 0 < theta < 90
        y_min = A_y;
        y_max = A_y + hct + wst;
        x_min = A_x - hst;
        x_max = A_x + wct;
    }
    else
    {
        // 90 <= theta <= 180
        y_min = A_y + hct;
        y_max = A_y + wst;
        x_min = A_x - hst + wct;
        x_max = A_x;
    }
}
else
{
    if ( theta > -90 )
    {
        // -90 < theta <= 0
        y_min = A_y + wst;
        y_max = A_y + hct;
        x_min = A_x;
        x_max = A_x + wct - hst;
    }
    else
    {
        // -180 <= theta <= -90
        y_min = A_y + wst + hct;
        y_max = A_y;
        x_min = A_x + hct;
        x_max = A_x - hst;
    }
}

This approach assumes that you have what you say you have i.e. point A and a value for theta that lies in the range [-180, 180]. I've also assumed that theta increases in the clockwise direction as that's what the rectangle that has been rotated by 30 degrees in your diagram seems to indicate you are using, I wasn't sure what the part on the right was trying to denote. If this is the wrong way around then just swap the symmetric clauses and also the sign of the st terms.

Answer 5

if you are using GDI+ , you can create a new GrpaphicsPath -> Add any points or shapes to it -> Apply rotate transformation -> use GraphicsPath.GetBounds() and it will return a rectangle that bounds your rotated shape.

Answer 6

Although Code Guru stated the GetBounds() method, I've noticed the question is tagged as3, flex, so here is an as3 snippet that illustrates the idea.

var box:Shape = new Shape();
box.graphics.beginFill(0,.5);
box.graphics.drawRect(0,0,100,50);
box.graphics.endFill();
box.rotation = 20;
box.x = box.y = 100;
addChild(box);

var bounds:Rectangle = box.getBounds(this);

var boundingBox:Shape = new Shape();
boundingBox.graphics.lineStyle(1);
boundingBox.graphics.drawRect(bounds.x,bounds.y,bounds.width,bounds.height);
addChild(boundingBox);

I noticed that there two methods that seem to do the same thing: getBounds() and getRect()

Answer 7

Hello MarkusQ

In your reply i have some doubts

x2 = x0+(x-x0)cos(theta)+(y-y0)sin(theta) y2 = y0-(x-x0)sin(theta)+(y-y0)cos(theta)

In thes two equations what is x2,y2 and x0,y0 and x,y?

As in the question we know only one point that is A(10,10)

Answer 8

is there any direct formula to calculate bounding box for gis related for example i have lat and lng and height of the 3d model and i want to know the bounding box coordinates.

Answer 9

Hi there.

I have a quite similar problem. I have a bounding box, given by all minima and maxima - and then the coordiate system is rotated by a principal component analysis (from some data within the box). Now I want to find out the resulting bounds on the principal components.

But this happens in an 166-dimensional space. So solutions based on the corner points are no option (as there are some 2^165 or so)

I'm looking for a direct formula or algorithm, that is linear or at least quardatic in the number of dimensions.

With regards M.

Answer 10

    fitRect: function( rw,rh,radians ){
            var x1 = -rw/2,
                x2 = rw/2,
                x3 = rw/2,
                x4 = -rw/2,
                y1 = rh/2,
                y2 = rh/2,
                y3 = -rh/2,
                y4 = -rh/2;

            var x11 = x1 * Math.cos(radians) + y1 * Math.sin(radians),
                y11 = -x1 * Math.sin(radians) + y1 * Math.cos(radians),
                x21 = x2 * Math.cos(radians) + y2 * Math.sin(radians),
                y21 = -x2 * Math.sin(radians) + y2 * Math.cos(radians), 
                x31 = x3 * Math.cos(radians) + y3 * Math.sin(radians),
                y31 = -x3 * Math.sin(radians) + y3 * Math.cos(radians),
                x41 = x4 * Math.cos(radians) + y4 * Math.sin(radians),
                y41 = -x4 * Math.sin(radians) + y4 * Math.cos(radians);

            var x_min = Math.min(x11,x21,x31,x41),
                x_max = Math.max(x11,x21,x31,x41);

            var y_min = Math.min(y11,y21,y31,y41);
                y_max = Math.max(y11,y21,y31,y41);

            return [x_max-x_min,y_max-y_min];
        }