在旋转的矩形内找到一个点

Question

Ok so, this should be super simple, but I'm not a smart man. Technically I want to know whether a point resides inside a rectangle, however the rectangle can be in different states. In my current context when I want to draw a rectangle rotated by, lets say, 45° clockwise, what I do is rotate the entire x,y axis centered at the top-left corner of the rectangle and then I just draw the rectangle as if nothing has happened. Same goes if I want to draw the rectangle at a random coordinate. Given that is the coordinate system who gets tossed and rotated, the rectangle always thinks it's being drawn at (0,0) with 0°, therefore, the best way to find if a given point is inside the rectangle would be to find the projection for the point based on the translation + rotation of the rectangle. But I have no idea how to do that.

This is what I currently do in order to find out if a point is inside a rectangle (not taking into consideration rotation):

bool Image::isPointInsideRectangle(int x, int y, const ofRectangle & rectangle){
    return x - xOffset >= rectangle.getX() && x - xOffset <= rectangle.getX() + rectangle.getWidth() &&
            y - yOffset >= rectangle.getY() && y - yOffset <= rectangle.getY() + rectangle.getHeight();
}

I already have angleInDegrees stored, as long as I could use it to project the (x,y) point I receive I should be able find out if the point is inside the rectangle.

Cheers!

Axel

Answer 1

The easiest way is to un-rotate x,y in the reverse direction relative to the origin and rotation of the rectangle.

For example, if angleInDegrees is 45 degrees, you would rotate the point to test -45 degrees (or 315 degrees if your rotation routine only allows positive rotations). This will plot the x,y on the same coordinate system as the unrotated rectangle.

Then, you can use the function you already provided to test whether the point is within the rectangle.

Note that prior to rotating x,y, you will probably need to adjust the x,y relative to the point of rotation - the upper-left corner of the rectangle. Since the rotation is relative to that point rather than the overall coordinate origin 0,0. You can compute the difference between x,y and the upper-left corner of your rectangle (which won't change during rotation), then simply rotate the adjusted point by -angleToRotate, then add the origin point difference back into the unrotated point to get absolute coordinates on your coordinate system.

Answer 2

Editted:

#include <cmath>

bool Image::isPointInsideRectangle(int x, int y, const ofRectangle & rectangle){
    return x*cosd(deg) - y*sin(deg) + xOffset >= rectangle.getX()
        && x*cosd(deg) - y*sin(deg) + xOffset <= rectangle.getX() + rectangle.getWidth() 
        && x*sind(deg) + y*cosd(deg) + yOffset >= rectangle.getY()
        && x*sind(deg) + y*cosd(deg) + yOffset <= rectangle.getY() + rectangle.getHeight();

Answer 3

Like you have already told, you could translate the coordinates of your point into the space of the rectangle. This is a common task in many software products which work with geometry. Each object have it own coordinate space and works as it would be at position (0, 0) without rotation. If your rectangle is at position v and rotated about b degree/radian , than you can translate your point P into the space of the rectangle with the following formula:

| cos(-b)  -sin(-b) |   | P_x - v_x |
|                   | ⋅ |           |
| sin(-b)   cos(-b) |   | P_y - v_y |

Many of the most important transformations can be represented as matrices. At least if you are using homogeneous coordinates . It is also very common to do that. Depending of the complexity and the goals of your program you could consider to use some math library like glm and use the transformations of your objects in form of matrices. Then you could write something like inverse(rectangle.transformation()) * point to get point translated into the space of rectangle .