Given an angle and dimensions, find a coordinate along the perimeter of a rectangle

Question

I'm writing a script where icons rotate around a given pivot (or origin). I've been able to make this work for rotating the icons around an ellipse but I also want to have them move around the perimeter of a rectangle of a certain width, height and origin.

I'm doing it this way because my current code stores all the coords in an array with each angle integer as the key, and reusing this code would be much easier to work with.

If someone could give me an example of a 100x150 rectangle, that would be great.

EDIT: to clarify, by rotating around I mean moving around the perimeter (or orbiting) of a shape.

Answer 1

You know the size of the rectangle and you need to split up the whole angle interval into four different, so you know if a ray from the center of the rectangle intersects right, top, left or bottom of the rectangle.

If the angle is: -atan(d/w) < alfa < atan(d/w) the ray intersects the right side of the rectangle. Then since you know that the x-displacement from the center of the rectangle to the right side is d/2, the displacement dy divided by d/2 is tan(alfa), so

dy = d/2 * tan(alfa)

You would handle this similarily with the other three angle intervals.

Ok, here goes. You have a rect with width w and depth d. In the middle you have the center point, cp. I assume you want to calculate P, for different values of the angle alfa.

I divided the rectangle in four different areas, or angle intervals (1 to 4). The interval I mentioned above is the first one to the right. I hope this makes sense to you.

First you need to calculate the angle intervals, these are determined completely by w and d. Depending on what value alfa has, calculate P accordingly, i.e. if the "ray" from CP to P intersects the upper, lower, right or left sides of the rectangle.

Cheers

Answer 2

This was made for and verified to work on the Pebble smartwatch, but modified to be pseudocode:

struct GPoint {
  int x;
  int y;
}

// Return point on rectangle edge.  Rectangle is centered on (0,0) and has a width of w and height of h
GPoint getPointOnRect(int angle, int w, int h) {
  var sine = sin(angle), cosine = cos(angle);   // Calculate once and store, to make quicker and cleaner
  var dy = sin>0 ? h/2 : h/-2;                  // Distance to top or bottom edge (from center)
  var dx = cos>0 ? w/2 : w/-2;                  // Distance to left or right edge (from center)
  if(abs(dx*sine) < abs(dy*cosine)) {           // if (distance to vertical line) < (distance to horizontal line)
    dy = (dx * sine) / cosine;                  // calculate distance to vertical line
  } else {                                      // else: (distance to top or bottom edge) < (distance to left or right edge)
    dx = (dy * cosine) / sine;                  // move to top or bottom line
  }
  return GPoint(dx, dy);                        // Return point on rectangle edge
}


Use:
rectangle_width  = 100;
rectangle_height = 150;
rectangle_center_x = 300;
rectangle_center_y = 300;
draw_rect(rectangle_center_x - (rectangle_width/2), rectangle_center_y - (rectangle_center_h/2), rectangle_width, rectangle_height);
GPoint point = getPointOnRect(angle, rectangle_width, rectangle_height);
point.x += rectangle_center_x;
point.y += rectangle_center_y;
draw_line(rectangle_center_x, rectangle_center_y, point.x, point.y);

Answer 3

One simple way to do this using an angle as a parameter is to simply clip the X and Y values using the bounds of the rectangle. In other words, calculate position as though the icon will rotate around a circular or elliptical path, then apply this:

(Assuming axis-aligned rectangle centered at (0,0), with X-axis length of XAxis and Y-axis length of YAxis):

if (X > XAxis/2)    
         X = XAxis/2;

if (X < 0 - XAxis/2)
         X = 0 - XAxis/2;

if (Y > YAxis/2)    
         Y = YAxis/2;

if (Y < 0 - YAxis/2)    
         Y = 0 - YAxis/2;

The problem with this approach is that the angle will not be entirely accurate and the speed along the perimeter of the rectangle will not be constant. Modelling an ellipse that osculates the rectangle at its corners can minimize the effect, but if you are looking for a smooth, constant-speed "orbit," this method will not be adequate.

Answer 4

If think you mean rotate like the earth rotates around the sun (not the self-rotation... so your question is about how to slide along the edges of a rectangle?)

If so, you can give this a try:

# pseudo coode
for i = 0 to 499
  if i < 100:  x++
  else if i < 250: y--
  else if i < 350: x--
  else y++

  drawTheIcon(x, y)

Update: (please see comment below)

to use an angle, one line will be

y / x = tan(th)       # th is the angle

the other lines are simple since they are just horizontal or vertical. so for example, it is x = 50 and you can put that into the line above to get the y. do that for the intersection of the horizontal line and vertical line (for example, angle is 60 degree and it shoot "NorthEast"... now you have two points. Then the point that is closest to the origin is the one that hits the rectangle first).

Answer 5

Use a 2D transformation matrix. Many languages (e.g. Java) support this natively (look up AffineTransformation); otherwise, write out a routine to do rotation yourself, once, debug it well, and use it forever. I must have five of them written in different languages.

Once you can do the rotation simply, find the location on the rectangle by doing line-line intersection. Find the center of the orbited icon by intersecting two lines:

1. A ray from your center of rotation at the angle you desire
2. One of the four sides, bounded by what angle you want (the four quadrants).

Answer 6

Draw yourself a sketch on a piece of paper with a rectangle and a centre of rotation. First translate the rectangle to centre at the origin of your coordinate system (remember the translation parameters, you'll need to reverse the translation later). Rotate the rectangle so that its sides are parallel to the coordinate axes (same reason).

Now you have a triangle with known angle at the origin, the opposite side is of known length (half of the length of one side of the rectangle), and you can now:

-- solve the triangle

-- undo the rotation

-- undo the translation