问题
Hey there guys, I'm learning processing.js, and I've come across a mathematical problem, which I can't seem to solve with my limited geometry and trigonometry knowledge or by help of Wikipedia.
I need to draw a rectangle. To draw this rectangle, I need to know the coordinate points of each corner. All I know is x and y for the midpoints of the top and bottom of the box, and the length of all four sides.
There is no guarantee on the orientation of the box.
Any help? This seems like it should be easy, but it is really stumping me.
回答1:
If this quadrilateral is a rectangle (all four angles are 90 degrees), then it can be solved. (if it could be any quadrilateral, then it is not solvable)
if the points are (x1,y1), and (x2, y2), and if the two points are not perfectly vertical (x1 = x2) or horizontal (y1 = y2), then the slope of one edge of the rectangle is
m1 = (y2-y1) / (x2-x1)
and the slope of the other edge is:
m2 = - 1 / m1
If you know the lengths of the sides, and the midpoints of two opposite sides, then the corrner points are easily determined by adding dx, dy to the midpoints: (if L is length of the sides that the midpoints are on)
dx = Sqrt( L^2 / (1 + m2^2) ) / 2
and
dy = m2 * dx
NOTE: if the points are vertically or horizontally aligned, this technique will not work, although the obvious solution for those degenerative cases is much simpler.
回答2:
If you know your quadrilateral is a rectangle, then you can use some simple vector maths to find the coordinates of the corners. The knowns are:
(x1,y1)- the coordinate of the midpoint on the top line(x2,y2)- the coordinate of the midpoint on the bottom linel1- the length of the top and bottom linesl2- the length of the other two lines
First, we find the vector between the two known points. This vector is parallel to the side lines:
(vx, vy) = (x2 - x1, y2 - y1)
We need to normalize this vector (i.e. make it length 1) so we can use it later as a basis to find our coordinates.
vlen = sqrt(vx*vx + vy*vy)
(v1x, v1y) = (vx / vlen, vy / vlen)
Next, we rotate this vector anticlockwise by 90 degrees. The rotated vector will be parallel to the top and bottom lines. 90 degree rotation turns out to just be swapping the coordinates and negating one of them. You can see this just by trying it out on paper. Or take at look at the equations for 2D rotations and substitute in 90 degrees.
(u1x, u1y) = (-v1y, v1x)
Now we have enough information to find the 'top-left' corner. We simply start at our point (x1, y1) and move back along that side by half the side length:
(p1x, p1y) = (x1 - u1x * l1 / 2, y1 - u1y * l1 / 2)
From here we can find the remaining points just by adding the appropriate multiples of our basis vectors. When implementing this you can obviously speed it up by only calculating each unique multiplication a single time:
(p2x, p2y) = (p1x + u1x * l1, p1y + u1y * l1)
(p3x, p3y) = (p1x + v1x * l2, p1y + v1y * l2)
(p4x, p4y) = (p3x + u1x * l1, p3y + u1y * l1)
回答3:
function getFirstPoint(x1,y1,x2,y2,l1,l2)
distanceV = {x2 - x1, y2 - y1}
vlen = math.sqrt(distanceV[1]^2 + distanceV[2]^2)
normalized = {distanceV[1] / vlen, distanceV[2] / vlen}
rotated = {-normalized[2], normalized[1]}
p1 = {x1 - rotated[1] * l1 / 2, y1 - rotated[2] * l1 / 2}
p2 = {p1[1] + rotated[1] * l1, p1[2] + rotated[2] * l1}
p3 = {p1[1] + normalized[1] * l2, p1[2] + normalized[2] * l2}
p4 = {p3[1] + rotated[1] * l1, p3[2] + rotated[2] * l1}
points = { p1 , p2 , p3 , p4}
return p1
end
回答4:
It's definitely a rectangle? Then you know the orientation of the short sides (they're parallel to the line between your points), and hence the orientation of the long sides.
You know the orientation and length of the long sides, and you know their midpoints, so it's straightforward to find the corners from there.
Implementation is left as an exercise to the reader.
回答5:
This means that there will be two lines parallel to the line between the two points you have. Get the corners by translating the line you have 1/2 the length of the top side in each direction perpendicular to the line you have.
回答6:
If you know the midpoint for the top, and the length of the top, then you know that the y will stay the same for both top corners, and the x will be the midpoint plus/minus the width of the rectangle. This will also be true for the bottom.
Once you have the four corners, there is no need to worry about the side lengths, as their points are the same as those used for the top and bottom.
midpoint
x,10 10,10 x,10
*--------------------------------------------*
width = 30
mx = midpoint x.
top left corner = (w/2) - mx or 15 - 10
top left corner coords = -5,10
mx = midpoint x.
top right corner = (w/2) + mx or 15 + 10
top left corner coords = 25,10
回答7:
There's a difference between a "quadrilateral" and a "rectangle".
If you have the midpoint of the top and bottom, and the sides lengths, the rest is simple.
Given:
(x1, y1) -- (top_middle_x, top_middle_y) -- (x2, y1)
(x1, y2) -- (btm_middle_x, btm_middle_y) -- (x2, y2)
and top/bottom length along with right/left length.
x1 = top_middle_x - top/bottom_length / 2; x2 = x1 + top/bottom_length;
y1 = top_middle_y y2 = bottom_middle_y
Obviously, that's the simplest case and assuming that the line of (tmx, tmy) (bmx, bmy) is solely along the Y axis.
We'll call that line the "mid line".
The next trick is to take the mid line, and calculate it's rotational offset off the Y axis.
Now, my trig is super rusty.
dx = tmx - bmx, dy = tmy - bmy.
So, the tangent of the angle is dy / dx. The arctangent(dy / dx) is the angle of the line.
From that you can get your orientation.
(mind, there's some games with quadrants, and signs, and stuff to get this right -- but this is the gist of it.)
Once you have the orientation, you can "rotate" the line back to the Y axis. Look up 2D graphics for the math, it's straight forward.
That gets you your normal orientation. Then calculate the rectangles points, in this new normal form, and finally, rotate them back.
Viola. Rectangle.
Other things you can do is "rotate" a line that's half the length of the "top" line to where it's 90 deg of the mid line. So, say you have a mid line that's 45 degrees. You would start this line at tmx, tmy, and rotate this line 135 degrees (90 + 45). That point would be your "top left" corner. Rotate it -45 (45 - 90) to get the "top right" point. Then do something similar with the lower points.
回答8:
Calculate the angle of the line joining the two midpoints using an arc-tangent function applied to the vector you get between them.
Subtract 90 degrees from that angle to get the direction of the top edge
Starting from the top-center point, move relative (1/2 top width x sin(angle), 1/2 top width x cos(angle)) - that gets the top right corner point.
Continue around the rectangle using the sin and cos of the angles and widths as appropriate
As a test: Check you made it back to the starting point
回答9:
/* rcx = center x rectangle, rcy = center y rectangle, rw = width rectangle, rh = height rectangle, rr = rotation in radian from the rectangle (around it's center point) */
function toRectObjectFromCenter(rcx, rcy, rw, rh, rr){
var a = {
x: rcx+(Math.sin((rr-degToRad(90))+Math.asin(rh/(Math.sqrt(rh*rh+rw*rw)))) * (Math.sqrt(rh*rh+rw*rw)/2)),
y: rcy-(Math.cos((rr-degToRad(90))+Math.asin(rh/(Math.sqrt(rh*rh+rw*rw)))) * (Math.sqrt(rh*rh+rw*rw)/2))
};
var b = {
x: a.x+Math.cos(rr)*rw,
y: a.y+Math.sin(rr)*rw
};
var c = {
x: b.x+Math.cos(degToRad(radToDeg(rr)+90))*rh,
y: b.y+Math.sin(degToRad(radToDeg(rr)+90))*rh
};
var d = {
x: a.x+Math.cos(degToRad(radToDeg(rr)+90))*rh,
y: a.y+Math.sin(degToRad(radToDeg(rr)+90))*rh
};
return {a:a,b:b,c:c,d:d};
}
来源：https://stackoverflow.com/questions/2219108/knowing-two-points-of-a-rectangle-how-can-i-figure-out-the-other-two