Lets say I have a point within a circle(not necessarily the origin) moving at a given vector how would I calculate the x and y coordinate of the point where it hits the edge of the circle.
Shift all coordinates by -cx, -cy. Now circle is centered at origin and has equation
x^2+y^2=R^2
Point coordinate (px, py), unit direction vector is (dx,dy). Equation of ray:
x = px + t * dx
y = py + t * dy
Substitute these variables into the circle equation, solve equation, find parameter t>0, then find intersection point (x,y), shift it back by (cx, cy).
I have a Vector2 in my 2D Game and what I would like to do now is set my vector2 x and y by calculating them using rotation in degrees
Do I need to use PI to calculate new X and Y coordinates then add move distance per second in order to get the correct coordinates?
Example : Lets say degree is 90, which means my gameobject would move forward,at 5 floating units per second, then Y would be 5,10,15 and if degree would be 180 then X would increase by 5 every second, this is simple, but how to do it for other degrees such as 38,268 etc?
The usual convention is that 0 degrees points in the positive X direction and as the angle increases you rotate the direction anti-clockwise. Your convention seems to be that 0 degrees points in the negative X direction and the angle increases clockwise, so first of all you must translate your angle, say alpha, into one with the usual convention, say beta
beta = 180.0 - alpha
Next, trigonometric functions assume radians which run from 0 to 2π rather than from 0 to 360, so you must translate beta into an angle in radians, say theta
theta = 2.0*PI*beta/360.0
Finally, cos(theta) gives the change in X for a move of 1 unit in the direction given by theta and sin(theta) gives the change in Y. So you need
X = X + D * cos(theta)
Y = Y + D * sintheta)
for a distance D. Using your convention this translates to
X = X + D * cos(2.0*PI*(180.0-alpha)/360.0)
Y = Y + D * sin(2.0*PI*(180.0-alpha)/360.0)
I have a unit sphere (radius 1) that is drawn centred in orthogonal projection.
The sphere may rotate freely.
How can I determine the point on the sphere that the user clicks on?
Given:
the height and width of the monitor
the radius of the projected circle, in pixels
the coordinates of the point the user clicked on
And assuming that the top-left corner is (0,0), the x value increases as you travel to the right, and the y value increases as you travel down.
Translate the user's click point into the coordinate space of the globe.
userPoint.x -= monitor.width/2
userPoint.y -= monitor.height/2
userPoint.x /= circleRadius
userPoint.y /= circleRadius
Find the z coordinate of the point of intersection.
//solve for z
//x^2 + y^2 + z^2 = 1
//we know x and y, from userPoint
//z^2 = 1 - x^2 - y^2
x = userPoint.x
y = userPoint.y
if (x^2 + y^2 > 1){
//user clicked outside of sphere. flip out
return -1;
}
//The negative sqrt is closer to the screen than the positive one, so we prefer that.
z = -sqrt(1 - x^2 - y^2);
Now that you know the (x,y,z) point of intersection, you can find the lattitude and longitude.
Assuming that the center of the globe facing the user is 0E 0N,
longitude = 90 + toDegrees(atan2(z, x));
lattitude = toDegrees(atan2(y, sqrt(x^2 + z^2)))
If the sphere is rotated so that the 0E meridian is not directly facing the viewer, subtract the angle of rotation from the longitude.
One possible approach is to generate the sphere from triangles, consisting of rows and columns. They can be invisible too. And then hit-testing those triangles with a mouse pick ray.
See this picture's latitude/longitude grid, but apply it much denser. For each grid cell, you need 2 triangles.
I'm using rectangles defined in terms of their x y coordinates and their width and height. I figured out how to rotate them in terms of coordinates (x = cos(deg) * x - sin(deg) * y y = sin(deg) * x + cos(deg) * y) but I'm stuck on the height and width. I'm sure there's an obvious solution that I'm missing. If it matters, I'm using Python.
edit Sorry for the confusing description. My intention is to get the width and height either reversed or negated due to whatever the angle is. For example, in a 90 degree rotation the values would switch. In a 180 degree rotation the width would be negative. Also, I only intend to use multiples of 90 in my script. I could just use if statements, but I assumed there would be a more "elegant" method.
Just calculate four corners of Your rectangle:
p1 = (x, y)
p2 = (x + w, y)
p3 = (x, y + h)
and rotate each by angle You want:
p1 = rotate(p1, angle)
# and so on...
and transform back to Your rectangle representation:
x, y = p1
w = dist(p1, p2) # the same as before rotation
h = dist(p1, p3)
where dist calculates distance between two points.
Edit: Why don't You try apply formula You have written to (width, height) pair?
x1 = cos(deg) * x - sin(deg) * y
y2 = sin(deg) * x + cos(deg) * y
It is easy to see that if deg == 90 the values will switch:
x1 = -y
y2 = x
and if deg == 180 they will be negated:
x1 = -x
y2 = -y
and so on... I think this is what You are looking for.
Edit2:
Here comes fast rotation function:
def rotate_left_by_90(times, x, y):
return [(x, y), (-y, x), (-x, -y), (y, -x)][times % 4]
The proper way would be to resort to transformation matrices. Also, judging from your question I suppose that you want to rotate with respect to (x=0,y=0), but if not you will need to take this into account and translate your rectangle to the center of the plan first (and then translate it back when the rotation is carried out).
M = Matrix to translate to the center
R = Rotation Matrix
Transformation Matrix = M^(-1) * R * M
But to give you an easy answer to your question, just take the two other corners of your rectangle and apply the same transformation on them.
To learn more about transformation matrices :
http://en.wikipedia.org/wiki/Transformation_matrix
From the way you describe only rotating by 90 degrees, and the way you seem to be defining width and height, perhaps you are looking for something like
direction = 1 // counter-clockwise degrees
// or
direction = -1 // clockwise 90 degrees
new_height = width * direction
new_width = -height * direction
width = new_width
height = new_height
Not sure why you want to have negative values for width and height, though .. because otherwise each 90 degree rotation effectively just swaps width and height, regardless which way you rotate.
Rotation should not change width and height. Your equation is correct if you want to rotate (x,y) about (0,0) by deg, but note that often cos and sin functions expect arguments in radians rather than degrees, so you may need to multiply deg by pi/180 (radians per degree).
If you need to find the locations of other rectangle vertices besides (x,y) after rotating, then you should either store and rotate them along with (x,y) or keep some information about the rectangle's orientation (such as deg) so you can recompute them as e.g. x+widthcos(deg), y+heightsin(deg).
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.
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.
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 line
l1 - the length of the top and bottom lines
l2 - 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)
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
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.
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.
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
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.
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
/* 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};
}