I have a simple maths/physics problem here: In a Cartesian coordinate system, I have a point that moves in time with a known velocity. The point is inside a box, and bounces orthognally on its walls.
Here is a quick example I did on paint:
What we know: The red point position, and its velocity which is defined by an angle θ and a speed. Of course we know the dimensions of the green box.
On the example, I've drawn in yellow its approximate trajectory, and let's say that after a determined period of time which is known, the red point is on the blue point. What would be the most efficient way to compute the blue point position?
I've tought about computing every "bounce point" with trigonometry and vector projection, but I feel like it's a waste of resources because trigonometric functions are usually very processor hungry. I'll have more than a thousand points to compute like that so I really need to find a more efficient way to do it.
If anyone has any idea, I'd be very grateful.
Apart from programming considerations, it has an interesting solution from geometric point of view. You can find the position of the point at a specific time T without considering its temporal trajectory during 0<t<T
For one minute, forget the size and the boundaries of the box; and assume that the point can move on a straight line for ever. Then the point has constant velocity components vx = v*cos(θ), vy = v*sin(θ) and at time T its virtual porition will be x' = x0 + vx * T, y' = y0 + vy * T
Now you need to map the virtual position (x',y') into the actual position (x,y). See image below
You can recursively reflect the virtual point w.r.t the borders until the point comes back into the reference (initial) box. And this is the actual point. Now the question is how to do these mathematics? and how to find (x,y) knowing (x',y')?
Denote by a and b the size of the box along x and y respectively. Then nx = floor(x'/a) and ny = floor(y'/b) indicates how far is the point from the reference box in terms of the number of boxes. Also dx = x'-nx*a and dy = y'-ny*b introduces the relative position of the virtual point inside its virtual box.
Now you can find the true position (x,y): if nx is even, then x = dx else x = a-dx; similarly if ny is even, then y = dy else y = b-dy. In other words, even number of reflections in each axis x and y, puts the true point and the virtual point in the same relative positions, while odd number of reflections make them different and complementary.
You don't need to use trigonometric function all the time. Instead get normalized direction vector as (dx, dy) = (cos(θ), sin(θ))
After bouncing from vertical wall x-component changes it's sign dx = -dx, after bouncing from horizontal wall y-component changes it's sign dy = -dy. You can see that calculations are blazingly simple.
If you (by strange reason) prefer to use angles, use angle transformations from here (for ball with non-zero radius)
if ((ball.x + ball.radius) >= window.width || (ball.x - ball.radius) <= 0)
ball.theta = M_PI - ball.theta;
else
if ((ball.y + ball.radius) >= window.height || (ball.y - ball.radius) <= 0)
ball.theta = - ball.theta;
To get point of bouncing:
Starting point (X0, Y0)
Ray angle Theta, c = Cos(Theta), s = Sin(Theta);
Rectangle coordinates: bottom left (X1,Y1), top right (X2,Y2)
if c >= 0 then //up
XX = X2
else
XX = X1
if s >= 0 then //right
YY = Y2
else
YY = Y1
if c = 0 then //vertical ray
return Intersection = (X0, YY)
if s = 0 then //horizontal ray
return Intersection = (XX, Y0)
tx = (XX - X0) / c //parameter when vertical edge is met
ty = (YY - Y0) / s //parameter when horizontal edge is met
if tx <= ty then //vertical first
return Intersection = (XX, Y0 + tx * s)
else //horizontal first
return Intersection = (X0 + ty * c, YY)
#Problem
Hey I wondered how we can find the intersection between a point with a vector direction, assuming that the point is in the circle.
In Other words, how can I find where the particle will hit the circle circumference when I know the following: Circle position, radius, Particle Position and direction (velocity).
#Implementation
I am currently creating a flight radar, all flights are stored in a Queue and I need to sort the queue based on the time untill the flight leaves the cirlce (Radar Radius)
So I need to return a Point, thus I can just get the distance and calculate the time based on distance and velocity.
At first subtract center cordinates to simplify calculations (now circle center is coordinate origin)
x11 = x1 - x0
y11 = y1 - y0
Point position against time is
x = x11 + vx * t
y = y11 + vy * t
where vx, vy are components of velocity
Point is at circumference when
x^2 + y^2 = R^2
so
(x11 + vx * t)^2 + (y11 + vy * t)^2 = R^2
(vx^2+vy^2)*t^2 + (2*vx*x11+2*vy*y11)*t + (x11^2+y11^2-R^2) = 0
This is quadratic equation for unknown t. Solve it, find roots - none for outside moving, one for single touch case, two roots for regular intersection - larger one is needed time of intersection t_int
Substitute t into point position equations and get coordinates:
x_int = x1 + t_int * vx
y_int = y1 + t_int * vy
I have two points in a 3d space, one point is (x,0,z) and the other one is the origin (0,0,0), through those points there is a passing line with length L that is starting from the first point and continuing after the origin point, in the end of this line there is a perpendicular (to the line) flat board with dimensions of W x H, the line ends in the middle of this board.
Assume that x,z,L,H,W are given, I need a way to find all the 3D points coordinates where those points forming a pixels image over the board (meaning each point has a distance of 1 from its left, right, top, bottom neighboring points).
Attached a pretty ugly drawing :) I made trying to illustrate the problem (I marked the pixels points with two question marks but I need them all).
Thanks.
It is possible to define that plane. But there is no selected direction to build a grid uniquelly.
Let we choose OY direction as base (because normal has zero Y-component).
So we have:
Normal vector N = (xx, 0, zz) //I renamed values to avoid confusion with coordinate
variables
Unit normal vector n = (nx, 0, nz), where
nx = xx / Sqrt(xx*xx+zz*zz)
nz = zz / Sqrt(xx*xx+zz*zz)
Base point
B = (bx, 0, bz) = (xx - nx * L, 0, zz - nz * L)
Unit base vector in the plane
dy = (0, 1, 0)
Another base vector
dc = dy x n //vector product
= (-bz, 0, bx)
Now it is possible to generate a grid, using integer indexes i, j in ranges (-W/2..W/2) and (-H/2.. H/2). Grid nodes coordinates:
x(i, j) = bx - j * bz
y(i, j) = 0 + i
z(i, j) = bz + j * bx
I want to construct a triangle in the real world to represent a 2D "viewing frustum" using the user's coordinates, heading (degrees currently facing from true north), and fixed distances that represent how far they can see.
I was imagining drawing a line of K1 distance from the user's point in the direction of the heading and marking a temporary point, then drawing a perpendicular line at that point to the previous line and marking 2 points on each side of the perpendicular line K2 distance away from the point.
This would give me the 3 points that I need. For those who are great at math, first is this possible and second can you give me some pointers on how to approach this? Thanks.
In cartesian co-ordinates:
Assume:
+Y axis is north.
K2 is distance from "temp point" to the two points you're creating
Current position is (Cx, Cy)
Heading (H) is angle clockwise from the Y-axis.
Temporary point is (Tx, Ty)
Remaining two points are (Px, Py) and (Qx, Qy)
Then:
Tx = Cx + K1 * sin(H)
Ty = Cy + K1 * cos(H)
Px = Tx - K2 * cos(H)
Py = Ty + K2 * sin(H)
Qx = Tx + K2 * cos(H)
Qy = Ty - K2 * sin(H)
When computing (Tx, Ty), you use sin(H) with the x-coord and cos(H) with the y-coord because the angle is being measured from the Y-axis. When computing (Px, Py) and (Qx, Qy) you use the fact that if (a, b) is some vector, then any multiple of (-a, b) is a vector perpendicular to the first. Hence the (sin(H), cos(H)) turns into (-sin(H), cos(H)) and (cos(H), -sin(H)). This falls out of the definition of dot product in 2-D cartesian space and the coordinate-free fact that the dot product of perpendicular vectors is zero.
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};
}