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I'm looking for an algorithm (or pseudo code) that can calculate the maximum number of (smaller) circles with diameter "s" that can be squeezed into the circumference of another (larger) circle with radius "r" ...
Image: http://teasy.space/images/terracolony-squeezingcircles2.jpg
You can alternate between radius/diameter etc if you wish -- as these are the only 2 parameters (other than the center (large circle) coordinate) that i have, i.e. that are known ...
The outer circles may not overlap but can fit "snug" together ...
After various upgrades to my routine through the years, I'm currently using an algorithm that is not perfect (and it needs to be accurate or the galaxy breaks down lol)
which does a broad interpolation between small outside circle diameter and large inside circle circumference, to somewhat accurately plot the circle count in a polygon style fitting pattern, which causes problems (i.e. overlaps) when using larger outside circles ...
; try to fit a random number of circles
num_Circles = Rand( min,max )
; check if the number of circles exceed the maximum that can fit
If num_Circles * SmallCircle_Diameter > LargeCircle_Circumference
; adjust the amount accordingly
num_Circles = LargeCircle_Circumference / SmallCircle_Diameter
End If
Another assumption is that the size of the smaller outer circles never exceeds that of the larger inner circle ...
something less to worry about ;)
I'm using this algorithm for one of my projects called Terra Colony, based on Gravity Well, a 2D space/gravity realtime colonization simulation game with moons, planets, stars, black/white holes, etc
Image: http://teasy.space/images/terracolony-squeezingcircles1.jpg
This is an issue that has plagued this project for over a decade!
Hopefully you can point me in the right direction :D
I have previously done many experiments and wrote different programs to find a solution, and have traveled the internet looking for formulas and solutions which in the end are very close, but not close enough! :P
Thank you! <3
Teasy
P.S. I tried to add the tag "circumference" but it apparently requires "1500 reputation" (points i guess, maybe to prevent spam)
There is formula that establishes relation between radius of big circle R, radius of small circle r and number of (touching) small circles N
R = r / Sin(Pi/N)
So maximum number of small circles might be found as
Sin(Pi/N) = r / R
Pi / N = arcsin(r / R)
and finally
N = Pi / arcsin(r / R)
Example:
R=5
r=2.5
so
N = Pi / arcsin(1/2) =
Pi / (Pi/6) =
6
Given the diam. of the small circle 'd' and the number of them 'c'
then the dia. of the large circle 'D' is
D=d/sin(180/c)
I'm interested in a fast way to calculate the rotation-independent center of a simple, convex, (non-intersecting) 2D polygon.
The example below (on the left) shows the mean center (sum of all points divided by the total), and the desired result on the right.
Some options I've already considered.
bound-box center (depends on rotation, and ignores points based on their relation to the axis).
Straight skeleton - too slow to calculate.
I've found a way which works reasonably well, (weight the points by the edge-lengths) - but this means a square-root call for every edge - which I'd like to avoid.(Will post as an answer, even though I'm not entirely satisfied with it).
Note, I'm aware of this questions similarity with:What is the fastest way to find the "visual" center of an irregularly shaped polygon?
However having to handle convex polygons increases the complexity of the problem significantly.
The points of the polygon can be weighted by their edge length which compensates for un-even point distribution.
This works for convex polygons too but in that case the center point isn't guaranteed to be inside the polygon.
Psudo-code:
def poly_center(poly):
sum_center = (0, 0)
sum_weight = 0.0
for point in poly:
weight = ((point - point.next).length +
(point - point.prev).length)
sum_center += point * weight
sum_weight += weight
return sum_center / sum_weight
Note, we can pre-calculate all edge lengths to halve the number of length calculations, or reuse the previous edge-length for half+1 length calculations. This is just written as an example to show the logic.
Including this answer for completeness since its the best method I've found so far.
There is no much better way than the accumulation of coordinates weighted by the edge length, which indeed takes N square roots.
If you accept an approximation, it is possible to skip some of the vertices by curve simplification, as follows:
decide of a deviation tolerance;
start from vertex 0 and jump to vertex M (say M=N/2);
check if the deviation along the polyline from 0 to M exceeds the tolerance (for this, compute the height of the triangle formed by the vertices 0, M/2, M);
if the deviation is exceeded, repeat recursively with 0, M/4, M/2 and M/2, 3M/4, M;
if the deviation is not exceeded, assume that the shape is straight between 0 and M.
continue until the end of the polygon.
Where the points are dense (like the left edge on your example), you should get some speedup.
I think its easiest to do something with the center of masses of the delaunay triangulation of the polygon points. i.e.
def _centroid_poly(poly):
T = spatial.Delaunay(poly).simplices
n = T.shape[0]
W = np.zeros(n)
C = 0
for m in range(n):
sp = poly[T[m,:],:]
W[m] = spatial.ConvexHull(sp).volume
C += W[m] +np.mean(sp, axis = 0)
return C / np.sum(W)
This works well for me!
Basically I am creating an algorithm that plays the game of Snake independently.
I have managed to make it work out which direction to head to within the square of the game, but the version of Snake that I made has toroidal space, namely that when the snake's head goes off one side of the square, it reappears on the opposite side.
On that basis, it is not important for me to calculate what the distance between two points (namely the snake's head and the food block), but rather what direction the food is in relation to the head.
I.e. on a grid of 50 pixels square, when the head is at [row 49, column 49] and the food is at [1,1], I don't want the snake to have to travel 48 squares up and to the left. I'd rather it just travel 2 squares down and 2 right.
What formula would I then use to calculate which direction will result in the shortest travel time between the head and the food?
PS please let me know if I should post this in another Stack Exchange site and I will do so.
It's probably easier to consider one dimensional case. Basically a circular buffer, see pic:
You have the player "P" at position 3 and your "Goal" at position 9, and total play space Width = 10. The direction of shortest motion towards goal could then be given by:
if(P.coord < Goal.coord)
{
int dist1 = Goal.coord - P.coord;
int dist2 = (Width - Goal.coord) + P.coord;
if(dist1 < dist2)
{
//move forwards
}
else
{
//move backwards
}
}
else
{
int dist1 = P.coord - Goal.coord;
int dist2 = (Width - P.coord) + Goal.coord;
if(dist1 < dist2)
{
//move backwards
}
else
{
//move forwards
}
}
Similarly handle the 'else' part if p>coord > Goal.coord. Now extend it to 2 dimensions.
So, I've been struggling with a frankly now infuriating problem all day today.
Given a set of verticies of a triangle on a plane (just 3 points, 6 free parameters), I need to calculate the area of intersection of this triangle with the unit square defined by {0,0} and {1,1}. (I choose this because any square in 2D can be transformed to this, and the same transformation can move the 3 vertices).
So, now the problem is simplified down to only 6 parameters, 3 points... which I think is short enough that I'd be willing to code up the full solution / find the full solution.
( I would like this to run on a GPU for literally more than 2 million triangles every <0.5 seconds, if possible. as for the need for simplification / no data structures / libraries)
In terms of my attempt at the solution, I've... got a list of ways I've come up with, none of which seem fast or ... specific to the nice case (too general).
Option 1: Find the enclosed polygon, it can be anything from a triangle up to a 6-gon. Do this by use of some intersection of convex polygon in O(n) time algorithms that I found. Then I would sort these intersection points (new vertices, up to 7 of them O(n log n) ), in either CW or CCw order, so that I can run a simple area algorithm on the points (based on Greens function) (O(n) again). This is the fastest i can come with for an arbitrary convex n-gon intersecting with another m-gon. However... my problem is definitely not that complex, its a special case, so it should have a better solution...
Option 2:
Since I know its a triangle and unit square, i can simply find the list of intersection points in a more brute force way (rather than using some algorithm that is ... frankly a little frustrating to implement, as listed above)
There are only 19 points to check. 4 points are corners of square inside of triangle. 3 points are triangle inside square. And then for each line of the triangle, each will intersect 4 lines from the square (eg. y=0, y=1, x=0, x=1 lines). that is another 12 points. so, 12+3+4 = 19 points to check.
Once I have the, at most 6, at fewest 3, points that do this intersection, i can then follow up with one of two methods that I can think of.
2a: Sort them by increasing x value, and simply decompose the shape into its sub triangle / 4-gon shapes, each with an easy formula based on the limiting top and bottom lines. sum up the areas.
or 2b: Again sort the intersection points in some cyclic way, and then calculate the area based on greens function.
Unfortunately, this still ends up being just as complex as far as I can tell. I can start breaking up all the cases a little more, for finding the intersection points, since i know its just 0s and 1s for the square, which makes the math drop out some terms.. but it's not necessarily simple.
Option 3: Start separating the problem based on various conditions. Eg. 0, 1, 2, or 3 points of triangle inside square. And then for each case, run through all possible number of intersections, and then for each of those cases of polygon shapes, write down the area solution uniquely.
Option 4: some formula with heaviside step functions. This is the one I want the most probably, I suspect it'll be a little... big, but maybe I'm optimistic that it is possible, and that it would be the fastest computationally run time once I have the formula.
--- Overall, I know that it can be solved using some high level library (clipper for instance). I also realize that writing general solutions isn't so hard when using data structures of various kinds (linked list, followed by sorting it). And all those cases would be okay, if I just needed to do this a few times. But, since I need to run it as an image processing step, on the order of >9 * 1024*1024 times per image, and I'm taking images at .. lets say 1 fps (technically I will want to push this speed up as fast as possible, but lower bound is 1 second to calculate 9 million of these triangle intersection area problems). This might not be possible on a CPU, which is fine, I'll probably end up implementing it in Cuda anyways, but I do want to push the limit of speed on this problem.
Edit: So, I ended up going with Option 2b. Since there are only 19 intersections possible, of which at most 6 will define the shape, I first find those 3 to 6 verticies. Then i sort them in a cyclic (CCW) order. And then I find the area by calculating the area of that polygon.
Here is my test code I wrote to do that (it's for Igor, but should be readable as pseudocode) Unfortunately it's a little long winded, but.. I think other than my crappy sorting algorithm (shouldn't be more than 20 swaps though, so not so much overhead for writing better sorting)... other than that sorting, I don't think I can make it any faster. Though, I am open to any suggestions or oversights I might have had in chosing this option.
function calculateAreaUnitSquare(xPos, yPos)
wave xPos
wave yPos
// First, make array of destination. Only 7 possible results at most for this geometry.
Make/o/N=(7) outputVertexX = NaN
Make/o/N=(7) outputVertexY = NaN
variable pointsfound = 0
// Check 4 corners of square
// Do this by checking each corner against the parameterized plane described by basis vectors p2-p0 and p1-p0.
// (eg. project onto point - p0 onto p2-p0 and onto p1-p0. Using appropriate parameterization scaling (not unit).
// Once we have the parameterizations, then it's possible to check if it is inside the triangle, by checking that u and v are bounded by u>0, v>0 1-u-v > 0
variable denom = yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]*xPos[2]+yPos[1]*xPos[2]+xPos[0]*yPos[2]-xPos[1]*yPos[2]
//variable u00 = yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]*Xx+yPos[1]*Xx+xPos[0]*Yx-xPos[1]*Yx
//variable v00 = -yPos[2]*Xx+yPos[0]*(Xx-xPos[2])+xPos[0]*(yPos[2]-Yx)+yPos[2]*Yx
variable u00 = (yPos[0]*xPos[1]-xPos[0]*yPos[1])/denom
variable v00 = (yPos[0]*(-xPos[2])+xPos[0]*(yPos[2]))/denom
variable u01 =(yPos[0]*xPos[1]-xPos[0]*yPos[1]+xPos[0]-xPos[1])/denom
variable v01 =(yPos[0]*(-xPos[2])+xPos[0]*(yPos[2]-1)+xPos[2])/denom
variable u11 = (yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]+yPos[1]+xPos[0]-xPos[1])/denom
variable v11 = (-yPos[2]+yPos[0]*(1-xPos[2])+xPos[0]*(yPos[2]-1)+xPos[2])/denom
variable u10 = (yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]+yPos[1])/denom
variable v10 = (-yPos[2]+yPos[0]*(1-xPos[2])+xPos[0]*(yPos[2]))/denom
if(u00 >= 0 && v00 >=0 && (1-u00-v00) >=0)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
if(u01 >= 0 && v01 >=0 && (1-u01-v01) >=0)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
if(u10 >= 0 && v10 >=0 && (1-u10-v10) >=0)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
if(u11 >= 0 && v11 >=0 && (1-u11-v11) >=0)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
// Check 3 points for triangle. This is easy, just see if its bounded in the unit square. if it is, add it.
variable i = 0
for(i=0; i<3; i+=1)
if(xPos[i] >= 0 && xPos[i] <= 1 )
if(yPos[i] >=0 && yPos[i] <=1)
if(!((xPos[i] == 0 || xPos[i] == 1) && (yPos[i] == 0 || yPos[i] == 1) ))
outputVertexX[pointsfound] = xPos[i]
outputVertexY[pointsfound] = yPos[i]
pointsfound+=1
endif
endif
endif
endfor
// Check intersections.
// Procedure is: loop over 3 lines of triangle.
// For each line
// Check if vertical
// If not vertical, find y intercept with x=0 and x=1 lines.
// if y intercept is between 0 and 1, then add the point
// Check if horizontal
// if not horizontal, find x intercept with y=0 and y=1 lines
// if x intercept is between 0 and 1, then add the point
for(i=0; i<3; i+=1)
variable iN = mod(i+1,3)
if(xPos[i] != xPos[iN])
variable tx0 = xPos[i]/(xPos[i] - xPos[iN])
variable tx1 = (xPos[i]-1)/(xPos[i] - xPos[iN])
if(tx0 >0 && tx0 < 1)
variable yInt = (yPos[iN]-yPos[i])*tx0+yPos[i]
if(yInt > 0 && yInt <1)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = yInt
pointsfound+=1
endif
endif
if(tx1 >0 && tx1 < 1)
yInt = (yPos[iN]-yPos[i])*tx1+yPos[i]
if(yInt > 0 && yInt <1)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = yInt
pointsfound+=1
endif
endif
endif
if(yPos[i] != yPos[iN])
variable ty0 = yPos[i]/(yPos[i] - yPos[iN])
variable ty1 = (yPos[i]-1)/(yPos[i] - yPos[iN])
if(ty0 >0 && ty0 < 1)
variable xInt = (xPos[iN]-xPos[i])*ty0+xPos[i]
if(xInt > 0 && xInt <1)
outputVertexX[pointsfound] = xInt
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
endif
if(ty1 >0 && ty1 < 1)
xInt = (xPos[iN]-xPos[i])*ty1+xPos[i]
if(xInt > 0 && xInt <1)
outputVertexX[pointsfound] = xInt
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
endif
endif
endfor
// Now we have all 6 verticies that we need. Next step: find the lowest y point of the verticies
// if there are multiple with same low y point, find lowest X of these.
// swap this vertex to be first vertex.
variable lowY = 1
variable lowX = 1
variable m = 0;
for (i=0; i<pointsfound ; i+=1)
if (outputVertexY[i] < lowY)
m=i
lowY = outputVertexY[i]
lowX = outputVertexX[i]
elseif(outputVertexY[i] == lowY)
if(outputVertexX[i] < lowX)
m=i
lowY = outputVertexY[i]
lowX = outputVertexX[i]
endif
endif
endfor
outputVertexX[m] = outputVertexX[0]
outputVertexY[m] = outputVertexY[0]
outputVertexX[0] = lowX
outputVertexY[0] = lowY
// now we have the bottom left corner point, (bottom prefered).
// calculate the cos(theta) of unit x hat vector to the other verticies
make/o/N=(pointsfound) angles = (p!=0)?( (outputVertexX[p]-lowX) / sqrt( (outputVertexX[p]-lowX)^2+(outputVertexY[p]-lowY)^2) ) : 0
// Now sort the remaining verticies based on this angle offset. This will orient the points for a convex polygon in its maximal size / ccw orientation
// (This sort is crappy, but there will be in theory, at most 25 swaps. Which in the grand sceme of operations, isn't so bad.
variable j
for(i=1; i<pointsfound; i+=1)
for(j=i+1; j<pointsfound; j+=1)
if( angles[j] > angles[i] )
variable tempX = outputVertexX[j]
variable tempY = outputVertexY[j]
outputVertexX[j] = outputVertexX[i]
outputVertexY[j] =outputVertexY[i]
outputVertexX[i] = tempX
outputVertexY[i] = tempY
variable tempA = angles[j]
angles[j] = angles[i]
angles[i] = tempA
endif
endfor
endfor
// Now the list is ordered!
// now calculate the area given a list of CCW oriented points on a convex polygon.
// has a simple and easy math formula : http://www.mathwords.com/a/area_convex_polygon.htm
variable totA = 0
for(i = 0; i<pointsfound; i+=1)
totA += outputVertexX[i]*outputVertexY[mod(i+1,pointsfound)] - outputVertexY[i]*outputVertexX[mod(i+1,pointsfound)]
endfor
totA /= 2
return totA
end
I think the Cohen-Sutherland line-clipping algorithm is your friend here.
First off check the bounding box of the triangle against the square to catch the trivial cases (triangle inside square, triangle outside square).
Next check for the case where the square lies completely within the triangle.
Next consider your triangle vertices A, B and C in clockwise order. Clip the line segments AB, BC and CA against the square. They will either be altered such that they lie within the square or are found to lie outside, in which case they can be ignored.
You now have an ordered list of up to three line segments that define the some of the edges intersection polygon. It is easy to work out how to traverse from one edge to the next to find the other edges of the intersection polygon. Consider the endpoint of one line segment (e) against the start of the next (s)
If e is coincident with s, as would be the case when a triangle vertex lies within the square, then no traversal is required.
If e and s differ, then we need to traverse clockwise around the boundary of the square.
Note that this traversal will be in clockwise order, so there is no need to compute the vertices of the intersection shape, sort them into order and then compute the area. The area can be computed as you go without having to store the vertices.
Consider the following examples:
In the first case:
We clip the lines AB, BC and CA against the square, producing the line segments ab>ba and ca>ac
ab>ba forms the first edge of the intersection polygon
To traverse from ba to ca: ba lies on y=1, while ca does not, so the next edge is ca>(1,1)
(1,1) and ca both lie on x=1, so the next edge is (1,1)>ca
The next edge is a line segment we already have, ca>ac
ac and ab are coincident, so no traversal is needed (you might be as well just computing the area for a degenerate edge and avoiding the branch in these cases)
In the second case, clipping the triangle edges against the square gives us ab>ba, bc>cb and ca>ac. Traversal between these segments is trivial as the start and end points lie on the same square edges.
In the third case the traversal from ba to ca goes through two square vertices, but it is still a simple matter of comparing the square edges on which they lie:
ba lies on y=1, ca does not, so next vertex is (1,1)
(1,1) lies on x=1, ca does not, so next vertex is (1,0)
(1,0) lies on y=0, as does ca, so next vertex is ca.
Given the large number of triangles I would recommend scanline algorithm: sort all the points 1st by X and 2nd by Y, then proceed in X direction with a "scan line" that keeps a heap of Y-sorted intersections of all lines with that line. This approach has been widely used for Boolean operations on large collections of polygons: operations such as AND, OR, XOR, INSIDE, OUTSIDE, etc. all take O(n*log(n)).
It should be fairly straightforward to augment Boolean AND operation, implemented with the scanline algorithm to find the areas you need. The complexity will remain O(n*log(n)) on the number of triangles. The algorithm would also apply to intersections with arbitrary collections of arbitrary polygons, in case you would need to extend to that.
On the 2nd thought, if you don't need anything other than the triangle areas, you could do that in O(n), and scanline may be an overkill.
I came to this question late, but I think I've come up with a more fully flushed out solution along the lines of ryanm's answer. I'll give an outline of for others trying to do this problem at least somewhat efficiently.
First you have two trivial cases to check:
1) Triangle lies entirely within the square
2) Square lies entirely within the triangle (Just check if all corners are inside the triangle)
If neither is true, then things get interesting.
First, use either the Cohen-Sutherland or Liang-Barsky algorithm to clip each edge of the triangle to the square. (The linked article contains a nice bit of code that you can essentially just copy-paste if you're using C).
Given a triangle edge, these algorithms will output either a clipped edge or a flag denoting that the edge lies entirely outside the square. If all edges lie outsize the square, then the triangle and the square are disjoint.
Otherwise, we know that the endpoints of the clipped edges constitute at least some of the vertices of the polygon representing the intersection.
We can avoid a tedious case-wise treatment by making a simple observation. All other vertices of the intersection polygon, if any, will be corners of the square that lie inside the triangle.
Simply put, the vertices of the intersection polygon will be the (unique) endpoints of the clipped triangle edges in addition to the corners of the square inside the triangle.
We'll assume that we want to order these vertices in a counter-clockwise fashion. Since the intersection polygon will always be convex, we can compute its centroid (the mean over all vertex positions) which will lie inside the polygon.
Then to each vertex, we can assign an angle using the atan2 function where the inputs are the y- and x- coordinates of the vector obtained by subtracting the centroid from the position of the vertex (i.e. the vector from the centroid to the vertex).
Finally, the vertices can be sorted in ascending order based on the values of the assigned angles, which constitutes a counter-clockwise ordering. Successive pairs of vertices correspond to the polygon edges.
With the help of the Stack Overflow community I've written a pretty basic-but fun physics simulator.
You click and drag the mouse to launch a ball. It will bounce around and eventually stop on the "floor".
My next big feature I want to add in is ball to ball collision. The ball's movement is broken up into a x and y speed vector. I have gravity (small reduction of the y vector each step), I have friction (small reduction of both vectors each collision with a wall). The balls honestly move around in a surprisingly realistic way.
I guess my question has two parts:
What is the best method to detect ball to ball collision?
Do I just have an O(n^2) loop that iterates over each ball and checks every other ball to see if it's radius overlaps?
What equations do I use to handle the ball to ball collisions? Physics 101
How does it effect the two balls speed x/y vectors? What is the resulting direction the two balls head off in? How do I apply this to each ball?
Handling the collision detection of the "walls" and the resulting vector changes were easy but I see more complications with ball-ball collisions. With walls I simply had to take the negative of the appropriate x or y vector and off it would go in the correct direction. With balls I don't think it is that way.
Some quick clarifications: for simplicity I'm ok with a perfectly elastic collision for now, also all my balls have the same mass right now, but I might change that in the future.
Edit: Resources I have found useful
2d Ball physics with vectors: 2-Dimensional Collisions Without Trigonometry.pdf
2d Ball collision detection example: Adding Collision Detection
Success!
I have the ball collision detection and response working great!
Relevant code:
Collision Detection:
for (int i = 0; i < ballCount; i++)
{
for (int j = i + 1; j < ballCount; j++)
{
if (balls[i].colliding(balls[j]))
{
balls[i].resolveCollision(balls[j]);
}
}
}
This will check for collisions between every ball but skip redundant checks (if you have to check if ball 1 collides with ball 2 then you don't need to check if ball 2 collides with ball 1. Also, it skips checking for collisions with itself).
Then, in my ball class I have my colliding() and resolveCollision() methods:
public boolean colliding(Ball ball)
{
float xd = position.getX() - ball.position.getX();
float yd = position.getY() - ball.position.getY();
float sumRadius = getRadius() + ball.getRadius();
float sqrRadius = sumRadius * sumRadius;
float distSqr = (xd * xd) + (yd * yd);
if (distSqr <= sqrRadius)
{
return true;
}
return false;
}
public void resolveCollision(Ball ball)
{
// get the mtd
Vector2d delta = (position.subtract(ball.position));
float d = delta.getLength();
// minimum translation distance to push balls apart after intersecting
Vector2d mtd = delta.multiply(((getRadius() + ball.getRadius())-d)/d);
// resolve intersection --
// inverse mass quantities
float im1 = 1 / getMass();
float im2 = 1 / ball.getMass();
// push-pull them apart based off their mass
position = position.add(mtd.multiply(im1 / (im1 + im2)));
ball.position = ball.position.subtract(mtd.multiply(im2 / (im1 + im2)));
// impact speed
Vector2d v = (this.velocity.subtract(ball.velocity));
float vn = v.dot(mtd.normalize());
// sphere intersecting but moving away from each other already
if (vn > 0.0f) return;
// collision impulse
float i = (-(1.0f + Constants.restitution) * vn) / (im1 + im2);
Vector2d impulse = mtd.normalize().multiply(i);
// change in momentum
this.velocity = this.velocity.add(impulse.multiply(im1));
ball.velocity = ball.velocity.subtract(impulse.multiply(im2));
}
Source Code: Complete source for ball to ball collider.
If anyone has some suggestions for how to improve this basic physics simulator let me know! One thing I have yet to add is angular momentum so the balls will roll more realistically. Any other suggestions? Leave a comment!
To detect whether two balls collide, just check whether the distance between their centers is less than two times the radius. To do a perfectly elastic collision between the balls, you only need to worry about the component of the velocity that is in the direction of the collision. The other component (tangent to the collision) will stay the same for both balls. You can get the collision components by creating a unit vector pointing in the direction from one ball to the other, then taking the dot product with the velocity vectors of the balls. You can then plug these components into a 1D perfectly elastic collision equation.
Wikipedia has a pretty good summary of the whole process. For balls of any mass, the new velocities can be calculated using the equations (where v1 and v2 are the velocities after the collision, and u1, u2 are from before):
If the balls have the same mass then the velocities are simply switched. Here's some code I wrote which does something similar:
void Simulation::collide(Storage::Iterator a, Storage::Iterator b)
{
// Check whether there actually was a collision
if (a == b)
return;
Vector collision = a.position() - b.position();
double distance = collision.length();
if (distance == 0.0) { // hack to avoid div by zero
collision = Vector(1.0, 0.0);
distance = 1.0;
}
if (distance > 1.0)
return;
// Get the components of the velocity vectors which are parallel to the collision.
// The perpendicular component remains the same for both fish
collision = collision / distance;
double aci = a.velocity().dot(collision);
double bci = b.velocity().dot(collision);
// Solve for the new velocities using the 1-dimensional elastic collision equations.
// Turns out it's really simple when the masses are the same.
double acf = bci;
double bcf = aci;
// Replace the collision velocity components with the new ones
a.velocity() += (acf - aci) * collision;
b.velocity() += (bcf - bci) * collision;
}
As for efficiency, Ryan Fox is right, you should consider dividing up the region into sections, then doing collision detection within each section. Keep in mind that balls can collide with other balls on the boundaries of a section, so this may make your code much more complicated. Efficiency probably won't matter until you have several hundred balls though. For bonus points, you can run each section on a different core, or split up the processing of collisions within each section.
Well, years ago I made the program like you presented here.
There is one hidden problem (or many, depends on point of view):
If the speed of the ball is too
high, you can miss the collision.
And also, almost in 100% cases your new speeds will be wrong. Well, not speeds, but positions. You have to calculate new speeds precisely in the correct place. Otherwise you just shift balls on some small "error" amount, which is available from the previous discrete step.
The solution is obvious: you have to split the timestep so, that first you shift to correct place, then collide, then shift for the rest of the time you have.
You should use space partitioning to solve this problem.
Read up on
Binary Space Partitioning
and
Quadtrees
As a clarification to the suggestion by Ryan Fox to split the screen into regions, and only checking for collisions within regions...
e.g. split the play area up into a grid of squares (which will will arbitrarily say are of 1 unit length per side), and check for collisions within each grid square.
That's absolutely the correct solution. The only problem with it (as another poster pointed out) is that collisions across boundaries are a problem.
The solution to this is to overlay a second grid at a 0.5 unit vertical and horizontal offset to the first one.
Then, any collisions that would be across boundaries in the first grid (and hence not detected) will be within grid squares in the second grid. As long as you keep track of the collisions you've already handled (as there is likely to be some overlap) you don't have to worry about handling edge cases. All collisions will be within a grid square on one of the grids.
A good way of reducing the number of collision checks is to split the screen into different sections. You then only compare each ball to the balls in the same section.
One thing I see here to optimize.
While I do agree that the balls hit when the distance is the sum of their radii one should never actually calculate this distance! Rather, calculate it's square and work with it that way. There's no reason for that expensive square root operation.
Also, once you have found a collision you have to continue to evaluate collisions until no more remain. The problem is that the first one might cause others that have to be resolved before you get an accurate picture. Consider what happens if the ball hits a ball at the edge? The second ball hits the edge and immediately rebounds into the first ball. If you bang into a pile of balls in the corner you could have quite a few collisions that have to be resolved before you can iterate the next cycle.
As for the O(n^2), all you can do is minimize the cost of rejecting ones that miss:
1) A ball that is not moving can't hit anything. If there are a reasonable number of balls lying around on the floor this could save a lot of tests. (Note that you must still check if something hit the stationary ball.)
2) Something that might be worth doing: Divide the screen into a number of zones but the lines should be fuzzy--balls at the edge of a zone are listed as being in all the relevant (could be 4) zones. I would use a 4x4 grid, store the zones as bits. If an AND of the zones of two balls zones returns zero, end of test.
3) As I mentioned, don't do the square root.
I found an excellent page with information on collision detection and response in 2D.
http://www.metanetsoftware.com/technique.html (web.archive.org)
They try to explain how it's done from an academic point of view. They start with the simple object-to-object collision detection, and move on to collision response and how to scale it up.
Edit: Updated link
You have two easy ways to do this. Jay has covered the accurate way of checking from the center of the ball.
The easier way is to use a rectangle bounding box, set the size of your box to be 80% the size of the ball, and you'll simulate collision pretty well.
Add a method to your ball class:
public Rectangle getBoundingRect()
{
int ballHeight = (int)Ball.Height * 0.80f;
int ballWidth = (int)Ball.Width * 0.80f;
int x = Ball.X - ballWidth / 2;
int y = Ball.Y - ballHeight / 2;
return new Rectangle(x,y,ballHeight,ballWidth);
}
Then, in your loop:
// Checks every ball against every other ball.
// For best results, split it into quadrants like Ryan suggested.
// I didn't do that for simplicity here.
for (int i = 0; i < balls.count; i++)
{
Rectangle r1 = balls[i].getBoundingRect();
for (int k = 0; k < balls.count; k++)
{
if (balls[i] != balls[k])
{
Rectangle r2 = balls[k].getBoundingRect();
if (r1.Intersects(r2))
{
// balls[i] collided with balls[k]
}
}
}
}
I see it hinted here and there, but you could also do a faster calculation first, like, compare the bounding boxes for overlap, and THEN do a radius-based overlap if that first test passes.
The addition/difference math is much faster for a bounding box than all the trig for the radius, and most times, the bounding box test will dismiss the possibility of a collision. But if you then re-test with trig, you're getting the accurate results that you're seeking.
Yes, it's two tests, but it will be faster overall.
This KineticModel is an implementation of the cited approach in Java.
I implemented this code in JavaScript using the HTML Canvas element, and it produced wonderful simulations at 60 frames per second. I started the simulation off with a collection of a dozen balls at random positions and velocities. I found that at higher velocities, a glancing collision between a small ball and a much larger one caused the small ball to appear to STICK to the edge of the larger ball, and moved up to around 90 degrees around the larger ball before separating. (I wonder if anyone else observed this behavior.)
Some logging of the calculations showed that the Minimum Translation Distance in these cases was not large enough to prevent the same balls from colliding in the very next time step. I did some experimenting and found that I could solve this problem by scaling up the MTD based on the relative velocities:
dot_velocity = ball_1.velocity.dot(ball_2.velocity);
mtd_factor = 1. + 0.5 * Math.abs(dot_velocity * Math.sin(collision_angle));
mtd.multplyScalar(mtd_factor);
I verified that before and after this fix, the total kinetic energy was conserved for every collision. The 0.5 value in the mtd_factor was the approximately the minumum value found to always cause the balls to separate after a collision.
Although this fix introduces a small amount of error in the exact physics of the system, the tradeoff is that now very fast balls can be simulated in a browser without decreasing the time step size.
Improving the solution to detect circle with circle collision detection given within the question:
float dx = circle1.x - circle2.x,
dy = circle1.y - circle2.y,
r = circle1.r + circle2.r;
return (dx * dx + dy * dy <= r * r);
It avoids the unnecessary "if with two returns" and the use of more variables than necessary.
After some trial and error, I used this document's method for 2D collisions : https://www.vobarian.com/collisions/2dcollisions2.pdf
(that OP linked to)
I applied this within a JavaScript program using p5js, and it works perfectly. I had previously attempted to use trigonometrical equations and while they do work for specific collisions, I could not find one that worked for every collision no matter the angle at the which it happened.
The method explained in this document uses no trigonometrical functions whatsoever, it's just plain vector operations, I recommend this to anyone trying to implement ball to ball collision, trigonometrical functions in my experience are hard to generalize. I asked a Physicist at my university to show me how to do it and he told me not to bother with trigonometrical functions and showed me a method that is analogous to the one linked in the document.
NB : My masses are all equal, but this can be generalised to different masses using the equations presented in the document.
Here's my code for calculating the resulting speed vectors after collision :
//you just need a ball object with a speed and position vector.
class TBall {
constructor(x, y, vx, vy) {
this.r = [x, y];
this.v = [0, 0];
}
}
//throw two balls into this function and it'll update their speed vectors
//if they collide, you need to call this in your main loop for every pair of
//balls.
function collision(ball1, ball2) {
n = [ (ball1.r)[0] - (ball2.r)[0], (ball1.r)[1] - (ball2.r)[1] ];
un = [n[0] / vecNorm(n), n[1] / vecNorm(n) ] ;
ut = [ -un[1], un[0] ];
v1n = dotProd(un, (ball1.v));
v1t = dotProd(ut, (ball1.v) );
v2n = dotProd(un, (ball2.v) );
v2t = dotProd(ut, (ball2.v) );
v1t_p = v1t; v2t_p = v2t;
v1n_p = v2n; v2n_p = v1n;
v1n_pvec = [v1n_p * un[0], v1n_p * un[1] ];
v1t_pvec = [v1t_p * ut[0], v1t_p * ut[1] ];
v2n_pvec = [v2n_p * un[0], v2n_p * un[1] ];
v2t_pvec = [v2t_p * ut[0], v2t_p * ut[1] ];
ball1.v = vecSum(v1n_pvec, v1t_pvec); ball2.v = vecSum(v2n_pvec, v2t_pvec);
}
I would consider using a quadtree if you have a large number of balls. For deciding the direction of bounce, just use simple conservation of energy formulas based on the collision normal. Elasticity, weight, and velocity would make it a bit more realistic.
Here is a simple example that supports mass.
private void CollideBalls(Transform ball1, Transform ball2, ref Vector3 vel1, ref Vector3 vel2, float radius1, float radius2)
{
var vec = ball1.position - ball2.position;
float dis = vec.magnitude;
if (dis < radius1 + radius2)
{
var n = vec.normalized;
ReflectVelocity(ref vel1, ref vel2, ballMass1, ballMass2, n);
var c = Vector3.Lerp(ball1.position, ball2.position, radius1 / (radius1 + radius2));
ball1.position = c + (n * radius1);
ball2.position = c - (n * radius2);
}
}
public static void ReflectVelocity(ref Vector3 vel1, ref Vector3 vel2, float mass1, float mass2, Vector3 intersectionNormal)
{
float velImpact1 = Vector3.Dot(vel1, intersectionNormal);
float velImpact2 = Vector3.Dot(vel2, intersectionNormal);
float totalMass = mass1 + mass2;
float massTransfure1 = mass1 / totalMass;
float massTransfure2 = mass2 / totalMass;
vel1 += ((velImpact2 * massTransfure2) - (velImpact1 * massTransfure2)) * intersectionNormal;
vel2 += ((velImpact1 * massTransfure1) - (velImpact2 * massTransfure1)) * intersectionNormal;
}