I am trying to solve the following problem (I am using Matlab, though pseudo-code / solutions in other languages are welcome):
I have two circles on a Cartesian plane defined by their centroids (p1, p2) and their radii (r1, r2). circle 1 (c1 = [p1 r1]) is considered 'dynamic': it is being translated along the vector V = [0 -1]. circle 2 (c2 = [p2 r2]) is considered 'static': it lies in the path of c1 but the x component of its centroid is offset from the x component of c2 (otherwise the solution would be trivial: the distance between the circle centroids minus the sum of their radii).
I am trying to locate the distance (d) along V at which circle 1 will 'collide' with circle 2 (see the linked image). I am sure that I can solve this iteratively (i.e. translate c1 to the bounding box of c2 then converge / test for intersection). However, I would like to know if there is a closed form solution to this problem.
Shift coordinates to simplify expressions
px = p1.x - p2.x
py = p1.y - p2.y
And solve quadratic equation for d (zero, one, or two solutions)
px^2 + (py - d)^2 = (r1 + r2)^2
(py - d)^2 = (r1 + r2)^2 - px^2
d = py +/- Sqrt((r1 + r2)^2 - px^2)
That's all.
As the question title does not match the question and accepted answer which is dependent on a fixed vector {0, -1}, or {0, 1} rather than an arbitrary vector I have added another solution that works for any unit vector.
Where (See diagram 1)
dx, dy is the unit vector of travel for circle c1
p1, p2 the centers of the moving circle c1 and static circle c2
r1, r2 the radius of each circle
The following will set d to the distance c1 must travel along dx, dy to collide with c2 if no collision the d will be set to Infinity
There are three cases when there is no solution
The moving circle is moving away from the static circle. u < 0
The moving circle never gets close enough to collide. dSq > rSq
The two circles are already overlapping. u < 0 luckily the math makes
this the same condition as moving away.
Note that if you ignore the sign of u (1 and 3) then d will be the distance to first (causal) contact going backward in time
Thus the pseudo code to find d
d = Infinity
rSq = (r1 + r2) ^ 2
u = (p1.x - p2.x) * dx + (p1.x - p2.x) * dy
if u >= 0
dSq = ((p2.x + dx * u) - p1.x) ^ 2 + ((p2.y + dy * u) - p1.y) ^ 2
if dSq <= rSq
d = u - (rSq - dSq) ^ 0.5
The point of contact can be found with
cpx = p1.x + dx * d;
cpy = p1.x + dy * d;
Diagram 1
I've created a mesh cone:
With points that are oriented as such:
p3
/
/
p1 ---- p2
I've calculate u = p2 - p1, v = p3 - p1
and then take the cross product u X v and normalize the vector before taking its perspective projection. But no matter which way I calculate u and v (e.g. u = p2 - p1, v = p3 - p2) my normals always face towards the inside of the cone.
I've looked at many other examples on here but none of them seems to work.
Any suggestions?
Edit - This is what I'm getting:
I've run the numbers through a vector calculator and all the calculations are correct.
One of the planes:
P1x: 0.866025
P1y: -0.500000
P1z: 9.000000
P2x: 0.939693
P2y: -0.342020
P2z: 9.000000
P3x: 0.000000
P3y: -0.000000
P3z: 10.000000
Vector U: 0.073667, 0.157980, 0.000000
Vector V: -0.866025, 0.500000, 1.000000
Normal: 0.157980, -0.073667, 0.173648
My code for calculating u, v, and n is:
//u = p2 - p1
vecU.m[1][1] = p2.m[1][1] - p1.m[1][1];
vecU.m[2][1] = p2.m[2][1] - p1.m[2][1];
vecU.m[3][1] = p1.m[3][1] - p1.m[3][1];
vecU.m[4][1] = 1;
printf("Vector U: %f, %f, %f\n", vecU.m[1][1], vecU.m[2][1], vecU.m[3][1]);
//v = p3 - p1
vecV.m[1][1] = p3.m[1][1] - p1.m[1][1];
vecV.m[2][1] = p3.m[2][1] - p1.m[2][1];
vecV.m[3][1] = p3.m[3][1] - p1.m[3][1];
vecV.m[4][1] = 1;
printf("Vector V: %f, %f, %f\n", vecV.m[1][1], vecV.m[2][1], vecV.m[3][1]);
//n = u X v
vecN.m[1][1] = (vecU.m[2][1] * vecV.m[3][1]) - (vecU.m[3][1] * vecV.m[2][1]);
vecN.m[2][1] = (vecU.m[3][1] * vecV.m[1][1]) - (vecU.m[1][1] * vecV.m[3][1]);
vecN.m[3][1] = (vecU.m[1][1] * vecV.m[2][1]) - (vecU.m[2][1] * vecV.m[1][1]);
vecN.m[4][1] = 1;
The easy answer ist switching u and v in the cross product, but I guess your question is rather something like "how do I determine if i just calculated the normal in the correct direction"? That should be determined by saving vertices counter-clockwise.
After seeing your edit:
I can't see an error in your calculations, but your drawing does not look like normals at all. It's not just about the sign of the normal, but you seem to just always point to the center of the ground surface of your cone.
I can't really say any more without seeing the actual code. Do all of your 'normals' in fact point to a single point?
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How do I calculate velocity for bouncing objects with mass?
f1 = ( (m1-m2)/(m1+m2) )*v1 + ( (m2*2 )/(m1+m2) )*v2
f2 = ( (m1*2 )/(m1+m2) )*v1 + ( (m2-m1)/(m1+m2) )*v2
Doesn't work if both objects collide while heading in the same direction
Any help will be appreiciated
For 1d objects
v1 = (u1(m1-m2) + 2m2u2)/(m1+m2)
v2 = (u2(m2-m1) + 2m1u1)/(m1+m2)
Where v1 and v2 are the velocities after and u1 and u2 are velocities before, with m1 and m2 being the mass of each.
Which is the same but simplified version of the formula you have given. BTW your the code block has an error on line 1 5th var is b2 should be m2
This will give the correct velocities if the bodies are moving in the same direction. All I can think that you may be doing wrong is not including the sign. If two bodies are moving in the same direction then the two velocities input u1 and u2 will have the same sign. For all other collisions they will each have a different sign.
If you are applying the above in a 2D situation you will need to modify the formula. This is the solution for circles where the point of contact is along the angle 'p' between them.
//
// u1 * cos(d1 - p) * (m1 - m2) + 2 * m2 * u2 * cos(d2 - p)
//V1x = -------------------------------------------------------- * cos(p) + u1 * sin(d1 - p) * cos(p + PI / 2)
// m1 + m2
// u1 * cos(d1 - p) * (m1 - m2) + 2 * m2 * u2 * cos(d2 - p)
//V1y = -------------------------------------------------------- * sin(p) + u1 * sin(d1 - p) * sin(p + PI / 2)
// m1 + m2
And do the same for the other object
v1x,v1y are the resulting x,y components of the velocity
u1,u2 is velocity.
m1, m2 is the mass
d1 , d2 us the direction of movement in radians.
p is the angle of contact. This is the angle from the center of first
object to the center of second object and the point of contact is on
the line this angle describes.
I've got a situation in which I have 2 circles (C1 and C2)
and i need to find the line equation for the line that is tangent to both of these circles.
So as far as i'm aware, given a single point (P1) and C2's point and radius it is possible to quite easily get 2 possible points of tangency for C2 and P1 to make 2 line equations. But as i don't have P1, only the knowledge that the point will be one of a possible 2 points on C1, i'm not sure how to calculate this.
I assume it will be something along the lines of getting the 2 tangent line equations of C1 that are equal to the same of C2.
Both circles can have any radius, they could be the same or they could be hugely different. They will also never overlap (they can still touch though). And I'm looking for the 2 possible internal tangents.
Oh, and also, visuals would be very helpful haha :)
Let O be the intersection point between the line through the centers and the tangent.
Let d be the distance between the centers and h1, h2 be the distances between O and the centers. By similarity, these are proportional to the radii.
Hence,
h1 / h2 = r1 / r2 = m,
h1 + h2 = d,
giving
h1 = m d / (1 + m),
h2 = d / (1 + m).
Then the coordinates of O are found by interpolating between the centers
xo = (h2.x1 + h1.x2) / d
yo = (h2.y1 + h1.y2) / d
and the angle of the tangent is that of the line through the centers plus or minus the angle between this line and the tangent,
a = arctan((y2 - y1)/(x2 - x1)) +/- arcsin(r1 / h1).
You can write the implicit equation of the tangent as
cos(a).y - sin(a).x = cos(a).yo - sin(a).xo.
(source: imag.fr)
So we are going to use a homothetic transformation. If the circles C and C' have respectively centres O and O', and radius r and r', then we know there exists a unique homothetic transformation with centre J and ratio a, such that :
a = |JO|/|JO'| = r/r'
Noting AB is the vector from A to B, and |z| the norm of a vector z.
Hence you get J, knowing that it is between O and O' which we both already know.
Then with u the projection of JR on JO', and v the decomposition on its orthogonal, and considering the sine s and cosine c of the angle formed by O'JR, we have
|u| = |JR| * c
|v| = |JR| * s
c^2 + s^2 = 1
And finally because the triangle JRO' is right-angled in R :
s = r' / |JO|'
Putting all of this together, we get :
J = O + OO' / |OO'| * a / (a+1)
if |OJ| == r and |O'J| == r' then
return the orthogonal line to (OO') passing through J
|JR| = √( |JO'|^ - r'^2 )
s = r' / |JO'|
c = √( 1 - s^2 )
u = c * |JR| * OO' / |OO'|
w = (-u.y, u.x) % any orthogonal vector to u
v = s * |JR| * w / |w|
return lines corresponding to parametric equations J+t*(u+v) and J+t*(u-v)
How can I find the line of intersection between two planes?
I know the mathematics idea, and I did the cross product between the the planes normal vectors
but how to get the line from the resulted vector programmatically
The equation of the plane is ax + by + cz + d = 0, where (a,b,c) is the plane's normal, and d is the distance to the origin. This means that every point (x,y,z) that satisfies that equation is a member of the plane.
Given two planes:
P1: a1x + b1y + c1z + d1 = 0
P2: a2x + b2y + c2z + d2 = 0
The intersection between the two is the set of points that verifies both equations. To find points along this line, you can simply pick a value for x, any value, and then solve the equations for y and z.
y = (-c1z -a1x -d1) / b1
z = ((b2/b1)*(a1x+d1) -a2x -d2)/(c2 - c1*b2/b1)
If you make x=0, this gets simpler:
y = (-c1z -d1) / b1
z = ((b2/b1)*d1 -d2)/(c2 - c1*b2/b1)
Finding the line between two planes can be calculated using a simplified version of the 3-plane intersection algorithm.
The 2'nd, "more robust method" from bobobobo's answer references the 3-plane intersection.
While this works well for 2 planes (where the 3rd plane can be calculated using the cross product of the first two), the problem can be further reduced for the 2-plane version.
No need to use a 3x3 matrix determinant,instead we can use the squared length of the cross product between the first and second plane (which is the direction of the 3'rd plane).
No need to include the 3rd planes distance,(calculating the final location).
No need to negate the distances.Save some cpu-cycles by swapping the cross product order instead.
Including this code-example, since it may not be immediately obvious.
// Intersection of 2-planes: a variation based on the 3-plane version.
// see: Graphics Gems 1 pg 305
//
// Note that the 'normal' components of the planes need not be unit length
bool isect_plane_plane_to_normal_ray(
const Plane& p1, const Plane& p2,
// output args
Vector3f& r_point, Vector3f& r_normal)
{
// logically the 3rd plane, but we only use the normal component.
const Vector3f p3_normal = p1.normal.cross(p2.normal);
const float det = p3_normal.length_squared();
// If the determinant is 0, that means parallel planes, no intersection.
// note: you may want to check against an epsilon value here.
if (det != 0.0) {
// calculate the final (point, normal)
r_point = ((p3_normal.cross(p2.normal) * p1.d) +
(p1.normal.cross(p3_normal) * p2.d)) / det;
r_normal = p3_normal;
return true;
}
else {
return false;
}
}
Adding this answer for completeness, since at time of writing, none of the answers here contain a working code-example which directly addresses the question.
Though other answers here already covered the principles.
Finding a point on the line
To get the intersection of 2 planes, you need a point on the line and the direction of that line.
Finding the direction of that line is really easy, just cross the 2 normals of the 2 planes that are intersecting.
lineDir = n1 × n2
But that line passes through the origin, and the line that runs along your plane intersections might not. So, Martinho's answer provides a great start to finding a point on the line of intersection (basically any point that is on both planes).
In case you wanted to see the derivation for how to solve this, here's the math behind it:
First let x=0. Now we have 2 unknowns in 2 equations instead of 3 unknowns in 2 equations (we arbitrarily chose one of the unknowns).
Then the plane equations are (A terms were eliminated since we chose x=0):
B1y + C1z + D1 = 0
B2y + C2z + D2 = 0
We want y and z such that those equations are both solved correctly (=0) for the B1, C1 given.
So, just multiply the top eq by (-B2/B1) to get
-B2y + (-B2/B1)*C1z + (-B2/B1)*D1 = 0
B2y + C2z + D2 = 0
Add the eqs to get
z = ( (-B2/B1)*D1 - D2 ) / (C2 * B2/B1)*C1)
Throw the z you find into the 1st equation now to find y as
y = (-D1 - C1z) / B1
Note the best variable to make 0 is the one with the lowest coefficients, because it carries no information anyway. So if C1 and C2 were both 0, choosing z=0 (instead of x=0) would be a better choice.
The above solution can still screw up if B1=0 (which isn't that unlikely). You could add in some if statements that check if B1=0, and if it is, be sure to solve for one of the other variables instead.
Solution using intersection of 3 planes
From user's answer, a closed form solution for the intersection of 3 planes was actually in Graphics Gems 1. The formula is:
P_intersection = (( point_on1 • n1 )( n2 × n3 ) + ( point_on2 • n2 )( n3 × n1 ) + ( point_on3 • n3 )( n1 × n2 )) / det(n1,n2,n3)
Actually point_on1 • n1 = -d1 (assuming you write your planes Ax + By + Cz + D=0, and not =-D). So, you could rewrite it as:
P_intersection = (( -d1 )( n2 × n3 ) + ( -d2 )( n3 × n1 ) + ( -d3 )( n1 × n2 )) / det(n1,n2,n3)
A function that intersects 3 planes:
// Intersection of 3 planes, Graphics Gems 1 pg 305
static Vector3f getIntersection( const Plane& plane1, const Plane& plane2, const Plane& plane3 )
{
float det = Matrix3f::det( plane1.normal, plane2.normal, plane3.normal ) ;
// If the determinant is 0, that means parallel planes, no intn.
if( det == 0.f ) return 0 ; //could return inf or whatever
return ( plane2.normal.cross( plane3.normal )*-plane1.d +
plane3.normal.cross( plane1.normal )*-plane2.d +
plane1.normal.cross( plane2.normal )*-plane3.d ) / det ;
}
Proof it works (yellow dot is intersection of rgb planes here)
Getting the line
Once you have a point of intersection common to the 2 planes, the line just goes
P + t*d
Where P is the point of intersection, t can go from (-inf, inf), and d is the direction vector that is the cross product of the normals of the two original planes.
The line of intersection between the red and blue planes looks like this
Efficiency and stability
The "robust" (2nd way) takes 48 elementary ops by my count, vs the 36 elementary ops that the 1st way (isolation of x,y) uses. There is a trade off between stability and # computations between these 2 ways.
It'd be pretty catastrophic to get (0,inf,inf) back from a call to the 1st way in the case that B1 was 0 and you didn't check. So adding in if statements and making sure not to divide by 0 to the 1st way may give you the stability at the cost of code bloat, and the added branching (which might be quite expensive). The 3 plane intersection method is almost branchless and won't give you infinities.
This method avoids division by zero as long as the two planes are not parallel.
If these are the planes:
A1*x + B1*y + C1*z + D1 = 0
A2*x + B2*y + C2*z + D2 = 0
1) Find a vector parallel to the line of intersection. This is also the normal of a 3rd plane which is perpendicular to the other two planes:
(A3,B3,C3) = (A1,B1,C1) cross (A2,B2,C2)
2) Form a system of 3 equations. These describe 3 planes which intersect at a point:
A1*x1 + B1*y1 + C1*z1 + D1 = 0
A2*x1 + B2*y1 + C2*z1 + D2 = 0
A3*x1 + B3*y1 + C3*z1 = 0
3) Solve them to find x1,y1,z1. This is a point on the line of intersection.
4) The parametric equations of the line of intersection are:
x = x1 + A3 * t
y = y1 + B3 * t
z = z1 + C3 * t
The determinant-based approach is neat, but it's hard to follow why it works.
Here's another way that's more intuitive.
The idea is to first go from the origin to the closest point on the first plane (p1), and then from there go to the closest point on the line of intersection of the two planes. (Along a vector that I'm calling v below.)
Given
=====
First plane: n1 • r = k1
Second plane: n2 • r = k2
Working
=======
dir = n1 × n2
p1 = (k1 / (n1 • n1)) * n1
v = n1 × dir
pt = LineIntersectPlane(line = (p1, v), plane = (n2, k2))
LineIntersectPlane
==================
#We have n2 • (p1 + lambda * v) = k2
lambda = (k2 - n2 • p1) / (n2 • v)
Return p1 + lambda * v
Output
======
Line where two planes intersect: (pt, dir)
This should give the same point as the determinant-based approach. There's almost certainly a link between the two. At least the denominator, n2 • v, is the same, if we apply the "scalar triple product" rule. So these methods are probably similar as far as condition numbers go.
Don't forget to check for (almost) parallel planes. For example: if (dir • dir < 1e-8) should work well if unit normals are used.
You can find the formula for the intersection line of two planes in this link.
P1: a1x + b1y + c1z = d1
P2: a2x + b2y + c2z = d2
n1=(a1,b1,c1); n2=(a2,b2,c2); n12=Norm[Cross[n1,n2]]^2
If n12 != 0
a1 = (d1*Norm[n2]^2 - d2*n1.n2)/n12;
a2 = (d2*Norm[n1]^2 - d1*n1.n2)/n12;
P = a1 n1 + a2 n2;
(*formula for the intersection line*)
Li[t_] := P + t*Cross[n1, n2];
The cross product of the line is the direction of the intersection line. Now you need a point in the intersection.
You can do this by taking a point on the cross product, then subtracting Normal of plane A * distance to plane A and Normal of plane B * distance to plane b. Cleaner:
p = Point on cross product
intersection point = ([p] - ([Normal of plane A] * [distance from p to plane A]) - ([Normal of plane B] * [distance from p to plane B]))
Edit:
You have two planes with two normals:
N1 and N2
The cross product is the direction of the Intersection Line:
C = N1 x N2
The class above has a function to calculate the distance between a point and a plane. Use it to get the distance of some point p on C to both planes:
p = C //p = 1 times C to get a point on C
d1 = plane1.getDistance(p)
d2 = plane2.getDistance(p)
Intersection line:
resultPoint1 = (p - (d1 * N1) - (d2 * N2))
resultPoint2 = resultPoint1 + C