Finding the intersection of the Circle and Infinite Cylinder. (all in 3D)
• Circle is defined by center, plane in which it lies and radius.
• Cylinder is defined by axis and radius.
how can i get the intersection of these two?
WLOG the cylinder has equation X² + Y² = 1 (if not, you can make it so by translation, rotation and scaling).
Then the parametric equation of the circle is
P = C + U cos t + V sin t
where C is the center point and U, V two orthogonal vectors in the circle plane, of length R.
You can rationalize with the substitution cos t = (1 - u²) / (1 + u²), sin t = 2u / (1 + u²).
Combining these equations,
(Cx (1 + u²) + Ux (1 - u²) + Vx 2u)² + (Cy (1 + u²) + Uy (1 - u²) + Vy 2u)² = (1 + u²)²
which is a quartic one. There is no particular simplification of the coefficients.
You can solve numerically or by the closed-form formulas. There can be up to four solutions.
I guess that this is strictly equivalent to finding the intersections between the torus formed by inflating the circle circumference and the straight line obtained by deflating the cylinder to its axis. This is well addressed in the ray-tracing literature.
You can also sse it as a circle/ellipse intersection problem in 2D.
Let's some base point of cylinder is A0, unit axis direction vector is AD, radius AR. Circle center is B0, circle plane unit normal is BN, radius BR.
Circle intersects cylinder, if distance from B0 to cyl. axis is smaller than sum of cylinder radius plus projection of circle radius onto normal to axis
Dist <= AR + BR * Abs(Cos(AD, BN)).
Cos(AD, BN) = DotProduct(AD, BN)
Distance(B0, cyl.axis) = Abs(VectorProduct(AD, B0-A0))
Related
I have the green x,y points, how would I get the missing red?
You can rotate the two known points of 90° around their midpoint.
In pseudo code:
// Evaluate the midpoint from the coordinates of points a and b,
h_x = (b_x - a_x) / 2;
h_y = (b_y - a_y) / 2;
m_x = a_x + h_x;
m_y = a_y + h_y;
// Apply a rotation of 90 degree around the midpoint to find c and d
c_x = m_x - h_y;
c_y = m_y + h_x;
d_x = m_x + h_y;
d_y = m_y - h_x;
This result can be formally derived in terms of homogeneous coordinates and transfomation matrices.
The midpoint m, expressed in homogeneous coordinates, can be calculated as
To rotate a vector around the origin of an angle α, we apply a rotation matrix like
If another center of rotation is needed (the midpoint, in our case), we need to translate from the original position to the origin, apply the rotation and translate back again. The translation matrices are
The complete transformation can be expressed as
Where
So that we can evaluate, let's say d, with
Q.e.d.
I have cone->p (vertex of the cone), cone->orient (axis vector), cone->k (half-angle tangent), cone->minm and cone->maxm (2 height values, for cone caps). Also I have point intersection which is on the cone. How do I find the cone (side surface) normal vector at intersection point using only these parameters?
Сame up with simpler method:
Find distance Dis from intersection point I to base P
Make axis orientation vector of length
D = Dis * sqrt(1+k^2)
and make point on axis at this distance
A = P + Normalized(Orient) * D
Now
Normal = I - A
Old answer:
Make orthogonal projection of point I (intersection) onto cone axis using vector `IP = I - P' and scalar (dot) product:
AxProj = P + Orient * dot(IP, Orient) / dot(Orient, Orient)
Vector from AxPr to I (perpendicular to axis):
AxPerp = I - AxProj
Vector, tangent to cone surface, using vector product:
T = IP x AxPerp
Vector, normal to cone surface:
N = T x IP
If I is the intersection point on the cone's surface and you know its coordinates, and P is the vertex of the cone, whose coordinates you also know, then this is enough:
Normal = (axis x PI) x PI
Normal = Normal / norm(Normal)
where axis is the vector aligned with the axis of the cone.
I have two normalized vectors:
A) 0,0,-1
B) 0.559055,0.503937,0.653543
I want to know, what rotations about the axes would it take to take the vector at 0,0,-1 to 0.559055,0.503937,0.653543?
How would I calculate this? Something like, rotate over X axis 40 degrees and Y axis 220 (that's just example, but I don't know how to do it).
Check this out. (google is a good thing)
This calculates the angle between two vectors.
If Vector A is (ax, ay, az) and
Vector B is (bx, by, bz), then
The cos of angle between them is:
(ax*bx + ay*by + az*bz)
--------------------------------------------------------
sqrt(ax*ax + ay*ay + az*az) * sqrt(bx*bx + by*by + bz*bz)
To calculate the angle between the two vectors as projected onto the x-y plane, just ignore the z-coordinates.
Cosine of Angle in x-y plane =
(ax*bx + ay*by)
--------------------------------------
sqrt(ax*ax + ay*ay) * sqrt(bx*bx + by*by
Similarly, to calculate the angle between the projections of the two vectors in the x-z plane, ignore the y-coordinates.
It sounds like you're trying convert from Cartesian coordinates (x,y,z) into spherical coordinates (rho,theta,psi).
Since they're both unit vectors, rho, the radius, will be 1. This means your magnitudes will also be 1 and you can skip the whole denominator and just use the dot-product.
Rotating in the X/Y plane (about the Z axis) will be very difficult with your first example (0,0,-1) because it has no extension in X or Y. So there's nothing to rotate.
(0,0,-1) is 90 degrees from (1,0,0) or (0,1,0). If you take the x-axis to be the 0-angle for theta, then you calculate the phi (rotation off of the X/Y plane) by applying the inverse cos upon (x,y,z) and (x,y,0), then you can skip dot-products and get theta (the x/y rotation) with atan2(y,x).
Beware of gimbal lock which may cause problems.
If i have a point (x,y,z) how to project it on to a sphere(x0,y0,z0,radius) (on its surface).
My input will be the coordinates of point and sphere.
The output should be the coordinates of the projected point on sphere.
Just convert from cartesian to spherical coordinates?
For the simplest projection (along the line connecting the point to the center of the sphere):
Write the point in a coordinate system centered at the center of the sphere (x0,y0,z0):
P = (x',y',z') = (x - x0, y - y0, z - z0)
Compute the length of this vector:
|P| = sqrt(x'^2 + y'^2 + z'^2)
Scale the vector so that it has length equal to the radius of the sphere:
Q = (radius/|P|)*P
And change back to your original coordinate system to get the projection:
R = Q + (x0,y0,z0)
Basically you want to construct a line going through the spheres centre and the point. Then you intersect this line with the sphere and you have your projection point.
In greater detail:
Let p be the point, s the sphere's centre and r the radius then x = s + r*(p-s)/(norm(p-s)) where x is the point you are looking for. The implementation is left to you.
I agree that the spherical coordinate approach will work as well but is computationally more demanding. In the above formula the only non-trivial operation is the square root for the norm.
It works if you set the coordinates of the center of the sphere as origin of the system (x0, y0, z0). So you will have the coordinates of the point referred to that origin (Xp', Yp', Zp'), and converting the coordinates to polar, you discard the radius (distance between the center of the sphere and the point) and the angles will define the projection.
Imagine a circle. Imagine a pie. Imagine trying to return a bool that determines whether the provided parameters of X, Y are contained within one of those pie pieces.
What I know about the arc:
I have the CenterX, CenterY, Radius, StartingAngle, EndingAngle, StartingPoint (point on circumference), EndingPoint (point on circumference).
Given a coordinate of X,Y, I'd like to determine if this coordinate is contained anywhere within the pie slide.
Check:
The angle from the centerX,centerY through X,Y should be between start&endangle.
The distance from centerX,centerY to X,Y should be less then the Radius
And you'll have your answer.
I know this question is old but none of the answers consider the placement of the arc on the circle.
This algorithm considers that all angles are between 0 and 360, and the arcs are drawn in positive mathematical direction (counter-clockwise)
First you can transform to polar coordinates: radius (R) and angle (A). Note: use Atan2 function if available. wiki
R = sqrt ((X - CenterX)^2 + (Y - CenterY)^2)
A = atan2 (Y - CenterY, X - CenterX)
Now if R < Radius the point is inside the circle.
To check if the angle is between StartingAngle (S) and EndingAngle (E) you need to consider two possibilities:
1) if S < E then if S < A < E the point lies inside the slice
2) if S > E then there are 2 possible scenarios
if A > S
then the point lies inside the slice
if A < E
then the point lies inside the slice
In all other cases the point lies outside the slice.
Convert X,Y to polar coordinates using this:
Angle = arctan(y/x);
Radius = sqrt(x * x + y * y);
Then Angle must be between StartingAngle and EndingAngle, and Radius between 0 and your Radius.
You have to convert atan2() to into 0-360 before making comparisons with starting and ending angles.
(A > 0 ? A : (2PI + A)) * 360 / (2PI)