As show in the image below, given the point O, A, B and the α, how to get the position of point P in the 3D-space?
Thanks advance!
Let's vectors
b = B - O
a = A - O
p = P - O
It seems that the simplest way is using of SLERP interpolation between vectors b and a.
At first find angle Omega between vectors b and a (for example, through dot product), then apply SLERP
p = Sin(Omega - Alpha) * b / Sin(Omega) + Sin(Alpha) * a / Sin(Omega)
Alternative way - you can find vector of rotation axis (normal to the circle plane) as
k = b x a //vector product
k = k.Normalized //unit vector
and then use Rodrigues' rotation formula to get p=P-O vector
Related
At the chapter 3 of "Eric Lengyel, Foundations Of Game Engine Development, Volume 1: Mathematics" i can not understand, how to solve exercise 10 ( on the "3.5 Plucker Coordinates" theme).
"Let {v|m1} and {v|m2} be parallel lines having the same direction but different moments. Find a formula for the distance d between theese two lines by considering the triangle formed by the origin and the closest point to the origin on each line"
It seemed simple to me at first, but after many attempts I can't figure out where to start. I've tried different ways, but couldn't get the right answer:
d = (| (v x (m2 - m1) |) / (sqr(v))
maybe someone know, what's the problem.
Thanks!
Table in that chapter (line G) shows "homogeneous point closest to the origin" as
p = (v x m) / v^2
Difference between such points for two lines with m1 and m2 moments is vector
diff = p1-p2 = (v x m1) / v^2 - (v x m2) / v^2 = (v x (m1-m2)) / v^2
Length of this vector is needed distance, because Op1 and Op2 vectors are perpendicular to direction v, points O, p1 and p2 belongs to a plane, perpendicular to lines, and diff=p1-p2 belonging that plane is perpendicular to the lines too, hence it's length is the distance between lines.
P.S. Op1p2 is triangle mentioned in the problem statement
P.P.S
Point p lies on Pluecker line if
p x v = m
multiply both parts by v
v x (p x v) = v x m
transform the left part by Lagrange identity
p * (v.dot.v) - v * (p.dot.v) = p * v^2
(because (p.dot.v)== 0 for perpendicular to the line)
p * v^2 = v x m
p = (v x m) / v^2
I have a normalized direction vector (from a 3d position to a light position) and I would like this vector to be rotated by some angle so I can create a "cone".
Id like to simulate cone tracing by using the direction vector as the center of the cone and create an X number of samples to create more rays to sample from.
What I would like to know is basically the math behind:
https://docs.unrealengine.com/latest/INT/BlueprintAPI/Math/Random/RandomUnitVectorinCone/index.html
Which seems to do exactly what Im looking for.
1) Make arbitrary vector P, perpendicular to your direction vector D.
You can choose component with max magnitude, exchange it with middle-magnitude component, negate it, and make min magnitude component zero.
For example, if z- component is maximal and y-component is minimal, you may make such P:
D = (dx, dy, dz)
p = (-dz, 0, dx)
P = Normalize(p) //unit vector
2) Make vector Q perpendicular both D and P through vector product:
Q = D x P //unit vector
3) Generate random point in the PQ plane disk
RMax = Tan(Phi) //where Phi is cone angle
Theta = Random(0..2*Pi)
r = RMax * Sqrt(Random(0..1))
V = r * (P * Cos(Theta) + Q * Sin(Theta))
4) Normalize vector V
Note that distribution of vectors is slightly non-uniform on the sphere segment.(it is uniform on the plane disk). There are methods to generate uniform distribution on the sphere but some work needed to apply them to segment (my first attempt before edit was wrong).
Edit: Modification to make sphere-uniform distribution (not checked thoroughly)
RMax = Tan(Phi) //where Phi is cone angle
Theta = Random(0..2*Pi)
u = Random(Cos(Phi)..1)
r = RMax * Sqrt(1 - u^2)
V = r * (P * Cos(Theta) + Q * Sin(Theta))
I have a line from (a, b) to (x, y), and I would like to draw a line starting at (x, y), with length ℓ, that makes an angle of θ with the original line.
How do I compute the coordinates of the endpoint of this new line? See the diagram:
It's nearly always simpler to use vector algebra for this kind of thing, rather than Cartesian coordinates. Let's start by labelling the points:
Let R(θ) be the matrix that rotates by θ radians counter-clockwise:
Then compute:
v = B − A (the vector from A to B)
v̂ = v / |v| (the unit vector in the direction of v)
ŵ = R(−θ) v̂ (the unit vector in the direction of BC; your rotation is clockwise, so we need R(−θ) here, not R(θ))
w = ℓ ŵ (the vector of length ℓ in the direction of BC)
C = B + w
This approach avoids the need to compute an arctangent, which would need some care (if done naïvely, it runs into trouble when B is vertically above or below A; but most languages have a function like atan2 for handling this case).
In any sensible programming language with a vector library you should be able to write this as a one-liner, perhaps like this:
C = B + (B - A).unit().rotate(-theta) * l
OK, so after a lot of scribbling, I came up with this:
The dashed lines represent lines parallel to the x- and y-axes.
m = x − a
n = y − b
α = tan−1 (n / m)
β = α − θ
p = ℓ cos β
q = ℓ sin β
c = x + p
d = y + q
So I was reading over something on this page (http://gamedeveloperjourney.blogspot.com/2009/04/point-plane-collision-detection.html)
The author mentioned
d = - D3DXVec3Dot(&vP1, &vNormal);
where vP1 is a point on the plane and vNormal is the normal to the plane. I'm curious as to how this gets you the distance from the world origin since the result will always be 0. In addition, just to be clear (since I'm still kind of hazy on the d part of a plane equation), is d in a plane equation the distance from a line through the world origin to the plane's origin?
In the generic case the distance between a point p and a plane can be computed by
<p - p0, normal>
where <a, b> is the dot product operation
<a, b> = ax*bx + ay*by + az*bz
and where p0 is a point on the plane.
When n is of unity length the dot product between a vector and it is the (signed) length of the projection of the vector on the normal
The formula you are reporting is just the special case when the point p is the origin. In this case
distance = <origin - p0, normal> = - <p0, normal>
This equality is formally wrong because the dot product is about vectors, not points... but still holds numerically. Writing down the explicit formula you get that
(0 - p0.x)*n.x + (0 - p0.y)*n.y + (0 - p0.z)*n.z
is the same as
- (p0.x*n.x + p0.y*n.y + p0.z*n.z)
Indeed a nice way to store a plane is to save the normal n and the value of k = <p0, n> where p0 is any point on the plane (the value of k is independent on which point you choose of the plane).
The result is not always zero. The result will only be zero if the plane goes through the origin. (Here let's assume the plane doesn't go through the origin.)
Basically, you are given a line from the origin to some point on the plane. (I.e. you have a vector from the origin to vP1). The problem with this vector is that most likely it's slanted and going to some far away place on the plane rather than to the closest point on the plane. So, if you simply took the length of vP1 you will get a distance that is too big.
What you need to do is get the projection of vP1 onto some vector that you know is perpendicular to the plane. That of course is vNormal. So take the dot product of vP1 and vNormal, and divide by the length of vNormal and you have the answer. (If they are kind enough to give you a vNormal that already is magnitude one, then no need to divide.)
You can work this out with Lagrange multipliers:
You know that the closest point on the plane must be of the form:
c = p + v
Where c is the closest point and v is a vector along the plane (which is thus orthogonal to n, the normal). You are trying for find the c with the smallest norm (or norm squared). So you are trying to minimized dot(c,c) subject to v being orthogonal to n (thus dot(v,n) = 0).
Thus, set up Lagrangian:
L = dot(c,c) + lambda * ( dot(v,n) )
L = dot(p+v,p+v) + lambda * ( dot(v,n) )
L = dot(p,p) + 2*dot(p,v) + dot(v,v) * lambda * ( dot(v,n) )
And take the derivative with respect to v (and set to 0) to get:
2 * p + 2 * v + lambda * n = 0
You can solve for lambda by in the equation above by dot producting both sides by n to get
2 * dot(p,n) + 2 * dot(v,n) + lambda * dot(n,n) = 0
2 * dot(p,n) + lambda = 0
lambda = - 2 * dot(p,n)
Note again that dot(n,n) = 1 and dot(v,n) = 0 (since v is in the plane and n is orthogonal to it). Then subtitute lambda back in to get:
2 * p + 2 * v - 2 * dot(p,n) * n = 0
and solve for v to get:
v = dot(p,n) * n - p
Then plug this back into c = p + v to get:
c = dot(p,n) * n
The length of this vector is |dot(p,n)| and the sign tells you whether the point is in the direction of the normal vector from the origin, or the reverse direction from the origin.
I tried using a raycasting-style function to do it but can't get any maintainable results. I'm trying to calculate the intersection between two tangents on one circle. This picture should help explain:
I've googled + searched stackoverflow about this problem but can't find anything similar to this problem. Any help?
Well, if your variables are:
C = (cx, cy) - Circle center
A = (x1, y1) - Tangent point 1
B = (x2, y2) - Tangent point 2
The lines from the circle center to the two points A and B are CA = A - C and CB = B - C respectively.
You know that a tangent is perpendicular to the line from the center. In 2D, to get a line perpendicular to a vector (x, y) you just take (y, -x) (or (-y, x))
So your two (parametric) tangent lines are:
L1(u) = A + u * (CA.y, -CA.x)
= (A.x + u * CA.y, A.y - u * CA.x)
L2(v) = B + v * (CB.y, -CB.x)
= (B.x + v * CB.y, B.x - v * CB.x)
Then to calculate the intersection of two lines you just need to use standard intersection tests.
The answer by Peter Alexander assumes that you know the center of the circle, which is not obvious from your figure http://oi54.tinypic.com/e6y62f.jpg.
Here is a solution without knowing the center:
The point C (in your figure) is the intersection of the tangent at A(x, y) with the line L perpendicular to AB, cutting AB into halves. A parametric equation for the line L can be derived as follows:
The middle point of AB is M = ((x+x2)/2, (y+y2)/2), where B(x2, y2). The vector perpendicular to AB is N = (y2-y, x-x2). The vector equation of the line L is hence
L(t) = M + t N, where t is a real number.