What is called a pure rotation in computer graphics? Does it mean the transformation applied on any matrix is only rotation?
If so, then how the angles between intersecting lines are preserved by pure rotation?
From a geometric perspective, angles are of course preserved.
From a mathematical prospective, you can show both length and dot product are unchanged under rotation. This means that if you have three points A, B, and C and let Ax, Bx, Cx be the location of these points after a rotation:
The distance from A to B is the same as the distance from Ax to Bx.
If you let V be the vector from A to C and and W be the vector from B to C, then V·W is the same as Vx·Wx.
This is just painful math.
Since the dot product of two vectors is the cosine of the angle between them times the length of each of the two vectors, you have that rotations are angle preserving.
Related
I need to offset a curve, which by the simplest way is just shifting the points perpendicularly. I can access each point to calculate angle of each line along given path, for now I use atan2. Then I take those two angle and make average of it. It returns the shortest angle, not what I need in this case.
How can I calculate angle of each connection? Concerning that I am not interested in the shortest angle but the one that would create parallel offset curve.
Assuming 2D case...
So do a cross product of direction vectors of 2 neighboring lines the sign of z coordinate of the result will tell you if the lines are CW/CCW
So if you got 3 consequent control points on the polyline: p0,p1,p2 then:
d1 = p1-p0
d2 = p2-p1
if you use some 3D vector math then convert them to 3D by setting:
d1.z=0;
d2.z=0;
now compute 3D cross:
n = cross(d1,d2)
which returns vector perpendicular to both vectors of size equals to the area of quad (parallelogram) constructed with d1,d2 as base vectors. The direction (from the 2 possible) is determined by the winding rule of the p0,p1,p2 so inspecting z of the result is enough.
The n.x,n.y are not needed so you can compute directly without doing full cross product:
n.z=(d1.x*d2.y)-(d1.y*d2.x)
if (n.z>0) case1
if (n.z<0) case2
if the case1 is CW or CCW depends on your coordinate system properties (left/right handness). This approach is very commonly used in CG fur back face culling of polygons ...
if n.z is zero it means that your vectors/lines are either parallel or at lest one of them is zero.
I think these might interest you:
draw outline for some connected lines
How can I create an internal spiral for a polygon?
Also in 2D you do not need atan2 to get perpendicular vector... You can do instead this:
u = (x,y)
v = (-y,x)
w = (x,-y)
so u is any 2D vector and v,w are the 2 possible perpendicular vectors to u in 2D. they are the result of:
cross((x,y,0),(0,0,1))
cross((0,0,1),(x,y,0))
I have two 3D points (x,y,z), namely A and B and a bunch of other 3D points. Point A is at (0,0,0).
I would like to set point B to (0,0,0) so that all other points including A and B are translated and rotated in a way that is appropriate (so that A is no longer at (0,0,0)).
I know that there are some translations and rotations involved, but nothing more than that.
UPGRADE:
Point B is also constrained by three vectors: x', y', z' that represent x, y, and z axis of B's coordinate system. I think these should be somehow considered for the rotation part.
As you have given two points, one (A) at the origin and one (B) somewhere else, and you want to shift (translate) B to the origin, I don't see the necessity for any rotation.
If you don't have any other contraints, just shift all coordinates by the initial coordinates of B.
You can construct a transformation matrix as given, e.g., https://en.wikipedia.org/wiki/Transformation_matrix#Affine_transformations for 2D, but if you simply translate, R' = R + T, where R' is the vector after transformation, R the vector before and T the translation vector.
For more general transformations including rotations, you have to specify the rotation angle and axis. Then, you can come up with more general transformation, see above link.
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.
I have a shape made out of several triangles which is positioned somewhere in world space with scale, rotate, translate. I also have a plane on which I would like to project (orthogonal) the shape.
I could multiply every vertex of every triangle in the shape with the objects transformation matrix to find out where it is located in world coordinates, and then project this point onto the plane.
But I don't need to draw the projection, and instead I would like to transform the plane with the inverse transformation matrix of the shape, and then project all the vertices onto the (inverse transformed) plane. Since it only requires me to transform the plane once and not every vertex.
My plane has a normal (xyz) and a distance (d). How do I multiply it with a 4x4 transformation matrix so that it turns out ok?
Can you create a vec4 as xyzd and multiply that? Or maybe create a vector xyz1 and then what to do with d?
You need to convert your plane to a different representation. One where N is the normal, and O is any point on the plane. The normal you already know, it's your (xyz). A point on the plane is also easy, it's your normal N times your distance d.
Transform O by the 4x4 matrix in the normal way, this becomes your new O. You will need a Vector4 to multiply with a 4x4 matrix, set the W component to 1 (x, y, z, 1).
Also transform N by the 4x4 matrix, but set the W component to 0 (x, y, z, 0). Setting the W component to 0 means that your normals won't get translated. If your matrix is composed of more that just translating and rotating, then this step isn't so simple. Instead of multiplying by your transformation matrix, you have to multiply by the transpose of the inverse of the matrix i.e. Matrix4.Transpose(Matrix4.Invert(Transform)), there's a good explanation on why here.
You now have a new normal vector N and a new position vector O. However I suppose you want it in xyzd form again? No problem. As before, xyz is your normal N all that's left is to calculate d. d is the distance of the plane from the origin, along the normal vector. Hence, it is simply the dot product of O and N.
There you have it! If you tell me what language you're doing this in, I'd happily type it up in code as well.
EDIT, In pseudocode:
The plane is vector3 xyz and number d, the matrix is a matrix4x4 M
vector4 O = (xyz * d, 1)
vector4 N = (xyz, 0)
O = M * O
N = transpose(invert(M)) * N
xyz = N.xyz
d = dot(O.xyz, N.xyz)
xyz and d represent the new plane
This question is a bit old but I would like to correct the accepted answer.
You do not need to convert your plane representation.
Any point lies on the plane if
It can be written as dot product :
You are looking for the plane transformed by your 4x4 matrix .
For the same reason, you must have
So and with some arrangements
TLDR : if p=(a,b,c,d), p' = transpose(inverse(M))*p
Notation:
n is a normal represented as a (1x3) row-vector
n' is the transformed normal of n according to transform matrix T
(n|d) is a plane represented as a (1x4) row-vector (with n the plane's normal and d the plane's distance to the origin)
(n'|d') is the transformed plane of (n|d) according to transform matrix T
T is a (4x4) (affine) column-major transformation matrix (i.e. transforming a column-vector t is defined as t' = T t).
Transforming a normal n:
n' = n adj(T)
Transforming a plane (n|d):
(n'|d') = (n|d) adj(T)
Here, adj is the adjugate of a matrix which is defined as follows in terms of the inverse and determinant of a matrix:
T^-1 = adj(T)/det(T)
Note:
The adjugate is generally not equal to the inverse of a transformation matrix T. If T includes a reflection, det(T) = -1, reversing the winding order!
Re-normalizing n' is mathematically not required (but maybe numerically depending on the implementation) since scaling is taken care off by the determinant. Thanks to Adrian Leonhard.
You can directly transform the plane without first decomposing and recomposing a plane (normal and point).
I have gone through all available study resources in the internet as much as possible, which are in form of simple equations, vectors or trigonometric equations.
I couldn't find the way of doing following thing:
Assuming Y is up in a 3D world.
I need to draw two 2D trajectories orthogonally (not the projections) for a 3D trajectory, say XY-plane for side view of the trajectory w.r.t. the trajectory itself and XZ-plane for top view for the same.
I have all the 3D points of the 3D trajectory, initial velocity, both the angles can be calculated by vector mathematics.
How should I proceed further?
refer:
Below a curve in different angles, which can loose its significance if projected along XY-plane. All I want is to convert the red curve along itself, the green curve along green curve and so on. and further how would I map side view to a plane. Top view is comparatively easy and done just by taking X and Z ordinates of each points.
I mean this the requirement. :)
I don't think I understand the question, but I'll answer my interpretation anyway.
You have a 3D trajectory described by a sequence of points p0, ..., pN. We are given an angle v for a plane P parallel to the Y-axis, and wish to compute the 2D coordinates (di, hi) of the points pi projected onto that plane, where hi is the height coordinate in the direction Y and di is the distance coordinate in the direction v. Assume p0 = (0, 0, 0) or else subtract p0 from all vectors.
Let pi = (xi, yi, zi). The height coordinate is hi = yi. Assume the angle v is given relative to the Z-axis. The vector for the direction v is then r = (sin(v), 0, cos(v)), and the distance coordinates becomes di = dot(pi, r).