Related
This is a big one for any math/3d geometry lovers. Thank you in advance.
Overview
I have a figure created by extruding faces around twisting spline curves in space. I'm trying to place a "loop" (torus) oriented along the spline path at a given segment of the curve, so that it is "aligned" with the spline. By that I mean the torus's width is parallel to the spline path at the given extrusion segment, and it's height is perpendicular to the face that is selected (see below for picture).
Data I know:
I am given one of the faces of the figure. From that I can also glean that face's centroid (center point), the vertices that compose it, the surrounding faces, and the normal vector of the face.
Current (Non-working) solution outcome:
I can correctly create a torus loop around the centroid of the face that is clicked. However, it does not rotate properly to "align" with the face. See how they look a bit "off" below.
Here's a picture with the material around it:
and here's a picture with it in wireframe mode. You can see the extrusion segments pretty clearly.
Current (Non-working) methodology:
I am attempting to do two calculations. First, I'm calculating the the angle between two planes (the selected face and the horizontal plane at the origin). Second, I'm calculating the angle between the face and a vertical plane at the point of origin. With those two angles, I am then doing two rotations - an X and a Y rotation on the torus to what I hope would be the correct orientation. It's rotating the torus at a variable amount, but not in the place I want it to be.
Formulas:
In doing the above, I'm using the following to calculate the angle between two planes using their normal vectors:
Dot product of normal vector 1 and normal vector 2 = Magnitude of vector 1 * Magnitude of vector 2 * Cos (theta)
Or:
(n1)(n2) = || n1 || * || n2 || * cos (theta)
Or:
Angle = ArcCos { ( n1 * n2 ) / ( || n1 || * || n2 || ) }
To determine the magnitude of a vector, the formula is:
The square root of the sum of the components squared.
Or:
Sqrt { n1.x^2 + n1.y^2 + n1.z^2 }
Also, I'm using the following for the normal vectors of the "origin" planes:
Normal vector of horizontal plane: (1, 0, 0)
Normal vector of Vertical plane: (0, 1, 0)
I've thought through the above normal vectors a couple times... and I think(?) they are right?
Current Implementation:
Below is the code that I'm currently using to implement it. Any thoughts would be much appreciated. I have a sinking feeling that I'm taking a wrong approach in trying to calculate the angles between the planes. Any advice / ideas / suggestions would be much appreciated. Thank you very much in advance for any suggestions.
Function to calculate the angles:
this.toRadians = function (face, isX)
{
//Normal of the face
var n1 = face.normal;
//Normal of the vertical plane
if (isX)
var n2 = new THREE.Vector3(1, 0, 0); // Vector normal for vertical plane. Use for Y rotation.
else
var n2 = new THREE.Vector3(0, 1, 0); // Vector normal for horizontal plane. Use for X rotation.
//Equation to find the cosin of the angle. (n1)(n2) = ||n1|| * ||n2|| (cos theta)
//Find the dot product of n1 and n2.
var dotProduct = (n1.x * n2.x) + (n1.y * n2.y) + (n1.z * n2.z);
// Calculate the magnitude of each vector
var mag1 = Math.sqrt (Math.pow(n1.x, 2) + Math.pow(n1.y, 2) + Math.pow(n1.z, 2));
var mag2 = Math.sqrt (Math.pow(n2.x, 2) + Math.pow(n2.y, 2) + Math.pow(n2.z, 2));
//Calculate the angle of the two planes. Returns value in radians.
var a = (dotProduct)/(mag1 * mag2);
var result = Math.acos(a);
return result;
}
Function to create and rotate the torus loop:
this.createTorus = function (tubeMeshParams)
{
var torus = new THREE.TorusGeometry(5, 1.5, segments/10, 50);
fIndex = this.calculateFaceIndex();
//run the equation twice to calculate the angles
var xRadian = this.toRadians(geometry.faces[fIndex], false);
var yRadian = this.toRadians(geometry.faces[fIndex], true);
//Rotate the Torus
torus.applyMatrix(new THREE.Matrix4().makeRotationX(xRadian));
torus.applyMatrix(new THREE.Matrix4().makeRotationY(yRadian));
torusLoop = new THREE.Mesh(torus, this.m);
torusLoop.scale.x = torusLoop.scale.y = torusLoop.scale.z = tubeMeshParams['Scale'];
//Create the torus around the centroid
posx = geometry.faces[fIndex].centroid.x;
posy = geometry.faces[fIndex].centroid.y;
posz = geometry.faces[fIndex].centroid.z;
torusLoop.geometry.applyMatrix(new THREE.Matrix4().makeTranslation(posx, posy, posz));
torusLoop.geometry.computeCentroids();
torusLoop.geometry.computeFaceNormals();
torusLoop.geometry.computeVertexNormals();
return torusLoop;
}
I found I was using an incorrect approach to do this. Instead of trying to calculate each angle and do a RotationX and a RotationY, I should have done a rotation by axis. Definitely was over thinking it.
makeRotationAxis(); is a function built into three.js.
Goal
I want to determine if a test point is within a defined quadrilateral. I'm probably going to implement the solution in Matlab so I only need pseudo-code.
Inputs
Corners of quadrilateral : (x1,y1) (x2,y2) (x3,y3) (x4,y4)
Test point : (xt, yt)
Output
1 - If within quadrilateral
0 - Otherwise
Update
It was pointed out that identifying the vertices of the quadrilateral is not enough to uniquely identify it. You can assume that the order of the points determines the sides of the quadrilateral (point 1 connects 2, 2 connects to 3, 3 connects to 4, 4 connects to 1)
You can test the Point with this condition. Also you can treat quadrilateral as 2 triangles to calculate its area.
Use inpolygon. Usage would be inpolygon(xt,yt,[x1 x2 x3 x4],[y1 y2 y3 y4])
Since it's a simple quadrilateral you can test for a point in triangle for each end and a point in rectangle for the middle.
EDIT Here is some pseudo code for point in triangle:
function SameSide(p1,p2, a,b)
cp1 = CrossProduct(b-a, p1-a)
cp2 = CrossProduct(b-a, p2-a)
if DotProduct(cp1, cp2) >= 0 then return true
else return false
function PointInTriangle(p, a,b,c)
if SameSide(p,a, b,c) and SameSide(p,b, a,c)
and SameSide(p,c, a,b) then return true
else return false
Or using Barycentric technique:
A, B, and C are the triangle end points, P is the point under test
// Compute vectors
v0 = C - A
v1 = B - A
v2 = P - A
// Compute dot products
dot00 = dot(v0, v0)
dot01 = dot(v0, v1)
dot02 = dot(v0, v2)
dot11 = dot(v1, v1)
dot12 = dot(v1, v2)
// Compute barycentric coordinates
invDenom = 1 / (dot00 * dot11 - dot01 * dot01)
u = (dot11 * dot02 - dot01 * dot12) * invDenom
v = (dot00 * dot12 - dot01 * dot02) * invDenom
// Check if point is in triangle
return (u > 0) && (v > 0) && (u + v < 1)
If the aim is to code your own test, then pick any classic point in polygon test to implement. Otherwise do what Jacob suggests.
assuming you the given coordinates are arranged s.t.
(x1,y1) = rightmost coordinate
(x2,y2) = uppermost coordinate
(x3,y3) = leftmost coordinate
(x4,y4) = botoom-most coordinate
You can do the following:
1. calculate the 4 lines of the quadrilateral (we'll call these quad lines)
2. calculate 4 lines, from the (xt, yt) to every other coordinate (we'll call these new lines)
3. if any new line intersects any of the quad lines, then the coordinate is outside of the quadrilateral, otherwise it is inside.
Assume A,B,C,D are the vertices of the quadrilateral and P is the point.
If P is inside the quadrilateral then all dot products dot(BP,BA), dot(BP,BC), dot(AP,AB), dot(AP,AD), dot(DP,DC), dot(DP,DA), dot(CP,CB) and dot(CP,CD) will be positive.
If P is outside the quadrilateral at least one of these products will be negative.
The solution I used to solve this problem was to get the angle of P (in the diagrams the OP posted) for each of the 4 triangles it makes with each side of the quadrilateral. Add the angles together. If they equal (or nearly equal, depending on the error tolerance of the code) 360, the point is inside the quadrilateral. If the sum is less than 360, the point is outside. However, this might only work with convex quadrilaterals.
I have coordinates for 4 vectors defining a quad and another one for it's normal. I am trying to get the rotation of the quad. I get good results for rotation on X and Y just using the normal, but I got stuck getting the Z, since I've used just 1 vector.
Here's my basic test using Processing and toxiclibs(Vec3D and heading methods):
import toxi.geom.*;
Vec3D[] face = {new Vec3D(1.1920928955078125e-07, 0.0, 1.4142135381698608),new Vec3D(-1.4142134189605713, 0.0, 5.3644180297851562e-07),new Vec3D(-2.384185791015625e-07, 0.0, -1.4142135381698608),new Vec3D(1.4142136573791504, 0.0, 0.0),};
Vec3D n = new Vec3D(0.0, 1.0, 0.0);
print("xy: " + degrees(n.headingXY())+"\t");
print("xz: " + degrees(n.headingXZ())+"\t");
print("yz: " + degrees(n.headingYZ())+"\n");
println("angleBetween x: " + degrees(n.angleBetween(Vec3D.X_AXIS)));
println("angleBetween y: " + degrees(n.angleBetween(Vec3D.X_AXIS)));
println("angleBetween z: " + degrees(n.angleBetween(Vec3D.X_AXIS)));
println("atan2 x: " + degrees(atan2(n.z,n.y)));
println("atan2 y: " + degrees(atan2(n.z,n.x)));
println("atan2 z: " + degrees(atan2(n.y,n.x)));
And here is the output:
xy: 90.0 xz: 0.0 yz: 90.0
angleBetween x: 90.0
angleBetween y: 90.0
angleBetween z: 90.0
atan2 x: 0.0
atan2 y: 0.0
atan2 z: 90.0
How can I get the rotation(around it's centre/normal) for Z of my quad ?
OK. I am frankly still not really clear on what it is you're looking for, but let me try to clarify the problem and then address my best guess as to what you really want and see if that helps.
As mentioned in the comment thread, a rotation is a transformation that maps one set of stuff (eg, vectors A, B, C...) to a different set of stuff (A', B', C'...). We can fully define this transformation in terms of an angle (call it θ) and an axis of rotation we'll call R.
Note that R is not a vector, it is a line. That means it has a location as well as a direction -- it is anchored somewhere in space -- and so you need either two points or a point and a direction vector to define it. For simplicity we might assume that the anchor point is the origin (0,0,0), since the question talks about the major axes X, Y and Z. In general, however, this need not be the case - if you want to determine rotation about arbitrary lines you will usually need to translate everything first so that the axis passes through the origin. (If all you care about is the orientation of your objects, rather than its position, then you can probably gloss over this issue.)
Given a start position A, end position A' and an axis R, it is conceptually straightforward to determine the angle θ (or an angle θ, since rotation is periodic and there are infinitely many θs that will take A to A'), though it can be a little fiddly for general R. In the simplest case, where R is one of the major axes, you can do something like this (for R = Z):
theta0 = atan2(A.x, A.y);
theta1 = atan2(A_prime.x, A_prime.y);
theta = theta1 - theta0;
In any case, it looks from your code as if you have the tools to do this already -- I'm not familiar with toxiclibs, but I would imagine the Vec3D angleBetween method ought to take you most of the way to the answer you want.
However, that presupposes that you know A, A' and R, and it seems like this is the real sticking point with your question. In the first place, you mention only a single set of points, defining an arbitrary quad. In the second, you talk about the normal as defining the centre of rotation. Both of these indicate that you haven't properly specified the problem.
As I have repeated tediously several times, a rotation is from one thing to another. A single set of quad vertices may define either the first state or the second, but not both (unless θ is 0, in which case the question is trivial). If you want to determine "the rotation of the quad", you need also to say "from an earlier position P" or "to a subsequent position Q", which you have not done.
Given that the particular quad in question is a square, you might think that there's an intuitive other position involved, to wit: with the sides axis-aligned. And we can indeed rather easily determine the angle of rotation required to get to that orientation, if we can assume that the quad is a rectangle:
// A and B are adjacent corners of the square
// B - A is the direction of the edge joining them
// theta is the angle between that side and the X axis
// (rotating by -theta around Z should align the square)
theta = atan2(B.x - A.x, B.y - A.y);
But, you made a point of stating that you might be looking at any arbitrary quad, for which there would be no "natural" base position to compare against. And even in the square case it is frankly not good practice to presume a baseline without explicitly declaring it.
Which brings us back to my original question: what do you mean? If you can actually pin that down properly I suspect you will find the problem itself relatively easy to solve.
EDIT: Based on your comments below, what you really want to do is to find a rotation that aligns your quad with one of the major planes. This is equivalent to rotating the quad's normal to align with the axis perpendicular to that plane: eg, to get the quad parallel to the XY plane, align its normal with the Z axis.
This can notionally be done with a single rotation about some calculated axis, but in practice you will decompose it into two rotations about major axes. The first rotates about the target axis until the vector is in the plane containing that axis and one of the others; then rotate around the third axis to get the normal to its final alignment. A verbal description is inevitably clunky, so let's formalise a bit:
Let's say you have a planar object Q, with vertices {v1, v2, v3, ...} (in your quad case there will be four of these, but it could be any number as long as all the points are coplanar), with unit normal n = (x y z)T. For the sake of explanation, let's arbitrarily assume that we want to align the object with the XY plane, and hence to rotate n to the Z axis -- the process would be essentially the same for XZ/Y or YZ/X.
Rotate around Z to get n into the XZ plane. We can calculate the angle required like this:
theta1 = -atan2(x,y);
However, we only need the sine and cosine to build a rotation matrix, and we can calculate these directly without knowing the angle:
hypoXY = sqrt(x*x + y*y);
c1 = x/hypoXY;
s1 = y/hypoXY;
(Obviously, if hypoXY is 0 this fails, but in that case n is already aligned with Z.)
Our first rotation matrix R1 looks like this:
[ c1 s1 0 ]
[ -s1 c1 0 ]
[ 0 0 1 ]
Next, rotate around Y to get n parallel to Z. Note that the previous rotation has moved x to a new position x' = sqrt(x2 + y2), so we need to account for this in calculating our second angle:
theta2 = -atan2(z, sqrt(x*x + y*y));
Again, we don't actually need theta2. And because we defined n to be a unit vector, our next calculations are easy:
c2 = z;
s2 = hypoXY;
Our second rotation matrix R2 looks like this:
[ c2 0 -s2 ]
[ 0 1 0 ]
[ s2 0 c2 ]
Compose the two together to get R = R2.R1:
[ c2c1 c2s1 -s2 ]
[ -s1 c1 0 ]
[ s2c1 s2s1 c2 ]
If you apply this matrix to n, you should get the normal aligned with the Z axis. (If not, check the signs first -- this is all a bit back of an envelope and I could easily have got some of the directions wrong. I don't have time to code it up and check right now, but will try to give it a go later. I'll also try to look over your sketch code then.)
Once that works, apply the same transformation to all the points in your object Q and it should become parallel to (although likely offset from) the XY plane.
Here is the rotation matrix for the z axis
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1
your vector
the result is the rotated vector
I have 3d mesh and I would like to draw each face a 2d shape.
What I have in mind is this:
for each face
1. access the face normal
2. get a rotation matrix from the normal vector
3. multiply each vertex to the rotation matrix to get the vertices in a '2d like ' plane
4. get 2 coordinates from the transformed vertices
I don't know if this is the best way to do this, so any suggestion is welcome.
At the moment I'm trying to get a rotation matrix from the normal vector,
how would I do this ?
UPDATE:
Here is a visual explanation of what I need:
At the moment I have quads, but there's no problem
converting them into triangles.
I want to rotate the vertices of a face, so that
one of the dimensions gets flattened.
I also need to store the original 3d rotation of the face.
I imagine that would be inverse rotation of the face
normal.
I think I'm a bit lost in space :)
Here's a basic prototype I did using Processing:
void setup(){
size(400,400,P3D);
background(255);
stroke(0,0,120);
smooth();
fill(0,120,0);
PVector x = new PVector(1,0,0);
PVector y = new PVector(0,1,0);
PVector z = new PVector(0,0,1);
PVector n = new PVector(0.378521084785,0.925412774086,0.0180059205741);//normal
PVector p0 = new PVector(0.372828125954,-0.178844243288,1.35241031647);
PVector p1 = new PVector(-1.25476706028,0.505195975304,0.412718296051);
PVector p2 = new PVector(-0.372828245163,0.178844287992,-1.35241031647);
PVector p3 = new PVector(1.2547672987,-0.505196034908,-0.412717700005);
PVector[] face = {p0,p1,p2,p3};
PVector[] face2d = new PVector[4];
PVector nr = PVector.add(n,new PVector());//clone normal
float rx = degrees(acos(n.dot(x)));//angle between normal and x axis
float ry = degrees(acos(n.dot(y)));//angle between normal and y axis
float rz = degrees(acos(n.dot(z)));//angle between normal and z axis
PMatrix3D r = new PMatrix3D();
//is this ok, or should I drop the builtin function, and add
//the rotations manually
r.rotateX(rx);
r.rotateY(ry);
r.rotateZ(rz);
print("original: ");println(face);
for(int i = 0 ; i < 4; i++){
PVector rv = new PVector();
PVector rn = new PVector();
r.mult(face[i],rv);
r.mult(nr,rn);
face2d[i] = PVector.add(face[i],rv);
}
print("rotated: ");println(face2d);
//draw
float scale = 100.0;
translate(width * .5,height * .5);//move to centre, Processing has 0,0 = Top,Lef
beginShape(QUADS);
for(int i = 0 ; i < 4; i++){
vertex(face2d[i].x * scale,face2d[i].y * scale,face2d[i].z * scale);
}
endShape();
line(0,0,0,nr.x*scale,nr.y*scale,nr.z*scale);
//what do I do with this ?
float c = cos(0), s = sin(0);
float x2 = n.x*n.x,y2 = n.y*n.y,z2 = n.z*n.z;
PMatrix3D m = new PMatrix3D(x2+(1-x2)*c, n.x*n.y*(1-c)-n.z*s, n.x*n.z*(1-c)+n.y*s, 0,
n.x*n.y*(1-c)+n.z*s,y2+(1-y2)*c,n.y*n.z*(1-c)-n.x*s,0,
n.x*n.y*(1-c)-n.y*s,n.x*n.z*(1-c)+n.x*s,z2-(1-z2)*c,0,
0,0,0,1);
}
Update
Sorry if I'm getting annoying, but I don't seem to get it.
Here's a bit of python using Blender's API:
import Blender
from Blender import *
import math
from math import sin,cos,radians,degrees
def getRotMatrix(n):
c = cos(0)
s = sin(0)
x2 = n.x*n.x
y2 = n.y*n.y
z2 = n.z*n.z
l1 = x2+(1-x2)*c, n.x*n.y*(1-c)+n.z*s, n.x*n.y*(1-c)-n.y*s
l2 = n.x*n.y*(1-c)-n.z*s,y2+(1-y2)*c,n.x*n.z*(1-c)+n.x*s
l3 = n.x*n.z*(1-c)+n.y*s,n.y*n.z*(1-c)-n.x*s,z2-(1-z2)*c
m = Mathutils.Matrix(l1,l2,l3)
return m
scn = Scene.GetCurrent()
ob = scn.objects.active.getData(mesh=True)#access mesh
out = ob.name+'\n'
#face0
f = ob.faces[0]
n = f.v[0].no
out += 'face: ' + str(f)+'\n'
out += 'normal: ' + str(n)+'\n'
m = getRotMatrix(n)
m.invert()
rvs = []
for v in range(0,len(f.v)):
out += 'original vertex'+str(v)+': ' + str(f.v[v].co) + '\n'
rvs.append(m*f.v[v].co)
out += '\n'
for v in range(0,len(rvs)):
out += 'original vertex'+str(v)+': ' + str(rvs[v]) + '\n'
f = open('out.txt','w')
f.write(out)
f.close
All I do is get the current object, access the first face, get the normal, get the vertices, calculate the rotation matrix, invert it, then multiply it by each vertex.
Finally I write a simple output.
Here's the output for a default plane for which I rotated all the vertices manually by 30 degrees:
Plane.008
face: [MFace (0 3 2 1) 0]
normal: [0.000000, -0.499985, 0.866024](vector)
original vertex0: [1.000000, 0.866025, 0.500000](vector)
original vertex1: [-1.000000, 0.866026, 0.500000](vector)
original vertex2: [-1.000000, -0.866025, -0.500000](vector)
original vertex3: [1.000000, -0.866025, -0.500000](vector)
rotated vertex0: [1.000000, 0.866025, 1.000011](vector)
rotated vertex1: [-1.000000, 0.866026, 1.000012](vector)
rotated vertex2: [-1.000000, -0.866025, -1.000012](vector)
rotated vertex3: [1.000000, -0.866025, -1.000012](vector)
Here's the first face of the famous Suzanne mesh:
Suzanne.001
face: [MFace (46 0 2 44) 0]
normal: [0.987976, -0.010102, 0.154088](vector)
original vertex0: [0.468750, 0.242188, 0.757813](vector)
original vertex1: [0.437500, 0.164063, 0.765625](vector)
original vertex2: [0.500000, 0.093750, 0.687500](vector)
original vertex3: [0.562500, 0.242188, 0.671875](vector)
rotated vertex0: [0.468750, 0.242188, -0.795592](vector)
rotated vertex1: [0.437500, 0.164063, -0.803794](vector)
rotated vertex2: [0.500000, 0.093750, -0.721774](vector)
rotated vertex3: [0.562500, 0.242188, -0.705370](vector)
The vertices from the Plane.008 mesh are altered, the ones from Suzanne.001's mesh
aren't. Shouldn't they ? Should I expect to get zeroes on one axis ?
Once I got the rotation matrix from the normal vector, what is the rotation on x,y,z ?
Note: 1. Blender's Matrix supports the * operator 2.In Blender's coordinate system Z point's up. It looks like a right handed system, rotated 90 degrees on X.
Thanks
That looks reasonable to me. Here's how to get a rotation matrix from normal vector. The normal is the vector. The angle is 0. You probably want the inverse rotation.
Is your mesh triangulated? I'm assuming it is. If so, you can do this, without rotation matrices. Let the points of the face be A,B,C. Take any two vertices of the face, say A and B. Define the x axis along vector AB. A is at 0,0. B is at 0,|AB|. C can be determined from trigonometry using the angle between AC and AB (which you get by using the dot product) and the length |AC|.
You created the m matrix correctly. This is the rotation that corresponds to your normal vector. You can use the inverse of this matrix to "unrotate" your points. The normal of face2d will be x, i.e. point along the x-axis. So extract your 2d coordinates accordingly. (This assumes your quad is approximately planar.)
I don't know the library you are using (Processing), so I'm just assuming there are methods for m.invert() and an operator for applying a rotation matrix to a point. They may of course be called something else. Luckily the inverse of a pure rotation matrix is its transpose, and multiplying a matrix and a vector are straightforward to do manually if you need to.
void setup(){
size(400,400,P3D);
background(255);
stroke(0,0,120);
smooth();
fill(0,120,0);
PVector x = new PVector(1,0,0);
PVector y = new PVector(0,1,0);
PVector z = new PVector(0,0,1);
PVector n = new PVector(0.378521084785,0.925412774086,0.0180059205741);//normal
PVector p0 = new PVector(0.372828125954,-0.178844243288,1.35241031647);
PVector p1 = new PVector(-1.25476706028,0.505195975304,0.412718296051);
PVector p2 = new PVector(-0.372828245163,0.178844287992,-1.35241031647);
PVector p3 = new PVector(1.2547672987,-0.505196034908,-0.412717700005);
PVector[] face = {p0,p1,p2,p3};
PVector[] face2d = new PVector[4];
//what do I do with this ?
float c = cos(0), s = sin(0);
float x2 = n.x*n.x,y2 = n.y*n.y,z2 = n.z*n.z;
PMatrix3D m_inverse =
new PMatrix3D(x2+(1-x2)*c, n.x*n.y*(1-c)+n.z*s, n.x*n.y*(1-c)-n.y*s, 0,
n.x*n.y*(1-c)-n.z*s,y2+(1-y2)*c,n.x*n.z*(1-c)+n.x*s, 0,
n.x*n.z*(1-c)+n.y*s,n.y*n.z*(1-c)-n.x*s,z2-(1-z2)*c, 0,
0,0,0,1);
face2d[0] = m_inverse * p0; // Assuming there's an appropriate operator*().
face2d[1] = m_inverse * p1;
face2d[2] = m_inverse * p2;
face2d[3] = m_inverse * p3;
// print & draw as you did before...
}
For face v0-v1-v3-v2 vectors v3-v0, v3-v2 and a face normal already form rotation matrix that would transform 2d face into 3d face.
Matrix represents coordinate system. Each row (or column, depending on notation) corresponds to axis coordinate system within new coordinate system. 3d rotation/translation matrix can be represented as:
vx.x vx.y vx.z 0
vy.x vy.y vy.z 0
vz.x vz.y vz.z 0
vp.x vp.y vp.z 1
where vx is an x axis of a coordinate system, vy - y axis, vz - z axis, and vp - origin of new system.
Assume that v3-v0 is an y axis (2nd row), v3-v2 - x axis (1st row), and normal - z axis (3rd row). Build a matrix from them. Then invert matrix. You'll get a matrix that will rotate a 3d face into 2d face.
I have 3d mesh and I would like to draw each face a 2d shape.
I suspect that UV unwrapping algorithms are closer to what you want to achieve than trying to get rotation matrix from 3d face.
That's very easy to achieve: (Note: By "face" I mean "triangle")
Create a view matrix that represents a camera looking at a face.
Determine the center of the face with bi-linear interpolation.
Determine the normal of the face.
Position the camera some units in opposite normal direction.
Let the camera look at the center of the face.
Set the cameras up vector point in the direction of the middle of any vertex of the face.
Set the aspect ratio to 1.
Compute the view matrix using this data.
Create a orthogonal projection matrix.
Set the width and height of the view frustum large enough to contain the whole face (e.g. the length of the longest site of a face).
Compute the projection matrix.
For every vertex v of the face, multiply it by both matrices: v * view * projection.
The result is a projection of Your 3d faces into 2d space as if You were looking at them exactly orthogonal without any perspective disturbances. The final coordinates will be in normalized screen coordinates where (-1, -1) is the bottom left corner, (0, 0) is the center and (1, 1) is the top right corner.
I'm writing a script where icons rotate around a given pivot (or origin). I've been able to make this work for rotating the icons around an ellipse but I also want to have them move around the perimeter of a rectangle of a certain width, height and origin.
I'm doing it this way because my current code stores all the coords in an array with each angle integer as the key, and reusing this code would be much easier to work with.
If someone could give me an example of a 100x150 rectangle, that would be great.
EDIT: to clarify, by rotating around I mean moving around the perimeter (or orbiting) of a shape.
You know the size of the rectangle and you need to split up the whole angle interval into four different, so you know if a ray from the center of the rectangle intersects right, top, left or bottom of the rectangle.
If the angle is: -atan(d/w) < alfa < atan(d/w) the ray intersects the right side of the rectangle. Then since you know that the x-displacement from the center of the rectangle to the right side is d/2, the displacement dy divided by d/2 is tan(alfa), so
dy = d/2 * tan(alfa)
You would handle this similarily with the other three angle intervals.
Ok, here goes. You have a rect with width w and depth d. In the middle you have the center point, cp. I assume you want to calculate P, for different values of the angle alfa.
I divided the rectangle in four different areas, or angle intervals (1 to 4). The interval I mentioned above is the first one to the right. I hope this makes sense to you.
First you need to calculate the angle intervals, these are determined completely by w and d. Depending on what value alfa has, calculate P accordingly, i.e. if the "ray" from CP to P intersects the upper, lower, right or left sides of the rectangle.
Cheers
This was made for and verified to work on the Pebble smartwatch, but modified to be pseudocode:
struct GPoint {
int x;
int y;
}
// Return point on rectangle edge. Rectangle is centered on (0,0) and has a width of w and height of h
GPoint getPointOnRect(int angle, int w, int h) {
var sine = sin(angle), cosine = cos(angle); // Calculate once and store, to make quicker and cleaner
var dy = sin>0 ? h/2 : h/-2; // Distance to top or bottom edge (from center)
var dx = cos>0 ? w/2 : w/-2; // Distance to left or right edge (from center)
if(abs(dx*sine) < abs(dy*cosine)) { // if (distance to vertical line) < (distance to horizontal line)
dy = (dx * sine) / cosine; // calculate distance to vertical line
} else { // else: (distance to top or bottom edge) < (distance to left or right edge)
dx = (dy * cosine) / sine; // move to top or bottom line
}
return GPoint(dx, dy); // Return point on rectangle edge
}
Use:
rectangle_width = 100;
rectangle_height = 150;
rectangle_center_x = 300;
rectangle_center_y = 300;
draw_rect(rectangle_center_x - (rectangle_width/2), rectangle_center_y - (rectangle_center_h/2), rectangle_width, rectangle_height);
GPoint point = getPointOnRect(angle, rectangle_width, rectangle_height);
point.x += rectangle_center_x;
point.y += rectangle_center_y;
draw_line(rectangle_center_x, rectangle_center_y, point.x, point.y);
One simple way to do this using an angle as a parameter is to simply clip the X and Y values using the bounds of the rectangle. In other words, calculate position as though the icon will rotate around a circular or elliptical path, then apply this:
(Assuming axis-aligned rectangle centered at (0,0), with X-axis length of XAxis and Y-axis length of YAxis):
if (X > XAxis/2)
X = XAxis/2;
if (X < 0 - XAxis/2)
X = 0 - XAxis/2;
if (Y > YAxis/2)
Y = YAxis/2;
if (Y < 0 - YAxis/2)
Y = 0 - YAxis/2;
The problem with this approach is that the angle will not be entirely accurate and the speed along the perimeter of the rectangle will not be constant. Modelling an ellipse that osculates the rectangle at its corners can minimize the effect, but if you are looking for a smooth, constant-speed "orbit," this method will not be adequate.
If think you mean rotate like the earth rotates around the sun (not the self-rotation... so your question is about how to slide along the edges of a rectangle?)
If so, you can give this a try:
# pseudo coode
for i = 0 to 499
if i < 100: x++
else if i < 250: y--
else if i < 350: x--
else y++
drawTheIcon(x, y)
Update: (please see comment below)
to use an angle, one line will be
y / x = tan(th) # th is the angle
the other lines are simple since they are just horizontal or vertical. so for example, it is x = 50 and you can put that into the line above to get the y. do that for the intersection of the horizontal line and vertical line (for example, angle is 60 degree and it shoot "NorthEast"... now you have two points. Then the point that is closest to the origin is the one that hits the rectangle first).
Use a 2D transformation matrix. Many languages (e.g. Java) support this natively (look up AffineTransformation); otherwise, write out a routine to do rotation yourself, once, debug it well, and use it forever. I must have five of them written in different languages.
Once you can do the rotation simply, find the location on the rectangle by doing line-line intersection. Find the center of the orbited icon by intersecting two lines:
A ray from your center of rotation at the angle you desire
One of the four sides, bounded by what angle you want (the four quadrants).
Draw yourself a sketch on a piece of paper with a rectangle and a centre of rotation. First translate the rectangle to centre at the origin of your coordinate system (remember the translation parameters, you'll need to reverse the translation later). Rotate the rectangle so that its sides are parallel to the coordinate axes (same reason).
Now you have a triangle with known angle at the origin, the opposite side is of known length (half of the length of one side of the rectangle), and you can now:
-- solve the triangle
-- undo the rotation
-- undo the translation