my question is how to get the second coordinate (in 2D) of a point that lies on a curve defined as a NURBS curve given the axial coordinate. I have the knot vector, control points, their weights and basis functions.
I looked through similar questions (How to find out Y coordinate of specific point in bezier curve in canvas?) but did not find a good answer so far.
Thanks,
M
This is not an easy task if you need to implement everything from scratch. Nevertheless, the codes will look like this:
for each non-null knot interval in the knot vector of the NURBS
{
extract Bezier curve B(t) from the NURBS for this knot interval [a, b];
compute the minimum and maximum X values of the Bezier curve's control points.
if ( X0 is within [Xmin, Xmax] )
{
t0 = a;
t1 = b;
epsilon = 1.0e-06; // a small value;
while ( (t1-t0) > epsilon )
{
Subdivide B(t) at t=0.5 to generate two Bezier curves: B1(t) and B2(t);
compute the [Xmin1, Xmax1] for B1(t) and [Xmin2, Xmax2] for B2(t);
if ( X0 is within [Xmin1, Xmax1] )
{
B(t) = B1(t);
t0 = a;
t1 = (a+b)/2;
}
else
{
B(t) = B2(t);
t0 = (a+b)/2;
t1 = b;
}
} // end while loop
return ((t0+t1)/2); // This is the parameter value you are looking for
} // end if()
} // end for loop
The (t0+t1)/2 is the parameter value you are looking for. Please note that you might find multiple solutions given the same X0 value.
Related
Background :
I am implementing an application where I need to do affine transformations on a selected line. Line is represented using 2 points. All the code is done in mouse events.
void mousePressEvent(){
lastPoint = event.point();
}
void mouseMoveEvent(){
currentPoint = event.point();
//Do transformations here.
translate_factor = currentPoint - lastPoint
scale_factor = currentPoint / lastPoint
rotation_angle = f(current_point,last_point) //FIND f
//Apply transformation
lastPoint = currentPoint
}
I have two points, p1 and p2. The X axis increases from left to right. Y axis increases top to bottom. I have a point d around which I want to rotate a set of points p_set .
How do I find the angle of rotation from p1 to p2 about d .
What I am doing is translate p1 , p2 by -d and calculate theta by using atan2f as shown in the following code.
Note that y is inverted : goes from top to bottom.
Here is the code
const auto about = stroke->getMidPoint();
Eigen::Affine2f t1(Eigen::Translation2f(0-about.x(),0-about.y()));
Eigen::Vector3f p1 = t1.matrix()*Eigen::Vector3f(lastPoint.x(),lastPoint.y(),1.0);
Eigen::Vector3f p2 = t1.matrix()*Eigen::Vector3f(currentPoint.x(),currentPoint.y(),1.0);
float theta = atan2f(p2[1]-p1[1],p2[0]-p1[0]);
When I transform using this, I get very weird rotations.
What I want is to find angle as a function of current and last points
It seems to me that you're finding the slope of the vector p2-p1.
If I understood correctly, you want the difference of the slopes of the two vectors p2-d and p1-d, so the rotation angle is something like atan((p2.y-d.y)/(p2.x-d.x)) - atan((p1.y-d.y)/(p1.x-d.x))
How to find leftmost/rightmost point of SVG C (bezier curve) path segment? I know there is getBoundingClientRect() and getBBox() but none of them apply since they return only single coordinate of the point.
Just to avoid XY problem - I want to split single path composed of bezier curves into several paths each monotonously going from left to right (or right to left). It means that on any single path should be no 2 points having equal X coordinate. I understand that required split point may potentially be inside the bounding box of a segment thus not being leftmost/rightmost, but I'm almost sure that way of finding such point should use same techniques as finding horizontally extreme point.
You would need to iterate through the path length with .getPointAtLength(i) method, and then find the limits. Seemed like a fun thing to do so I made a quick and dirty implementation, this is the important part:
function findLimits(path) {
var boundingPoints = {
minX: {x: dimensions.width, y: dimensions.height},
minY: {x: dimensions.width, y: dimensions.height},
maxX: {x: 0, y: 0},
maxY: {x: 0, y: 0}
}
var l = path.getTotalLength();
for (var p = 0; p < l; p++) {
var coords = path.getPointAtLength(p);
if (coords.x < boundingPoints.minX.x) boundingPoints.minX = coords;
if (coords.y < boundingPoints.minY.y) boundingPoints.minY = coords;
if (coords.x > boundingPoints.maxX.x) boundingPoints.maxX = coords;
if (coords.y > boundingPoints.maxY.y) boundingPoints.maxY = coords;
}
return boundingPoints
}
You can find the implementation here: https://jsfiddle.net/4gus3hks/1/
Paul LeBeau's comment and fancy animation on the wiki inspired me for the solution. It is based mostly on following terms:
Values of parameter t from [0, 1] can be matched to the curve
points.
For any parameter value point on the curve can be constructed
step-by-step by linearly combining pairs of adjacent control points
into intermediate control points of higher "depth". This operation
can be repeated until only single point left - point on the curve
itself.
Coordinates of the intermediate points can be defined by
t-polynoms of degree equal to point "depth". And coefficients of
those polynoms ultimately depend only on coordinates of initial
control points.
Penultimate step of construction gives 2 points that define tangent
to the curve at the final point, and coordinates of those points are
controlled by quadratic polynom.
Having direction of tangent in question as vector allows to
construct quadratic equation against t where curve has required
tangent.
So, in fact, finding required points can be performed in constant O(1) time:
tangentPoints: function(tx, ty){
var ends = this.getPolynoms(2);
var tangent = [ends[1][0].subtractPoly(ends[0][0]),
ends[1][1].subtractPoly(ends[0][1])];
var eq = tangent[0].multiplyScalar(ty).subtractPoly(tangent[1].multiplyScalar(tx));
return solveQuadratic(...eq.values).filter(t => t >= 0 && t <= 1);
}
Full code with assisting Polynom class and visual demo I placed into this repo and fiddle
I have a list of consecutive points and I need to find the coordinates of a polygon some size larger. I can calculate each of the points in the new polygon if it has convex angles, but I'm not sure how to adjust for when the angles are concave.
Concave angles can be treated in exactly the same way as convex ones: For each vertex you generate lines that are parallel to the two original segments but shifted by your offset value. Then the vertex is replaced with the intersection of these two lines.
The difficulty is that the resulting polygon can have intersections if the original one has one or more concave angles. There are different ways to handle these intersections. Generally they can produce inner contours (holes in the polygon) but maybe you are only interested in the outer contour.
In any case you have to find the intersection points first. If you don't find any, you are finished.
Otherwise find a start point of which you can be sure that it is on the outer contour. In many cases you can take the one with smallest X coordinate for that. Then trace the polygon contour until you get to the first intersection. Add the intersection to the polygon. If you are only interested in the outer contour, then skip all following vertices until you get back to the intersection point. Then continue to add the vertexes to the resulting polygon until you get to the next intersection and so on.
If you also need the inner contours (holes) it gets a bit more complicated, but I guess you can figure this out.
I also need to add that you should pe prepared for special cases like (almost) duplicate edges that cause numerical problems. Generally this is not a trivial task, so if possible, try to find a suitable polygon library.
For this problem I found a relatively simple solution for figuring out whether the calculated point was inside or outside the original polygon. Check to see whether the newly formed line intersects the original polygon's line. A formula can be found here http://www.geeksforgeeks.org/orientation-3-ordered-points/.
Suppose your polygon is given in counter-clockwise order. Let P1=(x1,y1), P2=(x2,y2) and P3=(x3,y3) be consecutive vertices. You want to know if the angle at P2 is “concave” i.e. more than 180 degrees. Let V1=(x4,y4)=P2-P1 and V2=(x5,y5)=P3-P2. Compute the “cross product” V1 x V2 = (x4.y5-x5.y4). This is negative iff the angle is concave.
Here is a code in C# that receives a list of Vector2D representing the ordered points of a polygon and returns a list with the angles of each vertex. It first checks if the points are clockwise or counterclockwise, and then it loops through the points calculating the sign of the cross product (z) for each triple of angles, and compare the value of the cross product to the clockwise function result to check if the calculated angle to that point needs to be the calculated angle or adjusted to 360-angle. The IsClockwise function was obtained in this discussion: How to determine if a list of polygon points are in clockwise order?
public bool IsClockwise(List<Vector2> vertices)
{
double sum = 0.0;
for (int i = 0; i < vertices.Count; i++)
{
Vector2 v1 = vertices[i];
Vector2 v2 = vertices[(i + 1) % vertices.Count];
sum += (v2.x - v1.x) * (v2.y + v1.y);
}
return sum > 0.0;
}
List<float> estimatePolygonAngles(List<Vector2> vertices)
{
if (vertices.Count < 3)
return null;
//1. check if the points are clockwise or counterclockwise:
int clockwise = (IsClockwise(vertices) ? 1 : -1);
List<float> angles = new List<float>();
List<float> crossProductsSigns = new List<float>();
Vector2 v1, v2;
//2. calculate the angles between each triple of vertices (first and last angles are computed separetely because index of the array):
v1 = vertices[vertices.Count - 1] - vertices[0];
v2 = vertices[1] - vertices[0];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
for (int i = 1; i < vertices.Count-1; i++)
{
v1 = vertices[i-1] - vertices[i];
v2 = vertices[i+1] - vertices[i];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
}
v1 = vertices[vertices.Count - 2] - vertices[vertices.Count - 1];
v2 = vertices[0] - vertices[vertices.Count - 1];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
//3. for each computed angle, check if the cross product is the same as the as the direction provided by the clockwise function, if dont, the angle must be adjusted to 360-angle
for (int i = 0; i < vertices.Count; i++)
{
if (crossProductsSigns[i] != clockwise)
angles[i] = 360.0f - angles[i];
}
return angles;
}
I've been looking at and trying to understand the following bit of code
float sdBox( vec3 p, vec3 b )
{
vec3 d = abs(p) - b;
return min(max(d.x,max(d.y,d.z)),0.0) +
length(max(d,0.0));
}
I understand that length(d) handles the SDF case where the point is off to the 'corner' (ie. all components of d are positive) and that max(d.x, d.y, d.z) gives us the proper distance in all other cases. What I don't understand is how these two are combined here without the use of an if statement to check the signs of d's components.
When all of the d components are positive, the return expression can be reduced to length(d) because of the way min/max will evaluate - and when all of the d components are negative, we get max(d.x, d.y, d.z). But how am I supposed to understand the in-between cases? The ones where the components of d have mixed signs?
I've been trying to graph it out to no avail. I would really appreciate it if someone could explain this to me in geometrical/mathematical terms. Thanks.
If you like to know how It works It's better do the following steps:
1.first of all you should know definitions of shapes
2.It's always better to consider 2D shape of them, because three dimensions may be complex for you.
so let me to explain some shapes:
Circle
A circle is a simple closed shape. It is the set of all points in a plane that are at a given distance from a given point, the center.
You can use distance(), length() or sqrt() to calculate the distance to the center of the billboard.
The book of shaders - Chapter 7
Square
In geometry, a square is a regular quadrilateral, which means that it has four equal sides and four equal angles (90-degree angles).
I describe 2D shapes In before section now let me to describe 3D definition.
Sphere
A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball.
Like a circle, which geometrically is an object in two-dimensional space, a sphere is defined mathematically as the set of points that are all at the same distance r from a given point, but in three-dimensional space.
Refrence - Wikipedia
Cube
In geometry, a cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex.
Refrence : Wikipedia
Modeling with distance functions
now It's time to understanding modeling with distance functions
Sphere
As mentioned In last sections.In below code length() used to calculate the distance to the center of the billboard , and you can scale this shape by s parameter.
//Sphere - signed - exact
/// <param name="p">Position.</param>
/// <param name="s">Scale.</param>
float sdSphere( vec3 p, float s )
{
return length(p)-s;
}
Box
// Box - unsigned - exact
/// <param name="p">Position.</param>
/// <param name="b">Bound(Scale).</param>
float udBox( vec3 p, vec3 b )
{
return length(max(abs(p)-b,0.0));
}
length() used like previous example.
next we have max(x,0) It called Positive and negative parts
this is mean below code is equivalent:
float udBox( vec3 p, vec3 b )
{
vec3 value = abs(p)-b;
if(value.x<0.){
value.x = 0.;
}
if(value.y<0.){
value.y = 0.;
}
if(value.z<0.){
value.z = 0.;
}
return length(value);
}
step 1
if(value.x<0.){
value.x = 0.;
}
step 2
if(value.y<0.){
value.y = 0.;
}
step 3
if(value.z<0.){
value.z = 0.;
}
step 4
next we have absolution function.It used to remove additional parts.
Absolution Steps
Absolution step 1
if(value.x < -1.){
value.x = 1.;
}
Absolution step 2
if(value.y < -1.){
value.y = 1.;
}
Absolution step 3
if(value.z < -1.){
value.z = 1.;
}
Also you can make any shape by using Constructive solid geometry.
CSG is built on 3 primitive operations: intersection ( ∩ ), union ( ∪ ), and difference ( - ).
It turns out these operations are all concisely expressible when combining two surfaces expressed as SDFs.
float intersectSDF(float distA, float distB) {
return max(distA, distB);
}
float unionSDF(float distA, float distB) {
return min(distA, distB);
}
float differenceSDF(float distA, float distB) {
return max(distA, -distB);
}
I figured it out a while ago and wrote about this extensively in a blog post here: http://fabricecastel.github.io/blog/2016-02-11/main.html
Here's an excerpt (see the full post for a full explanation):
Consider the four points, A, B, C and D. Let's crudely reduce the distance function to try and get rid of the min/max functions in order to understand their effect (since that's what's puzzling about this function). The notation below is a little sloppy, I'm using square brackets to denote 2D vectors.
// 2D version of the function
d(p) = min(max(p.x, p.y), 0)
+ length(max(p, 0))
---
d(A) = min(max(-1, -1), 0)
+ length(max([-1, -1], 0))
d(A) = -1 + length[0, 0]
---
d(B) = min(max(1, 1), 0)
+ length(max([1, 1], 0))
d(B) = 0 + length[1, 1]
Ok, so far nothing special. When A is inside the square, we essentially get our first distance function based on planes/lines and when B is in the area where our first distance function is inaccurate, it gets zeroed out and we get the second distance function (the length). The trick lies in the other two cases C and D. Let's work them out.
d(C) = min(max(-1, 1), 0)
+ length(max([-1, 1], 0))
d(C) = 0 + length[0, 1]
---
d(D) = min(max(1, -1), 0)
+ length(max([-1, 1], 0))
d(D) = 0 + length[1, 0]
If you look back to the graph above, you'll note C' and D'. Those points have coordinates [0,1] and [1,0], respectively. This method uses the fact that both distance fields intersect on the axes - that D and D' lie at the same distance from the square.
If we zero out all negative component of a vector and take its length we will get the proper distance between the point and the square (for points outside of the square only). This is what max(d,0.0) does; a component-wise max operation. So long as the vector has at least one positive component, min(max(d.x,d.y),0.0) will resolve to 0 leaving us with only the second part of the equation. In the event that the point is inside the square, we want to return the first part of the equation (since it represents our first distance function). If all components of the vector are negative it's easy to see our condition will be met.
This understanding should tranlsate back into 3D seamlessly once you wrap your head around it. You may or may not have to draw a few graphs by hand to really "get" it - I know I did and would encourage you to do so if you're dissatisfied with my explanation.
Working this implementation into our own code, we get this:
float distanceToNearestSurface(vec3 p){
float s = 1.0;
vec3 d = abs(p) - vec3(s);
return min(max(d.x, max(d.y,d.z)), 0.0)
+ length(max(d,0.0));
}
And there you have it.
I have an array of points (x0,y0)... (xn,yn) monotonic in x and wish to draw the "best" curve through these using Bezier curves. This curve should not be too "jaggy" (e.g. similar to joining the dots) and not too sinuous (and definitely not "go backwards"). I have created a prototype but wonder whether there is an objectively "best solution".
I need to find control points for all segments xi,y1 x(i+1)y(i+1). My current approach (except for the endpoints) for a segment x(i), x(i+1) is:
find the vector x(i-1)...x(i+1) , normalize, and scale it by factor * len(i,i+1) to give the vector for the leading control point
find the vector x(i+2)...x(i) , normalize, and scale it by factor * len(i,i+1) to give the vector for the trailing control point.
I have tried factor=0.1 (too jaggy), 0.33 (too curvy) and 0.20 - about right. But is there a better approach which (say) makes 2nd and 3nd derivatives as smooth as possible. (I assume such an algorithm is implemented in graphics packages)?
I can post pseudo/code if requested. Here are the three images (0.1/0.2/0.33). The control points are shown by straight lines: black (trailing) and red (leading)
Here's the current code. It's aimed at plotting Y against X (monotonic X) without close-ing. I have built my own library for creating SVG (preferred output); this code creates triples of x,y in coordArray for each curve segment (control1, xcontrol2, end). Start is assumed by last operation (Move or Curve). It's Java but should be easy to interpret (CurvePrimitive maps to cubic, "d" is the String representation of the complete path in SVG).
List<SVGPathPrimitive> primitiveList = new ArrayList<SVGPathPrimitive>();
primitiveList.add(new MovePrimitive(real2Array.get(0)));
for(int i = 0; i < real2Array.size()-1; i++) {
// create path 12
Real2 p0 = (i == 0) ? null : real2Array.get(i-1);
Real2 p1 = real2Array.get(i);
Real2 p2 = real2Array.get(i+1);
Real2 p3 = (i == real2Array.size()-2) ? null : real2Array.get(i+2);
Real2Array coordArray = plotSegment(factor, p0, p1, p2, p3);
SVGPathPrimitive primitive = new CurvePrimitive(coordArray);
primitiveList.add(primitive);
}
String d = SVGPath.constructDString(primitiveList);
SVGPath path1 = new SVGPath(d);
svg.appendChild(path1);
/**
*
* #param factor to scale control points by
* #param p0 previous point (null at start)
* #param p1 start of segment
* #param p2 end of segment
* #param p3 following point (null at end)
* #return
*/
private Real2Array plotSegment(double factor, Real2 p0, Real2 p1, Real2 p2, Real2 p3) {
// create p1-p2 curve
double len12 = p1.getDistance(p2) * factor;
Vector2 vStart = (p0 == null) ? new Vector2(p2.subtract(p1)) : new Vector2(p2.subtract(p0));
vStart = new Vector2(vStart.getUnitVector().multiplyBy(len12));
Vector2 vEnd = (p3 == null) ? new Vector2(p2.subtract(p1)) : new Vector2(p3.subtract(p1));
vEnd = new Vector2(vEnd.getUnitVector().multiplyBy(len12));
Real2Array coordArray = new Real2Array();
Real2 controlStart = p1.plus(vStart);
coordArray.add(controlStart);
Real2 controlEnd = p2.subtract(vEnd);
coordArray.add(controlEnd);
coordArray.add(p2);
// plot controls
SVGLine line12 = new SVGLine(p1, controlStart);
line12.setStroke("red");
svg.appendChild(line12);
SVGLine line21 = new SVGLine(p2, controlEnd);
svg.appendChild(line21);
return coordArray;
}
A Bezier curve requires the data points, along with the slope and curvature at each point. In a graphics program, the slope is set by the slope of the control-line, and the curvature is visualized by the length.
When you don't have such control-lines input by the user, you need to estimate the gradient and curvature at each point. The wikipedia page http://en.wikipedia.org/wiki/Cubic_Hermite_spline, and in particular the 'interpolating a data set' section has a formula that takes these values directly.
Typically, estimating these values from points is done using a finite difference - so you use the values of the points on either side to help estimate. The only choice here is how to deal with the end points where there is only one adjacent point: you can set the curvature to zero, or if the curve is periodic you can 'wrap around' and use the value of the last point.
The wikipedia page I referenced also has other schemes, but most others introduce some other 'free parameter' that you will need to find a way of setting, so in the absence of more information to help you decide how to set other parameters, I'd go for the simple scheme and see if you like the results.
Let me know if the wikipedia article is not clear enough, and I'll knock up some code.
One other point to be aware of: what 'sort' of Bezier interpolation are you after? Most graphics programs do cubic bezier in 2 dimensions (ie you can draw a circle-like curve), but your sample images look like it could be 1d functions approximation (as in for every x there is only one y value). The graphics program type curve is not really mentioned on the page I referenced. The maths involved for converting estimate of slope and curvature into a control vector of the form illustrated on http://en.wikipedia.org/wiki/B%C3%A9zier_curve (Cubic Bezier) would take some working out, but the idea is similar.
Below is a picture and algorithm for a possible scheme, assuming your only input is the three points P1, P2, P3
Construct a line (C1,P1,C2) such that the angles (P3,P1,C1) and (P2,P1,C2) are equal. In a similar fashion construct the other dark-grey lines. The intersections of these dark-grey lines (marked C1, C2 and C3) become the control-points as in the same sense as the images on the Bezier Curve wikipedia site. So each red curve, such as (P3,P1), is a quadratic bezier curve defined by the points (P3, C1, P1). The construction of the red curve is the same as given on the wikipedia site.
However, I notice that the control-vector on the Bezier Curve wikipedia page doesn't seem to match the sort of control vector you are using, so you might have to figure out how to equate the two approaches.
I tried this with quadratic splines instead of cubic ones which simplifies the selection of control points (you just choose the gradient at each point to be a weighted average of the mean gradients of the neighbouring intervals, and then draw tangents to the curve at the data points and stick the control points where those tangents intersect), but I couldn't find a sensible policy for setting the gradients of the end points. So I opted for Lagrange fitting instead:
function lagrange(points) { //points is [ [x1,y1], [x2,y2], ... ]
// See: http://www.codecogs.com/library/maths/approximation/interpolation/lagrange.php
var j,n = points.length;
var p = [];
for (j=0;j<n;j++) {
p[j] = function (x,j) { //have to pass j cos JS is lame at currying
var k, res = 1;
for (k=0;k<n;k++)
res*=( k==j ? points[j][1] : ((x-points[k][0])/(points[j][0]-points[k][0])) );
return res;
}
}
return function(x) {
var i, res = 0;
for (i=0;i<n;i++)
res += p[i](x,i);
return res;
}
}
With that, I just make lots of samples and join them with straight lines.
This is still wrong if your data (like mine) consists of real world measurements. These are subject to random errors and if you use a technique that forces the curve to hit them all precisely, then you can get silly valleys and hills between the points. In cases like these, you should ask yourself what order of polynomial the data should fit and ... well ... that's what I'm about to go figure out.