I'm developing a simulator gui in which the user clicks on different points of the map and the program connects these points to each other however the connection should be somehow curved (but preferably the curve should pass from the given points) I can't find a decent way to implement this.
A similar solution which I could not figure out
I have seen similar problems and often they are solved using QPainterPath or implementing a bezier curve. Or should I just compute the control points of the bezier curve (if so, how?) ?
Any help would be appreciated,
Thank you in advance
A cubic Bézier curve consists of 4 points: Start, End, Control1 and Control2. Depending on the two control points the curve can have different shapes. Since you don't have the control points you should calculate them in some way.
This gives a nice description of how to calculate the control points if you only know the start point, end point and one other point on the curve. In your case the point on the curve could be the mid point between start and end.
void Beziertest::Bezier2D(QList<QPoint> points)
{
QImage area(600,700,QImage::Format_RGB32);
int n=points.length()-1;
for(double u = 0.0 ; u <= 1.0 ; u += 0.001)
{
//calculate x coordinate
double xu=0.0;
for (int i = n; i >= 0; i--) {
xu+=points[i].x()*((factorial(n)/(factorial(i)*factorial((n-i))))*pow(u,i)*pow((1-u),(n-i)));
}
//calculate y coordinate
double yu=0.0;
for (int i = n; i >= 0; i--) {
yu+=points[i].y()*((factorial(n)/(factorial(i)*factorial((n-i))))*pow(u,i)*pow((1-u),(n-i)));
}
area.setPixel((int)xu , (int)yu , deger);
setPixmap(QPixmap::fromImage(res));//set image to label
}
}
Related
I have a QGraphicsView in my Qt application on which user can draw curves. Curves consist of QGraphicsEllipseItem's and QGraphicsPathItem's, which connect the adjacent ellipses.
I want to get a list of QPoint's which satisfy the given curve. I tried creating local QPainterPath for this procedure which would represent the whole curve and iterating over all the points from it's rectangle to see which ones satisfy this curve. The code looks like:
QPainterPath curvePath = edges[index]->at(0)->path();
qreal left, right, bottom, top;
for(int i=1;i<edges[index]->size();i++)
{
curvePath.connectPath(edges[index]->at(i)->path());
}
QRectF curveRect = curvePath.boundingRect();
left = curveRect.left();
right = curveRect.right();
top = curveRect.top();
bottom = curveRect.bottom();
for(qreal i = left;i<right;i++)
for(qreal j = top;j<bottom;j++)
{
QPointF pointToCheck(i, j);
if(curvePath.contains(pointToCheck))
list.append(pointToCheck);
}
where edges is QList of QLists of QGraphicsPathItem's. It works fine in case of calculations (the point of applying this is to increase precision of calculation), but it really slows down my application since those calculations are made quite often.
Is there more efficient way to implement this?
Following up from my original post Three.JS Object following a spline path - rotation / tangent issues & constant speed issue, I am still having the issue that the object flips at certain points along the path.
View this happening on this fiddle: http://jsfiddle.net/jayfield1979/T2t59/7/
function moveBox() {
if (counter <= 1) {
box.position.x = spline.getPointAt(counter).x;
box.position.y = spline.getPointAt(counter).y;
tangent = spline.getTangentAt(counter).normalize();
axis.cross(up, tangent).normalize();
var radians = Math.acos(up.dot(tangent));
box.quaternion.setFromAxisAngle(axis, radians);
counter += 0.005
} else {
counter = 0;
}
}
The above code is what moves my objects along the defined spline path (an oval in this instance). It was mentioned by #WestLangley that: "Warning: cross product is not well-defined if the two vectors are parallel.".
As you can see, from the shape of the path, I am going to encounter a number of parallel vectors. Is there anything I can do to prevent this flipping from happening?
To answer the why question in the title. The reason its happening is that at some points on the curve the vector up (1,0,0) and the tangent are parallel. This means their cross product is zero and the construction of the quaternion fails.
You could follow WestLangley suggestion. You really want the up direction to be the normal to the plane the track is in.
Quaternion rotation is tricky to understand the setFromAxisAngle function rotates around the axis by a given angle.
If the track lies in the X-Y plane then we will want to rotate around the Z-axis. To find the angle use Math.atan2 to find the angle of the tangent
var angle = Math.atan2(tangent.y,tangent.x);
putting this together set
var ZZ = new THREE.Vector3( 0, 0, 1 );
and
tangent = spline.getTangentAt(counter).normalize();
var angle = Math.atan2(tangent.y,tangent.x);
box.quaternion.setFromAxisAngle(ZZ, angle);
If the track leaves the X-Y plane things will get trickier.
The setup
Draw XY-coordinate axes on a piece of paper. Write a word on it along X-axis, so that the word's centerpoint is at origo (half on positive side of X/Y, the other half on negative side of X/Y).
Now, if you flip the paper upside down you'll notice that the word is mirrored in relation to both X- and Y-axis. If you look from behind the paper, it's mirrored in relation to Y-axis. If you look at it from behind and upside down, it's mirrored in relation to X-axis.
Ok, I have points in 2D-plane (vertices) that are created in similar way at the origo and I need to apply exactly the same rule for them. To make things interesting:
The 2D plane is actually 3D, each point (vertex) being (x, y, 0). Initially the vertices are positioned to the origo and their normal is Pn(0,0,1). => Correctly seen when looked at from point Pn towards origo.
The vertex-plane has it's own rotation matrix [Rp] and position P(x,y,z) in the 3D-world. The rotation is applied before positioning.
The 3D world is "right handed". The viewer would be looking towards origo from some distance along positive Z-axis but the world is also oriented by rotation matrix [Rw]. [Rw] * (0,0,1) would point directly to the viewer's eye.
From those I need to calculate when the vertex-plane should be mirrored and by which axis. The mirroring itself can be done before applying [Rp] and P by:
Vertices vertices = Get2DPlanePoints();
int MirrorX = 1; // -1 to mirror, 1 NOT to mirror
int MirrorY = 1; // -1 to mirror, 1 NOT to mirror
Matrix WorldRotation = GetWorldRotationMatrix();
MirrorX = GetMirrorXFactor(WorldRotation);
MirrorY = GetMirrorYFactor(WorldRotation);
foreach(Vertex v in vertices)
{
v.X = v.X * MirrorX * MirrorY;
v.Y = V.Y * MirrorY;
}
// Apply rotation...
// Add position...
The question
So I need GetMirrorXFactor() & ..YFactor() -functions that return -1 if the viewer's eyepoint is at greater "X/Y"-angle than +-90 degrees in relation to the vertex-plane's normal after the rotation and world orientation. I have already solved this, but I'm looking for more "elegant" mathematics. I know that rotation matrices somehow contain info about how much is rotated by which axis and I believe that can be utilized here.
My Solution for MirrorX:
// Matrix multiplications. Vectors are vertical matrices here.
Pnr = [Rp] * Pn // Rotated vertices's normal
Pur = [Rp] * (0,1,0) // Rotated vertices's "up-vector"
Wnr = [Rw] * (0,0,1) // Rotated eye-vector with world's orientation
// = vector pointing directly at the viewer's eye
// Use rotated up-vector as a normal some new plane and project viewer's
// eye on it. dot = dot product between vectors.
Wnrx = Wnr - (Wnr dot Pur) * Pur // "X-projected" eye.
// Calculate angle between eye's X-component and plane's rotated normal.
// ||V|| = V's norm.
angle = arccos( (Wnrx dot Pnr) / ( ||Wnrx|| * ||Pnr|| ) )
if (angle > PI / 2)
MirrorX = -1; // DO mirror
else
MirrorX = 1; // DON'T mirror
Solution for mirrorY can be done in similar way using viewer's up and vertex-plane's right -vectors.
Better solution?
if (([Rp]*(1,0,0)) dot ([Rw]*(1,0,0))) < 0
MirrorX = -1; // DO mirror
else
MirrorX = 1; // DON'T mirror
if (([Rp]*(0,1,0)) dot ([Rw]*(0,1,0))) < 0
MirrorY = -1; // DO mirror
else
MirrorY = 1; // DON'T mirror
Explaining in more detail is difficult without diagrams, but if you have trouble with this solution we can work through some cases.
I'm writing a function graph plotting application with Qt. And I need an algorithm to determine the domain of the function.
here is the part were i draw the function graph
QPainterPath p(QPointF(-m_w/2,f(-m_w/2)));
m_painter->setPen(m_functionPen);
for(double x=-m_w/2, y; x<m_w/2; x++)
{
y = f(x/100);
p.lineTo(x,y*100);
}
m_painter->drawPath(p);
i think that if i find the domain i would stop the progrma from drawing out of it
Plotting software usually doesn't bother determining the domain; it just evaluates the function at every visible position and skips drawing any lines if the result was "undefined"/"NaN"/etc. Here is your code modified to do that skipping (untested, and I didn't match your brace style because I can't stand it):
QPainterPath p();
double previousY = 1/0 /* NaN */;
m_painter->setPen(m_functionPen);
for(double x=-m_w/2, y; x<m_w/2; x++) {
y = f(x/100);
if (y == y /* not-NaN test */) {
if (previousY == previousY) {
p.lineTo(x,y*100);
} else {
p.moveTo(x,y*100);
}
}
previousY = y;
}
m_painter->drawPath(p);
(I'm assuming that QPainterPath p() will construct an empty path. I'm not familiar with the library you are using.) Note that this now treats the first point like the other points for simplicity of coding.
(Also, this strategy will not produce a correct graph if you are evaluating a function like f(x) = 1/(x + 0.00005), because the undefined point will just be skipped over and you'll get a vertical line. There is no simple general solution for this problem.)
On the other hand, if you're trying to find reasonable bounds for your graph (your m_w variable), then determining the domain is the problem. In this case, it will depend on what kinds of functions you have and how they are represented.
I render isosurfaces with marching cubes, (or perhaps marching squares as this is 2D) and I want to do set operations like set difference, intersection and union. I thought this was easy to implement, by simply choosing between two vertex scalars from two different implicit surfaces, but it is not.
For my initial testing, I tried with two spheres circles, and the set operation difference. i.e A - B. One circle is moving and the other one is stationary. Here's the approach I tried when picking vertex scalars and when classifying corner vertices as inside or outside. The code is written in C++. OpenGL is used for rendering, but that's not important. Normal rendering without any CSG operations does give the expected result.
void march(const vec2& cmin, //min x and y for the grid cell
const vec2& cmax, //max x and y for the grid cell
std::vector<vec2>& tri,
float iso,
float (*cmp1)(const vec2&), //distance from stationary circle
float (*cmp2)(const vec2&) //distance from moving circle
)
{
unsigned int squareindex = 0;
float scalar[4];
vec2 verts[8];
/* initial setup of the grid cell */
verts[0] = vec2(cmax.x, cmax.y);
verts[2] = vec2(cmin.x, cmax.y);
verts[4] = vec2(cmin.x, cmin.y);
verts[6] = vec2(cmax.x, cmin.y);
float s1,s2;
/**********************************
********For-loop of interest******
*******Set difference between ****
*******two implicit surfaces******
**********************************/
for(int i=0,j=0; i<4; ++i, j+=2){
s1 = cmp1(verts[j]);
s2 = cmp2(verts[j]);
if((s1 < iso)){ //if inside circle1
if((s2 < iso)){ //if inside circle2
scalar[i] = s2; //then set the scalar to the moving circle
} else {
scalar[i] = s1; //only inside circle1
squareindex |= (1<<i); //mark as inside
}
}
else {
scalar[i] = s1; //inside neither circle
}
}
if(squareindex == 0)
return;
/* Usual interpolation between edge points to compute
the new intersection points */
verts[1] = mix(iso, verts[0], verts[2], scalar[0], scalar[1]);
verts[3] = mix(iso, verts[2], verts[4], scalar[1], scalar[2]);
verts[5] = mix(iso, verts[4], verts[6], scalar[2], scalar[3]);
verts[7] = mix(iso, verts[6], verts[0], scalar[3], scalar[0]);
for(int i=0; i<10; ++i){ //10 = maxmimum 3 triangles, + one end token
int index = triTable[squareindex][i]; //look up our indices for triangulation
if(index == -1)
break;
tri.push_back(verts[index]);
}
}
This gives me weird jaggies:
(source: mechcore.net)
It looks like the CSG operation is done without interpolation. It just "discards" the whole triangle. Do I need to interpolate in some other way, or combine the vertex scalar values? I'd love some help with this.
A full testcase can be downloaded HERE
EDIT: Basically, my implementation of marching squares works fine. It is my scalar field which is broken, and I wonder what the correct way would look like. Preferably I'm looking for a general approach to implement the three set operations I discussed above, for the usual primitives (circle, rectangle/square, plane)
EDIT 2: Here are some new images after implementing the answerer's whitepaper:
1.Difference
2.Intersection
3.Union
EDIT 3: I implemented this in 3D too, with proper shading/lighting:
1.Difference between a greater sphere and a smaller sphere
2.Difference between a greater sphere and a smaller sphere in the center, clipped by two planes on both sides, and then union with a sphere in the center.
3.Union between two cylinders.
This is not how you mix the scalar fields. Your scalars say one thing, but your flags whether you are inside or not say another. First merge the fields, then render as if you were doing a single compound object:
for(int i=0,j=0; i<4; ++i, j+=2){
s1 = cmp1(verts[j]);
s2 = cmp2(verts[j]);
s = max(s1, iso-s2); // This is the secret sauce
if(s < iso) { // inside circle1, but not inside circle2
squareindex |= (1<<i);
}
scalar[i] = s;
}
This article might be helpful: Combining CSG modeling with soft blending using
Lipschitz-based implicit surfaces.