I'm trying to use the code here
http://www.iforce2d.net/b2dtut/raycasting
So I can know whether a line cross a box2d object or not. It half works, in that when the line crosses the object it does show as an intersection, however, if you click before the object, it still shows as an intersection, as if it doesn't know the line stops before the object. From reading about this code, it should not do this.
Here's a screen shot of the issue.
And here's the method I'm using for the check
-(b2Vec2)rayCheckWithInput:(b2Vec2)p1 andX:(b2Vec2)p2
{
b2RayCastInput input;
input.p1 = p1;
input.p2 = p2;
input.maxFraction = 1;
//check every fixture of every body to find closest
float closestFraction = 1; //start with end of line as p2
b2Vec2 intersectionNormal(0,0);
//for (b2Body* b = self.world.world->GetBodyList(); b; b = b->GetNext()) {
for (b2Fixture* f = box.body->GetFixtureList(); f; f = f->GetNext()) {
b2RayCastOutput output;
if ( ! f->RayCast( &output, input, 0 ) )
{
NSLog(#"Not Intersected");
continue;
}
if ( output.fraction < closestFraction ) {
closestFraction = output.fraction;
intersectionNormal = output.normal;
NSLog(#"Intersected");
NSLog(#"%f %f,", output.normal.x, output.normal.y);
NSLog(#"%f", output.fraction);
}
else{
NSLog(#"Intersected2");
NSLog(#"%f %f,", output.normal.x, output.normal.y);
NSLog(#"%f", output.fraction);
}
}
//}
b2Vec2 intersectionPoint = p1 + closestFraction * (p2 - p1);
NSLog(#"I point %f, %f", intersectionPoint.x, intersectionPoint.y);
return intersectionPoint;
}
I can't see how in that check I can tell if the click point is before (no intersection) or beyond (intersection) the bird, it seems to give the same result regardless of either of those possibilities.
Any ideas?
It's a problem of unit of measure: you should convert p1 and p2 coordinates (pixel or points) to box2d coordinates (meters).
Related
I am trying to get the left polygonal chain given a set of consecutive points. (NOTE: edges are non-intersecting.)
Image 1. Sample polygon and its bound.
What I did was:
Get the minY, maxY and minX. (Bound.)
Find the point that contains minY (or maxY) then save it as the first point.
Save any points until point with minY or maxY is found while checking for point with minX.
If the same Y is found first, save it as the new first point and repeat from #3.
If other Y is found first and the saved points has minX, this is the chain. Otherwise, save as the new first point and repeat from #3.
Image 2. The left chain of points.
But using this steps might give wrong result for some polygon, like this:
Since one point is (minX, maxY), either of the side will be returned.
EDIT:
With the idea of the left-bottom- and left-top-most points, here is the current code that I am using:
Get the min (left-bottom-most) and max (left-top-most) point.
std::vector<Coord> ret;
size_t i = 0;
Coord minCoord = poly[i];
Coord maxCoord = poly[i];
size_t minIdx = -1;
size_t maxIdx = -1;
size_t cnt = poly.size();
i++;
for (; i < cnt; i++)
{
Coord c = poly[i];
if (c.y < minCoord.y) // new bottom
{
minCoord = c;
minIdx = i;
}
else if (c.y == minCoord.y) // same bottom
{
if (c.x < minCoord.x) // left most
{
minCoord = c;
minIdx = i;
}
}
if (c.y > maxCoord.y) // new top
{
maxCoord = c;
maxIdx = i;
}
else if (c.y == maxCoord.y) // same top
{
if (c.x < maxCoord.x) // left most
{
maxCoord = c;
maxIdx = i;
}
}
}
Get the points connected to the max point.
i = maxIdx;
Coord mid = poly[i];
Coord ray1 = poly[(i + cnt - 1) % cnt];
Coord ray2 = poly[(i + 1) % cnt];
Get which has smallest angle. This will be the path we will follow.
double rad1 = Pts2Rad(mid, ray1);
double rad2 = Pts2Rad(mid, ray2);
int step = 1;
if (rad1 < rad2)
step = cnt - 1;
Save the points.
while (i != minIdx)
{
ret.push_back(poly[i]);
i = (i + step) % cnt;
}
ret.push_back(poly[minIdx]);
To be specific, I am assuming that no vertex is duplicated and define the "left chain" as the sequence of vertices from the original polygon loop that goes from the leftmost vertex in the top side of the bounding box, to the leftmost vertex in the bottom side of the bounding box. [In case the top and bottom sides coincide, these two vertices also coincide; I leave it to you what to return in this case.]
To obtain these, you can scan all vertices and keep the left-topmost so far and left-bottommost so far. Then compare to the next vertex. If above the left-topmost, becomes the new lef-topmost. If at the same level and to the left, becomes the new left-topmost. Similarly for the left-bottommost.
I am kinda stuck with my basic voxel physics right now. It's very, very choppy and I am pretty sure my maths is broken somewhere, but let's see what you have to say:
// SOMEWHERE AT CLASS LEVEL (so not being reinstantiated every frame, but persisted instead!)
glm::vec3 oldPos;
// ACTUAL IMPL
glm::vec3 distanceToGravityCenter =
this->entity->getPosition() -
((this->entity->getPosition() - gravityCenter) * 0.005d); // TODO multiply by time
if (!entity->grounded) {
glm::vec3 entityPosition = entity->getPosition();
if (getBlock(floorf(entityPosition.x), floorf(entityPosition.y), floorf(entityPosition.z))) {
glm::vec3 dir = entityPosition - oldPos; // Actually no need to normalize as we check for lesser, bigger or equal to 0
std::cout << "falling dir: " << glm::to_string(dir) << std::endl;
// Calculate offset (where to put after hit)
int x = dir.x;
int y = dir.y;
int z = dir.z;
if (dir.x >= 0) {
x = -1;
} else if (dir.x < 0) {
x = 1;
}
if (dir.y >= 0) {
y = -1;
} else if (dir.y < 0) {
y = 1;
}
if (dir.z >= 0) {
z = -1;
} else if (dir.z < 0) {
z = 1;
}
glm::vec3 newPos = oldPos + glm::vec3(x, y, z);
this->entity->setPosition(newPos);
entity->grounded = true; // If some update happens, grounded needs to be changed
} else {
oldPos = entity->getPosition();
this->entity->setPosition(distanceToGravityCenter);
}
}
Basic idea was to determine from which direction entityt would hit the surface and then just position it one "unit" back into that direction. But obviously I am doing something wrong as that will always move entity back to the point where it came from, effectively holding it at the spawn point.
Also this could probably be much easier and I am overthinking it.
As #CompuChip already pointed out, your ifs could be further simplified.
But what is more important is one logical issue that would explain the "choppiness" you describe (Sadly you did not provide any footage, so this is my best guess)
From the code you posted:
First you check if entity is grounded. If so you continue with checking if there is a collision and lastly, if there is not, you set the position.
You have to invert that a bit.
Save old position
Check if grounded
Set the position already to the new one!
Do collision detection
Reset to old position IF you registered a collision!
So basically:
glm::vec3 distanceToGravityCenter =
this->entity->getPosition() -
((this->entity->getPosition() - gravityCenter) * 0.005d); // TODO multiply by time
oldPos = entity->getPosition(); // 1.
if (!entity->grounded) { // 2.
this->fallingStar->setPosition(distanceToGravityPoint); // 3
glm::vec3 entityPosition = entity->getPosition();
if (getBlock(floorf(entityPosition.x), floorf(entityPosition.y), floorf(entityPosition.z))) { // 4, 5
this->entity->setPosition(oldPos);
entity->grounded = true; // If some update happens, grounded needs to be changed
}
}
This should get you started :)
I want to elaborate a bit more:
If you check for collision first and then set position you create an "infinite loop" upon first collision/hit as you collide, then if there is a collision (which there is) you set back to the old position. Basically just mathematic inaccuracy will make you move, as on every check you are set back to the old position.
Consider the if-statements for one of your coordinates:
if (dir.x >= 0) {
x = -1;
}
if (dir.x < 0) {
x = 1;
}
Suppose that dir.x < 0. Then you will skip the first if, enter the second, and x will be set to 1.
If dir.x >= 0, you will enter the first if and x will be set to -1. Now x < 0 is true, so you will enter the second if as well, and x gets set to 1 again.
Probably what you want is to either set x to 1 or to -1, depending on dir.x. You should only execute the second if when the first one was not entered, so you need an else if:
if (dir.x >= 0) {
x = -1;
} else if (dir.x < 0) {
x = 1;
}
which can be condensed, if you so please, into
x = (dir.x >= 0) ? -1 : 1;
I have a set of points that I'm trying to sort in ccw order or cw order from their angle. I want the points to be sorted in a way that they could form a polygon with no splits in its region or intersections. This is difficult because in most cases, it would be a concave polygon.
point centroid;
int main( int argc, char** argv )
{
// I read a set of points into a struct point array: points[n]
// Find centroid
double sx = 0; double sy = 0;
for (int i = 0; i < n; i++)
{
sx += points[i].x;
sy += points[i].y;
}
centroid.x = sx/n;
centroid.y = sy/n;
// sort points using in polar order using centroid as reference
std::qsort(&points, n, sizeof(point), polarOrder);
}
// -1 ccw, 1 cw, 0 collinear
int orientation(point a, point b, point c)
{
double area2 = (b.x-a.x)*(c.y-a.y) - (b.y-a.y)*(c.x-a.x);
if (area2 < 0) return -1;
else if (area2 > 0) return +1;
else return 0;
}
// compare other points relative to polar angle they make with this point
// (where the polar angle is between 0 and 2pi)
int polarOrder(const void *vp1, const void *vp2)
{
point *p1 = (point *)vp1;
point *p2 = (point *)vp2;
// translation
double dx1 = p1->x - centroid.x;
double dy1 = p1->y - centroid.y;
double dx2 = p2->x - centroid.x;
double dy2 = p2->y - centroid.y;
if (dy1 >= 0 && dy2 < 0) { return -1; } // p1 above and p2 below
else if (dy2 >= 0 && dy1 < 0) { return 1; } // p1 below and p2 above
else if (dy1 == 0 && dy2 ==0) { // 3-collinear and horizontal
if (dx1 >= 0 && dx2 < 0) { return -1; }
else if (dx2 >= 0 && dx1 < 0) { return 1; }
else { return 0; }
}
else return -orientation(centroid,*p1,*p2); // both above or below
}
It looks like the points are sorted accurately(pink) until they "cave" in, in which case the algorithm skips over these points then continues.. Can anyone point me into the right direction to sort the points so that they form the polygon I'm looking for?
Raw Point Plot - Blue, Pink Points - Sorted
Point List: http://pastebin.com/N0Wdn2sm (You can ignore the 3rd component, since all these points lie on the same plane.)
The code below (sorry it's C rather than C++) sorts correctly as you wish with atan2.
The problem with your code may be that it attempts to use the included angle between the two vectors being compared. This is doomed to fail. The array is not circular. It has a first and a final element. With respect to the centroid, sorting an array requires a total polar order: a range of angles such that each point corresponds to a unique angle regardless of the other point. The angles are the total polar order, and comparing them as scalars provides the sort comparison function.
In this manner, the algorithm you proposed is guaranteed to produce a star-shaped polyline. It may oscillate wildly between different radii (...which your data do! Is this what you meant by "caved in"? If so, it's a feature of your algorithm and data, not an implementation error), and points corresponding to exactly the same angle might produce edges that coincide (lie directly on top of each other), but the edges won't cross.
I believe that your choice of centroid as the polar origin is sufficient to guarantee that connecting the ends of the polyline generated as above will produce a full star-shaped polygon, however, I don't have a proof.
Result plotted with Excel
Note you can guess from the nearly radial edges where the centroid is! This is the "star shape" I referred to above.
To illustrate this is really a star-shaped polygon, here is a zoom in to the confusing lower left corner:
If you want a polygon that is "nicer" in some sense, you will need a fancier (probably much fancier) algorithm, e.g. the Delaunay triangulation-based ones others have referred to.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct point {
double x, y;
};
void print(FILE *f, struct point *p) {
fprintf(f, "%f,%f\n", p->x, p->y);
}
// Return polar angle of p with respect to origin o
double to_angle(const struct point *p, const struct point *o) {
return atan2(p->y - o->y, p->x - o->x);
}
void find_centroid(struct point *c, struct point *pts, int n_pts) {
double x = 0, y = 0;
for (int i = 0; i < n_pts; i++) {
x += pts[i].x;
y += pts[i].y;
}
c->x = x / n_pts;
c->y = y / n_pts;
}
static struct point centroid[1];
int by_polar_angle(const void *va, const void *vb) {
double theta_a = to_angle(va, centroid);
double theta_b = to_angle(vb, centroid);
return theta_a < theta_b ? -1 : theta_a > theta_b ? 1 : 0;
}
void sort_by_polar_angle(struct point *pts, int n_pts) {
find_centroid(centroid, pts, n_pts);
qsort(pts, n_pts, sizeof pts[0], by_polar_angle);
}
int main(void) {
FILE *f = fopen("data.txt", "r");
if (!f) return 1;
struct point pts[10000];
int n_pts, n_read;
for (n_pts = 0;
(n_read = fscanf(f, "%lf%lf%*f", &pts[n_pts].x, &pts[n_pts].y)) != EOF;
++n_pts)
if (n_read != 2) return 2;
fclose(f);
sort_by_polar_angle(pts, n_pts);
for (int i = 0; i < n_pts; i++)
print(stdout, pts + i);
return 0;
}
Well, first and foremost, I see centroid declared as a local variable in main. Yet inside polarOrder you are also accessing some centroid variable.
Judging by the code you posted, that second centroid is a file-scope variable that you never initialized to any specific value. Hence the meaningless results from your comparison function.
The second strange detail in your code is that you do return -orientation(centroid,*p1,*p2) if both points are above or below. Since orientation returns -1 for CCW and +1 for CW, it should be just return orientation(centroid,*p1,*p2). Why did you feel the need to negate the result of orientation?
Your original points don't appear form a convex polygon, so simply ordering them by angle around a fixed centroid will not necessarily result in a clean polygon. This is a non-trivial problem, you may want to research Delaunay triangulation and/or gift wrapping algorithms, although both would have to be modified because your polygon is concave. The answer here is an interesting example of a modified gift wrapping algorithm for concave polygons. There is also a C++ library called PCL that may do what you need.
But...if you really do want to do a polar sort, your sorting functions seem more complex than necessary. I would sort using atan2 first, then optimize it later once you get the result you want if necessary. Here is an example using lambda functions:
#include <algorithm>
#include <math.h>
#include <vector>
int main()
{
struct point
{
double x;
double y;
};
std::vector< point > points;
point centroid;
// fill in your data...
auto sort_predicate = [¢roid] (const point& a, const point& b) -> bool {
return atan2 (a.x - centroid.x, a.y - centroid.y) <
atan2 (b.x - centroid.x, b.y - centroid.y);
};
std::sort (points.begin(), points.end(), sort_predicate);
}
I have line that is defined as two points.
start = (xs,ys)
end = (xe, ye)
Drawing function that I'm using Only accepts lines that are fully in screen coordinates.
Screen size is (xSize, ySize).
Top left corner is (0,0). Bottom right corner is (xSize, ySize).
Some other funcions gives me line that that is defined for example as start(-50, -15) end(5000, 200). So it's ends are outside of screen size.
In C++
struct Vec2
{
int x, y
};
Vec2 start, end //This is all little bit pseudo code
Vec2 screenSize;//You can access coordinates like start.x end.y
How can I calculate new start and endt that is at the screen edge, not outside screen.
I know how to do it on paper. But I can't transfer it to c++.
On paper I'm sershing for point that belongs to edge and line. But it is to much calculations for c++.
Can you help?
There are many line clipping algorithms like:
Cohen–Sutherland wikipedia page with implementation
Liang–Barsky wikipedia page
Nicholl–Lee–Nicholl (NLN)
and many more. see Line Clipping on wikipedia
[EDIT1]
See below figure:
there are 3 kinds of start point:
sx > 0 and sy < 0 (red line)
sx < 0 and sy > 0 (yellow line)
sx < 0 and sy < 0 (green and violet lines)
In situations 1 and 2 simply find Xintersect and Yintersect respectively and choose them as new start point.
As you can see, there are 2 kinds of lines in situation 3. In this situation find Xintersect and Yintersect and choose the intersect point near the end point which is the point that has minimum distance to endPoint.
min(distance(Xintersect, endPoint), distance(Yintersect, endPoint))
[EDIT2]
// Liang-Barsky function by Daniel White # http://www.skytopia.com/project/articles/compsci/clipping.html
// This function inputs 8 numbers, and outputs 4 new numbers (plus a boolean value to say whether the clipped line is drawn at all).
//
bool LiangBarsky (double edgeLeft, double edgeRight, double edgeBottom, double edgeTop, // Define the x/y clipping values for the border.
double x0src, double y0src, double x1src, double y1src, // Define the start and end points of the line.
double &x0clip, double &y0clip, double &x1clip, double &y1clip) // The output values, so declare these outside.
{
double t0 = 0.0; double t1 = 1.0;
double xdelta = x1src-x0src;
double ydelta = y1src-y0src;
double p,q,r;
for(int edge=0; edge<4; edge++) { // Traverse through left, right, bottom, top edges.
if (edge==0) { p = -xdelta; q = -(edgeLeft-x0src); }
if (edge==1) { p = xdelta; q = (edgeRight-x0src); }
if (edge==2) { p = -ydelta; q = -(edgeBottom-y0src);}
if (edge==3) { p = ydelta; q = (edgeTop-y0src); }
r = q/p;
if(p==0 && q<0) return false; // Don't draw line at all. (parallel line outside)
if(p<0) {
if(r>t1) return false; // Don't draw line at all.
else if(r>t0) t0=r; // Line is clipped!
} else if(p>0) {
if(r<t0) return false; // Don't draw line at all.
else if(r<t1) t1=r; // Line is clipped!
}
}
x0clip = x0src + t0*xdelta;
y0clip = y0src + t0*ydelta;
x1clip = x0src + t1*xdelta;
y1clip = y0src + t1*ydelta;
return true; // (clipped) line is drawn
}
Here is a function I wrote. It cycles through all 4 planes (left, top, right, bottom) and clips each point by the plane.
// Clips a line segment to an axis-aligned rectangle
// Returns true if clipping is successful
// Returns false if line segment lies outside the rectangle
bool clipLineToRect(int a[2], int b[2],
int xmin, int ymin, int xmax, int ymax)
{
int mins[2] = {xmin, ymin};
int maxs[2] = {xmax, ymax};
int normals[2] = {1, -1};
for (int axis=0; axis<2; axis++) {
for (int plane=0; plane<2; plane++) {
// Check both points
for (int pt=1; pt<=2; pt++) {
int* pt1 = pt==1 ? a : b;
int* pt2 = pt==1 ? b : a;
// If both points are outside the same plane, the line is
// outside the rectangle
if ( (a[0]<xmin && b[0]<xmin) || (a[0]>xmax && b[0]>xmax) ||
(a[1]<ymin && b[1]<ymin) || (a[1]>ymax && b[1]>ymax)) {
return false;
}
const int n = normals[plane];
if ( (n==1 && pt1[axis]<mins[axis]) || // check left/top plane
(n==-1 && pt1[axis]>maxs[axis]) ) { // check right/bottom plane
// Calculate interpolation factor t using ratio of signed distance
// of each point from the plane
const float p = (n==1) ? mins[axis] : maxs[axis];
const float q1 = pt1[axis];
const float q2 = pt2[axis];
const float d1 = n * (q1-p);
const float d2 = n * (q2-p);
const float t = d1 / (d1-d2);
// t should always be between 0 and 1
if (t<0 || t >1) {
return false;
}
// Interpolate to find the new point
pt1[0] = (int)(pt1[0] + (pt2[0] - pt1[0]) * t );
pt1[1] = (int)(pt1[1] + (pt2[1] - pt1[1]) * t );
}
}
}
}
return true;
}
Example Usage:
void testClipLineToRect()
{
int screenWidth = 320;
int screenHeight = 240;
int xmin=0;
int ymin=0;
int xmax=screenWidth-1;
int ymax=screenHeight-1;
int a[2] = {-10, 10};
int b[2] = {300, 250};
printf("Before clipping:\n\ta={%d, %d}\n\tb=[%d, %d]\n",
a[0], a[1], b[0], b[1]);
if (clipLineToRect(a, b, xmin, ymin, xmax, ymax)) {
printf("After clipping:\n\ta={%d, %d}\n\tb=[%d, %d]\n",
a[0], a[1], b[0], b[1]);
}
else {
printf("clipLineToRect returned false\n");
}
}
Output:
Before clipping:
a={-10, 10}
b=[300, 250]
After clipping:
a={0, 17}
b=[285, 239]
I'm making a vector drawing application. I use this algorithm to generate outlines.
This algorthm works well, except it does not close the outline as seen here:
alt text http://img716.imageshack.us/img716/2633/noclosure.png
I'm not sure what I should do to ensure that it always closes the outline. I tried inserting the last vertex at position[0] in the std::vector but this did not help.
DOUBLEPOINT looks like:
struct DOUBLEPOINT {
double point[2];
};
How can I make it so it always properly closes the shape even on sharp corners?
Thanks
Try appending a copy of the first and second points to the end of the vector before plotting it. Your first and last line segment will overlap, but it should ensure that all the points are connected and all corners are rounded similarly.
I usually just use modulus:
nxt = input[(i+1) % input.size()];
How about:
for( size_t i = 0; i < input.size(); ++i )
{
POINTFLOAT cur;
POINTFLOAT nxt;
if( i == input.size() - 1 )
{
cur.x = input[i].point[0];
cur.y = input[i].point[1];
nxt.x = input[0].point[0];
nxt.y = input[0].point[1];
}
else
{
cur.x = input[i].point[0];
cur.y = input[i].point[1];
nxt.x = input[i+1].point[0];
nxt.y = input[i+1].point[1];
}