I want to randomly generate sprites on circumference of circle , but even after research of several hours , I couldn't come up with any solution.
That's what I can made till now
I've used this formula for it :
Sprite * pin = Sprite::create("pin.png");
pin->setPosition(Vec2((_circle->getContentSize().width/2)*(0.7/3), _circle->getContentSize().height*0.7));
Sprite * pin2 = Sprite::create("pin.png");
pin2->setPosition(Vec2((_circle->getContentSize().width/2)*(0.6/3), _circle->getContentSize().height*0.6));
Sprite * pin3 = Sprite::create("pin.png");
pin3->setPosition(Vec2((_circle->getContentSize().width/2)*(0.8/3), _circle->getContentSize().height*0.8));
Sprite * pin4 = Sprite::create("pin.png");
pin4->setPosition(Vec2((_circle->getContentSize().width/2)*(0.9/3), _circle->getContentSize().height*0.9));
Sprite * pin5 = Sprite::create("pin.png");
pin5->setPosition(Vec2((_circle->getContentSize().width/2)*(1/3), _circle->getContentSize().height));
_circle->addChild(pin);
_circle->addChild(pin2);
_circle->addChild(pin3);
_circle->addChild(pin4);
_circle->addChild(pin5);
But I want something like that(with correct angle which I couldn't do in sample image)
Please suggest some precise solution for it. Thanks for your time!
Basic trig stuff -- sin and cos are your friends.
Example:
const float circle_x = ...;
const float circle_y = ...;
const float circle_radius = ...;
const float angle = ...;
const float x = cos(angle)*circle_radius + circle_x;
const float y = sin(angle)*circle_radius + circle_y;
// Draw stuff at (x, y).
First, it's "circumference" not "circumstance" (that will help with your searches)
Second, you are using the size of the image, not the circle inside the image.
Third, you will need to use basic trigonometry for the solution. Determining points on a circle require the use of sin and cos functions. After you find the center of the circle and it's radius, these should be easy to calculate with just a little bit of research.
I'm drawing a point which reflects my current location. I've got some shapes on the map as well. However, due to changing GPS accuracy I need to show that I've crossed the line on map even if a LatLng point is still before it but the accuracy is ~20(m).
What I have:
- line's start and end points
- circle's center
- circle's radius (in meters)
I'm able to calculate the distance between line and circle's center point but that gives me the value like: 0.00987506668990474 which gives me really nothing because this value does not reflect the distance in meters and I can't really convert that to meters and compare with accuracy radius.
I'm not even sure if I'm on good target to get that information. Maybe there's some another method to get the information if there's an intersection or not.
thanks for any hints
update:
using the distance * 110km I'm getting much better results.
Happy face = intersection not detected - distance < radius, Sad face = intersection detected - distance > radius.
Yellow zone edge is mine line segment. As you can see it works when the intersection is on the top(/bottom), but not on the left(/right) side.
That's my calculation algorithm calculating distance from line segment to point. Forgive the mess... I'm still struggling with that, so it's not optimized yet:
public static double pointLineSegmentDistance(final List<Double> point, final List<List<Double>> line)
{
List<Double> v = line.get(0);
List<Double> w = line.get(1);
double d = pointPointSquaredDistance(v, w);
double t;
List<Double> calculateThis;
if (d > 0)
{
boolean test = (t = ((point.get(0) - v.get(0)) * (w.get(0) - v.get(0)) + (point.get(1) - v.get(1)) * (w.get(1) - v.get(1))) / d) < 0;
if (test)
{
calculateThis = v;
}
else
{
if (t > 1)
{
calculateThis = w;
}
else
{
calculateThis = new ArrayList<Double>();
calculateThis.add(v.get(0) + t * (w.get(0) - v.get(0)));
calculateThis.add(v.get(1) + t * (w.get(1) - v.get(1)));
}
}
}
else
{
calculateThis = v;
}
return Math.sqrt(pointPointSquaredDistance(point, calculateThis));
}
public static double pointPointSquaredDistance(final List<Double> v, final List<Double> w)
{
double dx = v.get(0) - w.get(0);
double dy = v.get(1) - w.get(1);
return dx * dx + dy * dy;
}
final List point - contains 2 elements - (0) latitude (1) longitude
List> line - contains 2 lists - (0) line segment start point (0/0)lat/(0/1)long (1) line segment end point (1/0)lat/(1/1)long
Finally I've solved my problem. It comes the solution is really very simple. Much simpler than I thought. There's maputil library available here:
https://github.com/googlemaps/android-maps-utils
That library privides a function named "isLocationOnPath":
boolean isLocationOnPath(LatLng point, List<LatLng> polyline, boolean geodesic, double tolerance)
Point - is a center or my point.
Polyline - for my needs is an edge of a polygon,
Geodesic - true (of course)
Tolerance - (in meters) is my accuracy taken from GPS accuracy - basicaly is a circle radius.
Details about the method:
http://googlemaps.github.io/android-maps-utils/javadoc/com/google/maps/android/PolyUtil.html#isLocationOnPath-LatLng-java.util.List-boolean-
I hope that helps.
You have two possibilities:
1) The cheap way:
You calcualte the distances between line start/end point and circle center with the android api (distanceBetween()). The result will be in meters, and correctby means of point to point distance calculation. Take the minimum of both start / end point to center circle distances.
If the line segment is very long related to the circle radius, the normal distance of the line to center, using this kind of caluclation is wrong
2) The better way, but more complex:
You transform the line start and end and the circle center to cartesian x,y space with unit = meter and calculate the distance to the line
segment using cartesian math. (see distance to line segment calculation)
First check option1 because it is just one line of code.
Your result of 0.00987506668990474 this looks like a distance in degrees, instead of meters. multiply it with 111km gives a value of about 1km. (at equator)
I am working on image editing using OPENGL in Android and I have applied filter to an image using photoshop curve now I want to reproduce the same in Android using glsl. Is there any formula to calculate single fragment color using photoshops curve output value?
EDIT
The math behind the photoshop curve has already been answered in this question
How to recreate the math behind photoshop curves but I am not very clear about how to reproduce the same in glsl fragment shader.
Screen Shot of my photoshop curve
You're after a function fragColour = curves(inColour, constants...). If you have just the one curve for red, green and blue you apply the same curve to each individually. This answer has a link (below) to code which plots points along the function. The key line is:
double y = ...
Which you'd return from curves. The variable x in the loop is your inColour. All you need now is the constants which come from the points and the second derivative sd arrays. These you'll have to pass in as uniforms. The function first has to figure out which point each colour x is between (finding cur, next, sd[i] and sd[i+1]), then evaluate and return y.
EDIT:
If you just want to apply some curve you've created in photoshop then the problem is much simpler. The easiest way is to create a simple function that gives a similar shape. I use these as a starting point. A gamma correction curve is also quite common.
This is overkill, but if you do need a more exact result, you could create an image with a linear ramp (e.g. 255 pixels from black to white), apply your filter to it in photoshop and the result becomes a lookup table. Passing in all 255 values to a shader is expensive so if it's a smooth curve you could try some curve fitting tools (for example).
Once you have a function, simply apply it to your colour in GLSL. Applying a gamma curve for example is done like this:
fragColour = vec4(pow(inColour.rgb, 1.0 / gamma.rgb), inColour.a);
EDIT2:
The curve you have looks very similar to this:
fragColour = vec4(pow(inColour.rgb, 1.0 / vec3(0.6)), inColour.a);
Or even simpler:
fragColour = vec4(inColour.rgb * inColour.rgb, inColour.a);
Just in case the link dies, I'll copy the code here (not that I've tested it):
Point[] points = /* liste de points, triés par "x" croissants */
double[] sd = secondDerivative(points);
for(int i=0;i<points.length-1;i++) {
Point cur = points[i];
Point next = points[i+1];
for(int x=cur.x;x<next.x;x++) {
double t = (double)(x-cur.x)/(next.x-cur.x);
double a = 1-t;
double b = t;
double h = next.x-cur.x;
double y= a*cur.y + b*next.y + (h*h/6)*( (a*a*a-a)*sd[i]+ (b*b*b-b)*sd[i+1] );
draw(x,y); /* ou tout autre utilisation */
}
}
And the second derivative:
public static double[] secondDerivative(Point... P) {
int n = P.length;
double yp1=0.0;
double ypn=0.0;
// build the tridiagonal system
// (assume 0 boundary conditions: y2[0]=y2[-1]=0)
double[][] matrix = new double[n][3];
double[] result = new double[n];
matrix[0][1]=1;
for(int i=1;i<n-1;i++) {
matrix[i][0]=(double)(P[i].x-P[i-1].x)/6;
matrix[i][1]=(double)(P[i+1].x-P[i-1].x)/3;
matrix[i][2]=(double)(P[i+1].x-P[i].x)/6;
result[i]=(double)(P[i+1].y-P[i].y)/(P[i+1].x-P[i].x) - (double)(P[i].y-P[i-1].y)/(P[i].x-P[i-1].x);
}
matrix[n-1][1]=1;
// solving pass1 (up->down)
for(int i=1;i<n;i++) {
double k = matrix[i][0]/matrix[i-1][1];
matrix[i][1] -= k*matrix[i-1][2];
matrix[i][0] = 0;
result[i] -= k*result[i-1];
}
// solving pass2 (down->up)
for(int i=n-2;i>=0;i--) {
double k = matrix[i][2]/matrix[i+1][1];
matrix[i][1] -= k*matrix[i+1][0];
matrix[i][2] = 0;
result[i] -= k*result[i+1];
}
// return second derivative value for each point P
double[] y2 = new double[n];
for(int i=0;i<n;i++) y2[i]=result[i]/matrix[i][1];
return y2;
}
From the man page for XFillPolygon:
If shape is Complex, the path may self-intersect. Note that contiguous coincident points in the path are not treated as self-intersection.
If shape is Convex, for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined.
If shape is Nonconvex, the path does not self-intersect, but the shape is not wholly convex. If known by the client, specifying Nonconvex instead of Complex may improve performance. If you specify Nonconvex for a self-intersecting path, the graphics results are undefined.
I am having performance problems with fill XFillPolygon and, as the man page suggests, the first step I want to take is to specify the correct shape of the polygon. I am currently using Complex to be on the safe side.
Is there an efficient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non-convex or complex?
You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.
In contrast, consider the case where the polygon is not self-intersecting, and it consists of a set of points in a list where the consecutive points form the boundary. In this case it is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):
For each consecutive pair of edges of the polygon (each triplet of points), compute the z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross product of these vectors:
given p[k], p[k+1], p[k+2] each with coordinates x, y:
dx1 = x[k+1]-x[k]
dy1 = y[k+1]-y[k]
dx2 = x[k+2]-x[k+1]
dy2 = y[k+2]-y[k+1]
zcrossproduct = dx1*dy2 - dy1*dx2
The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.
If there are N points, make sure you calculate N cross products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).
If the polygon is self-intersecting, then it fails the technical definition of convexity even if its directed angles are all in the same direction, in which case the above approach would not produce the correct result.
This question is now the first item in either Bing or Google when you search for "determine convex polygon." However, none of the answers are good enough.
The (now deleted) answer by #EugeneYokota works by checking whether an unordered set of points can be made into a convex polygon, but that's not what the OP asked for. He asked for a method to check whether a given polygon is convex or not. (A "polygon" in computer science is usually defined [as in the XFillPolygon documentation] as an ordered array of 2D points, with consecutive points joined with a side as well as the last point to the first.) Also, the gift wrapping algorithm in this case would have the time-complexity of O(n^2) for n points - which is much larger than actually needed to solve this problem, while the question asks for an efficient algorithm.
#JasonS's answer, along with the other answers that follow his idea, accepts star polygons such as a pentagram or the one in #zenna's comment, but star polygons are not considered to be convex. As
#plasmacel notes in a comment, this is a good approach to use if you have prior knowledge that the polygon is not self-intersecting, but it can fail if you do not have that knowledge.
#Sekhat's answer is correct but it also has the time-complexity of O(n^2) and thus is inefficient.
#LorenPechtel's added answer after her edit is the best one here but it is vague.
A correct algorithm with optimal complexity
The algorithm I present here has the time-complexity of O(n), correctly tests whether a polygon is convex or not, and passes all the tests I have thrown at it. The idea is to traverse the sides of the polygon, noting the direction of each side and the signed change of direction between consecutive sides. "Signed" here means left-ward is positive and right-ward is negative (or the reverse) and straight-ahead is zero. Those angles are normalized to be between minus-pi (exclusive) and pi (inclusive). Summing all these direction-change angles (a.k.a the deflection angles) together will result in plus-or-minus one turn (i.e. 360 degrees) for a convex polygon, while a star-like polygon (or a self-intersecting loop) will have a different sum ( n * 360 degrees, for n turns overall, for polygons where all the deflection angles are of the same sign). So we must check that the sum of the direction-change angles is plus-or-minus one turn. We also check that the direction-change angles are all positive or all negative and not reverses (pi radians), all points are actual 2D points, and that no consecutive vertices are identical. (That last point is debatable--you may want to allow repeated vertices but I prefer to prohibit them.) The combination of those checks catches all convex and non-convex polygons.
Here is code for Python 3 that implements the algorithm and includes some minor efficiencies. The code looks longer than it really is due to the the comment lines and the bookkeeping involved in avoiding repeated point accesses.
TWO_PI = 2 * pi
def is_convex_polygon(polygon):
"""Return True if the polynomial defined by the sequence of 2D
points is 'strictly convex': points are valid, side lengths non-
zero, interior angles are strictly between zero and a straight
angle, and the polygon does not intersect itself.
NOTES: 1. Algorithm: the signed changes of the direction angles
from one side to the next side must be all positive or
all negative, and their sum must equal plus-or-minus
one full turn (2 pi radians). Also check for too few,
invalid, or repeated points.
2. No check is explicitly done for zero internal angles
(180 degree direction-change angle) as this is covered
in other ways, including the `n < 3` check.
"""
try: # needed for any bad points or direction changes
# Check for too few points
if len(polygon) < 3:
return False
# Get starting information
old_x, old_y = polygon[-2]
new_x, new_y = polygon[-1]
new_direction = atan2(new_y - old_y, new_x - old_x)
angle_sum = 0.0
# Check each point (the side ending there, its angle) and accum. angles
for ndx, newpoint in enumerate(polygon):
# Update point coordinates and side directions, check side length
old_x, old_y, old_direction = new_x, new_y, new_direction
new_x, new_y = newpoint
new_direction = atan2(new_y - old_y, new_x - old_x)
if old_x == new_x and old_y == new_y:
return False # repeated consecutive points
# Calculate & check the normalized direction-change angle
angle = new_direction - old_direction
if angle <= -pi:
angle += TWO_PI # make it in half-open interval (-Pi, Pi]
elif angle > pi:
angle -= TWO_PI
if ndx == 0: # if first time through loop, initialize orientation
if angle == 0.0:
return False
orientation = 1.0 if angle > 0.0 else -1.0
else: # if other time through loop, check orientation is stable
if orientation * angle <= 0.0: # not both pos. or both neg.
return False
# Accumulate the direction-change angle
angle_sum += angle
# Check that the total number of full turns is plus-or-minus 1
return abs(round(angle_sum / TWO_PI)) == 1
except (ArithmeticError, TypeError, ValueError):
return False # any exception means not a proper convex polygon
The following Java function/method is an implementation of the algorithm described in this answer.
public boolean isConvex()
{
if (_vertices.size() < 4)
return true;
boolean sign = false;
int n = _vertices.size();
for(int i = 0; i < n; i++)
{
double dx1 = _vertices.get((i + 2) % n).X - _vertices.get((i + 1) % n).X;
double dy1 = _vertices.get((i + 2) % n).Y - _vertices.get((i + 1) % n).Y;
double dx2 = _vertices.get(i).X - _vertices.get((i + 1) % n).X;
double dy2 = _vertices.get(i).Y - _vertices.get((i + 1) % n).Y;
double zcrossproduct = dx1 * dy2 - dy1 * dx2;
if (i == 0)
sign = zcrossproduct > 0;
else if (sign != (zcrossproduct > 0))
return false;
}
return true;
}
The algorithm is guaranteed to work as long as the vertices are ordered (either clockwise or counter-clockwise), and you don't have self-intersecting edges (i.e. it only works for simple polygons).
Here's a test to check if a polygon is convex.
Consider each set of three points along the polygon--a vertex, the vertex before, the vertex after. If every angle is 180 degrees or less you have a convex polygon. When you figure out each angle, also keep a running total of (180 - angle). For a convex polygon, this will total 360.
This test runs in O(n) time.
Note, also, that in most cases this calculation is something you can do once and save — most of the time you have a set of polygons to work with that don't go changing all the time.
To test if a polygon is convex, every point of the polygon should be level with or behind each line.
Here's an example picture:
The answer by #RoryDaulton
seems the best to me, but what if one of the angles is exactly 0?
Some may want such an edge case to return True, in which case, change "<=" to "<" in the line :
if orientation * angle < 0.0: # not both pos. or both neg.
Here are my test cases which highlight the issue :
# A square
assert is_convex_polygon( ((0,0), (1,0), (1,1), (0,1)) )
# This LOOKS like a square, but it has an extra point on one of the edges.
assert is_convex_polygon( ((0,0), (0.5,0), (1,0), (1,1), (0,1)) )
The 2nd assert fails in the original answer. Should it?
For my use case, I would prefer it didn't.
This method would work on simple polygons (no self intersecting edges) assuming that the vertices are ordered (either clockwise or counter)
For an array of vertices:
vertices = [(0,0),(1,0),(1,1),(0,1)]
The following python implementation checks whether the z component of all the cross products have the same sign
def zCrossProduct(a,b,c):
return (a[0]-b[0])*(b[1]-c[1])-(a[1]-b[1])*(b[0]-c[0])
def isConvex(vertices):
if len(vertices)<4:
return True
signs= [zCrossProduct(a,b,c)>0 for a,b,c in zip(vertices[2:],vertices[1:],vertices)]
return all(signs) or not any(signs)
I implemented both algorithms: the one posted by #UriGoren (with a small improvement - only integer math) and the one from #RoryDaulton, in Java. I had some problems because my polygon is closed, so both algorithms were considering the second as concave, when it was convex. So i changed it to prevent such situation. My methods also uses a base index (which can be or not 0).
These are my test vertices:
// concave
int []x = {0,100,200,200,100,0,0};
int []y = {50,0,50,200,50,200,50};
// convex
int []x = {0,100,200,100,0,0};
int []y = {50,0,50,200,200,50};
And now the algorithms:
private boolean isConvex1(int[] x, int[] y, int base, int n) // Rory Daulton
{
final double TWO_PI = 2 * Math.PI;
// points is 'strictly convex': points are valid, side lengths non-zero, interior angles are strictly between zero and a straight
// angle, and the polygon does not intersect itself.
// NOTES: 1. Algorithm: the signed changes of the direction angles from one side to the next side must be all positive or
// all negative, and their sum must equal plus-or-minus one full turn (2 pi radians). Also check for too few,
// invalid, or repeated points.
// 2. No check is explicitly done for zero internal angles(180 degree direction-change angle) as this is covered
// in other ways, including the `n < 3` check.
// needed for any bad points or direction changes
// Check for too few points
if (n <= 3) return true;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
// Get starting information
int old_x = x[n-2], old_y = y[n-2];
int new_x = x[n-1], new_y = y[n-1];
double new_direction = Math.atan2(new_y - old_y, new_x - old_x), old_direction;
double angle_sum = 0.0, orientation=0;
// Check each point (the side ending there, its angle) and accum. angles for ndx, newpoint in enumerate(polygon):
for (int i = 0; i < n; i++)
{
// Update point coordinates and side directions, check side length
old_x = new_x; old_y = new_y; old_direction = new_direction;
int p = base++;
new_x = x[p]; new_y = y[p];
new_direction = Math.atan2(new_y - old_y, new_x - old_x);
if (old_x == new_x && old_y == new_y)
return false; // repeated consecutive points
// Calculate & check the normalized direction-change angle
double angle = new_direction - old_direction;
if (angle <= -Math.PI)
angle += TWO_PI; // make it in half-open interval (-Pi, Pi]
else if (angle > Math.PI)
angle -= TWO_PI;
if (i == 0) // if first time through loop, initialize orientation
{
if (angle == 0.0) return false;
orientation = angle > 0 ? 1 : -1;
}
else // if other time through loop, check orientation is stable
if (orientation * angle <= 0) // not both pos. or both neg.
return false;
// Accumulate the direction-change angle
angle_sum += angle;
// Check that the total number of full turns is plus-or-minus 1
}
return Math.abs(Math.round(angle_sum / TWO_PI)) == 1;
}
And now from Uri Goren
private boolean isConvex2(int[] x, int[] y, int base, int n)
{
if (n < 4)
return true;
boolean sign = false;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
for(int p=0; p < n; p++)
{
int i = base++;
int i1 = i+1; if (i1 >= n) i1 = base + i1-n;
int i2 = i+2; if (i2 >= n) i2 = base + i2-n;
int dx1 = x[i1] - x[i];
int dy1 = y[i1] - y[i];
int dx2 = x[i2] - x[i1];
int dy2 = y[i2] - y[i1];
int crossproduct = dx1*dy2 - dy1*dx2;
if (i == base)
sign = crossproduct > 0;
else
if (sign != (crossproduct > 0))
return false;
}
return true;
}
For a non complex (intersecting) polygon to be convex, vector frames obtained from any two connected linearly independent lines a,b must be point-convex otherwise the polygon is concave.
For example the lines a,b are convex to the point p and concave to it below for each case i.e. above: p exists inside a,b and below: p exists outside a,b
Similarly for each polygon below, if each line pair making up a sharp edge is point-convex to the centroid c then the polygon is convex otherwise it’s concave.
blunt edges (wronged green) are to be ignored
N.B
This approach would require you compute the centroid of your polygon beforehand since it doesn’t employ angles but vector algebra/transformations
Adapted Uri's code into matlab. Hope this may help.
Be aware that Uri's algorithm only works for simple polygons! So, be sure to test if the polygon is simple first!
% M [ x1 x2 x3 ...
% y1 y2 y3 ...]
% test if a polygon is convex
function ret = isConvex(M)
N = size(M,2);
if (N<4)
ret = 1;
return;
end
x0 = M(1, 1:end);
x1 = [x0(2:end), x0(1)];
x2 = [x0(3:end), x0(1:2)];
y0 = M(2, 1:end);
y1 = [y0(2:end), y0(1)];
y2 = [y0(3:end), y0(1:2)];
dx1 = x2 - x1;
dy1 = y2 - y1;
dx2 = x0 - x1;
dy2 = y0 - y1;
zcrossproduct = dx1 .* dy2 - dy1 .* dx2;
% equality allows two consecutive edges to be parallel
t1 = sum(zcrossproduct >= 0);
t2 = sum(zcrossproduct <= 0);
ret = t1 == N || t2 == N;
end
I am trying to pick objects in the bullet physics world but all I seem to be able to pick is the floor/ground plane!!! I am using the Vuforia SDK and have altered the ImageTargets demo code. I have used the following code to project my touched screen points to the 3d world:
void projectTouchPointsForBullet(QCAR::Vec2F point, QCAR::Vec3F &lineStart, QCAR::Vec3F &lineEnd, QCAR::Matrix44F &modelViewMatrix)
{
QCAR::Vec4F normalisedVector((2 * point.data[0] / screenWidth - 1),
(2 * (screenHeight-point.data[1]) / screenHeight - 1),
-1,
1);
QCAR::Matrix44F modelViewProjection;
SampleUtils::multiplyMatrix(&projectionMatrix.data[0], &modelViewMatrix.data[0] , &modelViewProjection.data[0]);
QCAR::Matrix44F inversedMatrix = SampleMath::Matrix44FInverse(modelViewProjection);
QCAR::Vec4F near_point = SampleMath::Vec4FTransform( normalisedVector,inversedMatrix);
near_point.data[3] = 1.0/near_point.data[3];
near_point = QCAR::Vec4F(near_point.data[0]*near_point.data[3], near_point.data[1]*near_point.data[3], near_point.data[2]*near_point.data[3], 1);
normalisedVector.data[2] = 1.0;//z coordinate now 1
QCAR::Vec4F far_point = SampleMath::Vec4FTransform( normalisedVector, inversedMatrix);
far_point.data[3] = 1.0/far_point.data[3];
far_point = QCAR::Vec4F(far_point.data[0]*far_point.data[3], far_point.data[1]*far_point.data[3], far_point.data[2]*far_point.data[3], 1);
lineStart = QCAR::Vec3F(near_point.data[0],near_point.data[1],near_point.data[2]);
lineEnd = QCAR::Vec3F(far_point.data[0],far_point.data[1],far_point.data[2]);
}
when I try a ray test in my physics world I only seem to be hitting the ground plane! Here is the code for the ray test call:
QCAR::Vec3F intersection, lineStart;
projectTouchPointsForBullet(QCAR::Vec2F(touch1.tapX, touch1.tapY), lineStart, lineEnd,inverseProjMatrix, modelViewMatrix);
btVector3 btRayFrom = btVector3(lineEnd.data[0], lineEnd.data[1], lineEnd.data[2]);
btVector3 btRayTo = btVector3(lineStart.data[0], lineStart.data[1], lineStart.data[2]);
btCollisionWorld::ClosestRayResultCallback rayCallback(btRayFrom,btRayTo);
dynamicsWorld->rayTest(btRayFrom, btRayTo, rayCallback);
if(rayCallback.hasHit())
{
char* pPhysicsData = reinterpret_cast<char*>(rayCallback.m_collisionObject->getUserPointer());//my bodies have char* messages attached to them to determine what has been touched
btRigidBody* pBody = btRigidBody::upcast(rayCallback.m_collisionObject);
if (pBody && pPhysicsData)
{
LOG("handleTouches:: notifyOnTouchEvent from physics world!!!");
notifyOnTouchEvent(env, obj,0,0, pPhysicsData);
}
}
I know I am predominantly looking top-down so I am bound to hit the ground plane, I at least know my touch is being correctly projected into the world, but I have objects lying on the ground plane and I can't seem to be able to touch them! Any pointers would be greatly appreciated :)
I found out why I wasn't able to touch the objects - I am scaling the objects up when they are drawn, so I had to scale the view matrix by the same value before I projected my touch point into the 3d world (EDIT I also had the btRayFrom and btRayTo input cooordinates reversed, it is now fixed):
//top of code
int kObjectScale = 100.0f
....
...
//inside touch handler method
SampleUtils::scalePoseMatrix(kObjectScale, kObjectScale, kObjectScale,&modelViewMatrix.data[0]);
projectTouchPointsForBullet(QCAR::Vec2F(touch1.tapX, touch1.tapY), lineStart, lineEnd,inverseProjMatrix, modelViewMatrix);
btVector3 btRayFrom = btVector3(lineStart.data[0], lineStart.data[1], lineStart.data[2]);
btVector3 btRayTo = btVector3(lineEnd.data[0], lineEnd.data[1], lineEnd.data[2]);
My touches are projected correctly now :)