I am interested in building a hexagonal Torus using a mesh of points?
I think I can start with a 2-d polygon, and then iterate 360 times (1 deg resolution) to build a complete solid.
Is this the best way to do this? What I'm really after is building wing profiles with variable cross section geometry over it's span.
In Your way You can do this with polyhedron(). Add an appropriate number of points per profile in defined order to a vector „points“, define faces by the indices of the points in a second vector „faces“ and set both vectors as parameter in polyhedron(), see documentation. You can control the quality of the surface by the number of points per profile and the distance between the profiles (sectors in torus).
Here an example code:
// parameter:
r1 = 20; // radius of torus
r2 = 4; // radius of polygon/ thickness of torus
s = 360; // sections per 360 deg
p = 6; // points on polygon
a = 30; // angle of the first point on Polygon
// points on cross-section
// angle = 360*i/p + startangle, x = r2*cos(angle), y = 0, z = r2*sin(angle)
function cs_point(i) = [r1 + r2*cos(360*i/p + a), 0, r2*sin(360*i/p + a)];
// returns to the index in the points - vector the section number and the number of the point on this section
function point_index(i) = [floor(i/p), i - p*floor(i/p)];
// returns the points x-, y-, z-coordinates by rotatating the corresponding point from crossection around the z-axis
function iterate_cs(i) = [cs[point_index(i)[1]][0]*cos(360*floor(i/p)/s), cs[point_index(i)[1]][0]*sin(360*floor(i/p)/s), cs[point_index(i)[1]][2]];
// for every point find neighbour points to build faces, ( + p: point on the next cross-section), points ordered clockwise
// to connect point on last section to corresponding points on first section
function item_add1(i) = i >= (s - 1)*p ? -(s)*p : 0;
// to connect last point on section to first points on the same and the next section
function item_add2(i) = i - p*floor(i/p) >= p-1 ? -p : 0;
// build faces
function find_neighbours1(i) = [i, i + 1 + item_add2(i), i + 1 + item_add2(i) + p + item_add1(i)];
function find_neighbours2(i) = [i, i + 1 + + item_add2(i) + p + item_add1(i), i + p + item_add1(i)];
cs = [for (i = [0:p-1]) cs_point(i)];
points = [for (i = [0:s*p - 1]) iterate_cs(i)];
faces1 = [for (i = [0:s*p - 1]) find_neighbours1(i)];
faces2 = [for (i = [0:s*p - 1]) find_neighbours2(i)];
faces = concat(faces1, faces2);
polyhedron(points = points, faces = faces);
here the result:
Since openscad 2015-03 faces can have more than 3 points, if all points of the face are on the same plane. So in this case faces could be build in one step too.
Are you building smth. like NACA airfoils? https://en.wikipedia.org/wiki/NACA_airfoil
There are a few OpenSCAD designs for those floating around, see e.g. https://www.thingiverse.com/thing:898554
BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.
I am experimenting a little with AR. I have got the angle of the direction I am looking to from a compass in degrees. I know my own position and the position of another object (POI), the position is giving in form of latitude and longitude.
Now I would like to know how I can calculate the angle between the direction I am looking to and the POI.
Dot Product:
a . b = ||a|| ||b|| cos(t)
t = acos( (a.b)/(||a|| ||b||) )
||vector|| = length of vector (magnitude)
t = angle between the two vectors
Your probably going to need to do this a couple times for each plane you have. (1x for 2D, 2x for 3D)
Or:
/|
/ |
h / | y
/ |
/____|
x
t is the lower left corner, which we'll assume is your object, the upper right corner is going to be your other object
x = obj2.x - obj1.x
y = obj2.y - obj1.y
h = sqrt( (x*x) + (y*y) )
sin(t) = y/h
cos(t) = x/h
tan(t) = y/x
t = asin(y/h)
t = acos(x/h)
t = atan(y/x)
What makes the first method better, is that it account's for you're current rotation. The second method (using atan, asin, and acos) doesn't.
Using this as a [reference][1]: Find a tangent point on circle?
cx = 0;
cy = 0;
px = -3;
py = -8;
dx = cx - px;
dy = cy - py;
a = asin(5 / ((dx*dx + dy*dy)^0.5));
b = atan2(dy, dx);
t_1 = deg2rad(180) + b - a;
t_2 = deg2rad(180) + b + a;
For a point (7,6) the angles are 7.9572/73.4434 and for (-3, -8) are 213.6264/285.2615. So for the first quadrant, the angles do not make sense, but for third quadrant they do. What I am doing wrong?
Your formula for a is wrong. You should use
a = acos(5 / ((dx*dx + dy*dy)^0.5))
instead of
a = asin(5 / ((dx*dx + dy*dy)^0.5))
i.e. use acos(...) instead of asin(...). The reason is shown in the image below. The formula for angle a is a=acos(r/H), where r is the radius of the circle and H is the length of the hypotenuse of the right angle triangle. So this has nothing to do with the fact that asin(...) has no way to know which of the two possible quadrants the value that is passed in lies. the argument of the asin is always positive, and you always want the answer in the range 0 to 90 degrees.
So the answer for the two angles that you want are b+a and b-a. Using acos instead of asin in your two cases produces 97.7592 & -16.5566 (or equivalently 343.4434) for your first quadrant example, and -164.7385 & -56.3736 (or equivalently 195.2615 and 303.6264) for your third quadrant example. (NB: instead of adding 180 degrees in the formula for t_1 and t-2, you could just switch the signs of dx and dy)
First -- I spent like 10 minutes figuring out what the heck you're trying to do (which in the end, I got from a comment in one of the answers), while solving your problem took 2 minutes. So, for future reference, please give a description of your problem as clear as you can first.
Now, I think you just have your signs messed up. Try the following:
%// difference vector
%// NOTE: these go the other way around for the atan2 to come out right
dx = px - cx;
dy = py - cy;
%// tip angle of the right triangle
a = asin( 5 / sqrt(dx*dx + dy*dy) );
%// angle between the (local) X-axis and the line of interest
b = atan2(dy, dx);
%// the third angle in the right triangle
%// NOTE: minus a here instead of plus b
g = pi/2 - a;
%// Angles of interest
%// NOTE1: signs are flipped; this automatically takes care of overshoots
%// NOTE2: don't forget to mod 360
t_1 = mod( rad2deg(b - g), 360)
t_2 = mod( rad2deg(b + g), 360)
Alternatively, you could skip computing the intermediate angle a by using acos instead of asin:
%// difference vector
dx = px - cx;
dy = py - cy;
%// Directly compute the third angle of the right triangle
%// (that is, the angle "at the origin")
g = acos( 5 / sqrt(dx*dx + dy*dy) );
%// angle between the (local) X-axis and the line of interest
b = atan2(dy, dx);
%// Angles of interest
t_1 = mod( rad2deg(b - g), 360)
t_2 = mod( rad2deg(b + g), 360)
Just another wayto re-discover the trigonometric identity acos(x) = pi/2 - asin(x) :)
This MathWorld entry is what you want: http://mathworld.wolfram.com/CircleTangentLine.html.
Alright, it looks like you are not accounting for the fact that asin, atan, ( any a-trig function ) has no way to know which of the two possible quadrants the value you passed in lies. To make up for that, a-trig function will assume that your point is in the first or fourth quadrant ( northeast / southeast ). Therefore, if you call atan function and your original point was in the second or third quadrant, you need to add 180 degrees / pi radians onto whatever value it returns.
See the documentation here stating that asin returns a value from [-pi/2, pi/2] :
http://www.mathworks.com/help/matlab/ref/asin.html
Hope that helps :)
EDIT
I misunderstood the situation originally
Here is what I think you have calculated :
t_1 and t_2 represent the angles you would travel at if you started on the circle from the tangent point and wanted to travel to your original starting point.
Viewed with this perspective your angles are correct.
For the point (7,6)
If you started on the circle at approx. (0,5) and traveled at 7 degrees, you would hit the point.
If you started on the circle at approx. (5,0) and traveled at 70 degrees, you would hit the point.
Now, what is going to be more useful and less confusing than angles, will be to know the slope of the line. To get this from the angle, do the following with angle in degrees:
angle = (angle + 90 + 360) % 180 - 90 // this gives us the angle as it would be in quad 1 or 4
slope = tan( deg2rad( angle ) )
I want to throw a ball that has a projectile motion. I have a monkey on centre of screen and onTouchBegin I am taking the starting point of the touch and onTouchEnded I am taking the ending points. From the starting and ending points I am taking the angle value between them. Like 30 degrees, 45 or 90 degree.
This is my code by which I have calculated angle of start to endpoint
float angleRadians = atan2(startTouchPoint.x - touchPoint.x, startTouchPoint.y - touchPoint.y);
float angleDegrees = CC_RADIANS_TO_DEGREES(angleRadians);
float cocosAngle = -1 * angleDegrees;
Now i am using Projectile motion formula to throw ball with angle i have calculated from above formula .
inside init method
gravity = 9.8; // metres per second square
X = 0;
Y = 0;
V0 = 50; // meters per second -- elevation
VX0 = V0 * cos(angle); // meters per second
VY0 = V0 * sin(angle); // meters per second
gameTime = 0;
and onTouchEnded i have called fire method which will throw ball .
-(void)fire:(ccTime) dt
{
CCLOG(#"Angle 1: %.2f",angle);
gameTime += dt*6;
// x = v0 * t * cos(angle)
X = (V0 * gameTime * cos(angle))/2+120;
// y = v0 * t * sin(angle) - 0.5 * g * t^2
Y = (V0 * gameTime * sin(angle) - 0.5 * gravity * pow(gameTime, 2))/2+255;
if (Y > 50)
{
sprite_webfire.position = ccp(X,Y);
flag = true;
}
else
{
//angleValue += 15;
angleValue = angle;
angle = [self DegreesToRadians:angleValue];
gravity = 9.8; // metres per second square
X = 0;
Y = 0;
V0 = 50; // meters per second -- elevation
VX0 = V0 * cos(angle); // meters per second
VY0 = V0 * sin(angle); // meters per second
gameTime = 0;
// [self pauseSchedulerAndActions];
}
if (Y < 50)
{
[self unschedule:#selector(fire:)];
}
NSLog(#"ball (%lf,%lf), dt = %lf angle value %d", X, Y, dt,angleValue);
}
this code is working . by this code i can throw ball in projectile motion but i cant throw it where i want to. i cant throw wrt to given angle from start to end point.
i can throw it like red mark but i want to throw it blue mark with swipe . but its not throwing like i am swiping screen.
I am not certain on what math you are using to do this, I find your documentation a bit confusing.
Generally, for project tile motion this is what you need to do:
Find out what the take off angle is relative to the horizontal. Then depending on whatever initial velocity you want the object to have, use that and you trig equations to put your initial velocities into rectangular components.
For example:
If initial velocity was 10, the initial velocity in the y direction would be 10sin(angle), and in the x direction it would be 10cos(angle).
Then in to update the position of the sprite you should use kinematics equations: http://www.physicsclassroom.com/class/1dkin/u1l6c.cfm
First update velocities:
Velocity in the Y direction: V = v(initial) + gravity*(Delta-time)
Velocity in the X direction is constant unless you want to factor in some sort of resistance to make things a lot more complicated.
then position y = oldPositionY + velocity(in Y direction)*(Delta-time) + 1/2(gravity)(delta-time)^2.
and position x = oldPositionX + Xvelocity*delta-time
I have done some projectile motion stuff, and I have found you need to make gravity a large constant, something around 500 to make it look life-like. Let me know if this is confusing or you don't know how to implement it.
I would suggest that you take a look at the following tutorial: http://www.raywenderlich.com/4756/how-to-make-a-catapult-shooting-game-with-cocos2d-and-box2d-part-1.
It shows you how to use a physics engine, so you don't need to do much of the math. All the 'bullets' in the tutorial are also moving with projectile motion.
I'll add a bit to what was already said (which was good). Firstly, you should not be wasting time computing any angles. Stick with vectors for your velocity. In other words, get the initial velocity vector from the touch start and end location, and that will be your (v0x, v0y). For example:
CGPoint initialVelocity = ccpSub(touchPoint, startTouchPoint);
float v0x = initialVelocity.x;
float v0y = initialVelocity.y;
If you wish to assign a different magnitude to the initial velocity vector, simply normalize it and then multiply it by a new magnitude.
CGPoint unitVelocity = ccpNormalize(initialVelocity);
float magnitude = 200; // or whatever you want it to be
CGPoint velocity = ccpMult(unitVelocity, magnitude);
Anyway, with this velocity set properly you can then use it in your position calculations as before, but without the added complexity of calculating the angles.
-(void) fire:(ccTime)dt
{
.
.
gameTime += dt;
// if x(t) = x0 + v0x*t, then dx = v0x*dt
x += v0x*dt;
// if y(t) = y0 + v0y*t - 0.5t^2, then dy = v0y*dt - g*t*dt
y += (v0y * dt - g*gameTime*dt);
.
.
}
Also you should not set v0 = 50. Calculate the velocity from the vector as I suggested.
Something important to consider is that you are calculating what the movement should be in a physical world based upon units of meters. The screen is operating in points, not meters, so you will probably have to apply a scaling factor to the new position (x,y) to get the look that you are going for.
Edit: my bad, I had to revisit my math in the position calculation. My differentials was a bit rusty.