Related
I'm trying to implement 2D Perlin noise to create Minecraft-like terrain (Minecraft doesn't actually use 2D Perlin noise) without overhangs or caves and stuff.
The way I'm doing it, is by creating a [50][20][50] array of cubes, where [20] will be the maximum height of the array, and its values will be determined with Perlin noise. I will then fill that array with arrays of cube.
I've been reading from this article and I don't understand, how do I compute the 4 gradient vector and use it in my code? Does every adjacent 2D array such as [2][3] and [2][4] have a different 4 gradient vector?
Also, I've read that the general Perlin noise function also takes a numeric value that will be used as seed, where do I put that in this case?
I'm going to explain Perlin noise using working code, and without relying on other explanations. First you need a way to generate a pseudo-random float at a 2D point. Each point should look random relative to the others, but the trick is that the same coordinates should always produce the same float. We can use any hash function to do that - not just the one that Ken Perlin used in his code. Here's one:
static float noise2(int x, int y) {
int n = x + y * 57;
n = (n << 13) ^ n;
return (float) (1.0-((n*(n*n*15731+789221)+1376312589)&0x7fffffff)/1073741824.0);
}
I use this to generate a "landscape" landscape[i][j] = noise2(i,j); (which I then convert to an image) and it always produces the same thing:
...
But that looks too random - like the hills and valleys are too densely packed. We need a way of "stretching" each random point over, say, 5 points. And for the values between those "key" points, you want a smooth gradient:
static float stretchedNoise2(float x_float, float y_float, float stretch) {
// stretch
x_float /= stretch;
y_float /= stretch;
// the whole part of the coordinates
int x = (int) Math.floor(x_float);
int y = (int) Math.floor(y_float);
// the decimal part - how far between the two points yours is
float fractional_X = x_float - x;
float fractional_Y = y_float - y;
// we need to grab the 4x4 nearest points to do cubic interpolation
double[] p = new double[4];
for (int j = 0; j < 4; j++) {
double[] p2 = new double[4];
for (int i = 0; i < 4; i++) {
p2[i] = noise2(x + i - 1, y + j - 1);
}
// interpolate each row
p[j] = cubicInterp(p2, fractional_X);
}
// and interpolate the results each row's interpolation
return (float) cubicInterp(p, fractional_Y);
}
public static double cubicInterp(double[] p, double x) {
return cubicInterp(p[0],p[1],p[2],p[3], x);
}
public static double cubicInterp(double v0, double v1, double v2, double v3, double x) {
double P = (v3 - v2) - (v0 - v1);
double Q = (v0 - v1) - P;
double R = v2 - v0;
double S = v1;
return P * x * x * x + Q * x * x + R * x + S;
}
If you don't understand the details, that's ok - I don't know how Math.cos() is implemented, but I still know what it does. And this function gives us stretched, smooth noise.
->
The stretchedNoise2 function generates a "landscape" at a certain scale (big or small) - a landscape of random points with smooth slopes between them. Now we can generate a sequence of landscapes on top of each other:
public static double perlin2(float xx, float yy) {
double noise = 0;
noise += stretchedNoise2(xx, yy, 5) * 1; // sample 1
noise += stretchedNoise2(xx, yy, 13) * 2; // twice as influential
// you can keep repeating different variants of the above lines
// some interesting variants are included below.
return noise / (1+2); // make sure you sum the multipliers above
}
To put it more accurately, we get the weighed average of the points from each sample.
( + 2 * ) / 3 =
When you stack a bunch of smooth noise together, usually about 5 samples of increasing "stretch", you get Perlin noise. (If you understand the last sentence, you understand Perlin noise.)
There are other implementations that are faster because they do the same thing in different ways, but because it is no longer 1983 and because you are getting started with writing a landscape generator, you don't need to know about all the special tricks and terminology they use to understand Perlin noise or do fun things with it. For example:
1) 2) 3)
// 1
float smearX = interpolatedNoise2(xx, yy, 99) * 99;
float smearY = interpolatedNoise2(xx, yy, 99) * 99;
ret += interpolatedNoise2(xx + smearX, yy + smearY, 13)*1;
// 2
float smearX2 = interpolatedNoise2(xx, yy, 9) * 19;
float smearY2 = interpolatedNoise2(xx, yy, 9) * 19;
ret += interpolatedNoise2(xx + smearX2, yy + smearY2, 13)*1;
// 3
ret += Math.cos( interpolatedNoise2(xx , yy , 5)*4) *1;
About perlin noise
Perlin noise was developed to generate a random continuous surfaces (actually, procedural textures). Its main feature is that the noise is always continuous over space.
From the article:
Perlin noise is function for generating coherent noise over a space. Coherent noise means that for any two points in the space, the value of the noise function changes smoothly as you move from one point to the other -- that is, there are no discontinuities.
Simply, a perlin noise looks like this:
_ _ __
\ __/ \__/ \__
\__/
But this certainly is not a perlin noise, because there are gaps:
_ _
\_ __/
___/ __/
Calculating the noise (or crushing gradients!)
As #markspace said, perlin noise is mathematically hard. Lets simplify by generating 1D noise.
Imagine the following 1D space:
________________
Firstly, we define a grid (or points in 1D space):
1 2 3 4
________________
Then, we randomly chose a noise value to each grid point (This value is equivalent to the gradient in the 2D noise):
1 2 3 4
________________
-1 0 0.5 1 // random noise value
Now, calculating the noise value for a grid point it is easy, just pick the value:
noise(3) => 0.5
But the noise value for a arbitrary point p needs to be calculated based in the closest grid points p1 and p2 using their value and influence:
// in 1D the influence is just the distance between the points
noise(p) => noise(p1) * influence(p1) + noise(p2) * influence(p2)
noise(2.5) => noise(2) * influence(2, 2.5) + noise(3) * influence(3, 2.5)
=> 0 * 0.5 + 0.5 * 0.5 => 0.25
The end! Now we are able to calculate 1D noise, just add one dimension for 2D. :-)
Hope it helps you understand! Now read #mk.'s answer for working code and have happy noises!
Edit:
Follow up question in the comments:
I read in wikipedia article that the gradient vector in 2d perlin should be length of 1 (unit circle) and random direction. since vector has X and Y, how do I do that exactly?
This could be easily lifted and adapted from the original perlin noise code. Find bellow a pseudocode.
gradient.x = random()*2 - 1;
gradient.y = random()*2 - 1;
normalize_2d( gradient );
Where normalize_2d is:
// normalizes a 2d vector
function normalize_2d(v)
size = square_root( v.x * v.x + v.y * v.y );
v.x = v.x / size;
v.y = v.y / size;
Compute Perlin noise at coordinates x, y
function perlin(float x, float y) {
// Determine grid cell coordinates
int x0 = (x > 0.0 ? (int)x : (int)x - 1);
int x1 = x0 + 1;
int y0 = (y > 0.0 ? (int)y : (int)y - 1);
int y1 = y0 + 1;
// Determine interpolation weights
// Could also use higher order polynomial/s-curve here
float sx = x - (double)x0;
float sy = y - (double)y0;
// Interpolate between grid point gradients
float n0, n1, ix0, ix1, value;
n0 = dotGridGradient(x0, y0, x, y);
n1 = dotGridGradient(x1, y0, x, y);
ix0 = lerp(n0, n1, sx);
n0 = dotGridGradient(x0, y1, x, y);
n1 = dotGridGradient(x1, y1, x, y);
ix1 = lerp(n0, n1, sx);
value = lerp(ix0, ix1, sy);
return value;
}
I'm writing my own basic physic engine and now I come to a problem I can't solve. Probably because I don't how to google this problem.
So here is my problem. I hope this image can explain it:
Collision response
I have two objects. The gray one is fixed and don't move and the green one which falls from the top.
The green object has three vectors: a force, the acceleration and the velocity. It collides with the fixed gray object.
The real question is how can I get the rotation of the green object when it falls down?
It sounds like you may not have an understanding of the fundamental physics underlying rigid body dynamics. I say that only because you don't mention any of the terminology commonly used when talking about this kind of problem. You'll need to introduce the idea of orientation and angular velocity (the rotational analogs of position and linear velocity) to each dynamic body in the system, and compute all kinds of intermediate quantities like moment of inertia, angular acceleration, and torque.
Perhaps the best introductory reference for this is Chris Hecker's series of articles for Game Developer Magazine. Assuming you already have non-rotational dynamics (covered in part 1) and collision detection (not covered by this series) solved, you should begin with part 2 and proceed to part 3. They'll give you a solid foundation in the physics and mathematics necessary for implementing rotational collision response.
You do as described below once, when the objects collide.
Let us call the green rectangle "a", and the other one "b".
1.
First you need the rectangles "rotational mass", mass of inertia.
a.i = 4/3 * width * height * (width^2 + height^2) * a.density
2.
Then you need the vector pointing from the rectangle's center of mass (average position of all corners) to the contact position (where the rectangles collide), let us call it "r".
3.
Then you need to find the collision normal. This normal is the direction of an impulse being applied to a from b. The normal is a vector with length 1 unit. In your example the normal would probably point upwards. Let us call the normal vector "n".
4.
Now you will need the velocity of the contact point on a. If a is not rotating, the formula would be:
vp = a.vel
If a is rotating the formula would be:
vp = a.vel + cross(a.r_vel, r)
a.r_vel is a's rotational velocity given in radians and positive direction is counter clockwise.
cross() means cross product, the function is:
cross (v,i) = [-i * v.y , i * v.x]
The expanded formula would be:
vp = a.v + [-r * a.r_vel.y , r * a.r_vel.x]
5.
Now you need to calculate whether the objects are moving towards each other. Project the vp onto n.
vp_p = dot(vp, n)
dot (v1, v2) = v1.x * v2.x + v1.y * v2.y
vp_p is a scalar (a value, not a vector).
If vp_p is negative the obejcts are moving towards each other, if it is > 0 they are moving apart.
6.
Now you need to calculate the impulse to stop a from moving into b, the impulse is:
j = -vp_p / (
1/a.mass + cross(r,n)^2 / a.i
)
The cross product between two vectors are:
cross(v1,v2) = v1.x * v2.y - v1.y * v2.x
It returns a scalar.
Multiply the impulse with the normal to get the impulse vector:
jn = j * n
7.
Now you need to apply the impulse to a:
a.new_vel = a.old_vel + jn / a.mass;
a.new_r_vel = a.old_r_vel + cross(r,jn) / a.i;
If you want the collision to be fully elastic, you must multiply the impulse by 2. Let us call this multiplier "e". e needs to be between 1 and 2. 1 means no energy is conserved, 2 means all energy is conserved.
Example code:
var vp = a.vel + cross(a.r_vel, r);
var vp_p = dot(vp,n); // negative val = moving towards each other
if (vp_p >= 0) { // do they move apart?
return false;
}
// normal impulse
var j = - e * vp_p / (
1/a.mass + cross(r,n)^2 / a.i
);
var jn = j * n;
//
a.vel = a.vel + jn / a.mass;
a.r_vel = a.r_vel + cross(r,jn) / a.i;
If b is not static the algorithm will be slightly different:
a.r = vector pointing from a's center of mass to the contact position
var vp = a.vel + cross(a.r_vel, a.r) - b.vel - cross(b.r_vel, b.r);
var vp_p = dot(vp,n); // negative val = moving towards each other
if (vp_p >= 0) { // do they move apart?
return false;
}
// normal impulse
var j = - e * vp_p / (
1/a.mass + cross(a.r,n)^2 / a.i +
1/b.mass + cross(b.r,n)^2 / b.i
);
var jn = j * n;
//
a.vel = a.vel + jp / a.mass;
a.r_vel = a.r_vel + cross(a.r,jn) / a.i;
b.vel = b.vel - jp / b.mass;
b.r_vel = b.r_vel - cross(b.r,jn) / b.i;
How the formulas work / sources:
http://www.myphysicslab.com/collision.html#resting_contact
Background: I have a bird view's JavaScript game where the player controls a space ship by touching a circle -- e.g. touch to the left of the circle center, and the ship will move left, touch the top right and it will move to the top right and so on... the further away from the circle center of pseudo joystick, the more speed in that direction. However, I'm not directly adjusting the ship's speed, but rather set a targetSpeed.x and targetSpeed.y value, and the ship will then adjust its speed using something like:
if (this.speed.x < this.targetSpeed.x) {
this.speed.x += this.speedStep;
}
else if (this.speed.x > this.targetSpeed.x) {
this.speed.x -= this.speedStep;
}
... and the same for the y speed, and speedStep is a small value to make it smoother and not too abrupt (a ship shouldn't go from a fast leftwards direction to an immediate fast rightwards direction).
My question: Using above code, I believe however that the speed will be adjusted quicker in diagonal directions, and slower along the horizontal/ vertical lines. How do I correct this to have an equal target speed following?
Thanks so much for any help!
var xdiff = targetSpeed.x - speed.x;
var ydiff = targetSpeed.y - speed.y;
var angle = Math.atan2(ydiff, xdiff);
speed.x += speedStep * Math.cos(angle);
speed.y += speedStep * Math.sin(angle);
Assuming you already checked that the touch is inside the circle, and that the edge of the circle represents max speed, and that the center of the circle is circleTouch == [0, 0]
In some C++-like pseudo code:
Scalar circleRadius = ...;
Scalar maxSpeed = ...;
Scalar acceleration = ...;
Vector calculateTargetSpeed( Vector circleTouch ) {
Vector targetSpeed = maxSpeed * circleTouch / circleRadius;
return targetSpeed;
}
Vector calculateNewSpeed( Vector currentSpeed, Vector targetSpeed ) {
Vector speedDiff = targetSpeed - currentSpeed;
Vector newSpeed = currentSpeed + acceleration * normalized(speedDiff);
return newSpeed;
}
// Divide v by its length to get normalized vector (length 1) with same x/y ratio
Vector normalized( Vector v ) {
return v / length(v);
}
// Pythagoras for the length of v
Scalar length( Vector v ) {
Scalar length = sqrt(v.x * v.x + v.y * v.y); // or preferably hypot(v.x, v.y)
return length;
}
This is just off the top of my head, and i haven't tested it. The other answer is fine, i just wanted to give an answer without trigonometry functions. :)
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.
Okay, this all takes place in a nice and simple 2D world... :)
Suppose I have a static object A at position Apos, and a linearly moving object B at Bpos with bVelocity, and an ammo round with velocity Avelocity...
How would I find out the angle that A has to shoot, to hit B, taking into account B's linear velocity and the speed of A's ammo ?
Right now the aim's at the current position of the object, which means that by the time my projectile gets there the unit has moved on to safer positions :)
I wrote an aiming subroutine for xtank a while back. I'll try to lay out how I did it.
Disclaimer: I may have made one or more silly mistakes anywhere in here; I'm just trying to reconstruct the reasoning with my rusty math skills. However, I'll cut to the chase first, since this is a programming Q&A instead of a math class :-)
How to do it
It boils down to solving a quadratic equation of the form:
a * sqr(x) + b * x + c == 0
Note that by sqr I mean square, as opposed to square root. Use the following values:
a := sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)
b := 2 * (target.velocityX * (target.startX - cannon.X)
+ target.velocityY * (target.startY - cannon.Y))
c := sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)
Now we can look at the discriminant to determine if we have a possible solution.
disc := sqr(b) - 4 * a * c
If the discriminant is less than 0, forget about hitting your target -- your projectile can never get there in time. Otherwise, look at two candidate solutions:
t1 := (-b + sqrt(disc)) / (2 * a)
t2 := (-b - sqrt(disc)) / (2 * a)
Note that if disc == 0 then t1 and t2 are equal.
If there are no other considerations such as intervening obstacles, simply choose the smaller positive value. (Negative t values would require firing backward in time to use!)
Substitute the chosen t value back into the target's position equations to get the coordinates of the leading point you should be aiming at:
aim.X := t * target.velocityX + target.startX
aim.Y := t * target.velocityY + target.startY
Derivation
At time T, the projectile must be a (Euclidean) distance from the cannon equal to the elapsed time multiplied by the projectile speed. This gives an equation for a circle, parametric in elapsed time.
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr(t * projectile_speed)
Similarly, at time T, the target has moved along its vector by time multiplied by its velocity:
target.X == t * target.velocityX + target.startX
target.Y == t * target.velocityY + target.startY
The projectile can hit the target when its distance from the cannon matches the projectile's distance.
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr(target.X - cannon.X) + sqr(target.Y - cannon.Y)
Wonderful! Substituting the expressions for target.X and target.Y gives
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr((t * target.velocityX + target.startX) - cannon.X)
+ sqr((t * target.velocityY + target.startY) - cannon.Y)
Substituting the other side of the equation gives this:
sqr(t * projectile_speed)
== sqr((t * target.velocityX + target.startX) - cannon.X)
+ sqr((t * target.velocityY + target.startY) - cannon.Y)
... subtracting sqr(t * projectile_speed) from both sides and flipping it around:
sqr((t * target.velocityX) + (target.startX - cannon.X))
+ sqr((t * target.velocityY) + (target.startY - cannon.Y))
- sqr(t * projectile_speed)
== 0
... now resolve the results of squaring the subexpressions ...
sqr(target.velocityX) * sqr(t)
+ 2 * t * target.velocityX * (target.startX - cannon.X)
+ sqr(target.startX - cannon.X)
+ sqr(target.velocityY) * sqr(t)
+ 2 * t * target.velocityY * (target.startY - cannon.Y)
+ sqr(target.startY - cannon.Y)
- sqr(projectile_speed) * sqr(t)
== 0
... and group similar terms ...
sqr(target.velocityX) * sqr(t)
+ sqr(target.velocityY) * sqr(t)
- sqr(projectile_speed) * sqr(t)
+ 2 * t * target.velocityX * (target.startX - cannon.X)
+ 2 * t * target.velocityY * (target.startY - cannon.Y)
+ sqr(target.startX - cannon.X)
+ sqr(target.startY - cannon.Y)
== 0
... then combine them ...
(sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)) * sqr(t)
+ 2 * (target.velocityX * (target.startX - cannon.X)
+ target.velocityY * (target.startY - cannon.Y)) * t
+ sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)
== 0
... giving a standard quadratic equation in t. Finding the positive real zeros of this equation gives the (zero, one, or two) possible hit locations, which can be done with the quadratic formula:
a * sqr(x) + b * x + c == 0
x == (-b ± sqrt(sqr(b) - 4 * a * c)) / (2 * a)
+1 on Jeffrey Hantin's excellent answer here. I googled around and found solutions that were either too complex or not specifically about the case I was interested in (simple constant velocity projectile in 2D space.) His was exactly what I needed to produce the self-contained JavaScript solution below.
The one point I would add is that there are a couple special cases you have to watch for in addition to the discriminant being negative:
"a == 0": occurs if target and projectile are traveling the same speed. (solution is linear, not quadratic)
"a == 0 and b == 0": if both target and projectile are stationary. (no solution unless c == 0, i.e. src & dst are same point.)
Code:
/**
* Return the firing solution for a projectile starting at 'src' with
* velocity 'v', to hit a target, 'dst'.
*
* #param ({x, y}) src position of shooter
* #param ({x, y, vx, vy}) dst position & velocity of target
* #param (Number) v speed of projectile
*
* #return ({x, y}) Coordinate at which to fire (and where intercept occurs). Or `null` if target cannot be hit.
*/
function intercept(src, dst, v) {
const tx = dst.x - src.x;
const ty = dst.y - src.y;
const tvx = dst.vx;
const tvy = dst.vy;
// Get quadratic equation components
const a = tvx * tvx + tvy * tvy - v * v;
const b = 2 * (tvx * tx + tvy * ty);
const c = tx * tx + ty * ty;
// Solve quadratic
const ts = quad(a, b, c); // See quad(), below
// Find smallest positive solution
let sol = null;
if (ts) {
const t0 = ts[0];
const t1 = ts[1];
let t = Math.min(t0, t1);
if (t < 0) t = Math.max(t0, t1);
if (t > 0) {
sol = {
x: dst.x + dst.vx * t,
y: dst.y + dst.vy * t
};
}
}
return sol;
}
/**
* Return solutions for quadratic
*/
function quad(a, b, c) {
let sol = null;
if (Math.abs(a) < 1e-6) {
if (Math.abs(b) < 1e-6) {
sol = Math.abs(c) < 1e-6 ? [0, 0] : null;
} else {
sol = [-c / b, -c / b];
}
} else {
let disc = b * b - 4 * a * c;
if (disc >= 0) {
disc = Math.sqrt(disc);
a = 2 * a;
sol = [(-b - disc) / a, (-b + disc) / a];
}
}
return sol;
}
// For example ...
const sol = intercept(
{x:2, y:4}, // Starting coord
{x:5, y:7, vx: 2, vy:1}, // Target coord and velocity
5 // Projectile velocity
)
console.log('Fire at', sol)
First rotate the axes so that AB is vertical (by doing a rotation)
Now, split the velocity vector of B into the x and y components (say Bx and By). You can use this to calculate the x and y components of the vector you need to shoot at.
B --> Bx
|
|
V
By
Vy
^
|
|
A ---> Vx
You need Vx = Bx and Sqrt(Vx*Vx + Vy*Vy) = Velocity of Ammo.
This should give you the vector you need in the new system. Transform back to old system and you are done (by doing a rotation in the other direction).
Jeffrey Hantin has a nice solution for this problem, though his derivation is overly complicated. Here's a cleaner way of deriving it with some of the resultant code at the bottom.
I'll be using x.y to represent vector dot product, and if a vector quantity is squared, it means I am dotting it with itself.
origpos = initial position of shooter
origvel = initial velocity of shooter
targpos = initial position of target
targvel = initial velocity of target
projvel = velocity of the projectile relative to the origin (cause ur shooting from there)
speed = the magnitude of projvel
t = time
We know that the position of the projectile and target with respect to t time can be described with some equations.
curprojpos(t) = origpos + t*origvel + t*projvel
curtargpos(t) = targpos + t*targvel
We want these to be equal to each other at some point (the point of intersection), so let's set them equal to each other and solve for the free variable, projvel.
origpos + t*origvel + t*projvel = targpos + t*targvel
turns into ->
projvel = (targpos - origpos)/t + targvel - origvel
Let's forget about the notion of origin and target position/velocity. Instead, let's work in relative terms since motion of one thing is relative to another. In this case, what we now have is relpos = targetpos - originpos and relvel = targetvel - originvel
projvel = relpos/t + relvel
We don't know what projvel is, but we do know that we want projvel.projvel to be equal to speed^2, so we'll square both sides and we get
projvel^2 = (relpos/t + relvel)^2
expands into ->
speed^2 = relvel.relvel + 2*relpos.relvel/t + relpos.relpos/t^2
We can now see that the only free variable is time, t, and then we'll use t to solve for projvel. We'll solve for t with the quadratic formula. First separate it out into a, b and c, then solve for the roots.
Before solving, though, remember that we want the best solution where t is smallest, but we need to make sure that t is not negative (you can't hit something in the past)
a = relvel.relvel - speed^2
b = 2*relpos.relvel
c = relpos.relpos
h = -b/(2*a)
k2 = h*h - c/a
if k2 < 0, then there are no roots and there is no solution
if k2 = 0, then there is one root at h
if 0 < h then t = h
else, no solution
if k2 > 0, then there are two roots at h - k and h + k, we also know r0 is less than r1.
k = sqrt(k2)
r0 = h - k
r1 = h + k
we have the roots, we must now solve for the smallest positive one
if 0<r0 then t = r0
elseif 0<r1 then t = r1
else, no solution
Now, if we have a t value, we can plug t back into the original equation and solve for the projvel
projvel = relpos/t + relvel
Now, to the shoot the projectile, the resultant global position and velocity for the projectile is
globalpos = origpos
globalvel = origvel + projvel
And you're done!
My implementation of my solution in Lua, where vec*vec represents vector dot product:
local function lineartrajectory(origpos,origvel,speed,targpos,targvel)
local relpos=targpos-origpos
local relvel=targvel-origvel
local a=relvel*relvel-speed*speed
local b=2*relpos*relvel
local c=relpos*relpos
if a*a<1e-32 then--code translation for a==0
if b*b<1e-32 then
return false,"no solution"
else
local h=-c/b
if 0<h then
return origpos,relpos/h+targvel,h
else
return false,"no solution"
end
end
else
local h=-b/(2*a)
local k2=h*h-c/a
if k2<-1e-16 then
return false,"no solution"
elseif k2<1e-16 then--code translation for k2==0
if 0<h then
return origpos,relpos/h+targvel,h
else
return false,"no solution"
end
else
local k=k2^0.5
if k<h then
return origpos,relpos/(h-k)+targvel,h-k
elseif -k<h then
return origpos,relpos/(h+k)+targvel,h+k
else
return false,"no solution"
end
end
end
end
Following is polar coordinate based aiming code in C++.
To use with rectangular coordinates you would need to first convert the targets relative coordinate to angle/distance, and the targets x/y velocity to angle/speed.
The "speed" input is the speed of the projectile. The units of the speed and targetSpeed are irrelevent, as only the ratio of the speeds are used in the calculation. The output is the angle the projectile should be fired at and the distance to the collision point.
The algorithm is from source code available at http://www.turtlewar.org/ .
// C++
static const double pi = 3.14159265358979323846;
inline double Sin(double a) { return sin(a*(pi/180)); }
inline double Asin(double y) { return asin(y)*(180/pi); }
bool/*ok*/ Rendezvous(double speed,double targetAngle,double targetRange,
double targetDirection,double targetSpeed,double* courseAngle,
double* courseRange)
{
// Use trig to calculate coordinate of future collision with target.
// c
//
// B A
//
// a C b
//
// Known:
// C = distance to target
// b = direction of target travel, relative to it's coordinate
// A/B = ratio of speed and target speed
//
// Use rule of sines to find unknowns.
// sin(a)/A = sin(b)/B = sin(c)/C
//
// a = asin((A/B)*sin(b))
// c = 180-a-b
// B = C*(sin(b)/sin(c))
bool ok = 0;
double b = 180-(targetDirection-targetAngle);
double A_div_B = targetSpeed/speed;
double C = targetRange;
double sin_b = Sin(b);
double sin_a = A_div_B*sin_b;
// If sin of a is greater than one it means a triangle cannot be
// constructed with the given angles that have sides with the given
// ratio.
if(fabs(sin_a) <= 1)
{
double a = Asin(sin_a);
double c = 180-a-b;
double sin_c = Sin(c);
double B;
if(fabs(sin_c) > .0001)
{
B = C*(sin_b/sin_c);
}
else
{
// Sin of small angles approach zero causing overflow in
// calculation. For nearly flat triangles just treat as
// flat.
B = C/(A_div_B+1);
}
// double A = C*(sin_a/sin_c);
ok = 1;
*courseAngle = targetAngle+a;
*courseRange = B;
}
return ok;
}
Here's an example where I devised and implemented a solution to the problem of predictive targeting using a recursive algorithm: http://www.newarteest.com/flash/targeting.html
I'll have to try out some of the other solutions presented because it seems more efficient to calculate it in one step, but the solution I came up with was to estimate the target position and feed that result back into the algorithm to make a new more accurate estimate, repeating several times.
For the first estimate I "fire" at the target's current position and then use trigonometry to determine where the target will be when the shot reaches the position fired at. Then in the next iteration I "fire" at that new position and determine where the target will be this time. After about 4 repeats I get within a pixel of accuracy.
I just hacked this version for aiming in 2d space, I didn't test it very thoroughly yet but it seems to work. The idea behind it is this:
Create a vector perpendicular to the vector pointing from the muzzle to the target.
For a collision to occur, the velocities of the target and the projectile along this vector (axis) should be the same!
Using fairly simple cosine stuff I arrived at this code:
private Vector3 CalculateProjectileDirection(Vector3 a_MuzzlePosition, float a_ProjectileSpeed, Vector3 a_TargetPosition, Vector3 a_TargetVelocity)
{
// make sure it's all in the horizontal plane:
a_TargetPosition.y = 0.0f;
a_MuzzlePosition.y = 0.0f;
a_TargetVelocity.y = 0.0f;
// create a normalized vector that is perpendicular to the vector pointing from the muzzle to the target's current position (a localized x-axis):
Vector3 perpendicularVector = Vector3.Cross(a_TargetPosition - a_MuzzlePosition, -Vector3.up).normalized;
// project the target's velocity vector onto that localized x-axis:
Vector3 projectedTargetVelocity = Vector3.Project(a_TargetVelocity, perpendicularVector);
// calculate the angle that the projectile velocity should make with the localized x-axis using the consine:
float angle = Mathf.Acos(projectedTargetVelocity.magnitude / a_ProjectileSpeed) / Mathf.PI * 180;
if (Vector3.Angle(perpendicularVector, a_TargetVelocity) > 90.0f)
{
angle = 180.0f - angle;
}
// rotate the x-axis so that is points in the desired velocity direction of the projectile:
Vector3 returnValue = Quaternion.AngleAxis(angle, -Vector3.up) * perpendicularVector;
// give the projectile the correct speed:
returnValue *= a_ProjectileSpeed;
return returnValue;
}
I made a public domain Unity C# function here:
http://ringofblades.com/Blades/Code/PredictiveAim.cs
It is for 3D, but you can easily modify this for 2D by replacing the Vector3s with Vector2s and using your down axis of choice for gravity if there is gravity.
In case the theory interests you, I walk through the derivation of the math here:
http://www.gamasutra.com/blogs/KainShin/20090515/83954/Predictive_Aim_Mathematics_for_AI_Targeting.php
I've seen many ways to solve this problem mathematically, but this was a component relevant to a project my class was required to do in high school, and not everyone in this programming class had a background with calculus, or even vectors for that matter, so I created a way to solve this problem with more of a programming approach. The point of intersection will be accurate, although it may hit 1 frame later than in the mathematical computations.
Consider:
S = shooterPos, E = enemyPos, T = targetPos, Sr = shooter range, D = enemyDir
V = distance from E to T, P = projectile speed, Es = enemy speed
In the standard implementation of this problem [S,E,P,Es,D] are all givens and you are solving either to find T or the angle at which to shoot so that you hit T at the proper timing.
The main aspect of this method of solving the problem is to consider the range of the shooter as a circle encompassing all possible points that can be shot at any given time. The radius of this circle is equal to:
Sr = P*time
Where time is calculated as an iteration of a loop.
Thus to find the distance an enemy travels given the time iteration we create the vector:
V = D*Es*time
Now, to actually solve the problem we want to find a point at which the distance from the target (T) to our shooter (S) is less than the range of our shooter (Sr). Here is somewhat of a pseudocode implementation of this equation.
iteration = 0;
while(TargetPoint.hasNotPassedShooter)
{
TargetPoint = EnemyPos + (EnemyMovementVector)
if(distanceFrom(TargetPoint,ShooterPos) < (ShooterRange))
return TargetPoint;
iteration++
}
Basically , intersection concept is not really needed here, As far as you are using projectile motion, you just need to hit at a particular angle and instantiate at the time of shooting so that you get the exact distance of your target from the Source and then once you have the distance, you can calculate the appropriate velocity with which it should shot in order to hit the Target.
The following link makes teh concept clear and is considered helpful, might help:
Projectile motion to always hit a moving target
I grabbed one of the solutions from here, but none of them take into account movement of the shooter. If your shooter is moving, you might want to take that into account (as the shooter's velocity should be added to your bullet's velocity when you fire). Really all you need to do is subtract your shooter's velocity from the target's velocity. So if you're using broofa's code above (which I would recommend), change the lines
tvx = dst.vx;
tvy = dst.vy;
to
tvx = dst.vx - shooter.vx;
tvy = dst.vy - shooter.vy;
and you should be all set.