# Kepler's Laws.py
# plots the orbit of a planet in an eccentric orbit to illustrate
# the sweeping out of equal areas in equal times, with sun at focus
# The eccentricity of the orbit is random and determined by the initial velocity
# program uses normalised units (G =1)
# program by Peter Borcherds, University of Birmingham, England
from vpython import *
from random import random
from IPython import display
import pandas as pd
def MonthStep(time, offset=20, whole=1): # mark the end of each "month"
global ccolor # have to make it global, since label uses it before it is updated
if whole:
Ltext = str(int(time * 2 + dt)) # end of 'month', printing twice time gives about 12 'months' in 'year'
else:
Ltext = duration + str(time * 2) + ' "months"\n Initial speed: ' + str(round(speed, 3))
ccolor = color.white
label(pos=planet.pos, text=Ltext, color=ccolor,
xoffset=offset * planet.pos.x, yoffset=offset * planet.pos.y)
ccolor = (0.5 * (1 + random()), random(), random()) # randomise colour of radial vector
return ccolor
scene = display(title="Kepler's law of equal areas", width=1000, height=1000, range=3.2)
duration = 'Period: '
sun = sphere(color=color.yellow, radius=0.1) # motion of sun is ignored (or centre of mass coordinates)
scale = 1.0
poss = vector(0, scale, 0)
planet = sphere(pos=poss, color=color.cyan, radius=0.02)
while 1:
velocity = -vector(0.7 + 0.5 * random(), 0, 0) # gives a satisfactory range of eccentricities
##velocity = -vector(0.984,0,0) # gives period of 12.0 "months"
speed = mag(velocity)
steps = 20
dt = 0.5 / float(steps)
step = 0
time = 0
ccolor = color.white
oldpos = vector(planet.pos)
ccolor = MonthStep(time)
curve(pos=[sun.pos, planet.pos], color=ccolor)
while not (oldpos.x > 0 and planet.pos.x < 0):
rate(steps * 2) # keep rate down so that development of orbit can be followed
time += dt
oldpos = vector(planet.pos) # construction vector(planet.pos) makes oldpos a varible in its own right
# oldpos = planet.pos makes "oldposs" point to "planet.pos"
# oldposs = planet.pos[:] does not work, because vector does not permit slicing
denom = mag(planet.pos) ** 3
velocity -= planet.pos * dt / denom # inverse square law; force points toward sun
planet.pos += velocity * dt
# plot orbit
curve(pos=[oldpos, planet.pos], color=color.red)
step += 1
if step == steps:
step = 0
ccolor = MonthStep(time)
curve(pos=[sun.pos, planet.pos], color=color.white)
else:
# plot radius vector
curve(pos=[sun.pos, planet.pos], color=ccolor)
if scene.kb.keys:
print
"key pressed"
duration = 'Duration: '
break
MonthStep(time, 50, 0)
label(pos=(2.5, -2.5, 0), text='Click for another orbit')
scene.mouse.getclick()
for obj in scene.objects:
if obj is sun or obj is planet: continue
obj.visible = 0 # clear the screen to do it again
I copied Kepler's Laws code in google and compiled it on pycharm.
But there is an error that
scene = display(title="Kepler's law of equal areas", width=1000, height=1000, range=3.2)
TypeError: 'module' object is not callable
I found some information on google that "pandas" library can improve this error so I tried it but I can't improve this error.
What should I do?
Replace "display" with "canvas", which is the correct name of this entity.
Related
I have a directed graph in networkx.
The nodes have a "height" label. Here is an example with heights 0, 1, 2, 3, 4, 5 and 6:
I would like to run spring layout (in two dimensions), but constrain the nodes to be of a fixed height. That is, I want to "constrain" spring layout so that the x coordinate of the nodes moves, by the y coordinate does not.
I am relatively new to networkx. What is the best way to accomplish this? Thanks in advance.
Following #Joe's request, I'm posting answer here.
This was just a matter of patching the code suggested above together. Thus absolutely no originality is claimed.
Your graph should have a "height" variable attached to each node. Thus, once you have added the code below, the following should work:
G = nx.Graph()
G.add_edges_from([[0,1],[1,2],[2,3]])
for g in G.nodes():
G.nodes()[g]["height"] = g
draw_graph_with_height(G,figsize=(5,5))
# Copyright (C) 2004-2015 by
# Aric Hagberg <hagberg#lanl.gov>
# Dan Schult <dschult#colgate.edu>
# Pieter Swart <swart#lanl.gov>
# All rights reserved.
# BSD license.
# import numpy as np
# taken from networkx.drawing.layout and added hold_dim
def _fruchterman_reingold(A, dim=2, k=None, pos=None, fixed=None,
iterations=50, hold_dim=None):
# Position nodes in adjacency matrix A using Fruchterman-Reingold
# Entry point for NetworkX graph is fruchterman_reingold_layout()
try:
nnodes, _ = A.shape
except AttributeError:
raise RuntimeError(
"fruchterman_reingold() takes an adjacency matrix as input")
A = np.asarray(A) # make sure we have an array instead of a matrix
if pos is None:
# random initial positions
pos = np.asarray(np.random.random((nnodes, dim)), dtype=A.dtype)
else:
# make sure positions are of same type as matrix
pos = pos.astype(A.dtype)
# optimal distance between nodes
if k is None:
k = np.sqrt(1.0 / nnodes)
# the initial "temperature" is about .1 of domain area (=1x1)
# this is the largest step allowed in the dynamics.
t = 0.1
# simple cooling scheme.
# linearly step down by dt on each iteration so last iteration is size dt.
dt = t / float(iterations + 1)
delta = np.zeros((pos.shape[0], pos.shape[0], pos.shape[1]), dtype=A.dtype)
# the inscrutable (but fast) version
# this is still O(V^2)
# could use multilevel methods to speed this up significantly
for _ in range(iterations):
# matrix of difference between points
for i in range(pos.shape[1]):
delta[:, :, i] = pos[:, i, None] - pos[:, i]
# distance between points
distance = np.sqrt((delta**2).sum(axis=-1))
# enforce minimum distance of 0.01
distance = np.where(distance < 0.01, 0.01, distance)
# displacement "force"
displacement = np.transpose(np.transpose(delta)*(k * k / distance**2 - A * distance / k))\
.sum(axis=1)
# update positions
length = np.sqrt((displacement**2).sum(axis=1))
length = np.where(length < 0.01, 0.1, length)
delta_pos = np.transpose(np.transpose(displacement) * t / length)
if fixed is not None:
# don't change positions of fixed nodes
delta_pos[fixed] = 0.0
# only update y component
if hold_dim == 0:
pos[:, 1] += delta_pos[:, 1]
# only update x component
elif hold_dim == 1:
pos[:, 0] += delta_pos[:, 0]
else:
pos += delta_pos
# cool temperature
t -= dt
pos = _rescale_layout(pos)
return pos
def _rescale_layout(pos, scale=1):
# rescale to (0,pscale) in all axes
# shift origin to (0,0)
lim = 0 # max coordinate for all axes
for i in range(pos.shape[1]):
pos[:, i] -= pos[:, i].min()
lim = max(pos[:, i].max(), lim)
# rescale to (0,scale) in all directions, preserves aspect
for i in range(pos.shape[1]):
pos[:, i] *= scale / lim
return pos
def draw_graph_with_height(g,highlighted_nodes=set([]),figsize=(15,15),iterations=150,title=''):
""" Try to draw a reasonable picture of a graph with a height feature on each node."""
pos = { p : (5*np.random.random(),2*data["height"]) for (p,data) in g.nodes(data=True)} # random x, height fixed y.
pos_indices = [i for i in pos.keys()]
pos_flat = np.asarray([pos[i] for i in pos.keys()])
A = nx.adjacency_matrix(g.to_undirected())
Adense = A.todense()
Adensefloat = Adense.astype(float)
new_pos = _fruchterman_reingold(Adensefloat, dim=2, pos=pos_flat, fixed=[0,len(pos_flat)-1], iterations=iterations, hold_dim=1)
pos_dict = { pos_indices[i] : tuple(new_pos[i]) for i in range(len(pos_indices))}
# for u,v,d in g.edges(data=True):
# d['weight'] = float(d['t'][1]-d['t'][0])
# edges,weights = zip(*nx.get_edge_attributes(g,'weight').items())
# print(weights)
fig, ax = plt.subplots(figsize=figsize)
if title: fig.suptitle(title, fontsize=16)
if highlighted_nodes:
nx.draw(g, pos=pos_dict, alpha=.1, font_size=14,node_color='b')
gsub = nx.subgraph(g,highlighted_nodes)
nx.draw(gsub, pos=pos_dict, node_color='r')
else:
nx.draw(g,pos=pos_dict)
plt.show()
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
I am working on basic distance calculation using accelerometer by dragging object on a single axis for physics class in MATLAB and have problem with plotting data.
My steps are:
1) After calibrating device to read zero g on every axis, I eliminate drift errors:
X_real = X_sample - X_calibrated;
if(X_real <= X_drift )
{
X_real = 0;
}
Where X_drift is 2 mg (From datasheet of accelerometer)
2) Calculate velocity:
velocity = 0; % On start
% Integration
v(i) = v(i-1) - x(i-1)-(x(i)+x(i-1)+x(i-2)+x(i-3))/4;
%Check if we stopped
if(x(i-1)==0 && x(i)==0)
v(i)=0;
end
%Check if velocity is under 0 (Not allowed)
if(v(i) < 0)
v(i)=0;
end
velocity = velocity + v(i);
3) Calculate distance:
distance = 0; % On start
%Integration
s(i) = s(i-1) + v(i-1) + (v(i)-v(i-1)-v(i-2)-v(i-3))/4;
distance = distance + s(i);
After testing this by dragging accelerometer on table 20 cm i got these results:
velocity = 0.09 m/s
distance = 0.21 m
time = 3.2s
Error of 1 cm is OK for classroom.
Chart tells something different:
I tried to plot distance after this:
s(i) = s(i)+s(i-1);
And got 21 cm on chart but after 6 s not after 4 s where it should be.
What am I doing wrong?
*UPDATE: Position y value is in mm not cm! Sorry
Im sorry for asking for help, i thought my formulas were ok, but they didn't. After step by step calculations my final solution is:
1) Velocity:
v(i) = v(i-1) - x(i-1)-(x(i)+x(i-1))/2;
2) Distance:
s(i) = s(i-1) + v(i-1)+(v(i)+v(i-1))/2;
And chart is:
Sorry once more time. I hope this will help someone calculating velocity and distance. It surely helped me as lesson to better study my code next time before asking for help.
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.
Question: A photon moving at speed 1 in the x–y plane starts at t = 0 at (x, y) = (0.5, 0.1) heading due east. Around every integer lattice point (i, j) in the plane, a circular mirror of radius 1/3 has been erected. How far from the origin is the photon at t = 10?
error:
Error: File: sodifficult.m Line: 1 Column: 28
Function definition is misplaced or improperly nested.
function [ output_args ] = function( input_args )
C=[1,-1,0,0,-1,0,-1,0,0,-1,0,-1,-1,-1;0,1,2,1,2,0,0,1,0,0,-1,-1,0,-1]
lx=zeros(1,15);
ly=zeros(1,15);
x=zeros(1,15);
y=zeros(1,15);
t=zeros(1,15);
ttotal=zeros(1,15);
ux=zeros(1,14);
uy=zeros(1,14);
vx=zeros(1,14);
vy=zeros(1,14);
x0=0.5;
y0=0.1;
lx(1)=1;
ly(1)=0;
x(1)=1-sqrt(0.91)/3;
y(1)=0.1;
t(1)=x(1)-x0;
ttotal=t(1);
for k=1:14
ux(k)=3*(x(k)-C(1,k))
uy(k)=3*(y(k)-C(2,k))
vx(k)=-uy(k)
vy(k)=ux(k)
i=k+1
lx(i) = -(lx(k)*ux(k)+ly(k)*uy(k))*ux(k)+(lx(k)*vx(k)+ly(k)*vy(k))*vx(k)
ly(i) = -(lx(k)*ux(k)+ly(k)*uy(k))*uy(k)+(lx(k)*vx(k)+ly(k)*vy(k))*vy(k)
if k<14
b = lx(i)*(x(k)-C(1,i)) + ly(i)*(y(k)-C(2,i))
c = (x(k)-C(1,i))^2 + (y(k)-C(2,i))^2 - 1/9
t(i) = -b-sqrt(b*b-c)
ttotal = ttotal + t(i)
else
t(i)=1-ttotal
ttotal=1
end
x(i)=x(k)+t(i)*lx(i)
y(i)=y(k)+t(i)*ly(i)
end
d = sqrt(x(15)^2+y(15)^2)
end
Don't have enough rep to comment yet, apparently... but horchler is right, the error you're getting is clearly on line 1. try function [ output_args ] = function_name( input_args ).