This should be an easy one. I am trying to find the coordinates of a point on a straight line. I am implementing in MATLAB. I know, the coordinates of the endpoints and the distance from one of the point.
I am using the following formula for calculating the coordinates (please note, I cannot use mid-point formula, as the distance can vary).
I am getting the wrong results when the slope is negative. Can you please suggest, what are the conditions, that needs to be considered for using this formula? Am not aware of any other formula as well.
That's too complicated solution for such a simple task. Use direct vector computations:
function P = point_on_line(A, B, AP)
D = B - A;
P = A + D / norm(D) * AP;
end
Call like this:
P = point_on_line([x1 y1], [x2 y2], len);
x = P(1);
y = P(2);
Ask if you need any clarifications.
Nothing wrong with your solution, but you need to take care of quadrant ambiguities when you take the arctangent to compute the angle θ.
There is a nice solution for that in most programming languages: atan2. Thus:
%// Your points (fill in any values)
A = [-10 0];
B = [-1 -1];
%// Use atan2!
th = atan2( B(2)-A(2) , B(1)-A(1) );
%// Distance from A to the point of interest
AP = sqrt( (B(2)-A(2))^2 + (B(1)-A(1))^2 ) / 2;
%// The point of interest
C = [
A(1) + AP*cos( th )
A(2) + AP*sin( th )];
%// Verify correctness with plots
figure(1), clf, hold on
line([A(1); B(1)], [A(2); B(2)])
plot(...
A(1), A(2), 'r.',...
B(1), B(2), 'b.',...
C(1), C(2), 'k.', 'markersize', 20)
In general, whenever and wherever you need to take an arctangent, use atan2 and not atan. The normal atan is only for cases where you don't know the individual components of the division y/x.
Note that your solution is not extensible to 3D, whereas the vector solutions proposed by the others here are. So in general I would indeed advise you to start working with vectors. Not only is it a lot simpler in many circumstances, it is also more versatile.
This is a wrong approach to the solution, as the solution is not unique. There are two points on you line with the same distance AP from the point A: one going left and another one going right.
The are infinite approaches to solve this, I prefer vector notation.
vector ab is a 2x1 matlab matrix:
ab = B-A
abN is the normalized vector
abN = ab/norm(ab)
stepping from A in the abN direction a distance of d (in your case AP) is:
A + abN*d
hope it helped.
Ohad
Related
I have data that look like this:
These are curves of the same process but with different parameters.
I need to find the index (or x value) for certain y values (say, 10).
For the blue curve, this is easy: I'm using min to find the index:
[~, idx] = min(abs(y - target));
where y denotes the data and target the wanted value.
This approach works fine since I know that there is an intersection, and only one.
Now what to do with the red curve? I don't know beforehand, if there will be two intersections, so my idea of finding the first one and then stripping some of the data is not feasible.
How can I solve this?
Please note the the curves can shift in the x direction, so that checking the found solution for its xrange is not really an option (it could work for the data I have, but since there are more to come, this solution is probably not the best).
Shameless steal from here:
function x0 = data_zeros(x,y)
% Indices of Approximate Zero-Crossings
% (you can also use your own 'find' method here, although it has
% this pesky difference of 1-missing-element because of diff...)
dy = find(y(:).*circshift(y(:), [-1 0]) <= 0);
% Do linear interpolation of near-zero-crossings
x0 = NaN(size(dy,1)-1,1);
for k1 = 1:size(dy,1)-1
b = [[1;1] [x(dy(k1)); x(dy(k1)+1)]] \ ...
[y(dy(k1)); y(dy(k1)+1)];
x0(k1) = -b(1)/b(2);
end
end
Usage:
% Some data
x = linspace(0, 2*pi, 1e2);
y = sin(x);
% Find zeros
xz = data_zeros1(x,y);
% Plot original data and zeros found
figure(1), hold on
plot(x, y);
plot(xz, zeros(size(xz)), '+r');
axis([0,2*pi -1,+1]);
The gist: multiply all data points with their consecutive data points. Any of these products that is negative therefore has opposite sign, and gives you an approximate location of the zero. Then use linear interpolation between the same two points to get a more precise answer, and store that.
NOTE: for zeros exactly at the endpoints, this approach will not work. Therefore, it may be necessary to check those manually.
Subtract the desired number from your curve, i.e. if you want the values at 10 do data-10, then use and equality-within-tolerance, something like
TOL = 1e-4;
IDX = 1:numel(data(:,1)); % Assuming you have column data
IDX = IDX(abs(data-10)<=TOL);
where logical indexing has been used.
I figured out a way: The answer by b3 in this question did the trick.
idx = find(diff(y > target));
Easy as can be :) The exact xvalue can then be found by interpolation. For me, this is fine since i don't need exact values.
I'm trying to find the line which best fits to the data. I use the following code below but now I want to have the data placed into an array sorted so it has the data which is closest to the line first how can I do this? Also is polyfit the correct function to use for this?
x=[1,2,2.5,4,5];
y=[1,-1,-.9,-2,1.5];
n=1;
p = polyfit(x,y,n)
f = polyval(p,x);
plot(x,y,'o',x,f,'-')
PS: I'm using Octave 4.0 which is similar to Matlab
You can first compute the error between the real value y and the predicted value f
err = abs(y-f);
Then sort the error vector
[val, idx] = sort(err);
And use the sorted indexes to have your y values sorted
y2 = y(idx);
Now y2 has the same values as y but the ones closer to the fitting value first.
Do the same for x to compute x2 so you have a correspondence between x2 and y2
x2 = x(idx);
Sembei Norimaki did a good job of explaining your primary question, so I will look at your secondary question = is polyfit the right function?
The best fit line is defined as the line that has a mean error of zero.
If it must be a "line" we could use polyfit, which will fit a polynomial. Of course, a "line" can be defined as first degree polynomial, but first degree polynomials have some properties that make it easy to deal with. The first order polynomial (or linear) equation you are looking for should come in this form:
y = mx + b
where y is your dependent variable and X is your independent variable. So the challenge is this: find the m and b such that the modeled y is as close to the actual y as possible. As it turns out, the error associated with a linear fit is convex, meaning it has one minimum value. In order to calculate this minimum value, it is simplest to combine the bias and the x vectors as follows:
Xcombined = [x.' ones(length(x),1)];
then utilized the normal equation, derived from the minimization of error
beta = inv(Xcombined.'*Xcombined)*(Xcombined.')*(y.')
great, now our line is defined as Y = Xcombined*beta. to draw a line, simply sample from some range of x and add the b term
Xplot = [[0:.1:5].' ones(length([0:.1:5].'),1)];
Yplot = Xplot*beta;
plot(Xplot, Yplot);
So why does polyfit work so poorly? well, I cant say for sure, but my hypothesis is that you need to transpose your x and y matrixies. I would guess that that would give you a much more reasonable line.
x = x.';
y = y.';
then try
p = polyfit(x,y,n)
I hope this helps. A wise man once told me (and as I learn every day), don't trust an algorithm you do not understand!
Here's some test code that may help someone else dealing with linear regression and least squares
%https://youtu.be/m8FDX1nALSE matlab code
%https://youtu.be/1C3olrs1CUw good video to work out by hand if you want to test
function [a0 a1] = rtlinreg(x,y)
x=x(:);
y=y(:);
n=length(x);
a1 = (n*sum(x.*y) - sum(x)*sum(y))/(n*sum(x.^2) - (sum(x))^2); %a1 this is the slope of linear model
a0 = mean(y) - a1*mean(x); %a0 is the y-intercept
end
x=[65,65,62,67,69,65,61,67]'
y=[105,125,110,120,140,135,95,130]'
[a0 a1] = rtlinreg(x,y); %a1 is the slope of linear model, a0 is the y-intercept
x_model =min(x):.001:max(x);
y_model = a0 + a1.*x_model; %y=-186.47 +4.70x
plot(x,y,'x',x_model,y_model)
I am trying to achieve 3d reconstruction from 2 images. Steps I followed are,
1. Found corresponding points between 2 images using SURF.
2. Implemented eight point algo to find "Fundamental matrix"
3. Then, I implemented triangulation.
I have got Fundamental matrix and results of triangulation till now. How do i proceed further to get 3d reconstruction? I'm confused reading all the material available on internet.
Also, This is code. Let me know if this is correct or not.
Ia=imread('1.jpg');
Ib=imread('2.jpg');
Ia=rgb2gray(Ia);
Ib=rgb2gray(Ib);
% My surf addition
% collect Interest Points from Each Image
blobs1 = detectSURFFeatures(Ia);
blobs2 = detectSURFFeatures(Ib);
figure;
imshow(Ia);
hold on;
plot(selectStrongest(blobs1, 36));
figure;
imshow(Ib);
hold on;
plot(selectStrongest(blobs2, 36));
title('Thirty strongest SURF features in I2');
[features1, validBlobs1] = extractFeatures(Ia, blobs1);
[features2, validBlobs2] = extractFeatures(Ib, blobs2);
indexPairs = matchFeatures(features1, features2);
matchedPoints1 = validBlobs1(indexPairs(:,1),:);
matchedPoints2 = validBlobs2(indexPairs(:,2),:);
figure;
showMatchedFeatures(Ia, Ib, matchedPoints1, matchedPoints2);
legend('Putatively matched points in I1', 'Putatively matched points in I2');
for i=1:matchedPoints1.Count
xa(i,:)=matchedPoints1.Location(i);
ya(i,:)=matchedPoints1.Location(i,2);
xb(i,:)=matchedPoints2.Location(i);
yb(i,:)=matchedPoints2.Location(i,2);
end
matchedPoints1.Count
figure(1) ; clf ;
imshow(cat(2, Ia, Ib)) ;
axis image off ;
hold on ;
xbb=xb+size(Ia,2);
set=[1:matchedPoints1.Count];
h = line([xa(set)' ; xbb(set)'], [ya(set)' ; yb(set)']) ;
pts1=[xa,ya];
pts2=[xb,yb];
pts11=pts1;pts11(:,3)=1;
pts11=pts11';
pts22=pts2;pts22(:,3)=1;pts22=pts22';
width=size(Ia,2);
height=size(Ib,1);
F=eightpoint(pts1,pts2,width,height);
[P1new,P2new]=compute2Pmatrix(F);
XP = triangulate(pts11, pts22,P2new);
eightpoint()
function [ F ] = eightpoint( pts1, pts2,width,height)
X = 1:width;
Y = 1:height;
[X, Y] = meshgrid(X, Y);
x0 = [mean(X(:)); mean(Y(:))];
X = X - x0(1);
Y = Y - x0(2);
denom = sqrt(mean(mean(X.^2+Y.^2)));
N = size(pts1, 1);
%Normalized data
T = sqrt(2)/denom*[1 0 -x0(1); 0 1 -x0(2); 0 0 denom/sqrt(2)];
norm_x = T*[pts1(:,1)'; pts1(:,2)'; ones(1, N)];
norm_x_ = T*[pts2(:,1)';pts2(:,2)'; ones(1, N)];
x1 = norm_x(1, :)';
y1= norm_x(2, :)';
x2 = norm_x_(1, :)';
y2 = norm_x_(2, :)';
A = [x1.*x2, y1.*x2, x2, ...
x1.*y2, y1.*y2, y2, ...
x1, y1, ones(N,1)];
% compute the SVD
[~, ~, V] = svd(A);
F = reshape(V(:,9), 3, 3)';
[FU, FS, FV] = svd(F);
FS(3,3) = 0; %rank 2 constrains
F = FU*FS*FV';
% rescale fundamental matrix
F = T' * F * T;
end
triangulate()
function [ XP ] = triangulate( pts1,pts2,P2 )
n=size(pts1,2);
X=zeros(4,n);
for i=1:n
A=[-1,0,pts1(1,i),0;
0,-1,pts1(2,i),0;
pts2(1,i)*P2(3,:)-P2(1,:);
pts2(2,i)*P2(3,:)-P2(2,:)];
[~,~,va] = svd(A);
X(:,i) = va(:,4);
end
XP(:,:,1) = [X(1,:)./X(4,:);X(2,:)./X(4,:);X(3,:)./X(4,:); X(4,:)./X(4,:)];
end
function [ P1,P2 ] = compute2Pmatrix( F )
P1=[1,0,0,0;0,1,0,0;0,0,1,0];
[~, ~, V] = svd(F');
ep = V(:,3)/V(3,3);
P2 = [skew(ep)*F,ep];
end
From a quick look, it looks correct. Some notes are as follows:
You normalized code in eightpoint() is no ideal.
It is best done on the points involved. Each set of points will have its scaling matrix. That is:
[pts1_n, T1] = normalize_pts(pts1);
[pts2_n, T2] = normalize-pts(pts2);
% ... code
% solution
F = T2' * F * T
As a side note (for efficiency) you should do
[~,~,V] = svd(A, 0);
You also want to enforce the constraint that the fundamental matrix has rank-2. After you compute F, you can do:
[U,D,v] = svd(F);
F = U * diag([D(1,1),D(2,2), 0]) * V';
In either case, normalization is not the only key to make the algorithm work. You'll want to wrap the estimation of the fundamental matrix in a robust estimation scheme like RANSAC.
Estimation problems like this are very sensitive to non Gaussian noise and outliers. If you have a small number of wrong correspondence, or points with high error, the algorithm will break.
Finally, In 'triangulate' you want to make sure that the points are not at infinity prior to the homogeneous division.
I'd recommend testing the code with 'synthetic' data. That is, generate your own camera matrices and correspondences. Feed them to the estimate routine with varying levels of noise. With zero noise, you should get an exact solution up to floating point accuracy. As you increase the noise, your estimation error increases.
In its current form, running this on real data will probably not do well unless you 'robustify' the algorithm with RANSAC, or some other robust estimator.
Good luck.
Good luck.
Which version of MATLAB do you have?
There is a function called estimateFundamentalMatrix in the Computer Vision System Toolbox, which will give you the fundamental matrix. It may give you better results than your code, because it is using RANSAC under the hood, which makes it robust to spurious matches. There is also a triangulate function, as of version R2014b.
What you are getting is sparse 3D reconstruction. You can plot the resulting 3D points, and you can map the color of the corresponding pixel to each one. However, for what you want, you would have to fit a surface or a triangular mesh to the points. Unfortunately, I can't help you there.
If what you're asking is how to I proceed from fundamental Matrix + corresponding points to a dense model then you still have a lot of work ahead of you.
relative camera locations (R,T) can be calculated from a fundamental matrix assuming you know the internal camera params (up to scale, rotation, translation). To get a full dense matrix there are a few ways to go. you can try using an existing library (PMVS for example). I'd look into OpenMVG but I'm not sure about matlab interface.
Another way to go, you can compute a dense optical flow (many available for matlab). Look for a epipolar OF (It takes a fundamental matrix and restricts the solution to lie on the epipolar lines). Then you can triangulate every pixel to get a depthmap.
Finally you will have to play with format conversions to get from a depthmap to VRML (You can look at meshlab)
Sorry my answer isn't more Matlab oriented.
I have a problem I've been trying to solve and I cannot come up with the answer. I have written a function in Matlab which, given two lat/lon points and two bearings, will return the two great circle points of intersection.
However, what I really need is the first great circle point of intersection along the two initial headings. I.e. if two airplanes begin at lat/lon points 1 and 2, with initial bearings of bearing1 and bearing2, which of the two great circle intersection points is the first one they encounter? There are many solutions (using haversine) which will give me the closer of the two points, but I don't actually care about which is closer, I care about which I will encounter first given specific start points and headings. There are many cases where the closer of the two intersections is actually the second intersection encountered.
Now, I realize I could do this with lots of conditional statements for handling the different cases, but I figure there's got to be a way to handle it with regard to the order I take all my cross products (function code given below), but I simply can't come up with the right solution! I should also mention that this function is going to be used in a large computationally intensive model, and so I'm thinking the solution to this problem needs to be rather elegant/speedy. Can anyone help me with this problem?
The following is not my function code (I can't list that here), but it is the pseudo code that my function was based off of:
%Given inputs of lat1,lon1,Bearing1,lat2,lon2,Bearing2:
%Calculate arbitrary secondary point along same initial bearing from first
%point
dAngle = 45;
lat3 = asind( sind(lat1)*cosd(dAngle) + cosd(lat1)*sind(dAngle)*cosd(Bearing1));
lon3 = lon1 + atan2( sind(Bearing1)*sind(dAngle)*cosd(lat1), cosd(dAngle)-sind(lat1)*sind(lat3) )*180/pi;
lat4 = asind( sind(lat2)*cosd(dAngle) + cosd(lat2)*sind(dAngle)*cosd(Bearing2));
lon4 = lon2 + atan2( sind(Bearing2)*sind(dAngle)*cosd(lat2), cosd(dAngle)-sind(lat2)*sind(lat4) )*180/pi;
%% Calculate unit vectors
% We now have two points defining each of the two great circles. We need
% to calculate unit vectors from the center of the Earth to each of these
% points
[Uvec1(1),Uvec1(2),Uvec1(3)] = sph2cart(lon1*pi/180,lat1*pi/180,1);
[Uvec2(1),Uvec2(2),Uvec2(3)] = sph2cart(lon2*pi/180,lat2*pi/180,1);
[Uvec3(1),Uvec3(2),Uvec3(3)] = sph2cart(lon3*pi/180,lat3*pi/180,1);
[Uvec4(1),Uvec4(2),Uvec4(3)] = sph2cart(lon4*pi/180,lat4*pi/180,1);
%% Cross product
%Need to calculate the the "plane normals" for each of the two great
%circles
N1 = cross(Uvec1,Uvec3);
N2 = cross(Uvec2,Uvec4);
%% Plane of intersecting line
%With two plane normals, the cross prodcut defines their intersecting line
L = cross(N1,N2);
L = L./norm(L);
L2 = -L;
L2 = L2./norm(L2);
%% Convert normalized intersection line to geodetic coordinates
[lonRad,latRad,~]=cart2sph(L(1),L(2),L(3));
lonDeg = lonRad*180/pi;
latDeg = latRad*180/pi;
[lonRad,latRad,~]=cart2sph(L2(1),L2(2),L2(3));
lonDeg2 = lonRad*180/pi;
latDeg2 = latRad*180/pi;
UPDATE: A user on the Mathworks forums pointed this out:
Actually they might each reach a different point. I might have misunderstood the question, but the way you worded it suggests that both trajectories will converge towards the same point, which is not true. If you image the first point being a little after one intersection and the second point being a little after the other intersection, you have a situation were each "plane" will travel towards the intersection that was the closest to the other plane at the beginning.
I didn't think about this. It is actually very possible that each of the planes intersects a different great circle intersection first. Which just made things much more complicated...
Presumably you have a velocity vector for the direction of the plane (the direction in which you want to look for your first intersection - the "bearing vector on the surface of the earth"). You didn't actually specify this. Let us call it v.
You also have the cartesian coordinates of two points, P1 and P2, as well as the initial position of the plane, P0. Each is assumed to be already on the unit sphere (length = 1).
Now we want to know which is the shorter distance - to P1, or P2 - so we need to know the angle "as seen from the direction of the normal vector". For this we need both the sin and the cos - then we can find the angle using atan2. We obtain sin from the cross product (remember all vectors were normalized), and cos from the dot product. To get the sin "as seen from the right direction", we take dot product with the normal vector.
Here is a piece of code that puts it all together - I am using very simple coordinates for the points and direction vector so I can confirm in my head that this gives the correct answer:
% sample points of P0...P2 and v
P0 = [1 0 0];
P1 = [0 1 0];
P2 = [0 -1 0];
v = [0 1 0];
% compute the "start direction normal":
n0 = cross(P0, v);
n0 = n0 / norm( n0 ); % unit vector
% compute cross and dot products:
cr01 = cross(P0, P1);
cr02 = cross(P0, P2);
cos01 = dot(P0, P1);
cos02 = dot(P0, P2);
% to get sin with correct sign, take dot product with direction normal:
sin01 = dot(cr01, n0);
sin02 = dot(cr02, n0);
% note - since P0 P1 and P2 are all in the same plane
% cr02 and cr02 are either pointing in the same direction as n0
% or the opposite direction. In the latter case we get a sign flip for the sin
% in the former case this does nothing
% Now get the angle, mapped from 0 to 2 pi:
ang1 = mod(atan2(sin01, cos01), 2*pi);
ang2 = mod(atan2(sin02, cos02), 2*pi);
if( ang1 < ang2 )
fprintf(1,'point 1 is reached first\n');
% point 1 is the first one reached
else
fprintf(1,'point 2 is reached first\n');
% point 2 is the first
end
When you change the direction of the velocity vector (pointing towards P2 instead of P1), the program correctly tells you "point 2 is reached first".
Let me know if this works for you!
I need to draw an elliptic curve over the finite field F17(in other words, I want to draw some specific dots on the curve), but somehow I don't get it right.
The curve is defined by the equation:
y^2 = x^3 +x + 1 (mod 17)
I tried the way below, but it can't work.
for x = 0:16, plot(x, mod(sqrt(x^3+x+1), 16),'r')', end
Can someone help ?
[Update]
According to Nathan and Bill's suggestions, here is a slightly modified version.
x = 0:18
plot(mod(x,16), mod(sqrt(x.^3+x+1), 16),'ro')
However, I feel the figure is WRONG , e.g.,y is not an integer when x=4 .
You have to test all points that fulfill the equation y^2 = x^3 +x + 1 (mod 17). Since it is a finite field, you cannot simply take the square root on the right side.
This is how I would go about it:
a=0:16 %all points of your finite field
left_side = mod(a.^2,17) %left side of the equation
right_side = mod(a.^3+a+1,17) %right side of the equation
points = [];
%testing if left and right side are the same
%(you could probably do something nicer here)
for i = 1:length(right_side)
I = find(left_side == right_side(i));
for j=1:length(I)
points = [points;a(i),a(I(j))];
end
end
plot(points(:,1),points(:,2),'ro')
set(gca,'XTick',0:1:16)
set(gca,'YTick',0:1:16)
grid on;
Matlab works with vectors natively.
your syntax was close, but needs to be vectorized:
x = 0:16
plot(x, mod(sqrt(x.^3+x+1), 16),'r')
Note the . in x.^3. This tells Matlab to cube each element of x individually, as opposed to raising the vector x to the 3rd power, which doesn't mean anything.
You can use this code if you want to plot on Real numbers:
syms x y;
v=y^2-x^3-x-1;
ezplot(v, [-1,3,-5,5]);
But, for plot in modulo, at first you can write below code;
X=[]; for x=[0:16], z=[x; mod(x^3+x+1,17)]; X=[X, z]; end, X,
Then, you can plot X with a coordinate matrix.