I have trouble deciphering the individual parameters of the NURBS formula in the shapes NURBSTo entry (used for splines - curved edges). MS Visio documentation didn't help too much.
The number of parameters is variable depending on the complexity of the curve. A more simple example is:
NURBS(0.4492,3,0,1,0,-0.1875,0,1,1,-0.1875,0,1)
where I found out the start and end coordinate parameters start is 5th for X and 6th for Y. End is 9th for X and 10th for Y. But the Y coordinates are still wrong, so I suppose they should be combined with another parameter. This Java code provided the best results so far in getting the control points of the spline:
int j = 0;
for (int i = 2; i + 4 < pointsS.length; i = i + 4)
{
mxPoint currPoint = new mxPoint();
currPoint.setX(startXY.getX() + (endXY.getX() - startXY.getX()) * pointsRaw[i + 2]);
currPoint.setY(startXY.getY() - (endXY.getY() - startXY.getY()) * pointsRaw[i + 3]);
pointList.add(currPoint);
j++;
}
Just an example for a more complex spline:
NURBS(2.9857,3,1,1,0.1875,0,0,1,0.1875,-0.8954,0,1,0.1875,-1.3431,0,1,0.1875,-1.7908,0.4521,1,-0.4936,-1.7908,1.049,1,-1.1747,-1.7908,1.424,1,-1.1747,-2.1799,1.902,1,-1.1747,-2.5689,2.3742,1)
The documentation say for param 2 only "degree". I suppose it's the degree of the polynomial that is used for approximation.
The wiki page about NURBS:
http://en.wikipedia.org/wiki/Non-uniform_rational_B-spline
Of course, it doesn't speak about Visio parameters :)
Are you accounting for the effect third and fourth parameters have on how you should interpret the x and y parameters?
From MSDN (http://msdn.microsoft.com/en-us/library/office/aa224197(v=office.11).aspx):
NURBS(knotLast, degree, xType, yType, x1, y1, knot1, weight1, ...)
knotLast The last knot.
degree The spline's degree.
xType Specifies how to interpret the x input data. If xType is 0, all x input data are interpreted as a percentage of Width. If xType is 1, all x input data are interpreted as local coordinates.
yType Specifies how to interpret the y input data. If yType is 0, all y input data are interpreted as a percentage of Height. If yType is 1, all y input data are interpreted as local coordinates.
x1 An x-coordinate.
y1 A y-coordinate.
knot1 A knot on the B-spline.
weight1 A weight on the B-spline.
This may help: Visio 2003 Developer's Survival Pack by Graham Wideman
http://www.amazon.com/Visio-2003-Developers-Survival-Pack/dp/1412011124
There's an extensive section on Visio NURBS. It's only $7 for the Kindle edition.
Related
I do not know what this error means or how to fix it. I am trying to perform an image rotation in a separate space of coordinates. When defining the reference space of the matrix to be at zero, I am getting the error that integers can only be comibined with integers of the same class or scalar doubles. the line is
WZcentered = WZ - [x0;yo]*ones(1,Ncols);
WZ is classified as a 400x299x3 unit 8, in the workspace. It is an image. x0 and y0 are set to 0 when the function is called. How can I fix this issue/what exactly is happening here?
Also, when I do the same thing yet make WZ to be equal to double(WZ) I get the error that 'matrix dimensions must agree.' I am not sure what the double function does however. Here is the whole code.
function [out_flag, WZout, x_final, y_final] = adopted_moveWZ(WZ, x0, y0);
%Initial Test of plot
[Nrows,Ncols]=size(WZ);
if Nrows ~= 2
if Ncols ==2
WZ=transpose(WZ); %take transpose
[Nrows,Ncols]=size(WZ); %reset the number of rows and columns
else
fprintf('ERROR: Input file should have 2-vectors for the input points.\n');
end
end
plot(WZ(1,:),WZ(2,:),'.')
title('These are the original points in the image');
pause(2.0)
%WZorig = WZ;
%centering
WZcentered = WZ - ([x0;y0] * ones(1,Ncols));
FigScale=400;
axis([-FigScale 2*FigScale -FigScale 2*FigScale])
disp('Hit any key to start the animation');
pause;
SceneCenter = zeros(Nrows,Ncols);
WZnew = WZcentered;
for ii=0:20
%rotate
R = [cos(pi/ii) -sin(pi/ii) 0; sin(pi/ii) cos(pi/ii) 0; 0 0 1];
WZnew = R * WZnew;
plot(WZnew(1,:),WZnew(2,:),'.')
%place WZnew at a different place in the scene
SceneCenter = (ii*[30;40])*ones(1,Ncols);
plot(SceneCenter(1,:) + WZnew(1,:), SceneCenter(2,:) + WZnew(2,:),'.')
axis([-FigScale 2*FigScale -FigScale 2*FigScale])
pause(1.0);
end
%Set final values for output at end of program
x_final = SceneCenter(1,1);
y_final = SceneCenter(2,1);
PPout = PPnew + SceneCenter;
This happens due to WZ and ([x0;y0] * ones(1,Ncols)) being of different data types. You might think MATLAB is loosely typed, and hence should do the right thing when you have a floating point type operated with an integer type, but this rule breaks every once in a while. A simpler example to demonstrate this is here:
X = uint8(magic(5))
Y = zeros(5)
X - Y
This breaks with the same error that you are reporting. One way to fix this is to force cast one of the operands to the other, typically up-casted to make sure the math works. When you do this, both the numbers you are working on are floating point (double precision), and so they are represented in the same byte formatting sequence in memory. This way, the '-' sign is valid, in the same way that you can say 3 apples + 4 apples = 7 apples, but 3 oranges (uint8) + 4 apples (double) = ?. The double(X) makes it clear that you really mean to use double precision arithmetic, and hence fixes the error. This is how it looks now:
double(X) - Y
After having identified this, the new error is 'matrix dimensions do not match'. This means exactly what it says. WZ is a 400x299x3 matrix, and the right hand side matrix is 2xnCols. Now can you subtract a 2D matrix from a 3D matrix of different sizes meaningfully?
Depending on what your code is really intending to do, you can pad the RHS matrix, or find out other ways to make the sizes equal.
All of this is why MATLAB includes routines to do image rotation, namely http://www.mathworks.com/help/images/ref/imrotate.html . This is part of the Image Processing Toolbox, though.
So I am working on my Thesis and I need to calculate geometric characteristics of an airfoil.
To do this, I need to interpolate the horizontal and vertical coordinates of an airfoil. This is used for a tool which will calculate the geometric characteristics automatically which come from random airfoil geometry files.
Sometime the Y values of the airfoil are non monotonic. Hence, the interp1 command gives an error since some values in the Y vector are repeated.
Therefore, my question is: How do I recognize and subsequently interpolate non monotonic increasing data automatically in Matlab.
Here is a sample data set:
0.999974 0.002176
0.994846 0.002555
0.984945 0.003283
0.973279 0.004131
0.960914 0.005022
0.948350 0.005919
0.935739 0.006810
0.923111 0.007691
0.910478 0.008564
0.897850 0.009428
0.885229 0.010282
0.872617 0.011125
0.860009 0.011960
0.847406 0.012783
0.834807 0.013598
0.822210 0.014402
0.809614 0.015199
0.797021 0.015985
0.784426 0.016764
0.771830 0.017536
0.759236 0.018297
0.746639 0.019053
0.734038 0.019797
0.721440 0.020531
0.708839 0.021256
0.696240 0.021971
0.683641 0.022674
0.671048 0.023367
0.658455 0.024048
0.645865 0.024721
0.633280 0.025378
0.620699 0.026029
0.608123 0.026670
0.595552 0.027299
0.582988 0.027919
0.570436 0.028523
0.557889 0.029115
0.545349 0.029697
0.532818 0.030265
0.520296 0.030820
0.507781 0.031365
0.495276 0.031894
0.482780 0.032414
0.470292 0.032920
0.457812 0.033415
0.445340 0.033898
0.432874 0.034369
0.420416 0.034829
0.407964 0.035275
0.395519 0.035708
0.383083 0.036126
0.370651 0.036530
0.358228 0.036916
0.345814 0.037284
0.333403 0.037629
0.320995 0.037950
0.308592 0.038244
0.296191 0.038506
0.283793 0.038733
0.271398 0.038920
0.259004 0.039061
0.246612 0.039153
0.234221 0.039188
0.221833 0.039162
0.209446 0.039064
0.197067 0.038889
0.184693 0.038628
0.172330 0.038271
0.159986 0.037809
0.147685 0.037231
0.135454 0.036526
0.123360 0.035684
0.111394 0.034690
0.099596 0.033528
0.088011 0.032181
0.076685 0.030635
0.065663 0.028864
0.055015 0.026849
0.044865 0.024579
0.035426 0.022076
0.027030 0.019427
0.019970 0.016771
0.014377 0.014268
0.010159 0.012029
0.007009 0.010051
0.004650 0.008292
0.002879 0.006696
0.001578 0.005207
0.000698 0.003785
0.000198 0.002434
0.000000 0.001190
0.000000 0.000000
0.000258 -0.001992
0.000832 -0.003348
0.001858 -0.004711
0.003426 -0.005982
0.005568 -0.007173
0.008409 -0.008303
0.012185 -0.009379
0.017243 -0.010404
0.023929 -0.011326
0.032338 -0.012056
0.042155 -0.012532
0.052898 -0.012742
0.064198 -0.012720
0.075846 -0.012533
0.087736 -0.012223
0.099803 -0.011837
0.111997 -0.011398
0.124285 -0.010925
0.136634 -0.010429
0.149040 -0.009918
0.161493 -0.009400
0.173985 -0.008878
0.186517 -0.008359
0.199087 -0.007845
0.211686 -0.007340
0.224315 -0.006846
0.236968 -0.006364
0.249641 -0.005898
0.262329 -0.005451
0.275030 -0.005022
0.287738 -0.004615
0.300450 -0.004231
0.313158 -0.003870
0.325864 -0.003534
0.338565 -0.003224
0.351261 -0.002939
0.363955 -0.002680
0.376646 -0.002447
0.389333 -0.002239
0.402018 -0.002057
0.414702 -0.001899
0.427381 -0.001766
0.440057 -0.001656
0.452730 -0.001566
0.465409 -0.001496
0.478092 -0.001443
0.490780 -0.001407
0.503470 -0.001381
0.516157 -0.001369
0.528844 -0.001364
0.541527 -0.001368
0.554213 -0.001376
0.566894 -0.001386
0.579575 -0.001398
0.592254 -0.001410
0.604934 -0.001424
0.617614 -0.001434
0.630291 -0.001437
0.642967 -0.001443
0.655644 -0.001442
0.668323 -0.001439
0.681003 -0.001437
0.693683 -0.001440
0.706365 -0.001442
0.719048 -0.001444
0.731731 -0.001446
0.744416 -0.001443
0.757102 -0.001445
0.769790 -0.001444
0.782480 -0.001445
0.795173 -0.001446
0.807870 -0.001446
0.820569 -0.001446
0.833273 -0.001446
0.845984 -0.001448
0.858698 -0.001448
0.871422 -0.001451
0.884148 -0.001448
0.896868 -0.001446
0.909585 -0.001443
0.922302 -0.001445
0.935019 -0.001446
0.947730 -0.001446
0.960405 -0.001439
0.972917 -0.001437
0.984788 -0.001441
0.994843 -0.001441
1.000019 -0.001441
First column is X and the second column is Y. Notice how the last values of Y are repeated.
Maybe someone can provide me with a piece of code to do this? Or any suggestions are welcome as well.
Remember I need to automate this process.
Thanks for your time and effort I really appreciate it!
There is quick and dirty method if you do not know the exact function defining the foil profile. Split your data into 2 sets, top and bottom planes, so the 'x' data are monotonic increasing.
First I imported your data table in the variable A, then:
%// just reorganise your input in individual vectors. (this is optional but
%// if you do not do it you'll have to adjust the code below)
x = A(:,1) ;
y = A(:,2) ;
ipos = y > 0 ; %// indices of the top plane
ineg = y <= 0 ; %// indices of the bottom plane
xi = linspace(0,1,500) ; %// new Xi for interpolation
ypos = interp1( x(ipos) , y(ipos) , xi ) ; %// re-interp the top plane
yneg = interp1( x(ineg) , y(ineg) , xi ) ; %// re-interp the bottom plane
y_new = [fliplr(yneg) ypos] ; %// stiches the two half data set together
x_new = [fliplr(xi) xi] ;
%% // display
figure
plot(x,y,'o')
hold on
plot(x_new,y_new,'.r')
axis equal
As said on top, it is quick and dirty. As you can see from the detail figure, you can greatly improve the x resolution this way in the area where the profile is close to the horizontal direction, but you loose a bit of resolution at the noose of the foil where the profile is close to the vertical direction.
If it's acceptable then you're all set. If you really need the resolution at the nose, you could look at interpolating on x as above but do a very fine x grid near the noose (instead of the regular x grid I provided as example).
if your replace the xi definition above by:
xi = [linspace(0,0.01,50) linspace(0.01,1,500)] ;
You get the following near the nose:
adjust that to your needs.
To interpolate any function, there must be a function defined. When you define y=f(x), you cannot have the same x for two different values of y because then we are not talking about a function. In your example data, neither x nor y are monotonic, so anyway you slice it, you'll have two (or more) "y"s for the same "x". If you wish to interpolate, you need to divide this into two separate problems, top/bottom and define proper functions for interp1/2/n to work with, for example, slice it horizontally where x==0. In any case, you would have to provide additional info than just x or y alone, e.g.: x=0.5 and y is on top.
On the other hand, if all you want to do is to insert a few values between each x and y in your array, you can do this using finite differences:
%// transform your original xy into 3d array where x is in first slice and y in second
xy = permute(xy(85:95,:), [3,1,2]); %// 85:95 is near x=0 in your data
%// lets say you want to insert three additional points along each line between every two points on given airfoil
h = [0, 0.25, 0.5, 0.75].'; %// steps along each line - column vector
%// every interpolated h along the way between f(x(n)) and f(x(n+1)) can
%// be defined as: f(x(n) + h) = f(x(n)) + h*( f(x(n+1)) - f(x(n)) )
%// this is first order finite differences approximation in 1D. 2D is very
%// similar only with gradient (this should be common knowledge, look it up)
%// from here it's just fancy matrix play
%// 2D gradient of xy curve
gradxy = diff(xy, 1, 2); %// diff xy, first order, along the 2nd dimension, where x and y now run
h_times_gradxy = bsxfun(#times, h, gradxy); %// gradient times step size
xy_in_3d_array = bsxfun(#plus, xy(:,1:end-1,:), h_times_gradxy); %// addition of "f(x)" and there we have it, the new x and y for every step h
[x,y] = deal(xy_in_3d_array(:,:,1), xy_in_3d_array(:,:,2)); %// extract x and y from 3d matrix
xy_interp = [x(:), y(:)]; %// use Matlab's linear indexing to flatten x and y into columns
%// plot to check results
figure; ax = newplot; hold on;
plot(ax, xy(:,:,1), xy(:,:,2),'o-');
plot(ax, xy_interp(:,1), xy_interp(:,2),'+')
legend('Original','Interpolated',0);
axis tight;
grid;
%// The End
And these are the results, near x=0 for clarity of presentation:
Hope that helps.
Cheers.
In order to find best fit (thru polyfit), i am getting negative p value but matlab is not accepting it (Subscript indices must either be real positive integers or logicals.). Is there any way that I can use it ? I can't think of alternative method. I'll always get negative values.
EDIT:
I am trying to flattening baseline of a curve, for that. I am running for loop to have fit from 1 to 3 order. And then I am using smallest normr s value to to find the best fit and then subtract it from the whole curve to get baseline straight. I tried with few curves it works well but not with all of the data because of above describes issue.
part of the code I am working on:
for i=1:3
[p,s]=polyfit(x,y,i);
a=s.normr;
b(i,1)=p(1);
normr(i,1)=a;
ind=find(b==min(b));
mn=b(ind,1);
Yflat=y-mn(1)*(x-mean(x));
ca{2,2}=Yflat;
clear a b normr p s rte ind ind2 Yflat
end
When I translate an image into negative coordinates,
I usually record an offset e.g.
offset = [ -5, -8.5 ]
and save the intensity values in matrix begin with (1, 1) as usual,
But when comes to calculation, let the coordinates array add up with the offset
e.g. [ actualX, actualY ] = [ x, y ] + offset ;
It may need extra efforts, but it works.
Good Luck!
The code below (your code from the comments + initialization of x, y) executes. What is the problem?
x = 1:50;
y = randn(size(x));
for i=1:3
[p,s]=polyfit(x,y,i);
a=s.normr;
b(i,1)=p(1);
normr(i,1)=a;
ind=find(b==min(b));
mn=b(ind,1);
Yflat=y-mn(1)*(x-mean(x));
ca{2,2}=Yflat;
end
I’m currently a Physics student and for several weeks have been compiling data related to ‘Quantum Entanglement’. I’ve now got to a point where I have to plot my data (which should resemble a cos² graph - and does) to a sort of “best fit” cos² graph. The lab script says the following:
A more precise determination of the visibility V (this is basically how 'clean' the data is) follows from the best fit to the measured data using the function:
f(b) = A/2[1-Vsin(b-b(center)/P)]
Granted this probably doesn’t mean much out of context, but essentially A is the amplitude, b is an angle and P is the periodicity. Hence this is also a “wave” like the experimental data I have found.
From this I understand, as previously mentioned, I am making a “best fit” curve. However, I have been told that this isn’t possible with Excel and that the best approach is Matlab.
I know intermediate JavaScript but do not know Matlab and was hoping for some direction.
Is there a tutorial I can read for this? Is it possible for someone to go through it with me? I really have no idea what it entails, so any feed back would be greatly appreciated.
Thanks a lot!
Initial steps
I guess we should begin by getting a representation in Matlab of the function that you're trying to model. A direct translation of your formula looks like this:
function y = targetfunction(A,V,P,bc,b)
y = (A/2) * (1 - V * sin((b-bc) / P));
end
Getting hold of the data
My next step is going to be to generate some data to work with (you'll use your own data, naturally). So here's a function that generates some noisy data. Notice that I've supplied some values for the parameters.
function [y b] = generateData(npoints,noise)
A = 2;
V = 1;
P = 0.7;
bc = 0;
b = 2 * pi * rand(npoints,1);
y = targetfunction(A,V,P,bc,b) + noise * randn(npoints,1);
end
The function rand generates random points on the interval [0,1], and I multiplied those by 2*pi to get points randomly on the interval [0, 2*pi]. I then applied the target function at those points, and added a bit of noise (the function randn generates normally distributed random variables).
Fitting parameters
The most complicated function is the one that fits a model to your data. For this I use the function fminunc, which does unconstrained minimization. The routine looks like this:
function [A V P bc] = bestfit(y,b)
x0(1) = 1; %# A
x0(2) = 1; %# V
x0(3) = 0.5; %# P
x0(4) = 0; %# bc
f = #(x) norm(y - targetfunction(x(1),x(2),x(3),x(4),b));
x = fminunc(f,x0);
A = x(1);
V = x(2);
P = x(3);
bc = x(4);
end
Let's go through line by line. First, I define the function f that I want to minimize. This isn't too hard. To minimize a function in Matlab, it needs to take a single vector as a parameter. Therefore we have to pack our four parameters into a vector, which I do in the first four lines. I used values that are close, but not the same, as the ones that I used to generate the data.
Then I define the function I want to minimize. It takes a single argument x, which it unpacks and feeds to the targetfunction, along with the points b in our dataset. Hopefully these are close to y. We measure how far they are from y by subtracting from y and applying the function norm, which squares every component, adds them up and takes the square root (i.e. it computes the root mean square error).
Then I call fminunc with our function to be minimized, and the initial guess for the parameters. This uses an internal routine to find the closest match for each of the parameters, and returns them in the vector x.
Finally, I unpack the parameters from the vector x.
Putting it all together
We now have all the components we need, so we just want one final function to tie them together. Here it is:
function master
%# Generate some data (you should read in your own data here)
[f b] = generateData(1000,1);
%# Find the best fitting parameters
[A V P bc] = bestfit(f,b);
%# Print them to the screen
fprintf('A = %f\n',A)
fprintf('V = %f\n',V)
fprintf('P = %f\n',P)
fprintf('bc = %f\n',bc)
%# Make plots of the data and the function we have fitted
plot(b,f,'.');
hold on
plot(sort(b),targetfunction(A,V,P,bc,sort(b)),'r','LineWidth',2)
end
If I run this function, I see this being printed to the screen:
>> master
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the function tolerance.
A = 1.991727
V = 0.979819
P = 0.695265
bc = 0.067431
And the following plot appears:
That fit looks good enough to me. Let me know if you have any questions about anything I've done here.
I am a bit surprised as you mention f(a) and your function does not contain an a, but in general, suppose you want to plot f(x) = cos(x)^2
First determine for which values of x you want to make a plot, for example
xmin = 0;
stepsize = 1/100;
xmax = 6.5;
x = xmin:stepsize:xmax;
y = cos(x).^2;
plot(x,y)
However, note that this approach works just as well in excel, you just have to do some work to get your x values and function in the right cells.
I have a tab separated XYZ file which contains 3 columns, e.g.
586231.8 2525785.4 15.11
586215.1 2525785.8 14.6
586164.7 2525941 14.58
586199.4 2525857.8 15.22
586219.8 2525731 14.6
586242.2 2525829.2 14.41
Columns 1 and 2 are the X and Y coordinates (in UTM meters) and column 3 is the associated Z value at the point X,Y; e.g. the elevation (z) at a point is given as z(x,y)
I can read in this file using dlmread() to get 3 variables in the workspace, e.g. X = 41322x1 double, but I would like to create a surface of size (m x n) using these variables. How would I go about this?
Following from the comments below, I tried using TriScatteredInterp (see commands below). I keep getting the result shown below (it appears to be getting some of my surface though):
Any ideas what is going on to cause this result? I think the problem lies with themeshgrid command, though I'm not sure where (or why). I am currently putting in the following set of commands to calculate the above figure (my X and Y columns are in meters, and I know my grid size is 8m, hence ti/tj going up in 8s):
F = TriScatteredInterp(x,y,z,'nearest');
ti = ((min(x)):8:(max(x)));
tj = ((min(y)):8:(max(y)));
[qx,qy] = meshgrid(ti,tj);
qz = F(qx,qy);
imagesc(qz) %produces the above figure^
I think you want the griddata function. See Interpolating Scattered Data in MATLAB help.
Griddata and tirscattteredinterp are extremely slow. Use the utm2deg function on the file exchange and from there a combination of both vec2mtx to make a regular grid and then imbedm to fit the data to the grid.
I.E.
for i = 1:length(X)
[Lat,Lon ] = utm2deg(Easting ,Northing ,Zone);
end
[Grid, R] = vec2mtx(Lat, Lon, gridsize);
Grid= imbedm(Lat, Lon,z, Grid, R);
Maybe you are looking for the function "ndgrid(x,y)" or "meshgrid(x,y)"