Finding the Y value corresponding to X value in a parametric plot

how do i find the value of Y when X= 1360 for every loop? X and Y gives a matrix. s is a random number for each loop

smax=0.4;
smin=0.1;
S=smin+rand(1,n)*(smax-smin);
S_cumulative=cumsum(S);
  n= 3200;
  R=200.04;
  V=30;                  
  v=20000;  
  i=1;
   while i<n
       t= linspace (0,50,100000);

         X=R*sind((v*t/(R))+(V*(t))+S_cumulative(i);
         Y=-R*cosd((v*t/R));
    end
    i=i+1;

[KL]

Hi Joseph, Can you describe what you're trying to achieve this code?

[Joseph Lee]

this plots 3200 parametric curves and im trying to find the minimum Y values among all the curves for X range {1360,1400}, eg. for position X=1360, i want to find the lowest Y value for that point given by all the curves. using find(X==1360) does not work. I can solve it on paper by finding t, time then substituting into Y equation but i cant put this into the code.

Answer 1

[KL]

If all your curves have the same timestep and same number of measurements why not put them all together in a matrix and then find the minimum? For example,

allY = rand(10,4); %say, you have 4 curves with 10 values (random data in this example)
X = (1:10)';
data_summary = [X allY] %put them together
intvl = 3:6; %in your case 1360 to 1400
minVals = min(data_summary(intvl,2:end))

[Joseph Lee]

i tried storing the Y values but Y is a matrix with 100000 columns for every loop because of the linspace and it says 'variable appears to change size every loop' Storing a matrix inside a matrix does not seem possible. what im attempting now is In loop:

j= 1360;
     while j<1400
     idx= find(round(X)== j);
     YY(j,:)=min(Y(idx));    
     end
     j=j+0.1;

i found out why find does not work, its because Y values are not exact and integers, they are a matrix of many decimal places. Still figuring out how to store YY for every loop and it takes a very long time, is there an alternative parametric equation coding method? If i decrease the linspace values, the values would not be accurate. Just ran for over an hour for a single i loop and is still running.

[KL]

Hi Joseph, I'm not able to understand what you're trying to do. Why on't you create a minimal example with a smaller size. let's say 100 columns?

P.S: "Storing a matrix inside a matrix does not seem possible" ... This is exactly why we have cell arrays in MATLAB.

[Joseph Lee]

it will affect the accuracy and the plot would be zig zagged instead of a curve, it took forever to run just a single matrix for n=1

[KL]

No, such minimal example is just for us to recreate your problem and suggest a solution which you could apply in your case. Do you get the idea? We don't even need your computations. Take the example I provided you in my answer.

allY = rand(10,4); %say, you have 4 curves with 10 values (random data in this example)
X = (1:10)';
data_summary = [X allY] %put them together
intvl = 3:6; %in your case 1360 to 1400
minVals = min(data_summary(intvl,2:end))

Here I create a dummy data for Y and X and then put them all together in one matrix and find the minimum value in a certain interval.

[Joseph Lee]

I tried your example but it doesnt work as intended, you are finding the minimum y value for each curve instead of finding minimum y value among the curves and X=1 does not mean y is located in first column since X is a sin function.

I tried cell array to store all the X and Y values now its about finding the minimum Y, I changed the X to range to find from X = 8 to 24, which I wonder how i can put them together... n= 3169 the total number of loops/ curves i is an increment of +1 from 1 to 3169

M = cell(n, 1) ;
N = cell(n, 1) ;
%for each loop
N{i}=X;
M{i}=Y;

[KL]

As far as I have understood, every column of the Y matrix is a curve and they are somehow associated with X along the row. So when I say find minimum value of Y when X = 6 (for example), we are trying to find the minimum along the row 6 across all columns (hence along all curves).

You're saying there are 100000 values of y for each iteration (or curve), well that means 100000 x, right? How about storing it vertically?

X       Y1
1      0.4
2      0.2
...    ...
100000 0.34

and then concatenate it horizontally for the next iteration,

X       Y1   Y2
1      0.4  0.8
2      0.2  0.4
...    ...  ...
100000 0.34 0.1

So, this how I could understand the problem from your descriptions. This is why we insist on creating a minimal example with a sample code.

[Joseph Lee]

 This worked, thanks for tip on arrays. Im looking for a way to find minimum for the 1st column of each cell, comparing the 1st,2nd columns and so on to get the minimum. 

   M = cell(n, 1) ;

j=8;
       N=1;
       while j<=24
           while N<=160

         idx=find(abs(X-j)<0.01);

         Ymin(N,:)=min(Y(idx));
         j=j+0.1;
         N=N+1;
             end
         end
         M{i}=Ymin;