Truncate matrix elements to a particular length

Ok. I am revising my previous question. I have two matrices that have the X and Y coordinates of a trajectory.
X= [ 0 1.1 1.9 2.6 3.01 4.56 5.04]
Y= [ 0 -0.013 0.008 0.008 -0.002 0.034 0.08]
So the trajectory comprises of 7 points and 6 segments. The length of each subsegment is given by the following matrix. The length is calculated by the distance formula
S=[ 1.1001 0.8003 0.7 0.4101 1.5504 0.4822]
The sum of all these subsegments equals sum(S)=5.0431. This is the length of my trajectory. To do further analysis, I want all my trajectories to have the same length (say 5 in this case). I want to interpolate the last point so that it falls on the trajectory and allows the trajectory to have a predefined length. Only the last point should be interpolated.
So my final output should look like this
X= [ 0 1.1 1.9 2.6 3.01 4.56 Xi]
Y= [ 0 -0.013 0.008 0.008 -0.002 0.034 Yi]
I need to find Xi and Yi using interpolation (interp1 probably) so that sum(S) becomes 5.
Can anyone help me write a code for this.
Thanks, Nancy

 采纳的回答

xy = [X;Y];
d = diff(xy(:,end-[1 0]),1,2);
k = 5 - sum(sqrt(sum(diff(xy(:,1:end-1),1,2).^2)))
xy(:,end)= k/sqrt(sum(d.^2))*d+xy(:,end-1)
OR
xy = [X;Y]
d = diff(xy,1,2)
D = hypot(d(1,:),d(2,:))
k = 5 - sum(D(1:end-1))
XYi = k/D(end)*d(:,end) + xy(:,end-1)

5 个评论

Thanks Andrei. What is d in the third line of the first solution. Is it the same as the one in the second solution.
And if possible could you explain your logic?
Hi Nancy! My typo! Corrected.
Could you explain your logic?
xy vertically concatenates x,y
d = is the horizontal numerical first derivative, e.g. a(i+1)-a(i)
D is the sqrt sum squares (doc hypot) of x and y
k is 5 minus the sum of all hypotenuses except the last
XYi is something.
Thanks Sean. I need the logic behind XYi.

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