How To Calculate Max Deviation On A Line
Asked by Kent on 29 Jun 2012
Latest activity Commented on by Kent on 29 Jun 2012
Hi there! I have a question regarding how to calculate maximum deviation on a line.
Given that I have one line (call it Line A), I'm interested in finding the MAX value at 1<POSITION<10 and MIN value at 50<POSITION<60, and then formed a line by this two points(call it Line B).
Next I would like to do is to calculate the max deviation (or distance) from Line A to the lowest point of Line B and output the value.
I've tried but kinda stuck at how to proceed with forming the lines from 2 points then calculating the max deviation from this lines to another line. Hope to get some input from you guys.
Thank you for your help.
Position,deltaP 1,0.79672217 2,0.7967096689 3,0.7967104044 4,0.7967030204 5,0.79669891 6,0.79670164 7,0.7966935311 8,0.7966885096 9,0.7966867596 10,0.7966908044 11,0.7966915659 12,0.7967002459 13,0.7966929348 14,0.7966808326 15,0.79668317 16,0.7966720107 17,0.7966774959 18,0.7966798385 19,0.7966586178 20,0.7966574996 21,0.7966417311 22,0.7966176296 23,0.7966008181 24,0.79657083 25,0.79653591 26,0.79651816 27,0.796498 28,0.79648868 29,0.796482303 30,0.79648864 31,0.7964957001 32,0.7965104026 33,0.79653118 34,0.79655322 35,0.79656086 36,0.7965753686 37,0.79657933 38,0.7965895335 39,0.796610771 40,0.7966209815 41,0.7966409801 42,0.7966490117 43,0.7966384365 44,0.7966236063 45,0.7966218648 46,0.7966473956 47,0.7966660522 48,0.7966582037 49,0.7966641863 50,0.7966611456 51,0.79665338 52,0.7966786663 53,0.7966801211 54,0.7966836322 55,0.7966905467 56,0.7966988863 57,0.79669877 58,0.7967126519 59,0.79669836 60,0.7967066396
P/S: I can send over the .ppt file for better illustrates about LineA and LineB. Please let me know your email.
~Kent
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1 Answer
Answer by Adam Filion on 29 Jun 2012
Accepted answer
Hi Kent, here's a quick solution I worked out. Assume you have your first column of data saved as 'x' and your second as 'y'.
% find max/min points and their location y1 = max(y(1:10)); idx1 = find(y==y1); y2 = min(y(50:60)); idx2 = find(y==y2); % plot new line hold on yp = (y2-y1)/(x(idx2)-x(idx1))*(x-x(idx1))+y1; plot(x,yp,'r') % max deviation and location maxdiff = max(abs(yp(idx1:idx2)-y(idx1:idx2))) idx_maxdiff = find((yp-y)==maxdiff)
The maxdiff is the maximum difference between the line and your data, and the idx_maxdiff is the index location in the vector where the maxdiff was found.
3 Comments
Kent on 29 Jun 2012
Hi Adam, thank you very much! I tried the code and it actually works pretty good. However, when trying to find the location of maximum difference (idx_maxdiff), it outputs
idx_deltamax =
Empty matrix: 0-by-1
I think it is because (yp-y) returns much smaller values and maxdiff has much larger value which doesn't coincide on the line. I'm not sure if this is true and would like to seek your opinion to work around this.
I've another question related to this which is to level or pivot the yp line and the Line A so that it becomes flat/parallel with the x-axis, before calculating for the maxdiff. Is there a line pivoting function in MATLAB that allows this kind of line pivoting?
~Kent
Adam Filion on 29 Jun 2012
Hi Kent, for your first question, this could be a mistake on my part. If you find the absolute difference in maxdiff, then you need to use abs again in finding idx_maxdiff otherwise the signs might not match. However when I was working with the data you provided, the code worked on my end. To fix that oversight change the following line
idx_maxdiff = find(abs(yp-y)==maxdiff)
Secondly, there is no built-in function in MATLAB to ‘pivot’ a line. However if you know the point you want to pivot around, then it is relatively simple. Let’s say you want to make Line B parallel to the x-axis at the max point you found in the first range of values. Then you can try:
yp=ones(size(y))*y(idx1); plot(x,yp,'r')
Kent on 29 Jun 2012
Hi Adam, regarding my first question, it was actually a mistake I made in the code that causes outputs "Empty Matrix". I rechecked my code and found that I had a typo in the maxdiff equation, y(idx1:idx2) became x(idx1:idx2), need to get my eyes check! But thanks for your help!
Regarding my 2nd question about pivoting, I will try that and let you know. Thanks!
~Kent
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