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I have two lines, one straight and one curvy. Both have an arbitrary number of x and y values defining the lines - the number of x and y values are not the same for either line. I am attempting to get separate distances of points between the curved line coordinates and the straight line coordinates. You can think of discrete integration to get a better picture of what I'm talking about, something along the lines of this: http://www.scientific-solutions.ch/tech/origin/products/images/calculus_integral.gif

By adding the different distances, I would get the area. The part on which I am stuck is the actual synchronization of the points. I can simply compare the x and y values of the straight and curve coordinates every ten indices for example because the curved coordinates are time dependent (as in the points do not change at a general rate). I need a way to synchronize the actual coordinates of the two sets of points. I thought about interpolating both sets of points to a specific number of points, but again, the time dependence of the curved set of points makes that solution void.

Could someone please suggest a good way of doing this, outlining the basics? I really appreciate the help.

Code to be tried (pseudo):

xLine = [value1 value2 ...]
yLine = [value1 value2 ...]
xCurve = [value1 value2 ...]
yCurve = [value1 value2 ...]

xLineInterpolate = %interpolate of every 10 points of x until a certain value. same with yLineInterpolate, xCurveInterpolate and yCurveInterpolate.

Then, I could just take the same index from each array and do some algebra to get the distance. My worry is that my line values increase at a constant rate whereas my curve values sometimes do not change (x and y values have different rates of change) and sometimes do. Would such an interpolation method be wrong then?

intl
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    Your question seems to be better suited to the [Math StackOverflow site](http://math.stackexchange.com/). But, that aside, if you've done any work towards your problem in Matlab, and are in need to working through it, you could post it here. – jrd1 Oct 09 '12 at 22:45
  • @jrd1 - apologies, didn't realise this "Software that mathematicians use" was in their FAQ. I've pinged the Maths mods to review the question and let me know if ok to migrate. – Kev Oct 10 '12 at 00:01
  • intl - please don't cross-post your question, and as jrd1 has rightly pointed out, any work (read code) you've done in Matlab for this question might be useful supplemental information. – Kev Oct 10 '12 at 00:03
  • @intl, we all make honest mistakes. That being said, the reason why I suggested that it be migrated to http://math.stackexchange.com/ is because you were asking how to approach the problem mathematically, which also implied you were attempting / or will attempt to code it after you've figured out how to do it - which is where you are now. That's why. Not because Matlab is the software that Mathematicians use - since its use is not exclusive to Math, it is used in a variety of fields (even Psychology). But, if you have attempted a solution, and need help with it, then here's the right place. – jrd1 Oct 10 '12 at 00:48
  • Would you like this migrated over to the Maths site, or are you happy enough here? One of the Maths mods also suggested reading this: http://mathoverflow.net/questions/40748/intuition-behind-the-notion-of-distance-between-curves/40774#40774 - which is on the the *other* maths website. @ ping me or flag if you'd like this. – Kev Oct 10 '12 at 21:24

2 Answers2

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If I understand correctly, you want to know the distance between a straight line and a curve. The easiest way is to perform a coordinate transformation such that the straight line is the new x-axis. In that frame, the y-values of the curved line are the distances you seek.

This coordinate transformation is equal to a rotation and a translation, as in the following: 

% estimate coefficients for straight line
sol = [x1 ones(size(x1))] \ y1;
m = sol(1); %# slope
b = sol(2); %# y-offset at x=0

% shift curved line down by b 
% (makes the straight line go through the origin)
y2 = y2 - b;

% rotate the curved line by -atan(m)
% (makes the straight line run parallel to the x-axis)
a = -atan(m);
R = [+cos(a) -sin(a)
     +sin(a) +cos(a)];    
XY = R*[x2; y2];

% the distances are then the values of y3. 
x3 = XY(1,:);
y3 = XY(2,:);
Rody Oldenhuis
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You need to use interpolation. I don't see how the time-dependence is relevant here - perhaps you are thinking of fitting a straight line to both curves? That's a bad idea.

You can do a simple interpolation for any curve just by assuming that every two adjacent points are connected by a straight line. This can be shown to be a reasonable approximation for the curve.

So, let's say you are looking at (x1,y1) and (x2,y2) which are adjacent to each other and now you choose an x3 that is between x1 and x2 (x1 < x2 < x3), and want to find the y3 value.

A simple way to find y3 is the following:

p=(x3-x1)/(x2-x1)
y3=y1+p*(y2-y1)

The idea is that p shows the relative position between x1 and x2 (0.5 would be the middle, for example), and then you use p as the relative position between y1 and y2.

Bitwise
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  • I meant time dependence as in some x and y values for the curves change more slowly and some other change more quickly. – intl Oct 10 '12 at 03:41
  • @intl the fact that the values change at different rates shouldn't matter, this is taken into account in the slope. – Bitwise Oct 10 '12 at 18:03