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put two plots in one pattern
- Thread starter tsonaki
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for Instance,we got a pattern that represent a curve with
(x,y)=(Wf,REICO) and another pattern that represent a different curve with (x,y)=(Wff,REICO) and I want the intersection of those curves.Extrapolating both curves to the point of intersection gives the bi-linear relationship.
Hopefully makes sense.If does not please let me know.
Also I am looking for the algorithm of least square method.I have tried to create the algorithm but it doesn't run and I got huge standard error.
Thanks a lot for your help
Regards
Maria
(x,y)=(Wf,REICO) and another pattern that represent a different curve with (x,y)=(Wff,REICO) and I want the intersection of those curves.Extrapolating both curves to the point of intersection gives the bi-linear relationship.
Hopefully makes sense.If does not please let me know.
Also I am looking for the algorithm of least square method.I have tried to create the algorithm but it doesn't run and I got huge standard error.
Thanks a lot for your help
Regards
Maria
If you're using least squares, you already have the data for the points.
1) Is there more than one crossing point?
2) Do the curves cross outside or inside your data values?
3) Are the values exponential? Maybe you need to take the log of one side first and do a curve fit on a log linear graph.
If they cross inside, a simple technique would be to find the two points on either side of the cross on both curves. With 2 points, you can create straight line graphs (y=mx+c) and solve the intersection that way.
If you want a more accurate result, take the next two points. It is simple enough to do a least squares on 4 points for a 2nd order curve. An then compute the intersection from that.
1) Is there more than one crossing point?
2) Do the curves cross outside or inside your data values?
3) Are the values exponential? Maybe you need to take the log of one side first and do a curve fit on a log linear graph.
If they cross inside, a simple technique would be to find the two points on either side of the cross on both curves. With 2 points, you can create straight line graphs (y=mx+c) and solve the intersection that way.
If you want a more accurate result, take the next two points. It is simple enough to do a least squares on 4 points for a 2nd order curve. An then compute the intersection from that.
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