Skip to main content

interpolation - Make an offset curve (parallel curve)


I have a polynomial curve that I got through interpolation.


pts1 = {{0, 0}, {12, 27}, {31, 52}, {58, 73}, {81, 85}};
y1 = pts1[[#, 1]] & /@ Range[Length[pts1]];
eq1 = Fit[pts1, {1, x, x^2, x^3, x^4}, x]//Chop;



$-2.74861*10^{-6}x⁴+0.00059554 x³-0.0516843 x²+2.7892 x$



eq1 /. x -> y1;


${0,27,52,73,85}$



pl1 = Plot[eq1, {x, Min[y1], Max[y1]}, Epilog -> {Blue, PointSize[0.02], Point[pts1]}, PlotRange -> {{-10, 100}, {-10, 100}}, AspectRatio -> 1, PlotStyle -> {Orange, Thick}]

I want to make another offset curve in 10 units. But I do not know how to proceed within Mathematica.



enter image description here


I made an offset through another software using multiple circles with a 10-unit radius to get the points I need.


enter image description here


Anyway, what would be the appropriate command to get these points?



Answer



This involves an algebraic curve so it can be done in closed form (one approach already shown does this, in the parametric form). We'll do the interpolation below at high precision in order to make some later computations more reliable.


pts1 = {{0, 0}, {12, 27}, {31, 52}, {58, 73}, {81, 85}};
f[x_] = Fit[N[pts1, 200], {1, x, x^2, x^3, x^4}, x];
N[f[x]]


(* Out[1252]= 0. + 2.78920381052 x - 0.0516843269209 x^2 + 
 0.000595539716825 x^3 - 2.74861498447*10^-6 x^4 *)

Here we find the parametric form of the (lower) offset curve).


offset[x_] = 
  With[{deriv = D[{x, f[x]}, x]},
    {grad = {-deriv[[2]], deriv[[1]]}},
    {x, f[x]} - grad/Sqrt[grad.grad]*10];

We check the plot.



ParametricPlot[{{t, f[t]}, offset[t]}, {t, 0, 80}]

plot


We can also implicitize. This part required the high precision interpolation. We could use Rationalize but that can get into round-off and cancellation error problems in plotting, since coefficients appear at very different scales.


imp = First[
   GroebnerBasis[Together[{x, y} - offset[t]], {x, y}, t, 
    MonomialOrder -> EliminationOrder]];
imp // N

(* Out[1260]= -6.73373194281*10^25 + 5.56558925951*10^24 x + 

 3.6738758353*10^23 x^2 - 4.34415318895*10^22 x^3 + 
 1.93344441642*10^21 x^4 - 5.36588154643*10^19 x^5 + 
 1.06609597248*10^18 x^6 - 1.60542472424*10^16 x^7 + 
 1.88490826277*10^14 x^8 - 1.74235671779*10^12 x^9 + 
 1.26230731848*10^10 x^10 - 7.01480518001*10^7 x^11 + 
 284576.519771 x^12 - 758.341572269 x^13 + 1. x^14 + 
 3.72787534059*10^24 y - 6.8504825964*10^23 x y + 
 9.94066556163*10^21 x^2 y + 7.41868236137*10^20 x^3 y - 
 4.34299189813*10^19 x^4 y + 1.19168781004*10^18 x^5 y - 
 2.1498342949*10^16 x^6 y + 2.77290008879*10^14 x^7 y - 

 2.63253515622*10^12 x^8 y + 1.82904347319*10^10 x^9 y - 
 8.95087000739*10^7 x^10 y + 280923.407204 x^11 y - 
 426.500272743 x^12 y - 3.17385078171*10^22 y^2 + 
 2.18287264042*10^22 x y^2 - 3.93259394774*10^20 x^2 y^2 - 
 1.00071010929*10^19 x^3 y^2 + 6.86935186234*10^17 x^4 y^2 - 
 1.75325222649*10^16 x^5 y^2 + 2.83739730618*10^14 x^6 y^2 - 
 3.21684582573*10^12 x^7 y^2 + 2.65009867828*10^10 x^8 y^2 - 
 1.59027982886*10^8 x^9 y^2 + 677296.69588 x^10 y^2 - 
 1950.02118583 x^11 y^2 + 3. x^12 y^2 - 1.48945282252*10^21 y^3 - 
 4.62703214764*10^20 x y^3 + 8.5373151706*10^18 x^2 y^3 + 

 6.50159149141*10^16 x^3 y^3 - 6.85046668404*10^15 x^4 y^3 + 
 1.58285949288*10^14 x^5 y^3 - 2.17546347848*10^12 x^6 y^3 + 
 1.98184395157*10^10 x^7 y^3 - 1.20720722952*10^8 x^8 y^3 + 
 465742.200007 x^9 y^3 - 853.000545485 x^10 y^3 + 
 6.2893710958*10^19 y^4 + 6.33039722844*10^18 x y^4 - 
 1.1302847025*10^17 x^2 y^4 + 4.83188192567*10^13 x^3 y^4 + 
 4.60017785742*10^13 x^4 y^4 - 9.9826564209*10^11 x^5 y^4 + 
 1.20048427437*10^10 x^6 y^4 - 9.37112169389*10^7 x^7 y^4 + 
 476873.182752 x^8 y^4 - 1625.01765486 x^9 y^4 + 3. x^10 y^4 - 
 1.09303426085*10^18 y^5 - 5.50342304678*10^16 x y^5 + 

 9.17636707233*10^14 x^2 y^5 - 4.3681823605*10^12 x^3 y^5 - 
 1.76328081245*10^11 x^4 y^5 + 3.57598551957*10^9 x^5 y^5 - 
 3.37083626949*10^7 x^6 y^5 + 184818.792803 x^7 y^5 - 
 426.500272743 x^8 y^5 + 1.03689348135*10^16 y^6 + 
 2.95870424611*10^14 x y^6 - 4.66457552293*10^12 x^2 y^6 + 
 3.02885768057*10^10 x^3 y^6 + 4.65196245499*10^8 x^4 y^6 - 
 1.00479219318*10^7 x^5 y^6 + 84153.0066438 x^6 y^6 - 
 433.338041296 x^7 y^6 + 1. x^8 y^6 - 5.50780682701*10^13 y^7 - 
 7.85681314181*10^11 x y^7 + 1.41189443972*10^10 x^2 y^7 - 
 1.57656872186*10^8 x^3 y^7 + 727639.19694 x^4 y^7 + 

 1.32364700231*10^11 y^8 *)

We can check the zero contour.


ContourPlot[imp == 0, {x, 0, 60}, {y, 0, 80}]

plot


One will notice we got both upper and lower offsets. This is due to the fact that GroebnerBasis internals will make polynomial relations out of radicals, in effect losing information about sign on square roots.


Comments

Post a Comment

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more