# How to fit a spline (not interpolate) through points in 3D?

I have coordinates of points in 3D and want to fit a spline through these points. I found a least square approach for 2D case: https://demonstrations.wolfram.com/GlobalBSplineCurveFittingByLeastSquares.

But I don't know how to use it for my 3D points. I would be grateful for any help.

points to fit:

{{-0.127862, 7.23797, 0.080385}, {-0.386039, 7.22523,
0.238487}, {-0.643736, 7.19972, 0.395825}, {-0.900618, 7.16136,
0.551879}, {-1.15633, 7.11002, 0.706108}, {-1.4105, 7.04555,
0.857945}, {-1.66271, 6.96778, 1.00679}, {-1.9125, 6.8765,
1.15199}, {-2.15936, 6.77151, 1.29287}, {-2.40271, 6.6526,
1.42869}, {-2.64192, 6.51958, 1.55865}, {-2.87627, 6.37229,
1.68191}, {-3.10494, 6.21063, 1.79756}, {-3.32704, 6.03455,
1.90464}, {-3.54155, 5.84411, 2.00217}, {-3.74738, 5.63948,
2.08911}, {-3.9433, 5.42099, 2.1644}, {-4.12798, 5.18913,
2.227}, {-4.30001, 4.94459, 2.27587}, {-4.45787, 4.68829,
2.31007}, {-4.59997, 4.42139, 2.3287}, {-4.72468, 4.14532,
2.33102}, {-4.83036, 3.8618, 2.31646}, {-4.91539, 3.57279,
2.28468}, {-4.9782, 3.28055, 2.23561}, {-5.01738, 2.98756,
2.16947}, {-5.03167, 2.69648, 2.08687}, {-5.02007, 2.41015,
1.98876}, {-4.98187, 2.13143, 1.87651}, {-4.9167, 1.86318,
1.75186}, {-4.82458, 1.60813, 1.61691}, {-4.70594, 1.36881,
1.47412}, {-4.56163, 1.14739, 1.32615}, {-4.39291, 0.945657,
1.17587}, {-4.2014, 0.764887, 1.02622}, {-3.98905, 0.605816,
0.880086}, {-3.75804, 0.4686, 0.740214}, {-3.51073, 0.352823,
0.609092}, {-3.24952, 0.257522, 0.488851}, {-2.97682, 0.181251,
0.381183}, {-2.6949, 0.122163, 0.287282}, {-2.40588, 0.0781093,
0.207801}, {-2.11161, 0.0467557, 0.142845}, {-1.81367, 0.0256998,
0.0919741}, {-1.51331, 0.0125879, 0.0542331}, {-1.21151, 0.00522497,
0.0282002}, {-0.908944, 0.00167311, 0.0120463}, {-0.60604,
0.00033535, 0.00360597}, {-0.303029, 0.0000222426, 0.000456223}, {0,
0, 0}, {0, 0, 0}, {0.303029, 0.000022046, -0.000449164}, {0.606041,
0.000333794, -0.00357979}, {0.908945,
0.00166793, -0.0119886}, {1.21151, 0.00521293, -0.028099}, {1.51332,
0.012565, -0.0540773}, {1.81368, 0.0256614, -0.0917535}, {2.11165,
0.0466967, -0.142551}, {2.40594, 0.0780248, -0.207425}, {2.69499,
0.122048, -0.286817}, {2.97694, 0.181102, -0.380626}, {3.24969,
0.257335, -0.488198}, {3.51095, 0.352596, -0.608342}, {3.75834,
0.468332, -0.739366}, {3.98943, 0.605506, -0.879142}, {4.20188,
0.764539, -1.02518}, {4.39349, 0.945273, -1.17475}, {4.56232,
1.14698, -1.32494}, {4.70675, 1.36837, -1.47283}, {4.82551,
1.60768, -1.61555}, {4.91775, 1.86271, -1.75043}, {4.98305,
2.13096, -1.87502}, {5.02137, 2.40969, -1.98722}, {5.0331,
2.69604, -2.08527}, {5.01892, 2.98713, -2.16782}, {4.97985,
3.28015, -2.23391}, {4.91715, 3.57242, -2.28293}, {4.83222,
3.86146, -2.31467}, {4.72663, 4.14502, -2.32918}, {4.602,
4.42112, -2.32682}, {4.45998, 4.68806, -2.30815}, {4.30219,
4.9444, -2.27392}, {4.13022, 5.18897, -2.225}, {3.94559,
5.42087, -2.16236}, {3.74972, 5.63939, -2.08703}, {3.54394,
5.84404, -2.00006}, {3.32946, 6.03451, -1.90249}, {3.1074,
6.21061, -1.79537}, {2.87875, 6.37229, -1.67969}, {2.64443,
6.51959, -1.5564}, {2.40524, 6.65262, -1.42641}, {2.1619,
6.77154, -1.29057}, {1.91506, 6.87653, -1.14966}, {1.66529,
6.96781, -1.00444}, {1.41309, 7.04558, -0.855578}, {1.15893,
7.11005, -0.703726}, {0.903221, 7.16138, -0.549484}, {0.646343,
7.19974, -0.393421}, {0.38865, 7.22524, -0.236076}, {0.130474,
7.23797, -0.077971}, {-0.127862, 7.23797, 0.080385}}

• Is f = BSplineFunction[big_list_of_points] is not enough? You can plot (ParametricPlot3D) or interpolate using parametrization: f[t], where t goes from 0 to 1. – Alx Nov 1 at 13:45
• It uses the points as control points, which resutantly gives a noisy curvature and torsion. – ARUN KUMAR Nov 1 at 15:46
• Have you tried like Show[ParametricPlot3D[f[t], {t, 0, 1}, PlotStyle -> Blue], ListPointPlot3D[data, PlotStyle -> Red]], it looks very smooth and accurate. Probably you confused with BezierFunction, that goes "around" points. – Alx Nov 1 at 15:57
• @ARUNKUMAR For curve fitting you need an initial guess of the model function. The coefficients inside the terms of model can be fitted by Mathematica. Do you have the model view? – Rom38 Nov 2 at 5:43
• @Rom38 I don't have specific form of function. Just want to fit a spline (for exmple : degree 3 or degree 5 spline) through points. Better the fitted spline be parameterized by arc length. If you see the ink in question that is exactly what I am looking but for 3D points. – ARUN KUMAR Nov 2 at 6:50