Fit 1/4 Circle with Sanche Besel curve

Sanche Besel curve Fitting 1/4 Circle

Based on the knowledge of the Bezier curve, we know that the parametric equation of the Sanche Besel curve is as follows, where a, B, C, D are four control point coordinates, and P (t) represents each point on the curve.

Because the 1/4 circle is to be simulated, the tangent direction of P (0) and P (1) should be placed in the position shown. Where AB is in the horizontal direction, DC is vertical and segment length | ab| = | dc| = h.

So this problem is actually converted to calculate a reasonable H value, so that the radius | oj| = 1, that is, the J point is just on the arc.

According to the symmetry of the Bezier curve, it is not difficult to think of the J Point at P (0.5), substituting the formula can be obtained:

The same conclusion can be inferred directly from the geometrical characteristics of the Bezier curve, namely:

So it is also possible to reconfirm that P (0.5) and J are the same point.

Substituting four control points coordinates a (0, 1), B (H, 1), C (1, h) and D (1, 0), you can solve the P (0.5) point coordinates as follows:

According to the definition of the circular equation, the following equation can be drawn:

The value of the H is then calculated as:

Therefore, the parametric equation P (t) of the Sanche Besel curve simulation 1/4 Circle can be finally defined as follows:

On the other hand, how much difference does the equation describe between the curve and the real 1/4 circle? The following is a numerical solution for this problem.

Use T = 0.0 to 1.0, step value 0.01, to solve the difference between the distance from each point to the origin and the radius 1.

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h>double bezier3 ( Double A, double b, double C, double D, double T) {double NT = 1.0-t;double Nt2 = NT * nt;double NT3 = NT * NT * NT;DOUBL E t2 = T * t;double t3 = t * t * T;return (A * nt3 + b * 3.0 * Nt2 * t + c * 3.0 * NT * t2 + d * T3);} int main () {        Double T, a;        Double D, e;        Dou ble max_e = 0.0, min_e = 1.0;        Double X, y;        Double h = (sqrt (2)-1.0 * 4.0/3.0;        for (t = 0.0; t < 1.01; t+=0.01)         {    &NB Sp           x = Bezier3 (0, H, 1, 1, t);                y = Bezier3 (1, 1, h, 0, T);                d = sqrt (x * x + y * y);    &nbsp ;           E = d-1.0;&nbsp                a = atan2 (y, x);                A = A * 180.0/3.1415926;                if (Max_e < e) Max_e = e;                if (Min_e > E) min_e = e;              &NBS P printf ("%4.1f,%f\n", A, E);       }        printf ("max_e =%f, min_e =%f\n", MA X_e, min_e); return 0;}

The output results are as follows:

90.0, 0.00000089.1, 0.00000388.1, 0.00001087.2, 0.00002286.2, 0.00003785.3, 0.00005484.4, 0.00007383.4, 0.00009282.5, 0.00011381.6, 0.00013380.7, 0.00015379.7, 0.00017278.8, 0.00019077.9, 0.00020677.0, 0.00022176.1, 0.00023475.2, 0.00024674.3, 0.00025573.4, 0.00026372.5, 0.00026871.6, 0.00027170.7, 0.00027369.8, 0.00027268.9, 0.00026968.0, 0.00026567.1, 0.00025966.2, 0.00025165.3, 0.00024264.4, 0.00023263.5, 0.00022062.6, 0.00020861.8, 0.00019460.9, 0.00018160.0, 0.00016659.1, 0.00015258.2, 0.00013757.3, 0.00012356.5, 0.00010855.6, 0.00009454.7, 0.00008153.8, 0.00006852.9, 0.00005652.0, 0.00004551.2, 0.00003550.3, 0.00002649.4, 0.00001848.5, 0.00001247.6, 0.00000746.8, 0.00000345.9, 0.00000145.0, 0.00000044.1, 0.00000143.2, 0.00000342.4, 0.00000741.5, 0.00001240.6, 0.00001839.7, 0.00002638.8, 0.00003538.0, 0.00004537.1, 0.00005636.2, 0.00006835.3, 0.00008134.4, 0.00009433.5, 0.00010832.7, 0.00012331.8, 0.00013730.9, 0.00015230.0, 0.00016629.1, 0.00018128.2, 0.00019427.4, 0.00020826.5,0.00022025.6, 0.00023224.7, 0.00024223.8, 0.00025122.9, 0.00025922.0, 0.00026521.1, 0.00026920.2, 0.00027219.3, 0.00027318.4, 0.00027117.5, 0.00026816.6, 0.00026315.7, 0.00025514.8, 0.00024613.9, 0.00023413.0, 0.00022112.1, 0.00020611.2, 0.00019010.3, 0.000172 9.3, 0.000153 8.4, 0.000133 7.5, 0.000113 6.6, 0.000092  5.6, 0.000073 4.7, 0.000054 3.8, 0.000037 2.8, 0.000022 1.9, 0.000010 0.9, 0.000003-0.0, 0.000000max_e = 0.000273, min_e = 0.000000

From the output analysis can be seen, the error is toward the arc outer convex, 0 degrees to 45 degrees, 45 degrees to 90 degrees.

At 0 degrees, 45 degrees and 90 degrees is the minimum error 0.000000, at 19.3 degrees and 70.7 degrees to achieve the maximum error of 0.000273, basically very close to 1/4 arc.

Above, that is, the Sanche Besel curve simulates the entire contents of the 1/4 arc.

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Fit 1/4 Circle with Sanche Besel curve