关于所附图像，我需要有一个计算算法来将轴A向下移动N英寸，并且轴B从左到右M英寸，以便组成圆D遵循抛物线的曲线；圆圈D不总是10英寸，并且可以更小。我不是数学专业的学生，所以这对我来说有点复杂我知道在轴A上有一个弧长，必须计算出来（我不知道怎么做）。然后我在轴B上也有一个弧长，并且圆弧相对于轴A的位置移动，与圆d的直径相关联的弧长将决定圆d与抛物线的交点在抛物线上的位置。为了沿着抛物线的曲线从左到右或从左到右-我需要一个公式来遵循抛物线考虑到尺寸的变化有人能提供一些关于如何做到这一点的答案吗？一个很好的公式和一些解释性的信息-至少足够的细节，我可以在这些部分和位上做一个搜索，以获得对该做什么的理解。
I have looked and found some information that may be helpful to me but does not answer my question at all:https://stackoverflow.com/questions/4039039/fastest-way-to-fit-a-parabola-to-set-of-points
最佳答案
似乎需要计算作为抛物线偏移的曲线。
在下面的C++程序中，我将展示如何首先找到抛物线的公式，然后如何计算从该曲线的偏移量，最后如何找到两个轴彼此形成的角度。
我认为点（0,0）是抛物线的左极端，其底部（顶点）在坐标（12，-8.75）处，而右极端在（24,0）处以这张照片为参考（抛物线为蓝色，圆心轨迹为橙色）：
注意，如果圆太大，一边相切，另一边可以与抛物线相交。我不确定12“是抛物线的总宽度还是只有一半，但在后一种情况下，10”的工具会太大：
程序将打印出代表刀具的圆与抛物线相切的点的坐标的一些样本（25）、圆心的中心坐标（工具的位置）和两轴（α和β）的角度。
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
using std::cout;
using std::setw;
using std::vector;
int main() {
// set number of steps or points of approximation
int n_steps = 25;
// declare vectors to store coordinates into
vector<double> x2(n_steps), y2(n_steps);
// calculate the parameters of the parabola expressed by the formula
// y = ax^2 + bx + c
// Knowing 2 points, one of which is the vertex.
// xv = -b/2a | b = -2axv
// y0 = ax0^2 + bx0 + c => | yv - y0 = a(xv^2 - x0^2) + b(xv - x0)
// yv = axv^2 + bxv + c | yv - y0 = a(xv - x0)(xv + x0) + b(xv - x0)
//
// a ((xv - x0)*(xv + x0) - 2xv(xv - x0)) = yv - y0
// a (xv - x0)*(xv + x0 - 2xv) = yv - y0
// Known coordinates
double xv = 12.0,
yv = -8.75,
x0 = 0.0,
y0 = 0.0;
double dx = xv - x0,
a = (y0 - yv) / ( dx * dx ),
b = - 2.0 * a * xv,
c = y0 - x0 * ( a * x0 + b );
cout << "Parabola formula:\n"
<< "y = " << a << "x^2 + " << b << "x + " << c << "\n\n"
<< "max acceptable diameter: " << 1.0 / a << "\n\n";
// Coordinates of rotating axes, extrapolated from your drawing
double r1 = 13,
r2 = 9,
x1 = xv - r1,
y1 = r2;
// some helper values (constant) I'll use later
double rad_to_deg = 180.0 / M_PI,
r1quad = r1 * r1,
r2quad = r2 * r2,
rdif = r1quad - r2quad,
rsum = r1quad + r2quad,
rden = 1.0 / ( 2.0 * r1 * r2 );
// radius of the circle (tool)
double diameter = 10,
radius = diameter / 2.0;
cout << "Diameter of tool (circle): " << diameter << "\n\n";
// calculate parabola points
cout << "\t\t\tTangent\t\t\t\tCenter of circle\t\t alpha\t\tbeta\n";
// xt[0] = x0 xt[n_steps] = x0 + 2*(xv - x0)
double step = 2.0 * dx / ( n_steps - 1 );
for ( int i = 0; i < n_steps; ++i ) {
// calculate the tangent points which lies on the parabola
double xt = x0 + i * step,
yt = xt * ( a * xt + b ) + c;
// calculate the offset points, coordinates of the center of the circle
// first derivative of the parabola
double delta = 2.0 * a * xt + b;
// point perpendicular to the tangent at distance equal to radius
double k = radius / sqrt(delta * delta + 1.0);
x2[i] = xt - k * delta;
y2[i] = yt + k;
// distance from x,y to x1,y1
double dx1 = x2[i] - x1,
dy1 = y2[i] - y1,
r3quad = dx1 * dx1 + dy1 * dy1,
r3 = sqrt(r3quad);
// Now that I know the coordinates of the vertices of the triangle
// and the lengths of its sides I can calculate the inner angles
// using Carnot teorem, for example: a^2 = b^2 + c^2 - 2bc*cos(alpha)
double alpha_Carnot = acos((rdif + r3quad) / (2.0 * r1 * r3)),
beta_Carnot = acos((rsum - r3quad) * rden);
// angle to the orizzontal of line from x1,y1 to x,y in radians
double gamma = atan2(dy1,dx1);
// angle of Axis A to the orizzontal in degrees
double alpha = (gamma + alpha_Carnot) * rad_to_deg;
// angle of Axis B to Axis A. beta = 0 if parallel
double beta = beta_Carnot * rad_to_deg - 180.0;
// output the coordinates
cout << std::fixed << setw(4) << i << setw(10) << xt << setw(10) << yt
<< setw(15) << x2[i] << setw(10) << y2[i]
<< setw(15) << alpha << setw(12) << beta << '\n';
}
return 0;
}
这是输出：
Parabola formula:
y = 0.0607639x^2 + -1.45833x + 0
max acceptable diameter: 16.4571
Diameter of tool (circle): 10
Tangent Center of circle alpha beta
0 0.000000 0.000000 4.123644 2.827642 -7.228866 -142.502245
1 1.000000 -1.397569 5.003741 1.597437 -7.151211 -132.856051
2 2.000000 -2.673611 5.860925 0.503378 -7.962144 -123.965745
3 3.000000 -3.828125 6.690144 -0.454279 -9.159057 -115.700562
4 4.000000 -4.861111 7.485392 -1.276137 -10.496232 -108.022957
5 5.000000 -5.772569 8.239777 -1.964178 -11.833367 -100.941141
6 6.000000 -6.562500 8.945861 -2.522462 -13.081185 -94.488527
7 7.000000 -7.230903 9.596439 -2.957906 -14.180211 -88.708034
8 8.000000 -7.777778 10.185964 -3.280939 -15.093523 -83.633631
9 9.000000 -8.203125 10.712644 -3.505588 -15.805504 -79.267662
10 10.000000 -8.506944 11.180897 -3.648397 -16.321003 -75.558850
11 11.000000 -8.689236 11.603201 -3.725755 -16.659970 -72.392050
12 12.000000 -8.750000 12.000000 -3.750000 -16.845543 -69.600878
13 13.000000 -8.689236 12.396799 -3.725755 -16.889440 -67.003199
14 14.000000 -8.506944 12.819103 -3.648397 -16.782985 -64.443028
15 15.000000 -8.203125 13.287356 -3.505588 -16.499460 -61.816277
16 16.000000 -7.777778 13.814036 -3.280939 -16.005878 -59.069041
17 17.000000 -7.230903 14.403561 -2.957906 -15.277444 -56.174060
18 18.000000 -6.562500 15.054139 -2.522462 -14.309495 -53.098717
19 19.000000 -5.772569 15.760223 -1.964178 -13.126180 -49.773973
20 20.000000 -4.861111 16.514608 -1.276137 -11.788732 -46.064496
21 21.000000 -3.828125 17.309856 -0.454279 -10.409278 -41.729113
22 22.000000 -2.673611 18.139075 0.503378 -9.184425 -36.337384
23 23.000000 -1.397569 18.996259 1.597437 -8.503984 -29.006402
24 24.000000 0.000000 19.876356 2.827642 -9.577076 -16.878208
这是一些图片（感谢excell）在不同的位置：
关于algorithm - 利用2轴运动计算抛物线路径的算法，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/36136175/