关于附图，我需要一个计算算法来将 A 轴向下移动 n 英寸，将 B 轴从左向右移动 m 英寸，以便组件圆 D 遵循抛物线的曲线；圆 D 并不总是 10 英寸，可以更小。我不是数学专业的，所以这对我来说有点复杂。我知道我在 A 轴上有一个必须计算的弧长(我不知道该怎么做。)，然后我在 B 轴上也有一个弧长并且弧正在引用的位置移动轴 A，与圆 D 的直径相关联的弧长将决定圆 D 和抛物线之间的交点在抛物线上的位置。为了从左到右跟随抛物线的曲线，反之亦然 - 我需要一个公式来跟随抛物线。考虑 D 大小的变化。有人可以提供一些关于如何做到这一点的答案吗？一个很好的公式，带有一些解释性信息 - 至少足够详细，我可以搜索这些部分和位以了解要做什么。
最佳答案
看来您需要计算作为抛物线偏移的曲线。
在接下来的 C++ 程序中，我将展示如何首先找到抛物线的公式，然后如何计算该曲线的偏移量，最后如何找到两个轴彼此形成的角度。
我将点 (0,0) 视为抛物线的左端，该点(顶点)的底部将位于坐标 (12,-8.75)，而右端位于 (24,0)。以这张图为引用(抛物线为蓝色，圆心轨迹为橙色):
请注意，如果圆太大，虽然它在一侧相切，但它可能会在另一侧与抛物线相交。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:
程序会打印出代表刀具的圆与抛物线相切点的坐标、对应的圆心坐标(刀具的位置)和角度的一些样本(25)两个轴(alpha 和 beta)。
#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
这些是不同位置的一些图片(感谢 excell):
关于利用 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/