从圆内任何点和角度到圆周的距离-解网

问：

从技术上讲，这是一个数学问题，但我不明白如何将其转换为代码。

我有一个圆的中心 C，它的半径 r，以及圆内的点 A 和 B。我需要找到距离 AD，所以 B 仅用于角度。因此，这意味着用 3 个点计算的角度和边来求解三角形 ACD。换句话说，我想用一个角度找到从圆内任何一点到圆周的距离。

我尝试过的伪代码：

float dis(float a_x, float a_y, float b_x, float b_y, float c_x, float c_y, float r) {
  float ab_x = a_x - b_x;
  float ab_y = a_y - b_y;

  float ac_x = a_x - c_x;
  float ac_y = a_y - c_y;

  float ab_a = atan(ab_y, ab_x);
  float ac_a = atan(ac_y, ac_x);
  float dc_a = pi - ac_a - ab_a;

  return (sin(dc_a) * r) / sin(ab_a);
}

这产生了错误的结果，可能是因为不止一个错误。如何正确地做到这一点？

我在数学交流中读到它，最终了解到它是一个简单的三角形，但是在编程上下文中，我找不到回答我的问题。

与数学语言无关的三角学

// Utility functions to work with 2D vectors:
function sub(u, v) {
    return [u[0] - v[0], u[1] - v[1]];
}

function mul(u, m) { // Scalar multiplication
    return [m * u[0], m * u[1]];
}

function size(u) {
    return Math.sqrt(u[0] * u[0] + u[1] * u[1]);
}

function norm(u) { // Resize vector to size 1
    return mul(u, 1 / size(u));
}

function solve(a, b, c) {
    // Translate problem to C at (0, 0): this does not influence the result
    a = sub(a, c);
    b = sub(b, c);
    // Scale problem to Unit circle
    const r = size(a);
    a = mul(a, 1/r);
    b = mul(b, 1/r);
    // Define W as the direction vector of the line A-B, with size 1.
    const w = norm(sub(b, a));
    // Solve equation to find m, such that D = A+mW and |D|² is 1, i.e.
    // |A+mW|² = 1, or 
    // (Ax+mWx)²+(Ay+mWy)² = 1, or 
    // (Ax)² + 2mAxWx + m²(Wx)² + (Ay)² + 2mAyWy + m²(Wy)² = 1, or
    // (Ax)²+(Ay)² - 1 + 2m(AxWx+AyWy) + m²((Wx)²+(Wy)²) = 0
    // As (Ax)²+(Ay)² = 1, we can eliminate (Ax)²+(Ay)² - 1.
    // m=0 is a solution. But we want to get the other root. 
    // So divide the equation by m, and solve by m:
    const m = -2 * (a[0] * w[0] + a[1] * w[1]) / (w[0] * w[0] + w[1] * w[1]);
    // Scale the solution back to the original size
    return m * r; // This is a signed number. For distance, take absolute value
}

// Demo 
const dist = solve([3, 5], [6, 6], [5, 10]);
console.log(dist);

这是一个交互式版本，其中 A 点和 C 点是固定的，但 B 对应于鼠标指针所在的位置。然后，您可以在移动鼠标指针时看到距离，以及相应的线段。

// Utility functions to work with 2D vectors:
function sub(a, b) {
    return [a[0] - b[0], a[1] - b[1]];
}

function add(a, b) {
    return [a[0] + b[0], a[1] + b[1]];
}

function mul(a, m) { // Scalar multiplication
    return [m * a[0], m * a[1]];
}

function size(a) {
    return Math.sqrt(a[0] * a[0] + a[1] * a[1]);
}

function norm(a) { // Resize vector to size 1
    return mul(a, 1 / size(a));
}

function solve(a, b, c) {
    // Translate problem to C at (0, 0): this does not influence the result
    a = sub(a, c);
    b = sub(b, c);
    // Scale problem to Unit circle
    const r = size(a);
    a = mul(a, 1/r);
    b = mul(b, 1/r);
    // Define G as the direction vector of the line A-B, with size 1.
    const w = norm(sub(b, a));
    // Solve equation to find m, such that D = A+mW and |D|² is 1, i.e.
    // |A+mW|² = 1, or 
    // (Ax+mWx)²+(Ay+mWy)² = 1, or 
    // (Ax)² + 2mAxWx + m²(Wx)² + (Ay)² + 2mAyWy + m²(Wy)² = 1, or
    // (Ax)²+(Ay)² - 1 + 2m(AxWx+AyWy) + m²((Wx)²+(Wy)²) = 0
    // As (Ax)²+(Ay)² = 1, we can eliminate (Ax)²+(Ay)² - 1.
    // m=0 is a solution. But we want to get the other root. 
    // So divide the equation by m, and solve by m:
    const m = -2 * (a[0] * w[0] + a[1] * w[1]) / (w[0] * w[0] + w[1] * w[1]);
    // Scale the solution back to the original size
    return m * r; // This is a signed number. For distance, take absolute value
}

// I/O handling and actual call of the solve function. 

const canvas = document.querySelector("canvas");
const ctx = canvas.getContext("2d");
const output = document.querySelector("span");

function drawCircle(c, a) {
    ctx.beginPath();
    ctx.arc(...c, size(sub(c, a)), 0, 2 * Math.PI);
    ctx.stroke();
}

function drawLine(a, d) {
    ctx.beginPath();
    ctx.moveTo(...a);
    ctx.lineTo(...d);
    ctx.stroke();
}

function clear() {
    ctx.clearRect(0, 0, canvas.width, canvas.height);
}

document.addEventListener("mousemove", refresh);

function refresh(e) {
    // For demo we fix points a and c. b is determined by pointer position
    const c = [200,  90];
    const a = [150,  30];
    clear();
    drawCircle(c, a);
    if (!e) return; // No coordinates for b provided.
    // Convert mouse pointer to coordinates for b (relative to the canvas)
    const b = [e.clientX - canvas.offsetLeft, e.clientY - canvas.offsetTop];
    const dist = solve(a, b, c); // <<<< EXECUTE THE ALGORITHM >>>>
    output.textContent = Math.abs(dist); // Output the result
    // For drawing the line segment, calculate d using a, b and dist
    const d = add(a, mul(norm(sub(b, a)), dist));
    drawLine(a, d);
}
refresh();

<canvas height="170" width="500"></canvas>
<div>Size of line segment: <span></span></div>

从圆内任何点和角度到圆周的距离

Distance to circumference of circle from any point and angle within the circle

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