如何编写函数实现turnRadius达±0.65后指数趋极并映射为sprite水平位置

云上老周

2026-10-1

实现代码

// 可调参数，按需修改
const THRESHOLD = 0.65; // 触发指数过渡的阈值
const EXPONENT = 2; // 指数曲线曲率，数值越大过渡越陡峭
const MAX_POS = 275;
const MIN_POS = 0;
const MID_POS = (MAX_POS + MIN_POS) / 2; // 也就是137.5

let turnRadius = user.input;
let spriteHorizontalPosition;

// 阈值点对应的位置，保证区间衔接无跳变
const rightThresholdPos = (1 - THRESHOLD) * MID_POS; // turnRadius=0.65对应48.125
const leftThresholdPos = MAX_POS - rightThresholdPos; // turnRadius=-0.65对应226.875

if (turnRadius >= THRESHOLD) {
  // 右侧阈值区间：指数衰减到最小位置0
  const normalized = (turnRadius - THRESHOLD) / (1 - THRESHOLD);
  const curve = Math.pow(normalized, EXPONENT);
  spriteHorizontalPosition = rightThresholdPos - curve * rightThresholdPos;
} else if (turnRadius <= -THRESHOLD) {
  // 左侧阈值区间：指数增长到最大位置275
  const normalized = (Math.abs(turnRadius) - THRESHOLD) / (1 - THRESHOLD);
  const curve = Math.pow(normalized, EXPONENT);
  spriteHorizontalPosition = leftThresholdPos + curve * (MAX_POS - leftThresholdPos);
} else {
  // 中间区间保持原有线性映射，操作跟手性不受影响
  spriteHorizontalPosition = (turnRadius * -1 + 1) * MID_POS;
}

// 边界兜底，避免极端输入超出范围
spriteHorizontalPosition = Math.max(MIN_POS, Math.min(MAX_POS, spriteHorizontalPosition));

playerSprite.transform.x = spriteHorizontalPosition;