欧拉–埃特尔维因方程

简化主干笔记（Summary）
Frenet 标架 → 力密度 → 张力 → 接触力 → 力平衡 → ODE → 解出张力 → 圆/椭圆算例

1–6 · 核心推导

  1. 1. Frenet frame and formula:
    1. $\mathbf{x}(s)$ be a curve with arc-length parametrization. $\mathbf{e}(s) = \mathbf{x}'(s)$ its tangent vector.
    2. $|\mathbf{x}'(s)|=1$ because arc-length parametrization means $s = L(s) := \int_{s_0}^{s}|\mathbf{x}'(t)|\,dt$ for some basepoint $s_0$ (the parameter is the length traveled); differentiating both sides, $1 = L'(s) = |\mathbf{x}'(s)|$ by the Fundamental Theorem of Calculus. So $\mathbf{e}(s)$ is a unit vector along $s$.
    3. $\boldsymbol\tau(s)$ be $\mathbf{e}(s)$ rotating counter-clockwise by $\frac{\pi}{2}$. The reason: this matches do Carmo's Remark 1 (§1-5) — orienting $\{\mathbf{e},\boldsymbol\tau\}$ the same way as the standard basis — which is exactly what makes $k(s)$ below come out in do Carmo's standard sign convention (positive for a CCW-traversed convex curve, and the classical, un-flipped formula for the ellipse).
    4. Frenet formula: Define the curvature $k(s) := \mathbf{e}'(s)\cdot\boldsymbol\tau(s)$ (equivalently $k(s) = \det(\mathbf{e}(s),\mathbf{e}'(s))$). Then
      1. $\mathbf{e}'(s) = k(s)\boldsymbol\tau(s)$.
      2. $\boldsymbol\tau'(s) = -k(s)\mathbf{e}(s)$.
  2. 2. Force:
    1. A force applied on a (measurable) subset $A$ on the rope is given by $\mathbf{F}(A) = \int_{A}\mathbf{f}(s)ds$, where $\mathbf{f}(s)$ is the density of the force.
  3. 3. Tension:
    1. The tension on the rope is always along the direction $\mathbf{e}(s)$. $\mathbf{T}(s)=T(s)\mathbf{e}(s)$. The question can be reformulated as:
      1. Find the dependence of quantity $T(s)$ on $s$.
      2. At point $s$, the pulling force (i.e. along direction $\mathbf{e}(s)$) is $\mathbf{T}(s)$; the opposing force is $-\mathbf{T}(s)$ (opposing the moving direction).
  4. 4. Contact force (pointwise):
    1. The contact force on the rope is defined to be the total force imposed from the bamboo to the rope (= support + friction).
    2. Let $\mathbf{f}(s)$ be the density of contact. Then at point $s$, let $\mathbf{n}(s)$ be the density of support force at point $s$, then $\mathbf{n}(s) = -n(s)\boldsymbol\tau(s)$, where $n(s)\ge 0$ is the magnitude of $\mathbf{n}(s)$ (support pushes away from the concave side, i.e. opposite $\boldsymbol\tau$, which points toward it by our choice above).
    3. Pointwise version of contact force: by the definitions we have
      1. $\mathbf{f}(s) = \mathbf{n}(s)-\mu n(s)\mathbf{e}(s) = -n(s)\boldsymbol\tau(s)-\mu n(s)\mathbf{e}(s)$. (key1)
  5. 5. Force on an interval:
    1. On an interval $[s_1,s_2]$, it has three forces: a pulling force $\mathbf{T}(s_2)$, an opposing force $-\mathbf{T}(s_1)$, and a contact force $\int_{s_1}^{s_2}\mathbf{f}(s)ds$.
    2. Since the interval $[s_1,s_2]$ is balanced, we have $\mathbf{T}(s_2) +(- \mathbf{T}(s_1)) + \int_{s_1}^{s_2}\mathbf{f}(s)ds = 0$.
    3. In particular, $\mathbf{T}(s)$ is differentiable with $\mathbf{T}'(s) = -\mathbf{f}(s)$. (key2)
  6. 6. The differential equation of $T(s)$:
    1. Plug in the definition of $T(s)$ and (key1) into (key2) we get
      1. LHS $= T'(s)\mathbf{e}(s)+T(s)\mathbf{e}'(s)$, by Frenet formula we get LHS $= T'(s)\mathbf{e}(s)+T(s)k(s)\boldsymbol\tau(s)$.
      2. RHS $= n(s)\boldsymbol\tau(s)+\mu n(s)\mathbf{e}(s)$.
      3. By LHS = RHS, and the fact that $\mathbf{e}(s),\boldsymbol\tau(s)$ is a frame, we get
        1. $T'(s) = \mu n(s)$
        2. $n(s) = T(s)k(s)$ (consistent with $n(s)\ge0$, since $T,k>0$).
          1. It follows that we have $T'(s) = \mu T(s)k(s)$.

Solving the ODE and concrete examples

  1. 7. Solving the ODE:
    1. Boundary condition: fix a basepoint $s_0$ with known tension $T(s_0)=G>0$ (e.g. where a weight $G$ hangs).
    2. Define $\Theta(s):=\int_{s_0}^s k(s')\,ds'$ (total turning of the tangent from $s_0$ to $s$).
    3. The ODE $T'(s)=\mu k(s)T(s)$ with $T(s_0)=G$ has unique solution $T(s)=G\,e^{\mu\Theta(s)}$ (integrating factor: $\varphi(s):=T(s)e^{-\mu\Theta(s)}$ has $\varphi'\equiv0$, so $\varphi\equiv\varphi(s_0)=G$).
    4. In particular $T(s)>0$ everywhere ($G>0$, $e^{(\cdot)}>0$), consistent with $n(s)=T(s)k(s)\ge0$ from item 6.

8 · Circle example

  1. Radius $R$: $k\equiv 1/R$ (constant).
  2. Parametrize by angle $\theta$ (arc length $s=R\theta$): $\Theta(\theta)=\int_0^\theta \frac1R\cdot R\,d\theta' = \theta$.
  3. $T(\theta)=G\,e^{\mu\theta}$ — the classical Euler–Eytelwein formula.
θ T(θ) G T(θ) = G e^{μθ}

9 · Ellipse example

  1. $\boldsymbol\delta(t)=(a\cos t,b\sin t)$, $k(t)=\dfrac{ab}{(a^2\sin^2t+b^2\cos^2t)^{3/2}}$, $v(t)=|\boldsymbol\delta'(t)|=\sqrt{a^2\sin^2t+b^2\cos^2t}$.
  2. $\Theta(t)=\int_0^t k(t')v(t')\,dt' = \int_0^t \dfrac{ab}{a^2\sin^2t'+b^2\cos^2t'}\,dt' = \arctan\!\big(\tfrac ab\tan t\big)$ for $t\in(-\tfrac\pi2,\tfrac\pi2)$; extends by $\Theta(k\tfrac\pi2+u)=k\tfrac\pi2+\arctan(\tfrac ab\tan u)$ for integer $k$ (each quarter of the ellipse turns the tangent by exactly $\pi/2$, by symmetry — consistent with the total-turning theorem: a full loop turns by $2\pi$ regardless of shape).
  3. So $T(t)=G\,e^{\mu\Theta(t)}$; at the symmetric points $t=\tfrac\pi2,\pi,\tfrac{3\pi}2,2\pi$: $\Theta=\tfrac\pi2,\pi,\tfrac{3\pi}2,2\pi$ exactly, same as the circle at those fractions — the ellipse only differs from the circle at non-symmetric wrap points, e.g. $t=\pi/4$ (with $a=6.25$cm, $b=4$cm): $\Theta(\pi/4)=\arctan(a/b)=\arctan(1.5625)\approx1.0015$ rad ($57.4°$), not $45°$ — curvature is higher near the major-axis tip, so the tangent turns faster there than a naive angle-guess assumes.

10 · 竹竿 + 登山扣 3:1 系统 (Z-drag)，物体 70kg

  1. 装置：绳一端固定在物体上（同一固定点也挂一个登山扣）；绳从固定点向上绕竹竿一圈（角度 $\Theta$），向下穿过登山扣（视为无摩擦，张力不变）；再向上绕竹竿第二圈（角度 $\Theta$），向下到拉绳人手中。
  2. 物体由三股绳拉住：直接固定的一股（张力 $T_0$）+ 穿过登山扣的两股（各 $T_1$）：$T_0+2T_1=70$（单位 kgf）。
  3. 张力沿绳传递：$T_1=T_0e^{\mu\Theta}$（第一圈绕竹竿），$T_{\text{final}}=T_1e^{\mu\Theta}=T_0e^{2\mu\Theta}$（第二圈绕竹竿，$T_{\text{final}}$ 即人手拉力）。
  4. 解出：$T_0=\dfrac{70}{1+2e^{\mu\Theta}}$，$T_{\text{final}}=\dfrac{70\,e^{2\mu\Theta}}{1+2e^{\mu\Theta}}$。
  5. 核对（$\mu\to0$，无摩擦极限）：$T_{\text{final}}\to 70/3\approx23.33$ kgf——经典 3:1 滑轮组（Z 字形省力装置）力学优势，这是"一个女生也能轻松拉动 70kg"的主要原因。
  6. 摩擦系数 $\mu$ 不确定，取几档比较（$\Theta=\pi$ 半圈，或 $3\pi/2$ 四分之三圈）：
    $\mu$$\Theta$$T_0$ (kgf)$T_1$ (kgf)$T_{\text{final}}$ (kgf)
    0.1$\pi$18.7325.6435.10
    0.1$3\pi/2$16.6526.6842.73
    0.2$\pi$14.7427.6351.79
    0.2$3\pi/2$11.4129.2975.18
    0.5$\pi$6.5931.71152.51
    0.5$3\pi/2$3.1733.42352.57
  7. 实际手感接近 $\mu\approx0.1$ 一档（$T_{\text{final}}\approx35$–$43$ kgf，比直接拎起一半重量稍多一点，量级上"一个女生能轻松拉动"是合理的）——但 $\mu$ 未经实测，仅为估计；注意结果对 $\mu$ 高度敏感（$\mu$ 从 0.1 到 0.5，$T_{\text{final}}$ 从 ~35 涨到 ~350 kgf，十倍差异）。
Z-drag 张力公式
$$T_{\text{final}} = \frac{70\,e^{2\mu\Theta}}{1+2e^{\mu\Theta}} \ \xrightarrow{\ \mu\to0\ }\ \frac{70}{3}\approx23.33\text{ kgf}$$