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Sag - downward vertical deflection

When a cable heats up, it gets slightly longer. If the endpoints stay fixed, that extra length mostly turns into more droop (sag) and lower tension. Engineers estimate the added sag by using a simple parabolic shape for small droop, or a catenary shape for better accuracy when droop is larger.

Posted 2026-01-14
Categories: Theory

Definition

We want to discuss how to calculate the sag relative to the straight line between two fixed points.

Temperature-driven cable elongation is commonly handled by: (1) computing added length from thermal expansion, then (2) converting that extra length into extra sag using a cable shape model (parabola for small sag, catenary for higher accuracy).

Heating increases cable length roughly proportional to original length and temperature rise:

$$\Delta L = \alpha L_0 \Delta T$$

That added length (with span fixed) shows up mainly as more sag and less tension.

To calculate the thermal elongation let:

  • S = span (horizontal distance between fixations)
  • L₀ = cable length at temperature T₀
  • α = linear thermal expansion coefficient (1/°F or 1/°C)
  • ΔT = T₁ − T₀
$$L_1 = L_0(1 + \alpha \Delta T), \qquad \Delta L = L_1 - L_0 = \alpha L_0 \Delta T$$

 

Methods

Now we want to discuss how to convert the extra length to sag.

To calculate the sag of a cable we have different options. Three of them we want to describe here.

  1. Parabola approximation
  2. Geometric approximation
  3. Catenary line

1) Parabolic approximation (small sag)

If sag is modest (common engineering approximation), model the cable as a parabola. The cable length is approximately:

$$L \approx S + \frac{8 f^2}{3S}$$

where f is the midspan sag (maximum deflection from the straight chord/axis).

Steps

  • Compute initial length from your known initial sag f₀:

    $$L_0 \approx S + \frac{8 f_0^2}{3S}$$

  • Apply thermal expansion:

    $$L_1 = L_0(1+\alpha\Delta T)$$

  • Solve for the new sag:

    $$f_1 \approx \sqrt{\frac{3S(L_1 - S)}{8}}$$

  • Sag increase:

    $$\Delta f = f_1 - f_0$$
Very small change shortcut (linearized about f₀)
$$\Delta f \approx \frac{3S}{16 f_0}\,\Delta L$$

2) Half-sine wave approximation (small sag)

This simplified relationship comes from a geometric approximation where thermal length increase is accommodated by bowing into a half sine wave, rather than using a full catenary / sag–tension model.

The formula

$$h = \frac{2L_0}{\pi}\sqrt{\alpha\,\Delta T}$$

is not the usual gravity sag equation; it is essentially a thermal-expansion → lateral bowing relationship under specific assumptions.

  • The distance between fixation points (the chord) is fixed and equals $L_0$.
  • When heated by $\Delta T$, the cable’s stress-free length would increase by:
    $$\Delta L = \alpha\,L_0\,\Delta T$$
  • Instead of creating axial thermal stress, the cable relieves the length increase purely by changing shape (no tension model, no weight/catenary, no stiffness effects).
  • The deflected shape is approximated as a half sine wave:
    $$y(x)=h\sin\left(\frac{\pi x}{L_0}\right),\quad x\in[0,L_0]$$
  • Slopes are small: $|y'|\ll 1$ (so arc length can be approximated).

A real cable between two supports typically has:

  • Gravity (weight per length) → a catenary/parabolic sag relation
  • Tension (initial pretension + temperature-dependent tension)
  • Sometimes creep / nonlinear conductor behavior (overhead lines)

This formula ignores weight and tension completely; it treats the cable like a member that simply “finds room” for extra length by bowing into a sine shape. It is best viewed as a geometric quick estimate under idealized conditions (or an analogy to a buckling-mode shape, noting a real cable cannot sustain compression).

3) True catenary (more accurate)

If sag is not “small” relative to span, use a catenary:

  • Shape parameter: $a = \dfrac{H}{w}$ (horizontal tension $H$, weight per horizontal length $w$)
  • Sag:
    $$f = a\left(\cosh\left(\frac{S}{2a}\right) - 1\right)$$
  • Length:
    $$L = 2a\sinh\left(\frac{S}{2a}\right)$$

Procedure

  • Compute/estimate $L_0$ (from known $a_0$ or measured sag/length).
  • Set $L_1 = L_0(1+\alpha\Delta T)$.
  • Solve the length equation for the new $a_1$ (usually numeric).
  • Compute $f_1$ from the sag equation.

 

New sag calculator tool

We implemented a new tool to calculate the sag for a cable being fixed between two points. This new tool will be released in the coming days.

 

Excursus in explanation

Where the $2/\pi$ comes from

Let's provide a derivation sketch to explain where the $2/\pi$ in the half-sine wave approximation comes from.

For $y(x)=h\sin(\pi x/L_0)$:

  • Derivative:

    $$y'(x)=h\frac{\pi}{L_0}\cos\left(\frac{\pi x}{L_0}\right)$$

  • Arc length:

    $$S=\int_0^{L_0}\sqrt{1+(y')^2}\,dx$$

For small slopes, use $\sqrt{1+u}\approx 1+\tfrac{1}{2}u$:

$$S\approx \int_0^{L_0}\left(1+\frac{1}{2}(y')^2\right)dx = L_0+\frac{1}{2}\int_0^{L_0}(y')^2dx$$

Compute the integral:

$$\int_0^{L_0}(y')^2dx =\left(h\frac{\pi}{L_0}\right)^2\int_0^{L_0}\cos^2\left(\frac{\pi x}{L_0}\right)dx =\left(h\frac{\pi}{L_0}\right)^2\cdot \frac{L_0}{2}$$

So the extra length beyond the straight span is:

$$\Delta L = S-L_0 \approx \frac{1}{2}\cdot \left(h\frac{\pi}{L_0}\right)^2\cdot \frac{L_0}{2} = \frac{\pi^2 h^2}{4L_0}$$

Equate geometric extra length to thermal expansion:

$$\alpha L_0\Delta T = \frac{\pi^2 h^2}{4L_0} \;\Rightarrow\; h=\frac{2L_0}{\pi}\sqrt{\alpha\Delta T}$$

References

  1. Linear Thermal Expansion: Formula, Coefficients & ...
  2. Thermal expansion of catenaries (PDF)
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