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Important Formulas in Sag and Tension Calculations

Byjaredcuchapin@gmail.com

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Quick Reference · Sag & Tension

Sag and Tension Formulas for Overhead Power Lines

A working cheat sheet, organized by the question you actually show up with — not just a formula dump. Each section opens with the geometry, then answers the specific things that come up on a real job: sag at a point, conductor length, ice and wind, blowout, and which elongation model to trust. Notation matches the catenary derivation post: S = span, D = sag, L = conductor length, w = weight per unit length, H = horizontal tension, a = distance from support A.

Running example used throughout this page
ConductorDrake 795 kcmil 26/7 ACSR — total area 468.6 mm² (Al 402.8 mm², St 65.8 mm²), OD 28.1 mm, bare weight w = 15.97 N/m, RTS = 140.1 kN, final composite modulus E = 73.9 GPa (CIGRE TB 324, Table 13)
Base conditionS = 300 m level span, installed at H₁ = 25% RTS = 35,025 N at t₁ = 15°C
Inclined variantSame span, supports differ in elevation by h = 10 m, used for Section 1–2 inclined-span questions
Loading variant12.5 mm radial ice, 380 Pa wind pressure, at t₂ = −10°C — used for Sections 3, 4, and 6

Every “Example” box below plugs into this same conductor and span, so results carry over between sections instead of resetting each time.

01

Catenary Curve

The exact shape a conductor takes under its own weight and tension. Everything else in this post — the parabola, the loading formulas, the elongation models — is either a simplification of this curve or something layered on top of it. Full derivation: The Role of Catenary Equations in Sag and Tension Calculations.

vertex reference line (y = 0) H/w a e(a) P S h A B
Catenary curve between supports A and B (shown with B lower, the general inclined-span case). The vertex is the true lowest point of the curve, sitting a height H/w above an abstract reference line — that reference line is a mathematical construct from the derivation, not a point you’d physically locate on site. Point P shows a general query at horizontal distance a from support A.

What is the catenary equation, and where does H/w come from?

Catenary coordinate equation, measured from the vertex.
\( y = \dfrac{H}{w}\cosh\left(\dfrac{wx}{H}\right) \)

Variables

  • H — horizontal component of conductor tension (N), constant along the span
  • w — conductor weight per unit length (N/m), where w = mg; m = unit mass (kg/m), g = 9.81 m/s²
  • x — horizontal distance from the vertex (m)
  • y — height above the reference line (m) — not the sag; see the note below
The vertex sits at y = H/w, not y = 0 — this is the single most common source of confusion when people first meet this formula. H/w is a scale constant of the curve (often called the catenary parameter), not sag. To get sag, you need the formulas below.
Example
H/w = 35,025 / 15.97 = 2193.2 m  →  vertex sits 2193.2 m above the reference line

How do I find sag in a level span?

Maximum sag, level span (occurs at the vertex, x = 0).
\( D_{max} = \dfrac{H}{w}\left[\cosh\left(\dfrac{wS}{2H}\right)-1\right] \)
Example — S = 300 m, H = 35,025 N, w = 15.97 N/m
D_max = 2193.2·[cosh(15.97·300 / (2·35025)) − 1]
= 2193.2·[cosh(0.06844) − 1] = 5.132 m

Where’s the point of maximum sag on a level span?

Exactly at midspan (x = 0, a = S/2) — but only because the span is level. That stops being true the moment the supports sit at different elevations (see below).

How do I find the conductor length in a level span?

Total conductor length, level span.
\( L = \dfrac{2H}{w}\sinh\left(\dfrac{wS}{2H}\right) \)
Slack — how much extra conductor is up there compared to a straight line.
\( \text{Slack} = L – S \)
Example — same S, H, w as above
L = 2(2193.2)·sinh(0.06844) = 300.234 m
Slack = 300.234 − 300 = 0.234 m (23.4 cm of extra conductor over 300 m)

How do I find sag when the supports are at different elevations?

Two steps: locate the vertex first, then compute sag relative to the chord line A–B.

Step 1 — locate the vertex. x_A is measured from the vertex to support A (negative on the A side); distance from A to the vertex is −x_A.
\( x_A = \dfrac{H}{w}\sinh^{-1}\left[\dfrac{h}{2\frac{H}{w}\sinh\left(\frac{wS}{2H}\right)}\right] – \dfrac{S}{2} \)

Variables

  • h — elevation of B above A (m); negative if B is lower
Step 2 — sag below the chord A–B, at distance a from support A.
\( D(a) = \dfrac{h}{S}a + \dfrac{H}{w}\left[\cosh\left(\dfrac{w\,x_A}{H}\right)-\cosh\left(\dfrac{w(x_A+a)}{H}\right)\right] \)
Example — h = 10 m, a = 100 m from A
x_A = 2193.2·sinh⁻¹[10 / (2·2193.2·sinh(0.06844))] − 150 = −223.03 m
→ vertex sits 223.03 m from A (i.e. 76.97 m from B side of midspan)
D(100) = (10/300)(100) + 2193.2[cosh(−223.03/2193.2) − cosh(−123.03/2193.2)]
= 4.563 m

How do I find the conductor’s elevation at distance a from support A?

This is the number you actually want for a clearance check — height relative to support A, not the drop below the sloped chord.

Elevation at a, relative to A (positive = above A).
\( e(a) = \dfrac{h}{S}a – D(a) \)
Example — continuing a = 100 m from A
e(100) = (10/300)(100) − 4.563 = 3.333 − 4.563 = −1.230 m
The conductor sits 1.23 m below support A’s level at that point — even though B is 10 m higher.

Where does the vertex actually sit on an inclined span, and where is max sag really located?

These are two different points once the span is inclined, and conflating them is a common hand-calculation error. The vertex (true lowest point of the curve) shifts toward the lower support — sometimes dramatically. The point of maximum sag-below-chord stays close to midspan, but not exactly at it.
Location of maximum sag-below-chord, measured from the vertex (exact).
\( x_c = \dfrac{H}{w}\sinh^{-1}\left(\dfrac{h}{S}\right) \)
Worked check — S = 300 m, our Drake running example (H = 35,025 N, w = 15.97 N/m)
h (m)Vertex, distance from AMax-sag point, distance from A
0 (level)150.0 m150.0 m
1077.0 m150.1 m
204.0 m150.1 m
This conductor is strung fairly taut (D_max is only 5.1 m on a 300 m span), which is exactly why it’s so sensitive to elevation difference — going from h = 10 m to h = 20 m pushes the vertex from 77 m almost all the way down to 4 m from A, while the max-sag point barely moves off midspan the whole time. Taut, low-sag lines are the case where this distinction matters most.

What’s the tension at any point along the span, not just at the supports?

Tension at any point, using the vertex-origin coordinate y.
\( T = w\,y = H\cosh\left(\dfrac{wx}{H}\right) \)

Useful because the highest tension in the span is always at the higher-elevation support attachment point — not at the vertex — which is the point you actually need to check against the conductor’s rated breaking strength.

Example — inclined span from above (h = 10 m)
T at A = 35,025·cosh(−223.03/2193.2) = 35,047 N
T at B = 35,025·cosh(76.97/2193.2) = 35,206 N
Tension at B (the higher support) is the largest in the span — 35,206 N, about 0.5% above H itself. On a taut span like this one the difference is small, but it grows fast on slacker, more inclined spans.
02

Parabolic Approximation

The small-slope limit of the catenary — the shortcut most engineers actually reach for by hand. Same questions as Section 1, parabolic answers.

exact catenary parabola gap exaggerated — real size in the worked example S A B
Both curves are forced through the same two support points, so they diverge only near the vertex — the gap here is stretched for visibility; the actual difference is centimeters on a span of hundreds of meters (see the worked example).

What’s the parabolic sag formula, and when is it accurate enough?

Parabolic approximation, measured from the vertex.
\( y \approx \dfrac{w}{2H}x^2 \)

Valid when the slope stays small — governed by wS/H, i.e. long slack spans and low tension push you toward the exact catenary; short, taut, everyday transmission spans are usually well inside where the parabola is fine. See the worked error check below for how “fine” scales with span length.

How do I find sag in a level span (parabola)?

Maximum sag, level span. Rearranges directly to solve for H given a target sag.
\( D_{max} \approx \dfrac{wS^2}{8H} \quad\longrightarrow\quad H \approx \dfrac{wS^2}{8D_{max}} \)
Example — S = 300 m, H = 35,025 N, w = 15.97 N/m
D_max ≈ 15.97·300² / (8·35025) = 5.130 m
Compare to the exact catenary value from Section 1: 5.132 m — 2 mm apart on a taut 300 m span.

How do I find the length of cable in a level span (parabola)?

Total conductor length, level span.
\( L \approx S\left(1+\dfrac{S^2w^2}{24H^2}\right) = S + \dfrac{S^3w^2}{24H^2} \)
Slack.
\( \text{Slack} \approx \dfrac{S^3w^2}{24H^2} \)
Example — same values
L ≈ 300(1 + 300²·15.97² / (24·35025²)) = 300.234 m
Matches the exact catenary length to the millimeter here — again, this is a taut span where the two methods barely differ.

How do I find sag at distance a from support A (parabola)?

Sag below the chord A–B, at distance a from A.
\( D(a) \approx \dfrac{w\,a(S-a)}{2H} \)
This formula is the same whether the span is level or inclined — h drops out entirely for sag-below-chord in the parabolic approximation. That’s a genuinely useful shortcut, not a coincidence worth double-checking every time.
Example — a = 100 m from A
D(100) ≈ 15.97·100·(300−100) / (2·35025) = 4.560 m
Compare to the exact catenary result in Section 1: 4.563 m.

How do I find the elevation at distance a from A, relative to A (parabola)?

Elevation at a, relative to A (positive = above A). For a level span, h = 0 and this reduces to −D(a).
\( e(a) \approx \dfrac{h}{S}a – \dfrac{w\,a(S-a)}{2H} \)
Example — h = 10 m, a = 100 m from A
e(100) ≈ (10/300)(100) − 4.560 = −1.226 m
Matches the exact result (−1.230 m) to 4 mm.

Where’s the vertex on an inclined span (parabola)?

Distance from support A to the vertex — closed form, no iteration needed.
\( x_1 = \dfrac{S}{2} – \dfrac{Hh}{wS} \)

This is the small-slope limit of the exact vertex formula in Section 1 — matches it closely for typical spans, diverges more as h/S grows (see the worked check in Section 1).

Example — h = 10 m
x₁ = 150 − (35025·10)/(15.97·300) = 76.89 m from A
Compare to the exact value from Section 1: 76.97 m — 8 cm apart.

How much error does the parabola actually introduce?

Relative error in sag from using the parabola instead of the exact catenary (Kießling).
\( \Delta\% = \dfrac{\cosh\left(\frac{S}{2C}\right)-1-\frac{1}{2}\left(\frac{S}{2C}\right)^2}{\frac{1}{2}\left(\frac{S}{2C}\right)^2}\times100,\quad C=\dfrac{H}{w} \)
Example — our running example, S = 300 m, C = 2193.2 m
Δ% = 0.039% — negligible, because this conductor is strung taut (low D_max/S ratio)
Worked example — a slacker case, H = 14 000 N, w = 12.12 N/m (C ≈ 1156 m)
SpanCatenary DParabola DAbs. error% error
400 m17.35 m17.31 m0.043 m0.25%
800 m69.92 m69.23 m0.694 m1.00%
Error grows with span and with slack (lower tension relative to weight). Kießling’s own rule of thumb: up to roughly 500 m span the parabola is generally sufficient; where the absolute error exceeds about 0.10 m, switch to the exact catenary.
03

Effects of Ice and Wind

Ice and wind don’t change the shape of the formulas — they change what you plug in for w. Everything from Sections 1–2 still applies once you’ve built the combined load.

How does ice change the effective conductor weight?

Weight of ice per unit length. Note: conductor outside diameter is written d_c here, kept separate from sag D to avoid the symbol clash in the original version of this post.
\( w_{ice} = \rho_{ice}\,g\,\pi\,t(d_c+t) \)

Variables

  • d_c — conductor outside diameter (m)
  • t — radial ice thickness (m)
  • ρ_ice — ice density, typically 900 kg/m³ (NESC / CIGRE)
  • g — 9.81 m/s²
Example — Drake (d_c = 0.0281 m), 12.5 mm radial ice
w_ice = 900·9.81·π·0.0125·(0.0281+0.0125) = 14.08 N/m

How does wind load combine with weight and ice?

Wind load per unit length — acts on the ice-covered diameter, not the bare conductor.
\( w_{wind} = P_{wind}(d_c+2t) \)

Variables

  • P_wind — wind pressure (N/m²)
Total resultant conductor weight under combined ice and wind.
\( w_{total} = \sqrt{(w+w_{ice})^2 + w_{wind}^2} \)
Example — continuing above, 380 Pa wind pressure
w_wind = 380·(0.0281+2·0.0125) = 20.18 N/m
w_total = √[(15.97+14.08)² + 20.18²] = 36.19 N/m
More than double the bare weight — this is the value that goes into H and the state change equation (Section 6).

How do I get sag under combined ice + wind loading?

Substitute w_total for w in the Section 1 / 2 formulas — the math doesn’t change, only the weight does.
\( D_{total} = \dfrac{H}{w_{total}}\left[\cosh\left(\dfrac{w_{total}S}{2H}\right)-1\right] \approx \dfrac{w_{total}S^2}{8H} \)
This D_total is measured along the direction of the resultant load, not straight down — the conductor plane itself has rotated to align with w_total. That’s the bridge to Section 4: you still need to split it into vertical and horizontal components before it’s useful for a clearance check.
Example — carried through to Section 4, using H = 65,073 N (solved in Section 6)
D_total = (65073/36.19)·[cosh(36.19·300/(2·65073)) − 1] = 6.261 m

How are ice and wind loading values actually chosen for a real design?

Briefly: from loading district maps and combined-loading cases (NESC) or the equivalent CIGRE TB 324 approach — not derived from first principles per project. That’s a big enough topic to live on its own; see this site’s NESC 2017 loading posts for the full treatment rather than repeating it here.

04

Vertical, Horizontal & Slant Sag

Wind doesn’t just increase sag — it rotates the plane the conductor hangs in. That rotation is why one sag number isn’t enough once wind is in the picture.

support θ wind D D_v D_h
Wind blows the conductor sideways off the vertical. D is the slant sag along the actual hang direction (what the loaded catenary/parabola formula gives you with w_total); D_v is the vertical drop that matters for ground clearance; D_h is the horizontal blowout that matters for clearance to structure or adjacent phases.

What’s the difference between these three, and why do you need all of them?

  • D — slant (total) sag, along the direction of the resultant load w_total. This is what the loaded catenary/parabola formula gives you directly.
  • D_v — vertical sag, the drop straight down. This is what matters for ground clearance.
  • D_h — horizontal blowout, the sideways displacement. This is what matters for clearance to the structure, to adjacent phases, or to anything beside the line.

What’s the blowout angle θ?

Angle of the resultant load (and therefore the conductor’s hanging plane) from vertical.
\( \theta = \tan^{-1}\left(\dfrac{w_{wind}}{w+w_{ice}}\right) \)
Example — continuing from Section 3 (w_wind = 20.18 N/m, w+w_ice = 30.05 N/m)
θ = tan⁻¹(20.18 / 30.05) = 33.9°

How do I compute vertical sag?

Vertical component of slant sag.
\( D_v = D\cos\theta \)
Example — D = 6.261 m (Section 3), θ = 33.9°
D_v = 6.261·cos(33.9°) = 5.198 m

How do I compute horizontal blowout?

Horizontal component of slant sag.
\( D_h = D\sin\theta \)
Example — same values
D_h = 6.261·sin(33.9°) = 3.490 m
Under this loading, the conductor swings almost 3.5 m sideways — a real number to check against structure and adjacent-phase clearance, not just an afterthought to the vertical sag.
05

Linear, SPE & EPE — Sag After Stringing

The catenary and parabola formulas above assume a fixed w and a fixed unstressed conductor length. Neither stays fixed over the life of a line — temperature changes elastically, and the conductor itself permanently stretches. These three models are the standard ways of tracking that stretch (CIGRE TB 324 terminology).

elongation stress LE SPE permanent set EPE (actual curve) fixed offset actual offset
Schematic only, not to scale. LE assumes no permanent stretch at all. SPE adds a single typical offset regardless of the actual loading event. EPE follows the conductor’s real non-linear curve, so the permanent set it predicts reflects the actual load applied — which may be larger or smaller than SPE’s fixed assumption.

Why can’t the catenary/parabola formulas alone handle a condition change after stringing?

Those formulas take H and w as given and return a shape. They don’t tell you what the new H is once temperature or load changes the conductor’s unstressed length. That’s a separate problem — solved by an elongation model plus the conductor state change equation in Section 6 — not something the catenary equation itself handles.

What is Linear Elongation (LE), and when is it good enough?

Treats the conductor as a single, purely elastic material — one E, one α — for its entire service life. Ignores permanent (plastic) stretch entirely, which means it always overstates final tension and understates how much sag will grow over time. Fine for preliminary sizing or low-consequence short lines; not something to trust for a final sag-tension table.

What is SPE (Simplified Plastic Elongation)?

Same elastic backbone as LE, but with a single fixed, typical permanent-elongation offset added on top — based on historical experience for that conductor family, not on the specific loading event or time in service. It’s the practical middle ground, and what most hand and spreadsheet sag-tension calculations actually use.

What is EPE (Experimental Plastic Elongation)?

Uses the conductor’s actual measured non-linear stress-strain curve, tracking plastic elongation as a function of the real loading history — the specific design ice/wind event — separately from time-based metallurgical creep, rather than one blanket allowance. Most accurate, most data-hungry (needs manufacturer/lab curves); this is the model behind this site’s EPE calculator and behind SAG10-class software.

Which one should you actually reach for?

  • Preliminary or rough sizing, no data on hand → LE
  • Everyday design work, standard conductor, no lab curve → SPE
  • Final design, critical or high-voltage lines, or initial-vs-final sag really matters → EPE

Worked examples: Linear and SPE Sag Tension Calculator and the EPE Sag and Tension Calculator.

06

Stringing States & the Conductor State Change Equation

A full sag-tension table is really just this equation solved repeatedly — once per loading case, and once more to move from “initial” to “final.”

What do initial, final-after-load, and final-after-creep actually mean?

  • Initial — the state right after stringing, sagging, and clipping. Reference condition; essentially no permanent stretch yet.
  • Final-after-load — the state after the conductor has experienced its worst design loading event (max ice + wind) at least once. Captures the permanent stretch that event induces — happens fast, not gradually.
  • Final-after-creep — the state after a specified sustained period, commonly 10 years, at everyday tension and temperature. Captures metallurgical creep — slow, continuous, and a separate mechanism from load-induced stretch (Section 7).

How do I get the effective modulus of elasticity and thermal expansion for a composite (ACSR) conductor?

Needed as inputs before you can run the state change equation below.

Composite modulus of elasticity, weighted by area fraction.
\( E_{cond} = E_{al}\dfrac{A_{al}}{A_{total}} + E_{st}\dfrac{A_{st}}{A_{total}} \)
Effective coefficient of linear thermal expansion, weighted by modulus × area fraction.
\( \alpha_{cond} = \alpha_{al}\dfrac{E_{al}}{E_{cond}}\dfrac{A_{al}}{A_{total}} + \alpha_{st}\dfrac{E_{st}}{E_{cond}}\dfrac{A_{st}}{A_{total}} \)

Variables

  • E_al, E_st — modulus of elasticity of aluminum and steel (Pa)
  • A_al, A_st, A_total — cross-sectional areas of aluminum, steel, and total (m²)
  • α_al, α_st — coefficient of linear thermal expansion of aluminum and steel (/°C)
Example — Drake (A_al = 402.8 mm², A_st = 65.8 mm²), typical component values E_al = 60.2 GPa, E_st = 162.0 GPa, α_al = 23.0×10⁻⁶/°C, α_st = 11.5×10⁻⁶/°C
E_cond = 60.2(402.8/468.6) + 162.0(65.8/468.6) = 74.49 GPa
CIGRE TB 324 lists a measured final modulus of 73.9 GPa for this exact conductor — close agreement; the small gap is real strand-interaction behavior a simple area-weighted average can’t capture.
α_cond = 23.0e-6(60.2/74.49)(402.8/468.6) + 11.5e-6(162.0/74.49)(65.8/468.6) = 19.5×10⁻⁶ /°C

How do I find the change in conductor length from tension or temperature alone?

Elastic stretch from a change in tension.
\( \dfrac{L_{final}-L_{initial}}{L_{initial}} = \dfrac{H}{E_{cond}A_{cond}} \)
Example — H = 35,025 N, E = 73.9 GPa, A = 468.6 mm²
Strain = 35025 / (73.9×10⁹ × 468.6×10⁻⁶) = 0.101%
Thermal expansion from a change in temperature.
\( L_{final} = L_{initial}\left[1+\alpha_{cond}(t_f-t_i)\right] \)
Example — L_initial = 300.234 m, cooling from 15°C to −10°C
L_final = 300.234·[1 + 19.5×10⁻⁶·(−25)] = 300.088 m
The conductor physically shrinks about 0.146 m from cooling alone — before the state change equation even accounts for the load increase.

Why does tension drop over time even with no load change?

Creep — a slow, continuous, permanent stretch under sustained stress that keeps happening even at everyday, unremarkable tension. As the unstressed length grows, the same span settles into more sag at lower tension for the same temperature. Full explanation in Section 7.

When do you actually need the conductor state change equation, and what does it solve for?

Any time you know the conductor’s state at one condition (H₁ at temperature t₁, weight w₁) and need the tension H₂ at a different condition — different temperature, different ice/wind load, or a different unstressed length after permanent stretch. Both tension and temperature changing together, plus a possible permanent-elongation offset, is exactly what the two length-change formulas above can’t handle separately — this equation solves all of it at once.

Conductor state change equation — cubic in H₂.
\( H_2^3 + H_2^2\left[\dfrac{(w_1S)^2AE}{24H_1^2}-H_1+AE\alpha\,\Delta t\right] – \dfrac{(w_2S)^2AE}{24} = 0 \)

Variables

  • H₁ — initial conductor tension, at the initial (reference) condition (N)
  • w₁, w₂ — unit weight of conductor at the initial and final conditions (N/m)
  • Δt = t₂ − t₁ — temperature change (°C)
  • A, E, α — cross-sectional area, modulus, and thermal coefficient (final/composite values)
  • S — ruling span, or single span length (m)

Solved once per row of a full sag-tension table (from a known reference state to each loading case), and again to move from “initial” to “final” — where the permanent-elongation offset comes from either SPE’s fixed value or EPE’s stress-strain-derived value.

What are the limitations of the conductor state change equation?

  • Linear-elastic only. It assumes a single, constant composite E over the whole range from state 1 to state 2. It has no idea about plastic or creep elongation on its own — that has to be fed in externally as a permanent-elongation offset from SPE or EPE (Section 5). Run it alone and you’ve implicitly assumed the LE model.
  • Assumes the conductor stays in tension throughout. Near or above the knee-point temperature, the aluminum layers of an ACSR conductor can go slack and the steel core alone carries the load — a real, documented behavior (Section 7) this equation doesn’t see coming.
  • Built on the parabolic length approximation, not the exact catenary — the (wS)²/24H² term is the same series term from Section 2. Same error behavior: negligible on typical taut transmission spans, growing on long or slack ones.
  • A, E, and α are treated as constants — real conductor stiffness and thermal behavior shift with stress and temperature, most noticeably for non-homogeneous ACSR near the knee point.
  • Solved per ruling span, not per physical span. It gives you one tension assumed common across a whole line section — actual tension in each individual suspension span can differ slightly from that.
  • The cubic can have more than one positive real root for some input combinations. You still need engineering judgment — usually “closest to H₁” or “within a sane %RTS range” — to pick the physical one.

How do you actually solve the conductor state change equation?

By hand, this cubic is almost never solved with the general cubic formula (Cardano’s) — it’s solved iteratively. Two practical routes, both standard practice:

  • Newton-Raphson iteration — fast, reliable, and what most sag-tension spreadsheets do under the hood (including a plain Excel Goal Seek / Solver call). Start from H₂ = H₁ as the first guess.
  • Graphical method — plot the loaded catenary curve and the elastic-modulus line on a stress-strain diagram and read off their intersection. This is the classic Varney/Alcoa technique, and it’s literally what the stress-strain picture in Section 5 is showing.
Newton-Raphson update step, with f(H) being the cubic above written as H³ + bH² + d (the linear term always drops out).
\( H_{n+1} = H_n – \dfrac{f(H_n)}{f'(H_n)}, \quad f(H)=H^3+bH^2+d,\ \ f'(H)=3H^2+2bH \)
Worked example — bare/everyday (H₁ = 35,025 N, w₁ = 15.97 N/m, t₁ = 15°C) → iced/windy (w₂ = 36.19 N/m, t₂ = −10°C)
With A = 468.6 mm², E = 73.9 GPa, α = 19.5×10⁻⁶/°C, S = 300 m: b = −24,900, d = −1.701×10¹⁴.
IterationH_n (N)H_(n+1) (N)
035,025116,476
1116,47685,752
285,75270,161
370,16165,487
465,48765,076
565,07665,073
Converges to H₂ ≈ 65,073 N (46.5% RTS) in about 5 iterations — the first jump overshoots because the starting guess is far from the answer, which is normal; it settles quickly after that. This is the H used in the Section 3 and 4 worked examples above.

How does this connect to ruling span?

The state change equation — and the whole sag-tension table — is solved using the ruling span, not each individual physical span, since tension is assumed common across all suspension spans in a section. See What is Ruling Span? and the step-by-step ruling span computation post.

07

Creep

The mechanism behind “final-after-creep” in Section 6 — worth its own section because it’s easy to under-account for.

What is creep, physically, and how is it different from elastic strain?

Creep is slow, continuous, permanent (non-recoverable) elongation of a material under sustained stress — distinct from elastic strain, which fully recovers when the load is removed. In ACSR, creep occurs almost entirely in the aluminum strands; the steel core creeps negligibly. That’s why, as a conductor ages, the steel core picks up a growing share of total tension while the aluminum layers “give.” Most creep happens in the first days to weeks after stringing, continuing at a decreasing rate for the rest of the line’s service life.

How is it normally handled in a sag-tension design?

Captured through the “final-after-creep” condition — commonly evaluated at 10 years — from either a typical creep allowance (SPE) or lab-measured long-term creep curves (EPE), folded into the state change equation as an equivalent permanent elongation or equivalent temperature shift.

Why does skipping it cause real problems years after construction?

Ignoring creep means your “final” condition is really still your initial one — sag will be under-predicted, and clearance that looked fine on paper can erode over the years as the line actually settles into more sag than the initial calculation showed. The opposite mistake is just as real: over-tightening a conductor at stringing to try to compensate for expected creep, without properly modeling it, risks over-tensioning before creep ever gets the chance to relax it. See Transmission Line Failure Due to Increase in Conductor Tension for a worked case of that second failure mode.

References: Kiessling, F.; Nefzger, P.; Kaintzyk, U.; Nefzger, F. — Overhead Power Lines: Planning, Design, Construction, Ch. 14. CIGRE Technical Brochure 324, WG B2-12, “Sag-Tension Calculation Methods for Overhead Lines.” See also this site’s catenary curve derivation for the full derivation behind Section 1.

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