# Sag Calculation: Segment Equilibrium Method

## Intro

In the previous article of this series, [Sag Calculation: Parabola vs. Hyperbolic](https://www.linkedin.com/pulse/sag-calculation-parabola-vs-hyperbolic-pavel-barysenka-kbrse), I compared two classical analytical models and showed where the shallow sag assumption (\\(dx \approx dL\\)) breaks down. Both curves are closed-form and convenient. Both also assume a smooth wire under a continuous load.

A real span of an overhead contact system (OCS) or a power line rarely looks like that. In practice we deal with:

- concentrated loads along the span: droppers, clamps, insulators, bird diverters, ice sleeves;
- inclined spans with supports at different heights (\\(y_A \neq y_B\\));
- uneven weight distribution when the sag is deep and the wire slope changes sharply between the support and mid-span.

Stitching a separate cosh curve between every pair of droppers quickly becomes unmanageable by hand. A simpler engineering route is to split the wire into a finite number of straight segments and check static equilibrium one segment at a time.

This article covers the piecewise-linear sag approximation, also known as the Segment Equilibrium or Local Displacement Method. I run it in standard Excel 365 in three successive approximations, including elastic elongation, with dynamic array formulas (SEQUENCE, MAP, LAMBDA) and no VBA.

## 1\. The Physical Model: From a Continuous Curve to Hinged Segments

Divide a span of horizontal length \\({L}\\) into \\({N}\\) straight rigid elements connected by frictionless hinges (nodes \\(i = 0, 1, 2, \dots, N\\)).

![01-discrete-span](https://cdn.hashnode.com/uploads/covers/6ad6959b6cdc3c187829e973/36897a2d-3502-4f19-b9c0-da587a6fb278.png align="center")

*Discretized wire model: N rigid straight segments connected by frictionless hinges under self-weight and concentrated loads*

The model rests on four assumptions:

- **Equal horizontal steps.** Every segment has the same horizontal projection \\(\Delta x_i = L/N\\).
- **Zero bending stiffness.** The nodes are ideal hinges with zero bending moment. For a flexible stranded conductor this is close to reality.
- **Constant horizontal tension.** As shown in the first article of the series, [State Equation of a Freely Suspended Wire](https://pavelbarysenka.hashnode.dev/state-equation-of-a-freely-suspended-wire), without horizontal external loads the horizontal component \\({H}\\) is the same in every segment.
- **Combined vertical load.** Each segment \\({i}\\) carries its own weight (projected, \\(q \cdot \Delta x_i\\), or along the true length, \\(q \cdot \Delta L_i\\)) plus any concentrated vertical force \\(F_i\\) at that location.

## 2\. Step-by-Step Equilibrium: No Matrices, No Iterations

With \\({H}\\) known, there is no stiffness matrix to invert and no solver to run. We march from the left support \\(A(0, y_A)\\) to the right support \\(B(L, y_B)\\).

### Step 1: Support Reactions Ra and Rb

Taking moments of all vertical loads about support \\({B}\\), and then about support \\({A}\\), with the height difference of the supports included, gives the vertical reactions:

$$R_a = \frac{1}{L}\left[\sum_k F_k (L-x_k) + W_{wire}\frac{L}{2} - H(y_B-y_A)\right]$$

$$R_b = \frac{1}{L}\left[\sum_k F_k x_k + W_{wire}\frac{L}{2} + H(y_B-y_A)\right]$$

Check:

$$R_a + R_b = \sum_k F_k + W_{wire}$$

where

- \\({L}\\): horizontal span length, m;
- \\(y_A\\), \\(y_B\\): heights of the left and right supports, m;
- \\({H}\\): horizontal component of tension, N;
- \\(F_k\\): concentrated vertical load \\({k}\\) at horizontal distance \\(x_k\\), N;
- \\(W_{wire}\\): total self-weight of the wire in the span, N.

Remember the \\(W_{wire} \cdot L/2\\) term. It assumes the wire's center of gravity is at mid-span, and Section 5 shows what that costs.

### Step 2: Equilibrium of the i-th Segment

Now isolate the \\({i}\\)\-th segment between nodes \\(\mathit{(i-1)}\\) and \\({i}\\).

![02-segment-equilibrium](https://cdn.hashnode.com/uploads/covers/6ad6959b6cdc3c187829e973/f99cada2-3c47-4cdc-8d80-fc08ad87ee0c.png align="center")

*Free-body diagram of the i-th wire segment and moment balance about node i*

At the left hinge (\\(\mathit{i-1}\\)), the rest of the wire pulls the segment with the horizontal force \\({H}\\) and the upward vertical force \\(R_{\mathit{(i-1)}}\\). The segment itself carries the total downward load \\(Q_i\\): its own weight plus any point load \\(F_i\\).

From node (\\(\mathit{i-1}\\)) to node \\({i}\\) the vertical internal force drops by \\(Q_i\\). To get the vertical drop \\(\Delta y_i\\), write the moment balance about the right hinge \\({i}\\), with \\(Q_i\\) applied at the segment midpoint (arm \\(\Delta x_i / 2\\)).

Vertical force balance:

$$R_i = R_{\mathit{(i-1)}} - Q_i$$

Moment balance about node \\({i}\\):

$$\sum M_i = 0: \quad H \Delta y_i - R_{\mathit{(i-1)}} \Delta x_i + Q_i \frac{\Delta x_i}{2} = 0$$

Vertical drop and node elevation:

$$\Delta y_i = \frac{\Delta x_i}{H}\left(R_{\mathit{(i-1)}} - \frac{Q_i}{2}\right), \qquad y_i = y_{\mathit{(i-1)}} - \Delta y_i$$

True length of the segment:

$$\Delta L_i = \sqrt{\Delta x_i^2 + \Delta y_i^2}$$

where

- \\({i}\\): segment and right-node index, from 1 to \\({N}\\); \\(y_0 = y_A\\), \\(R_0 = R_a\\);
- \\(\Delta x_i\\): horizontal step \\(L/N\\), m;
- \\(\Delta y_i\\): vertical drop across segment \\({i}\\), m;
- \\(Q_i\\): total vertical load on segment \\({i}\\) (self-weight plus \\(F_i\\)), N;
- \\(R_i\\): vertical component of wire tension at node \\({i}\\), N;
- \\(\Delta L_i\\): inclined length of segment \\({i}\\), m.

The algorithm is recursive. Starting from the left boundary (\\(y_0 = y_A\\), \\(R_0 = R_a\\)), each \\(y_i\\) comes directly from the previous node, so there is no simultaneous system to solve.

## 3\. Three Successive Approximations and Elastic Elongation

In the previous article, the difference between the parabola and the hyperbolic cosine came down to one question: is the wire weight distributed along the horizontal projection (\\({dx}\\)) or along the true arc length (\\({dL}\\))?

The segment method covers both cases with successive approximations:

- **0th approximation, projected weight.** The segment slopes are unknown at the start, so each segment weighs \\(q \cdot \Delta x_i\\): \\(Q_i^{(0)} = q \cdot \Delta x_i + F_i\\). One pass gives the drops \\(\Delta y_i^{(0)}\\) and lengths \\(\Delta L_i^{(0)}\\).
- **1st approximation, true length weight.** With the inclined lengths known, the segment weight becomes \\(Q_i^{(1)} = q \cdot \Delta L_i^{(0)} + F_i\\). Steep segments near the supports are longer than \\(\Delta x_i\\), so they are heavier. The same formulas give new \\(y_i^{(1)}\\) and \\(\Delta L_i^{(1)}\\).
- **2nd approximation, refined weight.** One more pass with \\(Q_i^{(2)} = q \cdot \Delta L_i^{(1)} + F_i\\) fixes the final geometry \\(y_i^{(2)}\\) and the strained wire length \\(L = \sum \Delta L_i^{(2)}\\).

A quick check for readers: set all point loads to zero (\\(F_k = 0\\)) in the workbook. The 0th approximation then reproduces the analytical parabola, and the 1st and 2nd approximations converge toward the cosh curve.

### Elastic Elongation and Unstressed Length L₀

Once the geometry is final, both tension components are known in every segment: the constant \\({H}\\) and the local vertical \\(R_i\\). The axial tension \\(T_i\\) and the unstressed (manufactured) length \\(\Delta L_{0,i}\\) of each segment follow from Hooke's law:

$$T_{i} = \sqrt{H^{2} + R_{i}^{2}}, \qquad \varepsilon_{i} = \frac{T_{i}}{E A}$$

$$\Delta L_{0,i} = \frac{\Delta L_{i}}{1 + \varepsilon_{i}}, \qquad L_{0} = \sum_{i=1}^{N} \Delta L_{0,i}$$

where

- \\(T_i\\): axial tension in segment \\({i}\\), N;
- \\(\varepsilon_i\\): elastic strain of segment \\({i}\\);
- \\({E}\\): equivalent Young's modulus of the conductor, N/mm²;
- \\({A}\\): cross-sectional area of the conductor, mm²;
- \\(\Delta L_{0,i}\\): unstressed length of segment \\({i}\\), m;
- \\(L_0\\): total unstressed length of the wire in the span, m.

## 4\. Stress-Testing the Model in Excel

To make the difference between the approximations visible, I pushed the model into a deep-sag, strongly asymmetric case on the "SEM" sheet of my workbook.

### Input Data (Sheet "SEM")

| Parameter | Value |
| --- | --- |
| Conductor | 243-AL1 (EN 50182) |
| Cross-section \\({A}\\) | 242.5 mm² |
| Young's modulus \\({E}\\) | 55,000 N/mm² |
| Unit weight \\({q}\\) | 6.5835 N/m |
| Span \\({L}\\) | 40 m, \\({N}\\) = 40 segments, \\(\Delta x_i\\) = 1.0 m |
| Support heights | \\(y_A\\) = 10.0 m (left), \\(y_B\\) = 15.0 m (right) |
| Horizontal tension \\({H}\\) | 100 N (deliberately low to get a deep sag) |

Point loads (table addForce):

| Load | \\(F_k\\), N | \\(x_k\\), m |
| --- | --: | --: |
| \\(F_1\\) | 50 | 10 |
| \\(F_2\\) | 10 | 20 |
| \\(F_3\\) | 0 | 30 |
| \\(F_4\\) | 30 | 35 |

### Excel Implementation Without Macros

The algorithm is purely sequential, so it needs neither VBA loops nor Solver. In Excel 365 the whole 40-segment calculation spills from dynamic array formulas built on SEQUENCE, FILTER, MAP and LAMBDA. Three examples.

Node coordinates \\(x_i\\):

```excel
=SEQUENCE(N, 1, L/N, L/N)
```

Point loads \\(F_i\\) mapped to the nodes:

```excel
=MAP(S5#, LAMBDA(x, SUM(FILTER(addForce[F, N], addForce[x, m]=x, 0))))
```

Cumulative vertical force \\(R_i\\):

```excel
=Ra - MAP(SEQUENCE(ROWS(U5#)), LAMBDA(i, SUM(INDEX(U5#, 1):INDEX(U5#, i))))
```

### Results of the Three Approximations

![03-sag-approximations](https://cdn.hashnode.com/uploads/covers/6ad6959b6cdc3c187829e973/a5903707-aa7f-4aea-904a-9d7fe09776b1.png align="center")

*Comparison of the 0th, 1st and 2nd approximations for a 40 m inclined span with four point loads and H = 100 N*

| Parameter | 0th approx. \\(q \cdot \Delta x_i\\) | 1st approx. \\(q \cdot \Delta L_i^{(0)}\\) | 2nd approx. \\(q \cdot \Delta L_i^{(1)}\\) |
| --- | --: | --: | --: |
| Total wire weight \\(W_{wire}\\), N | 263.34 | 368.37 | 396.93 |
| Left reaction \\(R_a\\), N | 165.42 | 217.93 | 232.21 |
| Right reaction \\(R_b\\), N | 187.92 | 240.43 | 254.71 |
| Strained wire length \\({L}\\), m | 55.9532 | 60.2914 | 61.1744 |
| Unstressed wire length \\(L_0\\), m |  |  | 61.1736 |
| Maximum sag \\(f_{max}\\), m | 17.17 | 20.74 | 20.99 |
| Computed right end \\(y_{40}\\), m | 15.45 | 13.76 | 14.17 |
| Right support error vs. \\(y_B\\) | +3.00% | −8.26% | −5.53% |

The gap between the 0th approximation (\\(f_{max}\\) = 17.17 m, \\({L}\\) = 55.95 m) and the 1st and 2nd (\\(f_{max}\\) = 20.99 m, \\({L}\\) = 61.17 m) is large. At \\({H}\\) = 100 N, distributing the weight over the horizontal projection (\\(q \cdot \Delta x_i\\)) underestimates the wire weight in the span by more than 33% (263.3 N vs. 396.9 N).

The method settles fast: between the 1st and 2nd approximations the maximum sag changes by only 0.25 m (20.74 m vs. 20.99 m).

## 5\. Why the Curve Misses the Right Support

The last two rows of the table (rows 44 and 45 on the "SEM" sheet) show a problem. The march starts at \\(y_0 = y_A\\) = 10.00 m, crosses \\({N}\\) = 40 segments and reaches \\({x}\\) = 40 m. The last node \\(y_{40}\\) should land on the right support, \\(y_B\\) = 15.00 m. It does not:

- 0th approximation: \\(y_{40}\\) = 15.45 m (+0.45 m, +3.0%);
- 1st approximation: \\(y_{40}\\) = 13.76 m (−1.24 m, −8.3%);
- 2nd approximation: \\(y_{40}\\) = 14.17 m (−0.83 m, −5.5%).

This is not an error in the workbook. Two simplifications in the textbook form of the method produce exactly these offsets.

### The +0.45 m Offset in the 0th Approximation: Midpoint vs. Node

In the reaction \\(R_a\\), the point loads \\(F_k\\) act at the nodes \\({x}\\) = 10, 20 and 35 m) with their exact arms \\(L-x_k\\). In the segment equation for \\(\Delta y_i\\), the same \\(F_i\\) is part of \\(Q_i = q \cdot \Delta x_i + F_i\\) and is halved. This moves it from the hinge \\(x_i\\) to the segment midpoint \\(x_i-\Delta x_i/2\\). Over the span the half-step shift adds up to:

$$\Delta y_{offset} = \frac{\Delta x \cdot \sum_{k} F_{k}}{2H}$$

With \\(\Delta x\\) = 1.0 m, \\({H}\\) = 100 N and loads of 50, 10 and 30 N:

$$\Delta y_{offset} = \frac{1.0}{2 \cdot 100}(50 + 10 + 30) = 0.45~\mathrm{m}$$

### The −0.83 m Offset in the 1st and 2nd Approximations: Center of Gravity

The 1st and 2nd approximations still use \\(W_{wire} \cdot L/2\\) in the reactions, so the wire's center of gravity stays at mid-span (\\(L/2\\) = 20 m). For a uniform load \\(q \cdot \Delta x_i\\) this is exact. With true segment lengths \\(q \cdot \Delta L_i\\) in an inclined, unevenly loaded span, the segments on the steeper side are longer and therefore heavier, and the center of gravity moves away from \\(L/2\\). Keeping \\(L/2\\) in \\(R_a\\) underestimates the starting reaction, and the curve ends below \\(y_B\\) = 15.00 m.

## 6\. Conclusions

1. **At working tensions the boundary error is negligible.** \\({H}\\) = 100 N is an extreme case: the sag exceeds half the span, which magnifies every nonlinear effect. With a realistic \\({H}\\) = 10,000 N (10 kN) on the "SEM" sheet, the right support error drops below 0.03% (about 4 mm), and the three approximations practically coincide.
2. **Where the method fits:**
   - spans with point loads (droppers, insulators, weights), inclined supports and a known horizontal tension \\({H}\\);
   - a fast, transparent spreadsheet model with no iterations, VBA or FEM software;
   - an independent check of complex spans.

The workbook is free to download: [Excel file, sheet "SEM"](https://drive.google.com/file/d/1VfJKbFGuSekAAyWV5TXwbNKNEFo7N0yh/view?usp=sharing). Two experiments worth trying on the "SEM" sheet:

- raise \\({H}\\) from 100 N to 5,000 N and watch the three approximations and the right support error converge;
- set all point loads in the addForce table (D3:D6) to 0 and compare the 0th and 2nd approximations with the parabola and cosh calculator from the previous article.

## Limits and the Next Step

The method handles uneven loading well, but it has two limits. It needs the horizontal tension \\({H}\\) in advance, and its simplified reaction formula drifts at the right support when the sag is deep and asymmetric.

In real design \\({H}\\) is often the unknown, while the unstressed wire length \\(L_0\\) is fixed. The next article in the series moves to the iterative nodal (tangent) method, which solves for exact equilibrium at every node for a given unstressed length.