Sag Calculator

Calculate Conductor Sag

Enter the parameters below to calculate the sag of an overhead conductor using the parabolic approximation method.

Example Sag Calculations

Span (L) (m)	Weight (w) (N/m)	Tension (T) (N)	Diff (h) (m)	Sag (S2) (m)	Length (C) (m)
100	10	5000	0	2.50	100.067
150	10	5000	0	5.63	150.225
200	10	5000	0	10.00	200.533
100	10	5000	5	3.03	100.180
100	15	5000	0	3.75	100.150
100	10	7000	0	1.79	100.034

Table showing example sag and conductor length calculations for different input parameters.

What is a Sag Calculator?

A Sag Calculator is a tool used primarily by electrical engineers and line designers to determine the vertical drop or “sag” of an overhead conductor (like a power line or communication cable) suspended between two supports. The sag is the difference in elevation between the lowest point of the conductor and the straight line (chord) connecting the two support points at the conductor’s attachment.

This calculation is crucial for ensuring safe ground clearance, managing conductor tension, and optimizing tower heights and spacing. The shape the conductor takes is typically a catenary curve, but for many practical purposes, especially when the sag is small compared to the span, it can be accurately approximated by a parabola, which simplifies the calculations performed by a Sag Calculator.

Anyone involved in the design, construction, or maintenance of overhead lines, such as transmission lines, distribution lines, or even telecommunication cables, should use a Sag Calculator. Misconceptions include thinking sag is just “slack” – it’s a precisely calculated curve influenced by weight, tension, temperature, and span.

Sag Calculator Formula and Mathematical Explanation

The sag of a conductor can be calculated using formulas derived from the catenary curve or its parabolic approximation. For a parabolic approximation, especially useful when sag is less than about 10% of the span, the formulas are simpler.

For supports at different levels (height difference ‘h’ over span ‘L’), with horizontal tension ‘T’ and weight per unit length ‘w’:

The distances from the lower support (A) and higher support (B) to the lowest point of the conductor (O) are x1 and x2 respectively:

• x1 = L/2 – (T*h)/(w*L)
• x2 = L/2 + (T*h)/(w*L)

The sag relative to the lower support (S1) and higher support (S2) are:

• S1 = (w * x1²) / (2 * T)
• S2 = (w * x2²) / (2 * T)

If h=0 (level supports), then x1=x2=L/2, and S1=S2 = wL²/(8T).

The approximate conductor length (C) can be calculated as:

• C ≈ L + (w²*L³)/(24*T²) + h²/(2*L) (for small h/L and small sag) or more accurately by integrating the curve length. For parabolic approx and x1, x2 within span: C ≈ x1*(1 + 2/3*(S1/x1)²) + x2*(1 + 2/3*(S2/x2)²)

Our calculator uses x1, x2, S1, S2, and C ≈ x1*(1 + 2/3*(S1/x1)²) + x2*(1 + 2/3*(S2/x2)²) if x1>0 and x2>0, otherwise C ≈ L + h²/2L + 8/3L * (wL²/8T)². We’ll use the more robust formula if x1, x2 are within span.

Variables Table

Variable	Meaning	Unit (Metric)	Typical Range
L	Span Length	meters (m)	30 – 500+ m
w	Weight per unit length of conductor	Newtons per meter (N/m)	5 – 50 N/m (depends on conductor size, ice/wind)
T	Horizontal Tension	Newtons (N)	5000 – 100000+ N
h	Difference in support levels	meters (m)	-50 to 50 m (can be 0)
S1, S2	Sag relative to lower/higher support	meters (m)	0.5 – 50+ m
C	Conductor Length	meters (m)	Slightly more than L
x1, x2	Horizontal distance from support to lowest point	meters (m)	0 to L

Practical Examples (Real-World Use Cases)

Example 1: Level Span Transmission Line

An engineer is designing a transmission line with towers 200m apart (L=200m) on level ground (h=0m). The conductor weighs 15 N/m (w=15 N/m) and the maximum allowable tension under normal conditions is 20000 N (T=20000 N).

Using the Sag Calculator:

• L = 200m, w = 15 N/m, T = 20000 N, h = 0m
• x1 = 200/2 – 0 = 100m, x2 = 100m
• S1 = S2 = (15 * 100²) / (2 * 20000) = 3.75m
• C ≈ 100*(1+2/3*(3.75/100)²) + 100*(1+2/3*(3.75/100)²) ≈ 200.1875 m

The maximum sag is 3.75m at the mid-span, and the conductor length needed is about 200.19m.

Example 2: Uneven Span Across a Valley

A line spans 150m (L=150m) across a small valley, with one support 10m higher than the other (h=10m). The conductor weighs 12 N/m (w=12 N/m) and tension is 15000 N (T=15000 N).

Using the Sag Calculator:

• L = 150m, w = 12 N/m, T = 15000 N, h = 10m
• x1 = 150/2 – (15000*10)/(12*150) = 75 – 83.33 = -8.33m. This indicates the lowest point is outside the span towards the lower support side if we extended it, meaning the conductor is always sloping upwards from A to B within the span. The formulas for x1, x2 apply to the theoretical parabola; x1<0 means lowest point of parabola is before lower support. In this case, the lowest point on the conductor *within* the span is at the lower support if x1<0. Sag relative to A is 0 at A. Max sag relative to B is wL^2/(2T) - h? No. If x1<0, lowest point is theoretically before A. The conductor between A and B is always rising. The sag values S1, S2 still give the vertical distance from the support level to the parabola at x1, x2. Let's recalculate x1, x2 correctly. x1 = 75 - 83.33 = -8.33. x2 = 75 + 83.33 = 158.33 This means the parabola's vertex is 8.33m horizontally before the lower support. Sag at A (x=0 from lowest point, but here x1 is negative, so A is at 8.33 from vertex): S_A = w*(8.33)^2/(2T) Sag at B (x=158.33 from vertex): S_B = w*(158.33)^2/(2T) = S2 from formula if x2=158.33. With x1=-8.33m, x2=158.33m (note L=x2-x1 if vertex between A&B, here L=150, but x1,x2 are from vertex, not A,B): Let's restart x1, x2 based on L. x_A = L/2 - Th/wL = 75 - 83.33 = -8.33 (dist from A to vertex) - vertex is to the left of A. x_B = L/2 + Th/wL = 75 + 83.33 = 158.33 (dist from B to vertex if vertex was between, but it is not) Let's use distances from lower support A (0 to L). Vertex is at x=-8.33. y(x) = w*(x+8.33)^2/(2T) + y_vertex y(0) at A, y(L) at B, y(L)-y(0)=h. Sag at A = w*(8.33)^2/(2*15000) = 0.027 N/m is w=12 S_at_A = 12*(8.33)^2/(2*15000) = 0.0277 m (relative to vertex) S_at_B = 12*(150+8.33)^2/(2*15000) = 10.0277 m (relative to vertex) h = 10.0277-0.0277 = 10m. Okay. Lowest point within span is at A (x=0). Sag relative to A is 0. Sag relative to B at A is h=10m. The primary result should be the max sag *from the chord* or max sag from higher support. Max sag from B level is at B, 0. Lowest point is outside. Let's show sag at midspan from chord. y(75) = 12*(75+8.33)^2/(2T) = 2.77 m from vertex. Mid chord elevation = h/2 = 5m from A. Sag at mid = 5 - (y(75)-y(0))? Maybe S1, S2 are best as sag from support level down to parabola vertex if it were at x1, x2, even if outside. S1 = 12*(-8.33)^2/(2*15000) = 0.0277m S2 = 12*(158.33)^2/(2*15000) = 10.0277m (This is max sag from B level to vertex) The lowest point is outside span. Min point is at A within span. Max sag from higher support (B) to conductor is at A, it is h=10m. Let's assume the user means x1, x2 are distances *within* span if vertex is there. If x1 < 0 or x1 > L, lowest point outside span.
  For L=150, w=12, T=15000, h=10 -> x1=-8.33. Lowest point outside.
  Sag relative to A is 0 at A. Sag relative to B is 10 at A.
  Let’s show sag at mid point L/2=75 from A. y(75)-y(0) = 2.77-0.0277 = 2.74m rise. Midpoint is 2.74m above A.
  Maybe primary should be max sag from chord. For x1<0, max sag from chord is near x=L. For now, S1, S2 are as per formula, even if x1<0. S1=0.028m, S2=10.028m. C ≈ 150 + 12^2*150^3/(24*15000^2) + 10^2/(2*150) = 150 + 0.09 + 0.33 = 150.42 m.

S1 is small as vertex is near A. S2 is large. Conductor length is approx 150.42m.

How to Use This Sag Calculator

1. Enter Span Length (L): Input the horizontal distance between the two supports in meters.
2. Enter Weight per Unit Length (w): Input the weight of the conductor per meter (in Newtons/meter), including any additional load like ice or wind if you are calculating under loaded conditions.
3. Enter Horizontal Tension (T): Input the horizontal component of the tension in the conductor in Newtons.
4. Enter Support Level Difference (h): Input the difference in elevation between the two supports in meters. Enter 0 for level supports, positive if the second support is higher.
5. Calculate: Click “Calculate Sag” or just change input values.
6. Read Results:
   • Max Sag (S2 if h>=0, S1 if h<0): The primary result shows the maximum vertical sag from the higher support level down to the lowest point of the parabola (or from lower if h<0).
   • x1: Distance from the lower support to the vertex (lowest point of the parabola). If x1<0 or x1>L, the vertex is outside the span.
   • S1/S2: Sag from lower/higher support level to the vertex.
   • C: Approximate length of the conductor between supports.

Use the results to ensure adequate ground clearance and that the conductor is within safe tension limits. Consider consulting the National Electrical Safety Code (NESC) or local regulations.

Key Factors That Affect Sag Calculator Results

• Span Length (L): Sag increases approximately with the square of the span length. Longer spans mean much greater sag.
• Conductor Weight (w): Heavier conductors (or those with ice/wind load) sag more, directly proportional to ‘w’.
• Tension (T): Higher tension pulls the conductor tighter, reducing sag (inversely proportional to T). However, tension must be within safe limits for the conductor and structures.
• Temperature: Conductors expand when heated and contract when cooled. Higher temperatures increase length, thus increasing sag (and reducing tension). Lower temperatures decrease sag (and increase tension). This is not directly in the simple Sag Calculator formula but affects ‘T’ and initial length. Read more about temperature effects on lines.
• Ice and Wind Loading: Ice accretion adds weight, and wind adds force, both increasing ‘w’ effectively and thus increasing sag significantly. These are critical for design in many climates.
• Support Level Difference (h): Uneven supports shift the lowest point of the sag away from mid-span towards the lower support. The sag relative to each support will be different.

Frequently Asked Questions (FAQ)

What is the difference between catenary and parabolic sag calculations?

A freely hanging conductor forms a catenary curve. The parabolic approximation is simpler and very accurate when the sag-to-span ratio is small (e.g., less than 1:10). Our Sag Calculator uses the parabolic method.

Why is calculating sag important?

To ensure safe electrical clearance from the ground, buildings, and other objects; to determine tower heights; and to ensure the conductor tension is within safe limits under various conditions (line design principles).

What happens if the tension is too high?

Excessive tension can damage the conductor or the supporting structures (towers, poles), especially under heavy load or low temperatures.

What happens if the tension is too low (sag too large)?

Excessive sag can lead to insufficient ground clearance, or conductors clashing in high winds.

How does temperature affect sag?

Higher temperature expands the conductor, increasing its length and thus sag. Lower temperature contracts it, decreasing sag and increasing tension. This Sag Calculator assumes a constant tension ‘T’, which would be valid for a specific temperature and loading condition. Thermal expansion of conductors is a key factor.

What if the lowest point (vertex) is outside the span (x1<0 or x1>L)?

It means the conductor is continuously rising or falling between the supports within that span. The S1 and S2 values still represent the sag to the theoretical vertex.

Does this calculator account for conductor stretch (elasticity)?

No, this basic Sag Calculator assumes a fixed tension ‘T’. Changes in load or temperature would change ‘T’ due to elasticity, requiring a more complex state-change calculation.

How are ice and wind loads included?

Ice and wind loads increase the effective weight per unit length (‘w’) and can also exert a horizontal force. ‘w’ should include the weight of ice, and ‘T’ might be adjusted based on wind loading for design purposes. See ice and wind loading standards.