--- title: "Position of maximum sag, unequal supports" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:max-sag-position-unequal-levels" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/max-sag-position-unequal-levels site: BCT Engineering --- # Position of maximum sag, unequal supports Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2078 Kartik, 6 marks; 2074 Bhadra, 6 marks; 2071 Bhadra, 6 marks; 2071 Magh, 6 marks (asked 4 times) - Video: [Lecture#30: Sag Calculation When Supports are at Equal Level and Unequal Levels | Dr. J. A. Laghari](https://www.youtube.com/watch?v=U6FkyPhCpXg) — Dr. J. A. LAGHARI - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/max-sag-position-unequal-levels - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Determine the position of maximum sag from either supporting towers of equal height at a different elevation from a common reference. Also, explain the condition of virtual sag. ## Short answer The lowest point is at x1 = L/2 - (Th)/(wL) from the lower support and x2 = L/2 + (Th)/(wL) from the higher. If h > wL^2/(2T), x1 becomes negative, the parabola's minimum lies outside the span, and the sag to that imaginary point is virtual sag; the conductor then exerts uplift on the lower tower. ## Answer When the two supports are at different heights, the lowest point of the conductor is not at mid-span. It shifts towards the lower support. Its position is found from the parabolic approximation. #### Derivation Let: - $L$ = span (horizontal distance between supports A and B) - $h$ = difference in level between the supports (B higher than A) - $w$ = weight of conductor per unit length, and $T$ = horizontal tension - $x_1$, $x_2$ = horizontal distances of the lowest point O from A (lower) and B (higher), so $x_1 + x_2 = L$ ```text B /| S2 / | A / | h |\ / | S1 | \_____ O _____/ .......| |<- x1 ->|<----- x2 ---->| ``` Taking O as the origin, the conductor is the parabola $y = \dfrac{w x^2}{2T}$. So: $$ S_1 = \frac{w x_1^2}{2T}, \qquad S_2 = \frac{w x_2^2}{2T} $$ $$ \begin{aligned} h &= S_2 - S_1 = \frac{w}{2T}(x_2^2 - x_1^2) \\ &= \frac{w}{2T}(x_2 + x_1)(x_2 - x_1) = \frac{wL}{2T}(x_2 - x_1) \\ \Rightarrow x_2 - x_1 &= \frac{2Th}{wL} \end{aligned} $$ Solving with $x_1 + x_2 = L$: $$ x_1 = \frac{L}{2} - \frac{Th}{wL}, \qquad x_2 = \frac{L}{2} + \frac{Th}{wL} $$ The lowest point (point of maximum sag) is at $x_1$ from the lower tower and $x_2$ from the higher tower. The sags are $S_1 = w x_1^2/2T$ below A and $S_2 = w x_2^2/2T$ below B. Under wind and ice, $w$ is replaced by the resultant loading $w_r$. If the supports are at equal height ($h = 0$), $x_1 = x_2 = L/2$ and $S = wL^2/8T$. #### Virtual sag - If $h > \dfrac{wL^2}{2T}$, then $x_1 = \dfrac{L}{2} - \dfrac{Th}{wL}$ becomes **negative**. The lowest point O of the parabola then lies **outside the span**, beyond the lower support. - The conductor rises all the way from A to B and has no real lowest point inside the span. The sag $S_1$ measured to the imaginary point O is called the **virtual sag**. - This happens on steep hill slopes, which are common in Nepal. The conductor pulls **upward** on the lower tower (uplift). That tower must be designed as a tension tower, or guyed, and the string must be a tension string, because a suspension string would be pulled up. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For supports with level difference h, the lowest point distance from the lower support is: A. x1 = Th/w B. x1 = L/2 - Th/(wL) C. x1 = L/2 D. x1 = L/2 + Th/(wL) Answer: B. The lowest point shifts toward the lower support. (Key point: Position of lowest point) 2. What gives x2 - x1 = 2Th/(wL)? A. h = wL^2/8T B. h = S2 - S1 = (w/2T)(x2^2 - x1^2) with x1 + x2 = L C. h = x1 x2 D. S = wL^2/8T Answer: B. Factorising the difference of squares gives the result. (Key point: Derivation step) 3. When does virtual sag occur? A. When h > wL^2/(2T), so x1 is negative and the lowest point lies outside the span B. When h = 0 C. When T is infinite D. When h < wL/2 Answer: A. The conductor then rises all the way from A to B. (Key point: Virtual sag condition) 4. What is a consequence of virtual sag? A. Sag increases at mid-span B. Conductor pulls upward on the lower tower; it needs a tension tower or guying C. No effect D. Tower carries only compression Answer: B. A suspension string would be lifted. 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