--- title: "Minimum clearance, towers at 45 m and 70 m" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:min-clearance-towers-45m-70m" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/min-clearance-towers-45m-70m site: BCT Engineering --- # Minimum clearance, towers at 45 m and 70 m Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2067 Mangsir (old course), 6 marks - Video: [Problem based on sag, two towers at different heights. Find the clearance between the conductor.](https://www.youtube.com/watch?v=Qpt5FUfdRTE) — BE Technical - 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/min-clearance-towers-45m-70m - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question A transmission line conductor of weight 0.844 kg/m and span 300 meters is supported from two towers TR1 and TR2 at the height of 45 m and 70 m respectively from a common reference. Find the minimum clearance of the conductor from reference. ## Short answer Minimum clearance is 45 m at lower tower TR1. With tension 4000 kg, the catenary vertex lies 244.9 m outside the span, so the conductor rises continuously and no sag occurs within the span. ## Answer For supports at unequal heights, first find where the lowest point of the conductor lies. If it falls outside the span, the lowest conductor point is at the lower tower. **Data:** $w = 0.844$ kg/m, $L = 300$ m, $h = 70-45 = 25$ m. The tension is not given. **Assume** the conductor is the same as in the companion question (UTS = 8000 kg, factor of safety 2), so $T = 4000$ kg. #### Position of the lowest point Measured from the lower tower TR1: $$ x_1 = \frac{L}{2} - \frac{Th}{wL} = 150 - \frac{4000\times25}{0.844\times300} = 150 - 394.94 = -244.94\ \text{m} $$ $x_1$ is negative, so the vertex lies 244.9 m **outside** the span, beyond TR1. Inside the span, the conductor rises continuously from TR1 (45 m) to TR2 (70 m). ```text TR2 o 70 m __..--` | TR1 __..--`` | vertex 45 m o | (outside) |<---- 300 m --->| ===== reference =================== ``` #### Generality of the result The vertex lies inside the span only if $$ T < \frac{wL^2}{2h} = \frac{0.844\times300^2}{2\times25} = 1519.2\ \text{kg} $$ Any practical working tension is above 1519 kg, since it is usually 20–50% of UTS. So the result does not depend on the exact tension assumed. **Answer: the minimum clearance of the conductor from the reference is 45 m, at tower TR1. The conductor has no dip within the span, because its vertex falls about 245 m outside it (for T = 4000 kg).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The tension was not given. How is it handled? A. The problem cannot be solved B. T is taken as 4000 kg, assuming UTS 8000 kg and factor of safety 2 C. T equals the conductor weight D. T is taken as the UTS, 8000 kg Answer: B. The companion conductor (UTS 8000 kg, FS 2) gives T = 4000 kg. (Key point: Missing data) 2. With x1 = L/2 - Th/(wL) = -244.94 m, what does the negative sign mean? A. The conductor is slack B. The vertex lies outside the span beyond TR1, the lower tower C. The tower TR2 is lower D. The vertex is at mid-span Answer: B. A negative x1 means the lowest point is outside the span, beyond the lower tower. (Key point: Vertex position) 3. What is the minimum clearance from the reference? A. 70 m at TR2 B. 57.5 m at mid-span C. 25 m D. 45 m at TR1 Answer: D. The conductor rises continuously from TR1, so the lowest point in the span is the lower tower, 45 m. (Key point: Minimum clearance) 4. The vertex is inside the span only if T < wL^2/(2h) = 1519 kg. Why does the result still hold? A. Because the span is short B. Practical working tensions (20-50% of UTS) exceed 1519 kg C. Because w is small D. Because h is small Answer: B. Any practical tension is above 1519 kg, so the answer does not depend on the assumed T. 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