--- title: "S-curve: 4-hr UH to 3-hr UH" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:scurve-4hr-to-3hr-300km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/scurve-4hr-to-3hr-300km2 site: BCT Engineering --- # S-curve: 4-hr UH to 3-hr UH Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2076 Asoj, 10 marks - Video: [Unit hydrograph of different duration using S-curve (4h to 2h) | Hydrology](https://www.youtube.com/watch?v=5d-M4ov836M) — The Study Sphere - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/scurve-4hr-to-3hr-300km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The ordinates of a 4 hr UH of a basin area of 300 km$^2$ measured at 1-hr intervals are 6, 36, 66, 91, 106, 93, 79, 68, 58, 49, 41, 34, 27, 23, 17, 13, 9, 6, 3 and 1.5 m$^3$/s respectively. Obtain the ordinates of a 3 hr UH using the S-curve technique. ## Short answer Construct S-curve by adding 4-h UH lagged every4 h, lag S-curve by3 h, multiply difference by4/3. The3-h UH peaks at113.3 m³/s at4 h; ordinates at1-h intervals:0,8.0,48.0,88.0,113.3,101.3,84.0,72.0,62.7,54.7,44.0,36.0,30.7,25.3,20.0,13.3,12.0,6.7,5.3,0,2.0 m³/s. ## Answer The 4-h UH is given at 1-h intervals (the ordinate is 0 at t = 0 and 6 m$^3$/s at t = 1 h). Check: the sum of the ordinates is 826.5 m$^3$/s, so the volume under the UH is $826.5\times3600 = 2.975\times10^6$ m$^3$. This is 0.99 cm over 300 km$^2$, so the UH is a 1 cm UH. #### Method 1. Build the S-curve: $S(t)=U_4(t)+S(t-4)$ (add the 4-h UH repeatedly, each lagged by 4 h). 2. Lag the S-curve by the new duration of 3 h: $S(t-3)$. 3. The difference is the response to 3 h of rain at the 4-h rate (1 cm per 4 h). Scale it by $4/3$: $$ U_3(t)=\frac{4}{3}\left[S(t)-S(t-3)\right] $$ | t (h) | $U_4$ | $S(t)$ | $S(t-3)$ | $S(t)-S(t-3)$ | $U_3$ (m$^3$/s) | |---|---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | 0.00 | | 1 | 6 | 6 | 0 | 6 | 8.00 | | 2 | 36 | 36 | 0 | 36 | 48.00 | | 3 | 66 | 66 | 0 | 66 | 88.00 | | 4 | 91 | 91 | 6 | 85 | 113.33 | | 5 | 106 | 112 | 36 | 76 | 101.33 | | 6 | 93 | 129 | 66 | 63 | 84.00 | | 7 | 79 | 145 | 91 | 54 | 72.00 | | 8 | 68 | 159 | 112 | 47 | 62.67 | | 9 | 58 | 170 | 129 | 41 | 54.67 | | 10 | 49 | 178 | 145 | 33 | 44.00 | | 11 | 41 | 186 | 159 | 27 | 36.00 | | 12 | 34 | 193 | 170 | 23 | 30.67 | | 13 | 27 | 197 | 178 | 19 | 25.33 | | 14 | 23 | 201 | 186 | 15 | 20.00 | | 15 | 17 | 203 | 193 | 10 | 13.33 | | 16 | 13 | 206 | 197 | 9 | 12.00 | | 17 | 9 | 206 | 201 | 5 | 6.67 | | 18 | 6 | 207 | 203 | 4 | 5.33 | | 19 | 3 | 206 | 206 | 0 | 0.00 | | 20 | 1.5 | 207.5 | 206 | 1.5 | 2.00 | The S-curve reaches an equilibrium value of about 207 m$^3$/s, in agreement with $Q_e=2.78A/T_r = 2.78\times300/4 = 208.5$ m$^3$/s. **Answer: The 3-h UH ordinates at 1-h intervals are 0, 8.0, 48.0, 88.0, 113.3, 101.3, 84.0, 72.0, 62.7, 54.7, 44.0, 36.0, 30.7, 25.3, 20.0, 13.3, 12.0, 6.7, 5.3, 0, 2.0 m$^3$/s (t = 0 to 20 h), with a peak of 113.3 m$^3$/s at 4 h. The small ripples at the tail come from rounding of the given ordinates.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The S-curve of the 4-h UH is built using: A. S(t) = U4(t) + S(t-4) B. S(t) = U4(t) - S(t-4) C. S(t) = U4(t) + S(t-3) D. S(t) = 4 U4(t) Answer: A. Lag by the original duration (4 h) and add. (Key point: S-curve rule) 2. The 3-h UH from the 4-h S-curve is: A. (3/4)[S(t) - S(t-3)] B. (4/3)[S(t) - S(t-3)] C. (4/3)[S(t) - S(t-4)] D. [S(t) - S(t-3)] Answer: B. S lagged by 3 h; the difference is 1 cm per 4 h for 3 h, scaled by 4/3. (Key point: 3-h UH formula) 3. At t = 4 h, S = 91 and S(1) = 6. U3(4) is: A. 113.3 B. 85.0 C. 63.8 D. 121.3 Answer: A. (4/3)(91 - 6) = 113.3 m^3/s. (Key point: Computation) 4. The S-curve equilibrium value should match: A. 2.78 A / T_r B. A x T_r C. A / 2.78 D. T_r / A Answer: A. Qe = 2.78 A/Tr = 2.78 x 300/4 = 208.5 m^3/s, close to the 207 obtained. 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