--- title: "Vertical curve 35 m headlight and momentum grade" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-80c-vc-headlight-35m" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-vc-headlight-35m site: BCT Engineering --- # Vertical curve 35 m headlight and momentum grade Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2080 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-vc-headlight-35m - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question A vertical curve of length 35 m is designed to connect a mild ascending gradient of 3% to a steep ascending gradient of 9.5% based on headlight sight distance condition (neglecting effect of gradient). i) Calculate the design speed of the highway section. ii) Determine the maximum allowable length of the steeper gradient to design it as a momentum grade with an exit speed of 15 Kmph. Assume the head light beam angle and height of the headlight from the road surface 1 degree and 0.75 m respectively and take coefficients of longitudinal friction and inertia as 0.35 and 0.85, respectively. ## Short answer i) Using headlight sight distance with L=35 m and N=0.065, S=39.7 m; solving SSD equation gives design speed ≈36 km/h. ii) Momentum grade length from v1=10 m/s to v2=4.17 m/s is 8 m. ## Answer **Given:** valley curve, $L = 35$ m, $g_1 = +3\%$, $g_2 = +9.5\%$, headlight height $h = 0.75$ m, beam angle $\alpha = 1^\circ$, $f = 0.35$, coefficient of inertia $\gamma = 0.85$, exit speed 15 km/h. $$ N = 0.095 - 0.03 = 0.065 $$ #### (i) Design speed The curve is designed for headlight sight distance (gradient effect neglected), reaction time $t = 2.5$ s. First find $S$. Try $S L$, so not valid. Hence $S > L$: $$ L = 2S - \frac{2\,(h + S\tan\alpha)}{N} $$ $$ 35 = 2S - \frac{2\,(0.75 + 0.017455\,S)}{0.065} = 1.4629\,S - 23.077 \ \Rightarrow\ S = 39.70\ \text{m} $$ $S = 39.7 > L = 35$ m, consistent. This is the stopping sight distance: $SSD = 0.278\,V\,t + \dfrac{V^2}{254 f}$ $$ 0.695\,V + \frac{V^2}{88.9} = 39.70\ \Rightarrow\ V = 36.07\ \text{km/h} $$ **Design speed $\approx$ 36 km/h** (check: SSD at 36 km/h $= 25.0 + 14.6 = 39.6$ m). #### (ii) Momentum grade The vehicle enters the steeper grade (9.5%) at $v_1 = 36/3.6 = 10$ m/s and must leave at $v_2 = 15/3.6 = 4.17$ m/s. Using $$ L = \frac{\gamma\,(v_1^2 - v_2^2)}{2g\,(G+f)} $$ with $G = 0.095$, $f = 0.35$ (taken as the resistance coefficient given): $$ L = \frac{0.85\,(10^2 - 4.17^2)}{2(9.81)(0.095+0.35)} = \frac{0.85\times82.6}{8.731} = 8.05\ \text{m} $$ **Answer:** design speed $\approx$ 36 km/h; maximum length of the steeper (9.5%) gradient as a momentum grade $\approx$ 8 m. (The value is small because the given friction coefficient 0.35 is large.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Grades +3% and +9.5% give which kind of vertical curve? A. Summit curve B. Valley curve, with N = 0.065 C. Reverse curve D. Transition curve Answer: B. The grade increases, so it is a valley curve, N = 0.095 - 0.03 = 0.065. (Key point: Type of curve) 2. Trial S < L gave S = 39.3 m, which is more than L = 35 m. What follows? A. Accept 39.3 m B. The assumption fails; use the S > L formula giving S = 39.7 m C. Use the comfort formula D. Increase L to 39.3 m Answer: B. S > L requires L = 2S - 2(h + S tan alpha)/N, giving S = 39.7 m. (Key point: S versus L check) 3. S = 39.7 m is set equal to the stopping sight distance. What is found? A. Design speed of about 36 km/h B. Design speed of about 80 km/h C. Design speed of about 15 km/h D. Design speed of about 60 km/h Answer: A. Solving 0.695V + V^2/88.9 = 39.7 gives V = 36 km/h. (Key point: Speed from SSD) 4. Which formula is used for the length of the momentum grade? A. L = gamma(v1^2 - v2^2)/(2g(G + f)) B. L = v^2/(2g f) C. L = N S^2/9.6 D. L = (v1 - v2)/G Answer: A. The kinetic-energy loss, with inertia coefficient gamma, is spent on climbing against G + f. (Key point: Momentum grade length formula) 5. With v1 = 10 m/s, v2 = 4.17 m/s, G = 0.095, f = 0.35, gamma = 0.85, the length is about A. 80 m B. 30 m C. 8 m D. 0.8 m Answer: C. L = 0.85 x 82.6/(2 x 9.81 x 0.445) = 8.05 m, small because f is large. 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