### Discussion :: Highway Engineering - Section 1 (Q.No.5)

Venkatesh said: (Sep 10, 2013) | |

Here N = 0.03+0.02 = 0.05. L = 0.38(NV^3)^0.5. L = 84.76MTS APPROX. |

Ram Kumar said: (Nov 18, 2013) | |

Length is 84.16 means, then what s the answer? option A or option C. |

Alik Biswas said: (Jan 21, 2014) | |

We Know, L= 2*(N*v^3/c)^0.5 (Length of valley curve). Here, N = 2%-(-3%) (Algebraic difference in grades). = 5% Or 0.05. v = 100 km/hr or 27.778 m/sec (Design Speed). c = 0.6 m/sec3 (Centrifugal acceleration). Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m. Say,84.60 m ([C] is the correct answer. |

Twizere Kagunga Elie said: (Sep 24, 2014) | |

I don't understand well if N is load or if is median. |

Nitin said: (Jan 2, 2015) | |

N is sum of upward & downward gradient. |

Manikanth said: (Jan 23, 2015) | |

Lenth of valley curve for comfort condition is 2(Nv^3/c)^0.5. |

Tarun said: (May 26, 2015) | |

V = 1km/hr*5/18 is m/s. |

Selvi said: (Aug 12, 2015) | |

L = 0.378(NV3)^1/2. = 0.378(.05*100*100*100)^.5. = 86.4 m. |

Rishabh said: (Oct 4, 2015) | |

L is the length of valley curve. And this formula is applicable in this case only 2(Nv^3/c)^0.5. |

Harpreet Singh said: (Dec 26, 2015) | |

I don't understand. Please explain. |

Nikhil said: (Jan 3, 2016) | |

What is mean by V? |

Santhi said: (Jan 27, 2016) | |

Here v is the design speed of the vehicle. |

Ayebare Junior Jeezy said: (Apr 27, 2016) | |

If V is the speed =100km/hr. Difference in grades = A = (m - n). = (-3 - 2) = 5% of acceleration = 0.6. But V =100 x 1000/60 x 60 = 27.8m/s. L = 2*(aV^3)^0.5. =((2)*0.05 x 27.8^3/0.6)^0.5). = 84.53m. |

Elsadig Ahmed Ali said: (Jul 14, 2016) | |

V = 1000 km/hr = 27.8 m/s. N = 5%. C = .6. L = 2 * ((N * V^3) / C)^.5, L = 2 * ((.05 * 27.8^3) / .6)^.5, L = 2 * ((.05 * 21484.952) / .6)^.5, L = 2 * (1074.2 / .6)^.5, L = 2 * (1790.3)^.5, L = 2 * 42.3. L = 84.6. |

Samm said: (Aug 21, 2016) | |

But N is not a difference in grade, it is deviation angle in radians. Please clarify this. |

Ravindra said: (Aug 21, 2016) | |

@Alik Biswas, Well. Here, 'c' means what? |

Elsadig Ahmed Ali said: (Sep 10, 2016) | |

@Samm. N = |n2 - n1| = |-3 - 2| =|-3 + -2| = |-5| = 5. @Ravindra. C = centrifugal acceleration = 0.6 m/s^2. |

Sarang Mote said: (Sep 25, 2016) | |

L = 2x(NV3/c)"0.5. L = 2*((0.05 * 27.78 * 27.78 * 27.78)/0.6)"0.5. L = 84.53m. |

Adikeri said: (Dec 1, 2016) | |

Not understanding. Explain in detail. |

Doha said: (May 3, 2017) | |

L = (A*V^2)/1300a, L = (5 * 100^2)/1300 * 0.6, L = 64.1m. |

Dame Abebe said: (May 22, 2017) | |

Yes, L =84.60 so option "C" is the answer. |

Sandi said: (Jun 7, 2017) | |

L = 2X(NV3/C)^.5, N = .05, V = 27.78, L = 84.498. |

Vamsi said: (Aug 19, 2017) | |

Answer is C. |

David Bielstein said: (Jan 29, 2018) | |

I thought the centrifugal acceleration at the beginning of a parabolic curve would be 0 m/sec^2, increasing to the maximum acceleration at the point directly above or below the PI, and then back to 0. Is 0.6 m/sec^2 assumed to be the maximum instantaneous centrifugal acceleration? |

Shahbaz Bashir said: (Jan 4, 2019) | |

N is the difference between raising gradient and falling gradient i.e N = n1 - n2 and also falling gradient = negative (-) similarly raising gradient =positive (+). |

Shaguftasultana said: (Jul 2, 2019) | |

N=|-3-2|=5, L=2* {(Nv^3)/C}, =2* 5*(100/3.6)/0.6, = 84.57. |

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