### Discussion :: Highway Engineering - Section 1 (Q.No.12)

Priya said: (Sep 30, 2013) | |

Directly proportional to the velocity and also inversely proportional to the radius of curvature. |

Sagar Shah said: (Oct 11, 2013) | |

Is it they are considering directly proportional to square of velocity? |

Srikanth Iiitbasar said: (Dec 23, 2013) | |

Actually it is not directly proportional to width of pavement. Because, Superelevation (e) = tan (tita) = sin (tita) = opp side /hyp= E/ width of pavement. |

Joce said: (Feb 15, 2014) | |

It is directly proportional to the velocity square and inversely proportional to the radius of curvature. |

Tanmoy Saha said: (Jul 1, 2014) | |

Superelevation(E) = Rate of superelevation(e)* Width of pavement(B). So E directly proportional to B, So the correct option will be 'B' . |

Pawan said: (Sep 24, 2014) | |

Equation e + f = v2/127 R. Thus e directly proportional to velocity and inversely proportional to radius of curvature which is right. So incorrect answer is option C inversely proportional to acceleration due to gravity. |

Bcr said: (Sep 30, 2014) | |

e+f=(v*v)/gR So directly prop to v. Indirectly prop to radius and gravity. Hence option A would be the answer. |

Bharathi said: (Nov 18, 2014) | |

Super elevation is directly proportional to the square of the velocity and inversely proportional to the acceleration due to gravity, radius of the curvature. |

Prasanna said: (Jun 11, 2015) | |

I think answer should be option (a), because question asking for incorrect statement. |

Padmini said: (Aug 13, 2015) | |

e = WV^2/gR. V^2 is direct proportional. |

Sabariii said: (Aug 24, 2015) | |

No @Mr. Prasanna, Total super elevation = e*W. Where e - rate of SE. W - Width of pavement. |

Bhagchad Golhani said: (Jul 6, 2016) | |

e is directly proportional to v^2 and also inversely proportional to R. |

Snehal said: (Aug 18, 2016) | |

e is the rate of superelevation. h = e * B is the super elevation (which is proportional to width). |

Pakeeza Jhan said: (Aug 22, 2016) | |

Super-elevation = v^2/2r. Hence, it should be directly proportional to square root of velocity and inversely proportional to the radius of curvature. |

Subeer Kumar said: (Sep 29, 2016) | |

Superelevation is directly proportional to velocity of the vehicle because for particular curve on the road radius of curve remains constant for that curve. |

Dambar said: (Oct 22, 2016) | |

Total super elevation E= e * W. |

Lucky Jaswal said: (Dec 22, 2016) | |

directly proportional to velocity and width. Inversely to the radius. It seems wrong in acceleration. |

Pramit said: (Jan 23, 2017) | |

According to me, It's directly proportional to the SQUARE of velocity. |

Engr Asad said: (Mar 10, 2017) | |

W V^2 _____ gr Directly proportional = width and sq of speed (not velocity). Inversely = acceleration due to gravity and radius of the curve. |

Lone Khursheed said: (Mar 14, 2017) | |

E = eB. Here in above relation; e= Rate of super elevation, And E = total super elevation. Therefore total super elevation is directly proportional to width B. But rate of super-elevation that is e is not proportional to width as the product of e and B is equal to E. So, option A is incorrect. |

Bikash Kabiraj said: (May 16, 2017) | |

V^2/gr. v-velocity of the vehicle, g-gravitational force, r-radius of the curve so right answer is option A. |

Surya said: (Jun 22, 2017) | |

I think option A is the correct answer. |

Sameer Naik said: (Jun 28, 2017) | |

1) e=E/w. 2) e+f=v^2/gR. Answer A is exactaly correct. |

Bicky Agarwal said: (Jul 21, 2017) | |

Here, e+f=v^2/gR. So, A will be answer. |

Sabin Acharya said: (Aug 2, 2017) | |

E=eB. e=v*v/gR. SO, E=v*v*B/gR. From final equation. E directly proportion to width, square of velocity and, inversely proportional to the g and R. SO, ans B IS RIGHT. |

Amit said: (Aug 3, 2017) | |

The correct answer is A. |

Amit said: (Aug 3, 2017) | |

A is the right answer. |

Kajal Das said: (Aug 11, 2017) | |

I think opt A is the actual incorrect statement. Because from the formulae E= e (rate of super elevation) * B (width of pavement) for designing super elevation, for a perticular pavement width B is same but rate of superelevation can be changed. So width is not proportional to value of superelevation (E). Again superelevation is provided for counteract centrifugal force which is ev^2/gR. So opt C is correct statement. So. Who answered opt A I will go with them. |

Satish said: (Aug 13, 2017) | |

Basically super elevation is provided to counter the centrifugal force and to reduce the tendency of fast moving vehicles to overturn. If super elevation is less or zero vehicles will have to reduce their speeds to safely turn it. Hence answer B is correct. |

Misganaw said: (Aug 15, 2017) | |

Actually, incorrect answer is opt A because when the width is extra large we need not provide super elevation and when the width is narrow we should increase super elevation to provide safe passage of vehicles. |

Rupam said: (Sep 15, 2017) | |

Please, anyone explain me how's answer B? |

Srinu said: (Sep 20, 2017) | |

e=v2/gR, So answer A is correct because question is asking incorrect statement, but superelevation is directly proportional to the square of velocity and inversely proportional to acceleration due to gravity and radius of curvature. |

Kshitij Mishra said: (Oct 6, 2017) | |

As super elevation e+f=v^2/ (g.r) so it is directly proportional to the design speed option b is incorrect because the speed of vehicles is mentioned there in the statement not the design speed of the vehicle will differ for the different vehicle so it is incorrect. Because s.e can be made on design speed. |

Vijay Rajput said: (Feb 9, 2018) | |

Super elevation =V*2/gR=F/W<=b/2h. So, the answer is b. |

Rudresh Meena said: (Apr 4, 2018) | |

A is correct because velocity of vehicle is directly proportional to super elevation. |

Rudresh Meena said: (Apr 4, 2018) | |

A is correct because velocity of vehicle is directly proportional to super elevation. |

Rudresh Meena said: (Apr 4, 2018) | |

A is correct because velocity of vehicle is directly proportional to super elevation. |

Santosh Kumar said: (Jun 3, 2018) | |

e=V2/gr. So, I think all option is true. But, e is directly proportional to the square of the velocity of the vehicle. |

Utsav said: (Dec 31, 2018) | |

Directly proportional to velocity square I guess that's why it's incorrect here! |

Swar said: (Jan 9, 2019) | |

e=tan (θ) = opp/adj = h/b = central rise/width. So super elevation is inversely proportional to width of pavement. But they have given as directly prop. So the answer is A. |

Pratibha Nishad said: (Feb 18, 2019) | |

Directly proportional to the velocity and also inversely proportional to the radius of curvature. |

Anjan said: (May 23, 2019) | |

e+f = (0.278V)'2/9.8R. So, option A is correct. |

Pari.. said: (Aug 11, 2019) | |

Super elevation is inversely proportional to the radius. E!=e B where e is the super-elevation. |

Jp Mahanta said: (Jan 18, 2020) | |

Super elevation doesn't depends on road width, The E value (height of outer edge of road at super-elevation) depends upon e (rate of super-elevation). |

Benjifanai said: (Mar 6, 2020) | |

Guys. Question is actually trick you. B is Right. E=eB. e=v*v/gR. SO, E=v*v*B/gR. From the final equation. E directly proportion to width, the square of the velocity and, inversely proportional to the g and R. here, superelevation does not directly proportional to the velocity of vehicles, but the only square of the velocity. SO, answer B IS RIGHT. |

Anomie said: (Jul 31, 2020) | |

e=v^2/Rg so directly proportional to the square of thÃ¨ velocity. |

Shiw Sanku Shaw said: (Aug 28, 2020) | |

How? |

M Ishfaq said: (Sep 15, 2020) | |

It is directly proportionel to quare of velocity not directly proportionl to single velocity. |

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