Discussion :: Highway Engineering - Section 1 (Q.No.41)
The absolute minimum radius of horizontal curve for a design speed 60 km ph is
Answer: Option D
No answer description available for this question.
|Gurdeep Singh said: (Aug 31, 2013)|
|R = 60*60/127(.07+.15) = 128.8.|
|Jagu said: (Jan 8, 2014)|
|e+f = V*V / 127 R.
R = V*V/ 127*(0.07+.15).
|Sohan Lal Jangra said: (Jul 8, 2014)|
|Please tell me formula for this?|
|Mukesh said: (Oct 18, 2014)|
|R = v*v/(e+f)*g.|
|Ihsan Baloch said: (Feb 13, 2015)|
|R = V^2/(e+f)g.|
|Rajnish said: (Jan 17, 2016)|
|Guys can you tell me where we have to use 127 and where 225?|
|Neeraj Yadav said: (Feb 10, 2016)|
|If we calculate for signal traffic, we will take 225.|
|Bhavya Jagyasi said: (Feb 13, 2016)|
|If we have to calculate the superelevation for only single vehicle then we use 127 and when we have to calculate superelevation for design speed (for all vehicle which are going to move on curve) we use 225.|
|Rajat said: (May 20, 2016)|
|R = V * V/27.5 = 131 m
So, the answer is A.
|Snehal said: (Aug 19, 2016)|
|You cannot use (e + f) = v^2/127R. It is for super-elevated curves not for horizontal curves.|
|Sahil Pathania said: (Dec 1, 2016)|
|We can use the formula (e+f)=v^2/127R. As curve is horizontal take e=0.
R = 60 * 60/127(0.15) = 189 approx.
Ans is (D) none of these.
|Johny Chauhan said: (Dec 14, 2016)|
|V = √ 27.94 R.
Underroot of (27.94 multiply by 27.94)
|Pramit said: (Jan 23, 2017)|
|Here, the value of V^2/Rg should be under 0.25.|
|Manoj Kabdwal said: (Feb 14, 2017)|
|The e equals v*v/225R put e equation 0.07 find R.
It is 228.57 as this question is saying about design speed so above formula will be used.
|Varsha said: (Mar 1, 2017)|
|0.07 for plain and rolling.
0.10 for the hilly area.
0.04 for urban roads and frequent intersections.
|Amrit said: (Mar 23, 2017)|
|R min= V * V/127(e+F).
= 60 * 60/ 127(0.07+0.15),
May be the correct answer 131m.
|Roop said: (Sep 27, 2017)|
|You are correct @Amrit.|
|Rahul said: (Nov 24, 2017)|
|What is 'F' and 'e'?|
|Suraj Giri said: (Nov 30, 2017)|
|R = 60*60/127(.07+.15) = 128.8.|
|Vijay said: (Dec 2, 2017)|
|Virender Paul said: (Jun 8, 2018)|
|Answer is in meter so we have to convert the speed in m/s.|
|Arsad Alam said: (Oct 31, 2018)|
|Value of radius (R) =226m correct answer. So option (D) right.|
|Muthu said: (Dec 13, 2018)|
|How we can use e is 0.07 and f is 0.15?|
|Tina said: (Feb 22, 2019)|
|e= 0.07 & f= 0.15 (both are constant).
e + f= V^2/127R.
Where: R= radius of horizontal Curve.
Note: here V is in kmph.
|Bazid said: (Mar 11, 2019)|
|For hilly road e = 0.10.
For plain e = 0.07 and the value of friction for cross slope f = 0.15.
|Kalke Patil said: (Mar 20, 2019)|
|Thanks for the given information.|
|Kinjal Patel said: (Mar 23, 2019)|
|We can Also used,
R =v * v/0.22 * g.
= 128.89 Approx.
|Anomie said: (Jun 11, 2019)|
|Thanks @Sahil Pathania.|
|Sujeet said: (Aug 11, 2019)|
|Absoulte min radius=( Vmin)^2/((127)(e(max)+.15).
Here Vmin=60, emax= .07 [7% for plain terrain].
So, D is correct.
|Mallesh said: (May 11, 2020)|
|A 2-lane horizontal curve in which radius varies from infinite to some finite value R,
constructed in hilly terrain. The radius R of curve was equal to the absolute minimum radius. If the design speed and minimum speed on the curve is 60 kmph and 40 kmph respectively.
Then calculate the length of this horizontal curve.
|Shyamlal said: (Jun 8, 2020)|
|Equation for super Elevation:
Take e=0.07 & f=0.15,
It get R=72.72m.
So the answer is option D.
|Rafeek said: (Jul 18, 2020)|
|Option D None of above.
The compensate gradients is 4 degree for, so the radius will be 60 x 4 = 120m.
|Sugumar said: (Jul 18, 2020)|
|How can we guess the weather it is a mixed traffic or single traffic. Can anyone explain based on this question?|
|Sanku said: (Nov 5, 2020)|
|The right value is 128.84 meter.|
|Pooja M said: (Feb 8, 2021)|
|Can we use e=(0.75V)*2/gr this formula?
r = (0.75*16.67) * 2/9.81 * 0.07,
r = 227.62.
|Akshay said: (Aug 10, 2021)|
|Factor 225 as kph, so answer 72.
|Deepankar said: (Aug 13, 2021)|
|For absolute minimum or ruling minimum radius we have to use V^2/127(e+f).
Where, V=speed in KMPH, e=given superelevation or according to given terrain condition (as there is none in this case so we HAVE TO assume plain for which it is 7% or 0.07) f = coefficient of lateral friction( 0.15).
R(absolute minimum)= (60)^2/127(0.07+0.15)= 128.84~129m.
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