Design“SagCurve”. Where; Y=thelowestpointofthecurve C=heightofPVCfromDatumwherex=0.Fromtheequationofparabolics,differentiationof.... 𝑌 = 𝑎𝑥! + 𝑏𝑥 + 𝑐 !"

!"= 2𝑎𝑥 + 𝑏 curvegradient

therefore,atPVC,x=0 !"

!"= 𝑏andG1=b(inpercentage)=initialgradient

Rateofslopechanges Fromthecurvegradient,wedothe2nddifferentiationtogettherateofslopechanges. !"

!"= 2𝑎𝑥 + 𝑏

PVC PVT

PVI

C‐G1

+G2

Y=ax2+bx+ccdifferentiation

ParabolicsEquation

Y

Datum

TT

!"!"= 2𝑎,wheretherateofslopechangesfromgeometricdesign

is!!!!!!

Therefore,2𝑎 =

!!!!!!,

a=!!!!!

!!

ExampleA 500-meter equal-tangent sag vertical curve has the PVC at station 100+00 with an elevation of 1000 m. The initial grade is -4% and the final grade is +2%.

Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve Solution WhereT(tangent)=500m

PVC PVT

PVI

1000m ‐4% +2%

Y=ax2+bx+ccdifferentiation

ParabolicsEquation

100+00

Y

x

a.DeterminethestationingatPVIandPVT Weknowthat,thecurvelength,Lforverticalcurvedesignis... L=2T, Sothat,PVI=PVC+tangent =(100+00)+500m =(100+00)+(5+00)=105+00stn. PVT=PVC+L =(100+00)+1000m =(100+00)+(10+00)=110+00stnb.DeterminetheelevationofPVIandPVTfromdatum

HeigthofPVI=1000 − !!

!"# 𝑥 5𝑠𝑡𝑛 = 980m

PVC

PVI

T=500m

‐4%1000m

???

Datum

HeigthofPVT=980 + !!!"# 

𝑥 5𝑠𝑡𝑛 = 990mc.Thelowestpointonthecurve Fromtheparabolicsequation,𝑌 = 𝑎𝑥! + 𝑏𝑥 + 𝑐 Find,a=? b=? c=ElevationofPVC=1000m b=G1=‐4%=−4 𝑠𝑡𝑛 a=!!!!!

!!= !!(!!)

!(!")!"#where,L=2T=2(500)

1000m=10stn=6 20𝑠𝑡𝑛=0.3/stn Fromtheequationofgradient,!"

!"= 2𝑎𝑥 + 𝑏

When!"

!"= 0,2ax+b=0

2(0.3/stn)x+(‐4)=0

PVT

PVI

T=500m

+2%

980m

???

Datum

2(0.3/stn)x+(‐4)=0

!.!!!"#

− 4 = 00.6x=4stnx=6.67stn=6.7stn=670m=6+70stn

So,thedistanceofthelowestpointofcurveis... DistpointfromPVC=PVC+x =(100+00)+(6+70) =106+70stn Thelowestpointofcurve,

Y =ax2+bx+c= 0.3 𝑠𝑡𝑛 6.7! 𝑠𝑡𝑛 + −4/𝑠𝑡𝑛 6.7𝑠𝑡𝑛 +  1000m=986.6m