# Learning Curve Example Problems

Curve Example problems here you will find all about curve with example and problems.

# Definition of Curve with Learning curve Example Problems

A road a railway line or canal etc is tried to make it straight. but due to some hindrance it is required to give it a turn. following can be cause for such turns:
Some in-habitation comes in its way
some obstacle like stream, drain, river or mountain is there
some area is to favored
to avoid much labor of cutting and filling of a project.
In such cases if a turn is required then a curve is given for easiness. It is very dangerous in respect of accidents if such a turn is not provided for a road or railways. This causes a dangerous obstacle. In the same way the water of the canal will act on its banks for such sharp turn and a good deal of oil will accurate inside. This will hinder the flew of water in the canal. This damage can be over powered by giving the canal a curved turn.
Must Read this post to know Super Elevation Formula Derivation And Explanation

## Learning curve Example Problems

First of all you have some data for solution
Like you have deflection angle ,change intersection point ( I.P)
Degree of curve if YOU HAVE THAT DATA YOU CAN SOLVE THE CURVE OK LETS START
If you have no Deflection angle you can find on this formula
Deflection angle = 0 = 180 – Intersection angle
WE HAVE :

GIVEN DATA :

Deflection angle = 400
chainage intersection point =I.P=17634.25
Degree of curve = D = 50
Required Data
Tangent point one= TP1=?
Tangent point two =TP2 =? SOLOUTION
Ist TP1 = change I.P –Tangent length

SO:

First we will find Tangent length
Formula we have
Tangent Length =T.L= R Tan 0/2
we have no “R”
Formula for finding “Radius” 5730/D
D= Degree of curve we have 50 5730/5 =1146
“1146” put this value in tangent length formula Tangent length = 1146 tan 40/2 =417.72
Now we have Tangent length “417.72”
Put this value in this formula
Ist TP1 = chainage I.P –Tangent length
= 17634.25 – 417.72 =17217.13
Tangent point one(TP1) = 17217.13 ft
Now for “TP2”

Formula we have
Tangent point = TP1 + Curve Length
First we find curve length
Formula we have
Curve Length = 3.14 R 0/180
3.14*1146/180 =800.16
Put this value in this ”800.16” formula
Tangent point = TP1 + Curve Length
=17217.13+800.16 = 18017.29 ft
Now we can find curve ordinate
On these formulas
For middle ordinate
For 1st ordinate we have formula
R = Radius of curve
X = R .D
L = Length of curve

I hope after reading this post , now you will completely understood the curve formula explanation . However if you any questions or suggestion then you can use the below commenting box .