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)
If you have no Deflection angle you can find on this formula
Deflection angle = 0 = 180 – Intersection angle


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 =?

Learning curve Example Problems

Ist TP1 = change I.P –Tangent length


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 .

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