# 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 .