Presented by: Civil Engineering Academy

Basic Circular Curve ElementsPresented by: Civil Engineering Academy

Circular curves are used in pipeline

alignments and for horizontal alignment of

highways.

C long chord (chord PC to PT)

L length of the curve (the length of the curve

from the PC to the PT)

T (semi-) tangent distance from V to PC or

from V to PT

R radius of the curve

PI point of intersection of back and forward

tangents

I intersection angle; central angle of curve;

deflection angle between back and

forward tangents; interior angle

M middle ordinate (the distance from the

midpoint of the curve to the long chord)

E external distance (the distance from the

vertex to the midpoint of the curve)

Fig. 79.1

arc definition, Eq. 79.1, CERM

chord definition, Eq. 79.2, CERM

Radius Of The Curve

Length Of The Curve

Eq. 79.3, CERM

Eq. 79.4, CERM

Tangent Distance

arc basis Eq. 79.8 CERM

chord basis Eq. 79.9 CERM

Eq. 79.7 CERM

Eq. 79.6 CERM

Eq. 79.5 CERM

Long Chord (Chord PC To PT)

Middle Ordinate

External Distance

Degree Of Curve

Tangent offsets are useful in laying out circular

curves.

The degree of curve is another parameter of

a circular curve that can be given in lieu of

radius.

For the arc definition:

For the chord definition:

Stationing along a circular curve is along its arc.

The angle of intersection of a circular curve is 53°40’ and its radius is

2600 ft, if PC is at sta. 35 + 590 , determine the right angle offset

distance from the tangent passing thru the PC to station 35 + 920 on

the curve.

Solution:Given :

R = 2600 ftI = 53°40’

P.C. = 35 + 590

Unknown tangent @

Sta. 35 + 920

Draw the problem

according to given

data.

2π(2600 ft)(53°40’)

360°

L = 2435.31 ft

l = 920 – 590 = 330ft

330 ft = 2π(2600 ft)(Ɵ)

360⁰

Ɵ = 7.27⁰

Cos(7.27 ) = 2600ft – t

2600t = 20.9 ft

=

The tangent of the simple curve has bearings on N75°12’E and

S78°36’E respectively. Determine the central angle of the curve.

Solution:

Draw the tangents and their bearings:From angle postulates:

I = 180 – 75°12’ - 78°36’ = 26°12’

A horizontal circular curve having an intersection angle of 28° is to

have a radius of 1200 ft. Determine the tangent distance, long

chord, middle ordinate, and external distance.

Solution:

T = 1200ft x tan(28°/2) = 299.2 ft

C= (2 x 299.2 ft) cos(28°/2) = 580.63 ft

For tangent distance, t:

For long chord , C:

For middle ordinate, M:

M= (580.61/2) tan (28°/4) = 35.65 ft

For external distance, E:

E= 1200 ft x tan(28°/2) x tan (28°/4) =36.74 ft

The figure represents a straight railroad spur that intersects the curved

highway route AB. Distances on the route are measured along the arc.

Applying the recorded data, determine the station of the intersection point P.

AV= T= 800 tan (108 °/2) = 1101.1 ft AM= 1101.1 - 220 = 881.1 ft AN=AM tan 28° = 468.5 ft ON= 800 - 468.5 = 331.5 ftOP = R = 800 ft ONP = 90° + 28° = 118°

Applying trigonometric relationships:

Solution:

Solve triangle ONP:sin (OPN)= sin ONP(ON)/(OP)OPN= 21°27.7‘ = 21.462°

AOP = 180° - (118° + 21.462°) =

40.538°

Arc Dist of AP = 2π(800)(40.538°)/360° = 566.02 ft

Station of P = (22 + 00) + (5 + 66) = 27 + 66.02 ft

Note: Law of Sines Review:Sin(a)/a = Sin(b)/b=Sin(c)/c

Eq. 6.57

Note: For any triangle there are 180°

CERM Ex. 79.2

More practice problems!

Next topic: Vertical Curves!