Part 5 Simple Curves

Prof. Moir D. Haug CE 271 Spring Survey Camp

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PART V

SIMPLE CIRCULAR CURVES

Figure 1: Simple Curve

Assume that AV and VD are two straight portions of a proposed highway and that the curve BC

is to be used as a gradual change of direction between them.

The curve has a constant radius R and is called a simple circular curve.

The curve starts at B and ends at D. ∴ ∡ AB0 = ∡ DCO = 90 o

Point V is referred to as the vertex or Point of Intersection (P.I.) .


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One of the values we must always know, or be able to determine, for any simple curve is the

deflection angle at the vertex. This angle is designated by either I or Δ . In the case of Figure 1,

the survey is assumed to be progressing from A towards D. If this is the case, the deflection

angle at the vertex is as shown.

Figure 2:

I n Figure 2 the quadrilateral BOCV has certain qualities of symmetry, as

follows :

OB = OC because both are radii

∡ VBO = ∡ VCO = 90o and, from this

∡ BVC = 180o - Δ and

∡ BOC = Δ

We can also say that BV = VC. BV and VC are referred to as the subtangent distances and are

abbreviated S.T. or, quite often, T.

Note also that, if we join OV, we create two right-angled triangles, BOV and COV, that are

similar in all respects. We also apply names to point B, where the curve begins, and point C,

where the curve ends. We have three choices and the various names are paired as shown.


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B C

Name Abb. Name Abb.

Beginning of Curve

Point of Curvature

Tangent-Curve Point

B.C.

P.C.

T.C.

End of Curve

Point of Tangency

Curve-Tangent Point

E.C.

P.T.

C.T.

1. Derivation of Formulae

Figure 3:

Referring to Figure 4, on the following page, there are certain dimensions associated with a

simple circular curve that are important and we will derive expressions for them that relate them

to R or T.

In Δ BOV

∡ VBO = 90 o

∡ BVO =

∡ BOV =

= Tan

Δ

BV = B0 TAN

But, BV = T, and

BO = R

…. (1)

T = R Tan


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Figure 4:

The straight line BC is called the Long Chord, abbreviated L.C.

In Δ BEO, ∡ BEO = 90o

= Sin

, or BE = BO Sin

But BE =

and B0 = R

∴

= R Sin

, OR

The line EF is called the Middle Ordinate, abbreviated M.O.

In Δ BE0 and sector BFO

M.O. = OF – OE, but OF = R and

= Cos

but OB = R

∴ OE = R Cos

L.C. = 2 R Sin

….. (2)


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Substituting the values of OF and OE in the original expression for the M.O. we get;

M.O. = R - R Cos

The expression, (1 - Cos

) i s known as the Versine of

and tables giving

values of the Versine are published. This is why we often see the above equation

in the form .

Figure 5:

In the right-triangle OBV and the Sector BFO

E = OV - OF but

= Sec

, or

M.O. = R (1 - Cos

) ..….. (3)

M.O. = R Versine

The length FV is referred

to as the External and is

abbreviated E


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OV = OB Sec

but OB = R, so

OV = R Sec

. also OF = R

The original expression was;

E = OV - OF; if we make the above substitutions then

E = R (Sec

- 1 ) ------------------ (a)

From Equation (1); T = R Ta n

, or

R =

. ,Substitute this value of R in (a) above

E = =

(Sec

- 1), or E = T

E = T

E = T

------------------ (b)

There i s an identity in trigonometry o f the form Ta n

=

I n the case of (b) above it would have the form Tan

=

From this;

E = Tan

..….. (4)


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2. Radius of Curvature and Degree of Curve

There are three methods by means of which we can designate the “sharpness” of curvature of a

curve.

a) Radius

The sharpness of curvature is inversely proportional to the radius, i.e. a reduction in

radius increases the sharpness.

b) Degree of Curve, chord definition – Dc

Degree of curve according to the chord definition, Dc, is defined as the angle subtended

by a chord having a length of one full station or 100 St. in the foot system and a 100

meters in the metric system. This method is followed in railroad practice. The radius of

such a curve may be computed by the following expression:

R =


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Note that the radius of curvature varies inversely as the degree of curve and the radius of

lo curve according to the chord definition is 5729.651 units of measurements.

c ) Degree o f Curve, arc definition – Da

The curvature is expressed by stating the "degree of curve" Da which has traditionally

been defined as the angle subtended a t the centre of the curve by an arc 100 ft. long. In

the metric system, Da , is defined as the angle by a 100 m arc.

Smaller values of D, decreases the sharpness o f curvature. The arc definition for degree

of curve is most frequently followed in highway practice.

When designing a curve, Da, is usually selected on the basis of design speed,

superelevation and road surface friction factor.

Figure 7

If a 100 ft or meter arc subtends an angle of lo , the radius of curve is 5729.578 units. We

refer to this as a "one degree curve” .

=

, where C = 100

Da =

=

; and

R =


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3. Length of Curve

a) Length of Curve Chord Definition - Lc

The length of curve on the chord basis, is the summation of chords which approximate

the curve and, is therefore, an inexact expression. The Lc obtained will always be less

than the true arc length. The difference will increase as Dc, increases.

Figure 8:

b) Length of Curve Arc Definition - La

The length of curve or arc length for corresponding radius varies directly to the central angle

subtended by the arc. See Figure 7.

=

=

;

Δ

This is an exact expression.

Note that the length of curve (on the chords basis) LC, is somewhat less than the actual arc

length La.

This difference will increase as Da or Dc increases.

=

, where C = 100

Δ = Deflection angle for the curve

Lc = 100 x


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The Arc Basis is Used for the Calculations Presented in This Chapter.

Actually both chords and arc methods are used in North America. For long gradual curves,

which are common in railroad practice, the chord basis (arc length considered to be same as

chords) is normally used. For highway curves and curved property boundaries, the arc basis is

more common.

In field practice, field measurements of the curve are laid out with tape, along the chord and not

along the arc, resulting in error. This error can be reduced when the arc basis is used

by using short arc ("chord”) length or applying the difference between the arc length and the

chord length. For example, for a 100 meter 2o curve, the chord length is 99.955 meters and for a

10' curve the chord length is 99.873 meters. The following general rules are suggested in the

metric system:

100 meter arcs "chords up to lo curves

30 meter arcs “chords” up to 4o curves

20 meter arcs "chords" up to 10o curves

10 meter arcs "chords” up to 25o curves

3 meter arcs “chords” up to 100o curves

4. Example Problem #1 – OAK HURST ROAD

Two tangents intersect at Station 4 + 016.770. The deflection angle to the right is 40o 00' 00". It

is decided to design the highway for a maximum speed of 90 km/hr, and using AASHTO*

recommendation for superelevation and friction a minimum radius of 270 meters and a

maximum degree of curve, Da is to be 22o.

Calculate T, La, R and the stationing of the P.C. and P.T. using 20 meter arc length.

* American Association of State Highway and Transportation Officials – “Green Book”

(AASHTO – 42,815 people killed in 38, 309 fatal crashes in 2002, 25% of these crashes occurred on horizontal curves)


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Explanation

When we speak of a station on a curve we are given a specific location from the start of the

survey. In the above problem, the P.I. is given as Stn. 4 + 316.770. What we are saying is that

The P.I. is 4,016.770 m from the start of the survey.

The problem also stated that is Δ is 40o to the right, this says that when we stand at the P.C. and

look in the direction in which the survey is progressing, the curve deflects to the right.

Solution

Given P . I . = Stn. 4 + 016.770

Δ = 40 o

20 meter arc

For simplification/safely Da may be rounded down to 20 o

.

Using arc definition, Recompute R and calculate T and La

R =

=

= 286.479 m

R = 286.479 m

La =

x 2 R = 100

Δ

= 100 x

= 200.000

La = 200.000 m

T = R tan Δ

= 286.479 tan

= 104.270

T = 104.270 m

P.I. = Stn. 4 + 016.770

T 0 - 104.270

P.C. Stn 3 + 912.500

La 0 + 200.000

P.T. Station 4 + 112.500


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Figure 9

5. Laying out a curve by Deflection Angles

Curves are staked out usually by the use of deflection angles turned a t the P.C. from the

tangent to stations along the curve together with the use of chords measured from station

to station along the curve.

In the past 100 ft. chord length could be laid out quite accurately, however, in the metric

system 100 meter chords will result in a large error. Most curves are presently laid out in

20 meter chords or less and the discussion presented here will use 20 meter chord length.


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First let us consider the first 20 meter past the P.C.

Note that the angle subtended by a 20 meter arc “chord” will be proportional to the

degree of curve Da.

Figure 10

To locate point B on the curve, we measure 20 m from the P.C.

With the theodolite set up at the P.C., and reading 0o when sighted on the vertex, we turn off an

angle D/2 to align the tape and establish point B.

In Isosceles triangle AOB,

∡ BAO = ∡ ABO

= 90 – D/2

∡ VAB = D/2


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Consider the second 20 metre chord BC

Measure 20 m from B to C. Join AC. In isosceles triangle OCA angle VAC = D. To locate point

C measure 20 m from point B. With the theodolite at the P.C., reading Oo on V, turn off angle D

to align tape and establish C.

Figure 11:


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Figure 12

From: Figure 10: Note t h a t deflection angle D/2 = 1/2 angle AOB at center

Figure 11: Note t h a t deflection angle D = 1/2 the angle at the center

Figure 12: Note that deflection angle 3D/2 = 1/2 angle AOD at center

From this we can deduce a general rule.

If we set a theodolite up at either the P.C. or the P.T., and sight on the vertex with the plates set

at zero, we can turn the correct angle to any point on a curve of constant radius by merely setting


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on the plates an angle equal to half the angle at the centre subtended by the chord joining the

instrument and the desired point.

Figure:

Figure 13 - Figure of Deflection Angles on a Simple Curve

On the above figure let points a, b, c, d represent station points on a simple curve. Point “a” is an

odd distance from P.C. and distance dB is also an odd increment. The deflection angles

are:

∡ VAa = d1/2

∡ VAb = d1/2 + D/2

∡VAc = d1/2 +D/2 + D/2

= d1/2 + D

∡ VAd = d1/2 + 3D/2

∡ VAB =d1/2 + 3D/2 + 2d2/2

= /2


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Usually the P.C. and P.T. do not fall at even

stations but we are usually required to place our

stakes at the even stations on the curve. The first

and last odd length of arc or odd station,

therefore, is usually less than 20 meters and the

first and last deflection angles are less than D/2.

In the case of an arc of a circle, the angle

subtended at the centre is directly proportional to

the length of the arc.

Da = Angle subtended by a full 100 meter station

D = Angle subtended by a full station (i.e. may to 100 or less)

d = Angle subtended by an odd length station

l = Length of arc of an odd station

S= Length of arc of a full station

C = Chord distance

The deflection angle for an odd station = d/2

OR in more general form for an arc length, S

D and d are in degrees

The degree of curve, Da is defined for a 100 m arc length, since we are using 20 m arcs the

angle, D subtended by the 20 m arc is 1/5 of Da namely 4o.


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The deflection angle, D/2 for a full 20.000 m station is thus 2o, The distance from P.C. to the

first full station is [(3 + 920.000) – (3 + 912.500)] = 7.50 m, and the distance from the last full

station on the curve to P.T, is [( 4 + 112.500) - (4 + 100.000)] = 12.500 m.

The deflection angles for the odd increments at the beginning and the end of the curve are:

d1 = (7.5/20.000)* (4o) = 1

o 30' (at beginning or curve), and

d1/2 = (7.5/20.000)* (4o /2

o) = 0

o 45' ( at beginning or curve).

d2/2 = (12.5/20.000)* (4o /2

o) = 1

o 15 ' (at end of curve )

Chord distances for the initial and final odd increments of arc and the full station are:

C1 = 2R Sin d1/2 = 2 x 286.479 Sin 0.75o = 7.500 meter

C2 = 2R Sin d2/2 = 2 x 286.479 Sin 1.25o = 12.499meter

C20 = 2R Sin D/2 = 2 x 286.479 Sin 2o = 19.996 meters

It can be seen that the chords lengths are nearly equal to the arc length and no correction would

have to be applied to account for the difference.

Let us now make up a set of field notes for the curve.

It should be emphasized that the form of field notes given is not necessarily THE form of field

notes though it satisfies our needs quite well. The notes are set up on the assumption that the

transit/theodolite is set up at the P.C, (Stn 3 + 912.500) and that the plates read 0' 00' 00'' with

the instrument sighted on the P.I. (Stn 4 + 016.770).


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The notes are designed to be absolutely complete so that the person who is going to lay out the

curve does not have to refer to any other source for any information he may require with regard

to the curve. The bottom and top lines in the notes relate our particular curve to the preceding

and following curves.

The column headed "Bearing' orients the curve in relation to the rest of the project as does the

sketch.

STATION POINT DEFLECTION

ANGLE BEARING CURVE DATA

9 + 918.570 P.C. S 25o 18' E

4 + 112.5 P.T. 20o 00' 00" = /2 = 40o 00' 00"

4 + 100 18o 45' 00

4 + 080 16o 45' 00" Da = 20

4 + 060 14o 45' 00"

4 + 040 12o 45' 00 La = 200.000

4 + 020 10o 45' 00

4 + 000 8o 45' 00 T = 104.270

2 + 980 6o 45' 00

2 + 960 4o 45' 00 D/2 = 2o 00' 00"

3 + 940 2o 45' 00 d/1 = 0o 45' 00"

3 + 920 0o 45' 00" d/2 = 1o 15' 00"

3 + 912.500 P.C. 0o 00' 00"

Figure 14

Note that the stationing increases from the bottom of the page towards the top so that the

instrument-person, standing at P.C. and looking at the notes, sees the notes going away from

him/her as does the curve.


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Figure 15: Right – hand Page

Show : Page No, (The right-hand and left-hand sheets have the same page no.)

Project (If it is separate from that on preceding and following pages)

Party; Date; Weather Conditions.

Sketch of curve (not necessary to scale)

Approximate orientation of curve (or correct values if known)


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We can check our curve by using latitudes and departures to calculate the theoretical closing

error a t the E.C. if a series of chords are laid out from the B.C,

Figure 16(a), not to scale.

Figure 16(b), not to scale.


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The actual, stacked E.C. {neglecting errors in layout) falls 0.012 m outside the forward tangent

and 0.035 m beyond the theoretical E.C. This error is not large and greater accuracy would be

acquired by having shorter chords lengths.


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6. Intermediate Set-up for Simple Circular Curve

Sometimes, because of the length of a curve or because of physical obstructions, the whole

curve cannot be run in from either the P.C. or the P.T. If this is the case it becomes necessary to

set the transit up at some station on the curve and to orient it in such a way that the field notes

can be used without modification or further calculations.

With reference to Figure 17 and using the field notes shown previously we will assume that,

with the instrument set up at the P.C., we have staked the curve to Station 4 + 000.000 but are

unable to see beyond that point . We move the transit to Station 4 + 000.000.

Consider the curve AOB (Fig. 17) as being a complete curve in itself,

The central angle AOB = 17o 30’

Angle BAO = Angle ABO = 81o'15'

Triangle AV,B is an isosceles because the two subtangent distances (T1 )* AV1 and V1B, are

equal.

Angle BAV1, = Angle V1 BA = 8o 45'

With the instrument set up at B, with the plates clamped at zero, sight back on A and clamp both

motions. Unclamp the upper motion and turn a clockwise angle of 8 o

45'. The line of sight now

lies along the line V1 BV2 and is tangent to the curve at B. If B was the beginning of a curve, the

deflection angle to Stn 4 + 020 would be 2o 00' If we plunge the telescope we will be sighting

along the line BV2 and if we add 2 o 00" to the 8

o 45' already set on the plates we will be sighted

on Stn 4 + 020 with 10 o

45 ' set on the plates which is the value of the notes for Stn 4 + 020.


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Let us now assume that we are able to stake to Stn 4 + 080 but cannot see beyond Point C. We

move our transit to Point C.

Let us consider curve BOC (Fig. 17) to be complete in itself. Angle BOC = 16 o 00'. Angle

OBC = Angle OCB = 82 o 00'

Triangle BV2 is an isosceles triangle because the two subtangent distances (T2) BV2 and V2C

are equal.

Angle V2BC = Angle V2CB = 8o 00'

With the instrument set up at C, with the plates clamped at 8 o 45’, sight back on B and clamp

both motions. Unclamp the upper motion and turn the instrument clockwise an additional 8o 00',

so that the plates read 16o 45' and the line of sight lies along the line V2 CV3 and is tangent to the

curve at Point C. If C was the beginning of a curve the deflection angle to Stn 4 + 100 would be

2o 00’. If we plunge the telescope the line of sight will be along the subtangent CV3 and

if we add 2 o 00’ to the 16

o 45' already set on the plates we will be sighted on Stn 4 + 100 with

18 o 45' set on the plates which is the value in the notes for Stn 4 +100.

Again, we can proceed to stake the next portion of the curve without changing the field notes.

The two statements allow us to formulate a general rule for intermediate setups on a simple

circular curve.

When we occupy an intermediate station on a simple circular curve and sight back on some

previously established station, for the purpose of orienting the transit, the plates of the transit

must be clamped at the value of the deflection angle for the station being sighted on. After

Backsighting and clamping the lower motion, if we change the value of the angle on the plates


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to that for the deflection angle of the station that is occupied by the instrument, the line of sight

will be tangent to the curve and we can plunge the telescope and continue staking the curve

without making any change in the field notes.

Figure 17 Intermediate backsighting setup on a Simple Circular Curve

(not too scale)


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On Fig. 18 there is a slightly different application of the rule. In this case, the curve has been

staked from the P.C. to Stn 0 + 300, but the instrument is oriented by sighting ahead to the P.T.

The principle is the same as that outlined above.

Figure 18, Intermediate foresighting set up on a Simple Circular Curve (not too scale) .

The curve has been staked from A t o B and the instrument has been set up a t B. You will orient

it by sighting ahead on the P.T. We will treat the sector BOC as an entity in itself. He deduce

the angle BOC to be 22o 30".

Angle OBC - Angle OCB = 78o 45'


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In a BV1,C1, B V1 = V1C - T1 So is isosceles

Angle V1BC = Angle V1CB = 11o 15'

Following the rule set out on the previous example with the instrument set up at B, s e t the

deflection angle for C (20o 00') on the plates. Sight on C and clamp the lower motion. Unclamp

the upper motion and set the plates the deflection angle for point B.

Note that 29 o00' - 8

o 45' = 11

o15' and the line of sight now lies along the line. BV1 and we can

continue staking without changing our notes.

7 . Field Procedure

a) Set up a t P.I. and layout tangents to establish P.C. and P. T.

b) Set up a t P.C. and sight back to P.I . Set plates to 00 o 00’ 00”.

c) Turn off first deflection angle on transit and measure out the first subchord distance, pound in

hub and write stationing on it.

d) Continue laying out chords and placing hubs. When the last station (P.T. Is reached the last

subchord measured out should be quite close t o the P.T. hub established earlier.

If it is not, an error has been made and the curve has to be run in all over again.

e) It is advisable to use intermediate s e t u p s when setting out the curve. They prevent

communication difficulties and tend to reduce the number of errors.


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PROBLEMS

1. Calculate the radius of the curve that will pass through Point P, using 1atitudes and

departures. Determine the stationing of the P.C. and P,T, and o f Point P on the curve.

2 . Solve for length of the curve intersection distance AB, based on the following diagram.


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3. Calculate BC, area ABC (shaded) and angle d:

Documents

Part 5 Simple Curves