Horizontal Curves Part 1

Lecture Outline:Simple CurveCompound CurveReverse Curve (Parallel Tangents)

curves our first topic here in finals

so horizontal curves is needed in

designing a road so introduce a horizontal curve

for safety purposes and shampoo to

to avoid an among inaccessible areas

so for safety reasons uh we avoid a designing roadway that that is

straight for too long so 200 meters 300 meters

400 meters at least there is a horizontal curve on that

on that roadway

[Music]

slope

[Music]

uh fundamentals of surveying will just focus on understanding the parts of a simple of a horizontal

curve rather than uh designing of a horizontal

of a overload wing which involves horizontal curve

we will just try to analyze what are the parts how to

what are the parts of a horizontal curve transportation engineering

design

this is just an introductory lesson with regards to

the transportation planning so

uh major suggestions

we have three horizontal curves

of intersection there's another point of intersection here so there is another

horizontal curve here and it is a partner this is just pi

horizontal curve so

so what are the horizontal curves

first one is the simple curve then the compound curve reverse curve

and the spiral curve so much start tires

which is the simple curve so from the

word itself it is simple so a simple curve is

derived from a sector of a circle so this is a

sector of a circle okay so this is sector of a circle and from

that uh computation

so what are the parts of a simple curve first one is we have pc or the point of

curvature so this is the starting point of a simple curve

and we have bt or the point of tangency which is the n point

of the simple curve now at easy npt uh we have tangent lines

from that point which is yeah uh

we have a line is tangent to that we call that

the back tangent which is this one the forward dungeon in the man's apartment

now the point of intersection of these tangents is called p i or p point of

intersection and the angle formed

by this uh point by by the intersection of the

tangents is called the angle of intersection which is i

so an angle of intersection i is also equivalent to the central angle

of the uh of the of the simple curve

so so other parts are we have

tangentia t so the distance from pc to pi

has the same distance from pi to pt

so that's tangent also we also have a chord length of chord l

this a line measured from b c to b

which is s l c or length of curve or the pathway

of the road itself so young curves

simple curve so we have a external distance

or uh or the distance from the point of intersection to the midpoint of

of the arc of the symbol curve also have a middle ordinate measured

from the midpoint of the curve to the midpoint

or halfway the distance of the chord of the

simple curve so we have the also the reduce here okay and this is the center of the

simple curve now we also have uh

we have a offset distance so at any point of the curve

there is an equivalent offset distance which is measured from the tangent

and then perpendicular to it measure a distance from the tangent line

to that point we are determining so that's what you call

the offset distance x and it also have a

corresponding equivalent angle that is measured from

the tangent to that point so we assign that as theta or

the offset again so that is the offset angle subtended

at pc between pi at any point of the curve

now so if if the point we determine is at pt

the equivalent offset uh offset angle

is this one okay by the way the

offset angle of a point when measured

at the equivalent central angle of an offset angle is equivalent to twice its value on the

central

if we are measuring the offset distance at a point where located at pt

so from there to here so if this is the

central angle of the curve which is it which it ends at pt from pc to pt

is i over 2. so

okay so uh

what are the formulas that we will use here in simple group

soma formula um

we have this triangle okay we can use that

t equals foreign r tangent i over 2 e equals to r second i over 2 minus r

we have m times r times minus r cosine i over 2 we have l 2 r sine i over 2

and lc pi ri over 180 formulas

formula trigonometric functions

that is used in a right triangle so sine

cosine so just find it here and because

you can derive it on your own so just use this right triangle also use this right triangle from that you can derive

those formulas for e m l tangent and so on

so for the length of curve this is just the formula for length of arc

of a sector okay this is net of arc

so latin again if you can memorize it memorize it

so one factor in how sharp a curve is on a horizontal curve

is by determining its degree of curve so

the smaller the degree of curve the flatter is the curve and vice versa

so the sharpness of simple curve is also determined by its radius

large radius are flat whereas smaller edges are sharp so

smaller ranges on a horizontal curve on a simple curve it means

it has a sharper curve or only

curve so how do we determine the degree of curve

we have two methods or uh we have two basis

in determining the degree of curve the first one is by arc basis

second is true chord bases so in arc basis

we have a one station then we will measure

that equivalent central angle of that of of that sector

okay so if we have a given one station distance

which is measured along the arc okay we will find the equivalent

central angle of that or the degree of curve so here the philippines we use one

station equivalent to 20 meters so for every 20 meters we have one

station so uh

we take the ratio of one station

over the equivalent uh central angle which is d

is equals to a proportion attention is a whole circle

which is

now the equivalent central angle of that is of course the whole angle which is 360 degrees one

revolution so substitute now one station is

20 meters divided by d so it is equals to 2 pi r

over 360. now to determine d uh

cross multiply nothing we will have

20 times 360 equals to 2 pi r

t so divide both sides by 2 pi r we have 20 times 360 over 2 pi r equals

d now let's evaluate 20 times 360 over 2 pi

when we solve this that is equivalent to 1 1 4 5 0.916

over r that is the solution in determining the degree of curve

basing on the ranges of the curve so just substitute the radius and

on this value on this equation so that you will have the degree of

curve so for chord bases one station is measured along

its cord so 20 meters

okay so i think 20 meters nothing

so uh

so take the trigonometric function of sine so sine d over 2

take the half of the station because so 10

over the hypotenuse which is r so sine d over 2 equals to 10 over r

so what happened you degree of curve take the arc sine

of 10 over r then multiply it by 2.

so this is the solution in determining the degree

of curve by chord pieces

so example two tangents of a simple curve have azimuth of 120 degrees and 156 degrees

30 minutes respectively with a reduce of 400 meters

that remain the following first degree of curve using arc basis

then degree of curve through chord bases tangent distance external distance

middle ordinate long length of long chord then of curve

so before we start uh

clear

foreign

let's answer that problem so we have

here two tangents so usually some problem uh my counter not tense horizontal curve

is that there's no given figure so why you miss

120 and 156 degrees 30 minutes so forgiving your tangents

is nothing so your starting point down 120 degrees so first make sure and as you move from

the south [Music]

then after that draw another line which is measured from the azimuth 156

degrees 30 minutes again from south make sure 156

so then

yes so

given your reduce which is 400 meters

now basically someone given direction of tangents

from that we can compute the angle of intersection of the curve

opening tangents

120 degrees so therefore if i subtract

this asimo 156 degrees 30 minutes to the azimuth of the back tangent

which is 120 degrees how much

156 degrees minus 120 degrees is equals to

36 degrees and 30 minutes

okay angle of intersection equivalent

central angle okay

so annoying required what is the degree of curve through our basis

and chord bases for a and b

for the degree of curve by arc basis

t is equals to

over r so just substitute r which is 400

so the degree of curve is two point eight

six five degrees so read it in chiang mai degrees minutes seconds

i'm using chord bases so inside of that equation

we have 2 arc sine 10 over r

so t is equals to 2 arc sine of 10

over r which is 400 so the degree of curve

to our basis is also equivalent

2.865 degrees so but yes but

this a lot of magnesium degree of curve through arc bases and corn bases so

careful in following uh to what in following the instruction

or what is required on the problem basis

for the tangent distance the tangent distance has a formula of

r tangent i over 2

so substitute r 400 then tangent of

i over 2. so um i over topola divided that into 36 degrees

30 minutes divides it two i equivalently 18 degrees 15 minutes

okay okay t is equals to 131.9

meters okay so get it direct substitution that's a

formula external distance external distance

e is equals to our second i

over 2 minus one

so e is equals to 400 times second of

uh 18 degrees 15 minutes minus one okay substitution

class uh

21.186 meters middle ordinate

middle coordinate is r 1 minus cosine i over 2

so 400 times 1 minus cosine 18 degrees 15

minutes okay so the middle ordinate is

20.12 meters

length of long chord formula base then long chord two r sine

i over two so two times four hundred times

sine of eighty degrees 15 minutes so the length of long cord is

250.531

meters okay

how about g length of curve for the length of curve

length of r pi r i over 180 so pi r is 400

degrees 30 minutes

36 degrees 30 minutes

problem but as a simple curve a simple curve given a middle ordinate

of 4.5 and a deflection of gel from pc to pt of 27 degrees determine the ranges of the

first

a pc to pt which is 27 degrees so deflection angle from pc to

pt deflection angle is this one so your angle subtended from

so from 27 degrees multiplied by 2 that's equivalent to 54

degrees equivalent

curve that's 54 degrees which is that's also

the eye so we say

so what is required here is to determine the ratios since given is the middle ordinate we

can use the formula m equals to r one minus cosine i over

two now regarding r so divide both sides by

one minus cosine i over 2 to eliminate the quantity on the right

hand side of the equation material r so substitute the given 4.5

1 minus cosine i over 2 and at the moment 54 q evaluates 27 degrees

so r r is

41.287 meters

okay next

proceed

the line connecting the pc and pt on the simple curve is 120.4

meters the radius is 213.6 meters

if the stationing of the pc is 9 plus 346.1

just draw a simple curve sample then the new s i human given

the stationing which is this one nine plus three four six point two one

and we have a ranges of 213.6

another given is the line connecting pc and pt or should

you length of chord from pc to pt that's the horde of the

of a sector which measures 120.4

so required is to determine the stationing of pt

so in measuring the stationing and determining the stationing of pt all we need to do is

add the stationing from pc to the length of curve

station

okay yeah when we determine the length of curve just add that to the station of a pc

now we can determine the stationing of pt but before that uh madame

unknown the angel of intersection is unknown

knowing that we don't have angle of intersection yet so sanated vedic derived the angle of

intersection so based on the given we have length of chord

and we have a radius so from the formula of the length of chord

uh we can substitute l and r from that formula so that we will be able to

determine the unknown angle of intersection so to determine i

so divide both sides by 2r

paramagne sine i over 2 then take the arc sine

of l over 2 r which is equals to i over 2 and multiply

it by 2 r sine of l over 2 r

equals to i so substitute that

substitute l 120.4 to r 213.6

[Music]

32.74 degrees or in degrees milliseconds we have 32 degrees

44 minutes point 22.8

seconds okay

now we can find the length of curve

just pi r i over 180 degrees so pi r i over 180

we have pi r is 213.6

times i

over 180 degrees so the length of curve is

120.054 meters

okay so the stationing of pt

the stationing of ptsd was the stationing of pc which is equivalent to

nine plus three four six point one meters

plus 122.054 meters

so in adding station a class uh again i'm nine plus three four six point

one is means that the station is nine kilometers and three thousand three

hundred four hundred nine kilometers and three hundred forty six point one

meters so the time that there's a calculator it is 9346.1

then at 122.054 [Music] so when you add that

current stationing is 9 plus four six eight point one five four meters so that's the

stationary of pt okay

lappo let's proceed to

compound curve so when it comes to compound curve

uh this is just two adjacent simple curve

so makata big simple curve the parentheses

sharpness of the curve due to the design of the roadway which is

a compound curve which is which has a different which has two different uh ranges

of the simple curve of the compound curve rather so

all we need to do is extend that tangent

here and the tangent of the second curve

section

angular intersection just add the two central angle or angle of intersection on the lower

simple curve so i want plus i now mathematically uncommon tangent or

hormone tangent this one okay this is the common tangent

another we have the long chord long chord is a line measured from pc

guidance vertex to point of intersection x

so essentially

so later

[Music]

by considering this triangle formed by the chords

of each simple curve

[Music]

find the stationing of pcc and pt

station

and just add it to the station in your pc so you let of curve number

curve will be added on pc but what if

you give an seo

that's why we need to subtract this x and the tangent one

then length

lc1 and the length of curve of the second

station a given subtract x subtract t1 then add

lc1 nlc

so let's start answering this problem

okay

component curve laid on the dungeons have the following data

i 131 degrees i2 36 degrees

t13 degrees t2 4 degrees so

so give a new degree of curve

basis okay use our basis

using our bases uh

parasite ranges one so can you derive that degree of square 0.145.916

over r so simultaneously or whether it be nothing you are in terms of the

degree of curve so 101 145

0.916 over t so yeah solution

okay uh one one four five point nine one six over three degrees

the first reduce is 72 meters

r2 is 1145.916 over d equals to 1145.916

over four is four degrees

so the second ranges is 286.479

meters so this

2. so from that we can draw the figure so it wasn't my sample throwing now

but uh

then

foreign

[Music]

is tangent one plus tangent two the tangent one is r one tangent i one

over two okay you give it and you've seen all nothing

that

31 degrees divided by 2. so tangent one is

hundred five point nine three zero meters how about the tangent to

tangent two so again subscript

r2 is 286.479 tangent 36 over two

so tangent to 93 degrees 93.083

is 199.013

meters okay

[Music]

to solve the chord as i mentioned earlier

and pcc so

[Music]

okay

pcc

is equivalent to

2 times 381

972 sine 31 degrees divided by 2

the length of the first chord is 204.155

meters the length of the second chord is 2

to r1 r2 sine i

over 2 which is 2 286.479

sine 36 degrees over 2. so l2 is

177.054 meters

okay so we have l1 and l2 check now

[Music] so in order to determine the length of long word we can use law of cosines

now going back here it's adding compound curve

so the equivalent deflection angle then from the tangent to the chord

is one half the central

so when we add the three angles which is i one over two beta and i two over two that is equals

to 180 degrees

180 degrees minus i one over two minus i two

over two so solve the netting class

beta 180 degrees i one over 2 minus i 2 over 2

so 180 degrees minus minus 31 degrees over 2

minus 36 degrees over 2. beta is

146.5 degrees okay

of cosines using law of cosines so your side not then is l

so l squared equals to l one squared plus l two squared minus 2 l 1 l 2

cosine beta so

substitution then take nothing square root of both sides

eliminate you exponent 2 so the square root of

l1 is 204.155 square

plus zero five 177.054 square minus two times two oh four point one

five five times uh

one seven seven point zero five four cosine of angle

beta 146.5 degrees okay

computer calculator so i got i 365

meters so you know length

of longhorn okay angle pc pcc and pt

next question what is the length of the first curve

the length of the first curve length of arc so substitute

31 degrees over 180

so lc1 is 206.667

meters the length of the second curve can announce it

by r2 is 286 point

asana 479 times 36 degrees

over 180 degrees now lc2 is

180 meters okay 180 meters

that's the answer for letter d

[Music]

206.667 plus 180

so the total length of curve is 386.667

meters

from bi vertex so parama compute

angle of intersection which is 31 plus 36 degrees which is equal to 67

67 degrees

180 minus 67 is 130 degrees

is law of sine so x over sine of i two

is equals to give me nothing common tangent over sine of 130 degrees

over sine 130 degrees so you see 10999.013

times sine i to 36 degrees

over sine of 113 degrees

so x is 127.079

meters you

right station pt equals to station pi

minus x minus t1 plus lc1 plus lc2

so substitute stationing now

p i eight plus 750

minus x 127.079 minus t one computer now 105.930

plus lc1

206.6 plus lc 280.

so the stationing of pts station 8 plus

903.658 meters so that's

your answer for stationing of beating

okay

okay next problem for compound curve

given

the tangents of a compound curve intersect at an angle of 104 degrees so ion

those first curve has a radius of 420 meters so r1 420

the central angle of the first curve is 38 degrees so i1

if the common tangent is 200 meters long

ct 200 meters what is the radius of the second curve

can determine i2 i2 therefore is 104 degrees minus 38

degrees so i2 is 66 degrees so i'm going to

later on because

so the tangent one is r1 is 420

tangent of i1 is 38 over 2. so tangent 1 is

144 so

which is 200 substitutionary tangent 144.618

so 102 therefore is 200 minus 144.618

so subtract the two we have 555.382

meters so from tangent to the mediterranean

[Music] so divide both sides by tangent

i two over two that's the second ranges substitution

tangent to 55.382 over tangent 66

over 2 stored compute 85.281

meters so get it now no figure hidden

for this problem any questions uh compound curve

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Horizontal Curves Part 1

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