Show us why the public should always call a land surveyor

Where are you surveying?

Horizontal Curves Part 1

Land Surveyor
Views: 8
Get Embed Code

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
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
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
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
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
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
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
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
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
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
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
by considering this triangle formed by the chords
of each simple curve
find the stationing of pcc and pt
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
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
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
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
to solve the chord as i mentioned earlier
and pcc so
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
206.667 plus 180
so the total length of curve is 386.667
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 next problem for compound curve
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

You need to be a member of Land Surveyors United - Global Surveying Community to add thoughts!

Join Land Surveyors United - Global Surveying Community

E-mail me when people leave their comments –

Discover Videos by Type

Educate Future Land Surveyors

Comment as:

Votes: 0

You need to be a member of Land Surveyors United - Global Surveying Community to add thoughts!

Join Land Surveyors United - Global Surveying Community

Comments are closed.