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SIMPLE CURVES
RAILROAD AND HIGHWAY CURVES
In highway or railroad construction, the curves most generally used presently' are circular curves although parabolic and other curves are sometimes used. These types of curves are classified as Simple, Compound, Reversed or Spiral curves.
A. Simple Curve:
A simple curve is a circular are, extending from one tangent to the next. The point where the curve leaves the first tangent is called the "point of curvature" (P.C.) and the point where the curve joins the second tangent is called the "point of tangency" (P.T.). The P.C. and P. T. are often called the tangent points. If the tangent be produced, they will meet in a point of intersection called the "vertex". The distance from the vertex to the P.C. or P.T. is called the "tangent distance". The distance from the vertex to the curve is called the "external distance" (measured towards the center of curvature). While the line joining the middle of the curve and the middle of the chord line joining the P.C. and P.T. is called the "middle ordinate".
Geometry of the Circular Curves:
In the study of curves, the following geometric principles should be emphasized:
- An inscribed angle is measured by one half its intercepted arc.
LACB=~ ~AOB
- Inscribed angles having the same or equal intercepted arcs are equal.
@
.• B
D e C
LADB=LACB
- An angle formed by a tangent and a chord is measured by one half its intercepted arc.
l'
LBAC=- LADC
- Tangents from an extemal poiht a circle are equal.
AB=BC
- Angles whose sides are perpendicular each to each are either equal or supplementary.
B~AC
D F
LABC=LFED
SIMPLE CURVES
Sharpness of the curve is expressed in any of the three ways:
1. Degree of Curve: (Arc Basis) Degree of curve is the angle at the center subtended by an arc of 20 m. is the Metric system or 100 ft. in the English system. This is the method generally used in Highway practice.
a. Metric System:
By ratio and proportion:
20 2nR
D= 360(20)
2nR
D
b. English System:
100 2nR D= 360 D = 360(100) 2nR D _ 1145.916(5)
- R (5 times the metric system) D - 5729.
- R
S-
2. Degree of Curve:. (Chord Basis) Degree of curve is the angle subtended by a chord of 20 meters in Metric System or 100 ft. in English System.
a. Metrjc System:
b. English System:
S
. D 50
In-=- 2 R
R=-
S
. D
rn-
3. Radius = Length of radius is stated
Elements of a simple curve:
P. C. = poi nt of curvature P.T. = point oftangency
P.I. = point of intersection
R = radius of the curve
D = degree of the curve
T = tangent distance
I = -angle of intersection
E = external distance
M = middle ordinate Lc = length of curve C = long chord C1 and C2 =sl,lb-chord
. d 1 and d 2 = sub-angle
S-
SIMPlE CURVES
® Distance from mid point of CUNe to the mid point of long chord:
M= R (1 - Cos ~) M= 190.99 (1- Cos 18') M=9.35m.
I ,
, I ,
:R I
'"
, (^) "', ,I / " (^) '.. 1 I II I , (^) "';-1118",: / X"',r/ ex 111 ) • .....-:.,":~.1,~, .A »'t$ '<:."I
Solution:
ill Distance from mid point of CUNe to P.I.: R- 1145.
E=R(Sec~ -1) E = 190.99 (Sec 18' -1)
E= 9.83 m.
R R
Sjn~=~·
2 2R
Sin Q.=~ 2 2R
C
2R=-
S
In- 2
. d C, Sin~ Sin~=--
2 C
d C 1 Sin~. Sin ~ 2 =--- 20 (Metric I' (^) Il
d C 1 Sin~ Sin T= 100 (English)
20Sin~ C 1 =--- (Metric) S
10-
100 Sin~
C 1 =---
S
7. Sub-chords: (Chord b, ;is)
SIMPlE CURVES
@ Stationing of B: S=R
S = 190.99 (16) 1t
S= 5133m.
Sta. of B =(10 +020) + (53.33)
Sta. of B = 10 + 073.
® Stationing of D: S=R
S =286.48 (36) 1t
180 S= 180 m.
Sta. of D =(20 + 130.46) + 180
Sfa. of D = 20 + 310.
.
.^ \ ,
"'-_ \ \ IR
"'-- 12'~ \ /
R"<:i .. 30" , ffl
@ Distance DE:
Cos 36' =286.
OE
OE=354.11 m.
DE =354.11 - 286.
DE=67.63m.
Solution:
CD External distance:
·.rdTJtJI¥~.I~II-I~ii~fti~nt~:··
·.(1).••••. C9IllP~f~ ••• t~~E!xlE?rn~I.·9fsl<trce •• Qf.the 9J1'Y~.> •
- ~•••• C;oITlputetflttmj~¢I¢~dihaWfJf.thec;(lo/~ .•·•••.
(!)Cqrllpu!E!t~~~tatl(l6jn9PfJlqintAonme
•••••••• ,.~~~~,i~~~~&®.~f··ot6' •• frQrn
w.
j""- ' R <Y-,_ "-_ "- %n '-'''~- \
_A . 2}i1:~o PoC R 20+130.
Solution:
CD Long chord:
R= 1145.
R=286.48 m.
!:= RSin 25'
L = 2 (286.48) Sin 25' L = 242.14m.
SIMPlE CURIES
Solution:
<D Length of curve from P. C:
o
Angle x =24'40' + 6'13'
Angle x =30'53'
c
a=90·24'4O'
a = 65'20'
229.18 = OCCos 24'40'
OC = 252.20 m.
Sin 45'30' = Sin ~
l/l =180' ·45'30' -128'
l/l =6'13'
ro 229.
Sin 6'13' = Sin 45'30'
CD=34.80m,
@ Stationing of D:
R= 1145.
D
R
R=229.18m.
fa 8 =105.
n 229.
LC1 =Re
L
Cl - 180
LCl =98.68 m.
® Distance CD:
.
'', ' / '\ \1 ,I' R' 'I ', , /
'~t/
o
\ ' , I ', ' / ~'\ \ \ /
'{P'~~, /.
'\ '\ ' '\ / '\V' o
SIMPLE CURVES
Rxn
L~ = 180
L
_229.18(30'531n
L~ =148.24 m.
Station ofD =(2 +040) +(148.24)
Station of D =(2 + 188.24)
@ Length of curve from PC. to A: S=R
S - 560.13 (27'391n
® Length of long chord:
, ,^1 •
\ I ,I \ ' / , I ' _R_ ' / ! / \34" /
\V
.Thij~mm@lfJij~f9WItro.mpm~(
.9Q~Mfflp!OON~19qi:)~<t®!M9M~n~
..a4..n1..·Jfthe;d;@l(L~·frO#lilheiP·C;itijQOllffie •• .~~~H~.ZOOfu{>··· ....
.1•••• ·E~lr"..I·lrt!I~·.I~I·· ®•• lqh~~~9J~9fli'ltMli9fm~@W4~ ..·.•.•••••• ·.·§4·f@miwt~m~'laogthml§ng~@ff9@.eP.ri>p:tt)·· .. ... ....
Solution:
CD Radius of CUNe:
I,
!R I,
! / \201 /
\V
S· In^ 34'^ = 2 (560.13) L
L =928.74 m.
lan8=-
28 = 2T3g
R· Cos2T39'=-
R
R= 560.13m.
SIMPlE CURVES
@ Area bounded by the tangents and outside the central cUrve: T=Rtan24' T= 286.48 tan 24' T= 127. A 1R (2) 'It R2 I ",.... - 2 - 360 A_ _ 127.55 (286.48)(2) 'It (286.48)2(48) ",...a - 2 - 360
Alea = 2162.
® External distance:
E=R(sec~-1) E= 336.49 (sec 25' -1) E=34.79m. @ Length of long chord:
~=RSin25'
L =2 (336.49) Sin 25'
L = 284.41 In.
S· In^ e^ = 21.032.
e=T3T a= 90' -12'- a=70'23'
A
I_i.
Solution:
<D Radius of CUNe:
Sin 50' =12~
T= 156.91 m. T=Rtan 25' 156.91 =Rtan25' R=336.49m. D= 1145.
D= 3'24'
, / " ''., ' /~
,\ '
Ii'. 25' \ 25' /
7\7'. I,
1:lllilllBlijl
Solution:
<D Degree of curve:
SIMPLE CURVES
OB=R+E
OB = R+ R(sec~.1)
OB=R+R(SeC~.1)
OB= 1.0223 R
In /1 OPB
OB R
Sin IJ =Sin a.
1.0223R R
5iii'"'8 =Sin 70'23' LJ = 105'39' 0=180-a.-1J
. 21.03 _ R
Sin 3'58' - Sin70'23'
R= 286.36m..
® Length of chord:
o
@ Area bounded by the CUNe and the tangent lines:
A =. RT(2L 1t ~ (24') 2 360' T= Rtan 12'
T= 286.36 tan 12'
T=60.
A=286.36 (60.87) _1t (2~:~2 (24)
A = 256.26m
11lSillitillll; .!lPOl'dill~t*19tg91OQN~IJ~.2<l1l)Q·pV'1W~polllt ·~ri~I~~~~.h~ ••~9()r~i#t~~ .• §f.~Q~.~~ •• ~••
q)FjfJpth~di$tal1¥ofline~P') ~••• ®IY~J°rt'md~r~~Rfs.b!lple@f'I~tb~t
...... ·•••·,#icillliefangellt.lq.tfflj.thi'ee.lines.i\B,DI!Z
AM~P·>
@"p~lltlJisatM<itj(jn1t9S~.87tfetermilll'l
.··th¢sWIQl1ingofPT·< '.
Solution:
CD Distance of line BD:
A
Sin 4 '01' = 2 (2~.36) x= 40.12 m.
A c
SIMPlE CURVES
@ Chord distance:
C=2Rsin l
C=2(311) Sin 2" C= 139.92m.
@ Length of CUNe: 1t Lc =RI 180 1t
Lc = (311) 26 180
Lc = 141.13 m.
.li~'I~~I!~~~~~':~r:~i
@!'1®~@6~af~t:qli~6AAf@ffl*~ij@#
m~ll6~m!M~~~lp!@gl~9f~)
Willilli~tM9¢l\tfbffi@IHfjElgp;<
Solution:
CD Distance from mid point of CUNe to P.I.
R
- D
R
E= R(sec 112· 1)
E = 190.99 (sec 30' -1)
E=29,55m.
® Distance of mid point of CUNe to mid point of long chord:
M= R(1- Cos 1!2)
M= 190.99 (1 • Cos 30')
M=25.59m,
® Stationing of B:
S=Re
S = 190.99 (16) 1t
Sta. of B = 10 +(020) +(53.33)
Sta. of B = 10 + 073.
t~~ •• tans~~ltrr(ll~ep.g .• h~$.a(jlte¢ll«@ll~
n()rth•• and.the.ta~gerlt •• tht()U91'1 •• th~ •• ~rnt'l~s~.
b~rlng.()f.N,50· .•~, ....• lt.nll,$a.ra,dl~~pf2QOffi; •• lJ~in$ •• ·.·,;jrc ••• b<!llis.·•••..·.$tliltiQilitJ9 •••• pl••• P.q, ••• i§.
SIMPlE CURVES
Solution:
(j) Tangent distance:
S=R
S
- 200 (28) 1t
- 180 S=97.74m.
Sta. of B = (12 + 060) + (97.74)
Sta. ofB= 12+ 151,
Solution:
(j) Middle ordinate: T= Rtan 25'
T = 200 tan 25'
T=93.26m.
® Long chord:
'Sin 25'=1:... 2R
L = 2(200) Sin 25'
L = 169,05m.
@ Stationing of B:
p.e
Lc 2)
T=o 210 20
-I=4"
M= R (1- Cos 112)
. R= 11~.916 =286.
M = 286.48 (1- Cos 21') M= 19.03m.
S-
SIMPlE CURVES
1
\
. '\
R\ *" '* I R-60" \ ,I
28.0~
10+
® Degree of CUNe:
T= Rtan 20'
124.46 = Rtan 20'
R=341.
D =1145.
D=3.35'
@ Stationing of 8: S=RS
S =341.95 (16) 1t
S=95.
Sta. of 8 =(10 +060) +(95.49)
Sta. of8= 10+ 155.
IIIIII
1:11.""i~
Solution: CD Radius of CUNe:
80 = TSin 40'
T= 124.
Sin l= 70
2 R
S' 1 50
In 2= 95.
~ =31.5' 1=63' @ Tangent distance:
T= Rtan l
T= 95.67 tan 31.5'
T= 58.63 nI.
Ift.ili
Solution: CD Tangent dsistance:
SIMPLE CURVES
tane=-
e =14.04' 2e =28.08'
R - 60 = R Cos 28.08'
0.11nR= R=509,70m.
® Tangent distance: T=Rtan31' T= 509.70 tan 31' T=306.26m.
@ Stationing of painf x: S=Re
S = 509.70 (28.08) 1t
S=249.
Sta. of x= (10 + 080) + (249.80)
Sta. ofx= 10+ 329,
Solution:
<D Deflection angle at the P.C.:
Cos 2e = 219.
2e= 16.988'
e=8.49' (deflection angle)
@ Stationing at B: S= R(2e) S= 229.18 (16.988)' 1t 180 S=67.95m.
Sta. of B = (10 + 120.60) + 67.
Sta. ofB= 10+ 188.
@ Chord distance from P. C. to B:
Sin 8.49' = ~~ AB= 67.73 m.
S-
SIMPLE CURVES
I.'.
.tqt1~A~··)<·······.
Solution:
CD Central angle of 10' center curve:
OA=Ra SinQ=~ 2 2Ra
S,n-=- 2 Ra Ra= 143.47 m. '. 6' 10 SIn---
2 - Re
R 6 = 191.07 m.
Sin 10' =.1Q..
2 RlO R 10 = 114.74 m.
O'C= Re- RlO
O·C=191.07-114.
O·C=76.
OC=Re- Ra
OC = 191.07 -143.
OC=47.
Using Sine Law: 47.70 76.
S
. !.m. = Sin 136' In 2
ful=25'44'
@ Central angle of 6' end curves:
16 +~+ 136' = 180' 16 + 25'44' + 136' = 180'
@ stationing of P. T.:
L
Cl- 0
1 L
_20(18'16')
cl- 6'
LCl = 60.89 m.
L
_20/ 2
c2- ~
L
OL - 10'
LOL = 102.93 m.
P. T. = (10 + 185.42) + 60.89 + 102.93 + 60.
P. T. = (10 + 410,13)
mounth of tunn~
SIMPLE eURVES
~IIIIIIIIIIIIII~IIIIII
I :7OmI II
Il'lt.
Solution:
CD Stationing of the point of deviation:
j I , rt----...-.---,/.-------------....~ .. - .r.
al ,"^ ,/ ~11I2,' / ~ r;,' ,/;=163.
~ I '1/
U
~,,', r~ilway in ~y rhe runnel
Sin Q= 10 2 R
Sin 3.5' =~ R=163.80m.
C I
as = 163. 1= 55'04'
tan 55'04' = 70 x x=48.89m. I T tan-=- 2 R T = 163.80 tan 27'32' T=85.39 m.
Sta. of point of deviation (P. C.)
® Stationing of mouth of tunnel:
b.c._ 20
I - D
_ 55'04' (20)
4= 157.33m.
Sta. of mouth of tunnel = (7 + 677.72) + (157.33) Sta. of mouth of tunnel = 7 + 835,
@ Direction of railway in the tunnel:
Direction is S. 34'56' E.
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