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ADVANCE ENGINEERING SURVEYING (3+1) Lecture Lectu re 5: Reve Reverse rse Curve Curve Dr. Mohsin Mohsin Sidd Siddique ique Asst. Prof. Dept. of Civil Engineering FAST-NU 25/09/2012
Reverse curve • Rever verse Curve: rve: A reverse curve consists of two arc bending in opposite directions with common tangent at junction of arcs. • Their centers lie on opposite side sidess of the the curv curve. e. • Their radii may be either equal or different and they have one common common tangent. tangent. • Reverse curves are generally used to connect two parallel roads or railway lines or when two lines intersect at a very small small angle. angle.
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Reverse curve
• These curves are suitable for railways, city roads etc. • But they should be avoided as far as possible for important tracks or highways for the following reasons: ▫ 1. Supereve Superevelatio lations ns cann cannot ot be provided provided at the the point point of reverse reverse curvature ▫ 2. A sudden sudden change change of direction direction would would be dangerous dangerous for a vehicle vehicle ▫ 3. A sudden sudden chan change ge of supere superevelat velation ion cause causess discomfort discomfort to passengers ▫ 4. Carelessne Carelessness ss of the driver driver may cause the vehicle vehicle to overturn overturn over a reverse curve. • Setting out of reverse curve • They are generally short, and hence they are set out by the chain and tape method.
Reverse curve Notations • 1. AB and EF are the straight lines, • 2.BE is the common tangent • 3. C is the point of reverse curve • 4. T1 and T2 are the tangent points • Φ is the angle of intersection between the two straights straights • Φ1 and Φ2 are the deflection deflection angles of the common tangent • R1 and R2 are the radii of the arcs
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Reverse Curves Cases • Case I – When the straights straights are nonparallel • Suppose AB, BC and CD are lines of an open transverse along the alignment of a road as shown in figure. • AB and CD when produced meet at a point E, where Φ is the angle of intersection. • It is required to connect the lines AB and CD by a reverse curve with BC as the common tangent. • Let ▫ Φ1= angle angle of of deflecti deflection on for the first first arc arc ▫ Φ2= angle angle of of deflectio deflection n for the second second arc arc ▫ Φ= angle angle of interse intersection ction betwe between en AB and CD ▫ T1 and T2=t T2=tang angent ent poi points nts ▫ F=p F=point oint of of revers reversee curvat curvature ure ▫ R=c R=commo ommon n radiu radiuss for the the arcs arcs
Reverse Curves Cases • 1. Tangent length of first arc T 1 B = BF = R tan[φ 1 / 2] • 2. Tangent length of second arc T 2C = CF = R tan[φ 2 / 2] • 3. Length of common tangent BC = BF + CF = R tan
[φ 1 / 2] + R tan[φ 2 / 2]
• 4. Length of first curve o T 1 F = π Rφ 1 / 180 • 5. Length of second curve o T 2 F = π Rφ 2 / 180 • 6. Chainage Chainage of T1=chain T1=chainage age of B-T1B B-T1B • 7. Chainag Chainagee of F=chain F=chainage age of T1+1 T1+1st curve length • 8. Chainag Chainagee of T2=ch T2=chainag ainagee of F+2nd curve length
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Reverse Curves Cases • Case II – When the straight straight lines are parallel • In figure, PQ and RS are two parallel lines a distance y apart. It is required to connect the lines PQ and RS by a reverse curve having equal radii. Line AB is drawn parallel to PQ or RS through point C. • R= Common Radius
▫ C= poin pointt of reve reverse rse curv curvature ature ▫ T1 an and d T2=tan T2=tangen gentt points points ▫ Φ= angl anglee subte subtende nded d at the the centre by the curve ▫ T1T2=l =length =length of joining T1 and T2 ▫ x= perpe perpendi ndicul cular ar distanc distancee between T1 and T2 ▫ y=p y=para aralle llell distance distance betwe between en lines PQ and RS
Reverse Curves Cases • 1. Long chord for first curve T 1C = 2 R sin[φ / 2]
• 2. Long chord for second curve T 2C = 2 R sin[φ / 2]
• 3. Length T1T2= l
• From
l
=
2 R sin[φ / 2] + 2 R sin[φ / 2]
l
=
4 R sin[φ / 2]
k
∆T 1T 2 K
sin[φ / 2] = y / l l
=
l
=
4 R × y / l ⇒ l 2
=
4Ry
4Ry
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Reverse Curves Cases • 4. T 1 A = O1T 1 − O1 A = R − R cos φ T 2 B = O2T 2 − O2 B = R − R cos φ y
= T 1 A + T 2 B =
2 R(1 − cos φ )
• 5. x = AB = CA + CB = R sin φ + R sin φ =
2 R sin φ
Numerical 7 • While surveying along the the alignment of a road, the magnetic magnetic bearings of the line AB, BC and CD are measured as 80, 110 and 60 degrees respectively. respectively. The length of BC is 200m and the chainage chainage of B is 950m. Calculate the necessary data for setting out a reverse curve connecting connecting the lines AB and CD by taking BC as the common tangent. The radii of both curves may be assumed equal. • Solution φ 1
= 110 − 80 = 30
φ 2
= BB of BC -
φ 2
=
o
(180 + 60)
290 − 240 = 50
o
BC = 200m
Chainage of B = 950m Let F the point of reverse curvature
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Numerical 7 Now, BC = BF + FC = R tan(φ 1 / 2) + R tan(φ 2 / 2)
200 = R tan(30 / 2) + R tan(50 / 2) R = 272.4m
• 1. Tangent length T1B R tan[φ 1 / 2] = 72.97 m
• 2. Chaina Chainage ge of T1= 950 − 72.97 = 877.03m
• 3. First curve length π Rφ 1 / 180
o
= 142.62m
• 4. Chain Chainag age e of F 877.03 + 142.62 = 1019.25m
• 5. Second curve length π Rφ 2 / 180
o
=
237.71m
Numerical 7 • 6. Chaina Chainage ge of T2 1019.65 + 237.71 = 1257.36m
• 7. Length of long chord for small curve T 1 F = 2 R sin (φ 1 / 2 ) = 141.00m
• 8. Length of long chord for large curve T 2 F = 2 R sin (φ 2 / 2 ) = 230.24m
• Setting the first curve • The long chord T1F is divided into two equal parts, each of length 70.5m • Offsets for left half are calculated, taking peg interval of 10 m
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Numerical 7
Mid - ordinate = Oo 2
− x
2
R
2
−
( L / 2)2
= Oo =
272.40 − 272.40 2 − (70.5 / 2 )
2
= 9.28m
( R − OO )
O10
=
R
O10
=
272.40 2 − 10 2
−
O20
=
272.40 2 − 20 2
(272.40 − 9.28) = 8.54m − (272.40 − 9.28) = 7.62m
2
−
= R −
2
(272.40 − 9.28) = 9.09m
−
O30
=
272.40
O40
=
272.40 2 − 40 2
−
(272.40 − 9.28) = 6.32m
O50
=
272.40 2 − 50 2
−
(272.40 − 9.28) = 4.65m
O60
=
272.40 2 − 10 2
−
(272.40 − 9.28) = 2.58m
− 30
Numerical 7 • The offsets for right half are similar to those for the left half • Setting the second curve • The long chord T2F is divided into left and right halves. • The offsets for left half are calculated exactly as described above. The offsets of the right half are similar o those of the left half. • Then both the halves are set out in similar manner.
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Numerical 8 • A reverse is to be set set out to connect connect two parallel railway line 30m apart. The distance between the tangent points is 150m. Both the arcs have the same radius. The curve is to be set out by the method of ordinates from long chord, taking peg interval of 10m. Calculate necessary data for setting the curve. • Solution • AB and CD represent the parallel lines. The line line T1T2 is the distance between the tangent tangent points, which is given given as 150m.
Numerical 8 • We know that T 1T 2
= T 1 F + T 2 F =
T 1T 2
=
2 R sin[φ / 2] + 2 R sin[φ / 2]
4 R sin[φ / 2]
sin[φ / 2] = 30 / 150
150 = 4 R (30 / 150 ) R = 187.5m
• As sin[φ / 2] = 30 / 150 o
φ = 23
• Horizontal Distance between T1 and T2 GH = GF + FH = R sin =
[φ ] + R sin[φ ]
2 ×187.5 sin [23
o
] = 146.52m
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Numerical 8 • Setting out of curve • The reverse curve is symmetrical, so the long chord for each curve =T1F=T2F=75m. The long chord of the first curve is divided into two halves (left and right) and ordinates are calculated for he left half. The ordinates for the right half will be similar to those for the left half. Taking peg interval of 10m, the ordinates for the left half are calculated as follows Mid - ordinate = Oo 2
− x
2
−
= R −
R
2
−
( L / 2 )2
= Oo = 187.5 −
187.52 − (37.5 / 2 )
2
= 3.78m
( R − OO )
O10
=
R
O10
=
187.52 − 10 2
O20
=
187.52 − 20 2
−
(187.5 − 3.78) = 2.71m
O30
=
187.52 − 30 2
−
(187.5 − 3.78) = 1.36m
−
(187.5 − 3.78) = 3.51m
Numerical 9 • While surveying along the the alignment of a road, the magnetic magnetic bearings of the line AB, BC and CD are measured as 70, 120 and 50 degrees respectively. The length of BC is 250m and the chaina chainage ge of B is 1050m 1050m.. Calculate the necessary data for setting out a reverse curve connecting connecting the lines AB and CD by taking BC as the common tangent. The radii of both curves may be assumed equal.
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THANK YOU • Questions…. • Comments… • Suggestions…
• Feel free to contact
[email protected]
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