BUREAU OF DESIGN AND ENVIRONMENT
SURVEY MANUAL
May 2001
Illinois CURVES May 2001
A-1
APPENDIX - A
CURVES
The radius can be used to define a curve in lieu of the degree of curve definition to describe
horizontal curves.
For projects designed or constructed in English units of measure the arc definition of degree of
curve is primarily used. The arc definition is defined as the change of direction of the central
angle per an arc length of 100 feet. The relationship between the degree of curve and the
radius in feet is stated by the following equation:
R = 5729.58
D
Some of the various definitions and relationships of a simple curve are illustrated in Figure A.1,
page A-9 and given as follows:
(Δ) = Deflection Angle
Tangent distance (T) = R(Tan Δ/2)
External distance (E) = R(Sec Δ/2 -1) = R(Exsec Δ/2) = T (Sin Δ/2) - M
Middle Ordinate (M) = R(1-Cos Δ/2) = R(Vers Δ/2) = R – R (Cos Δ/2)
Long Chord (L.C.) = 2R(Sin Δ/2) = 2T(Cos Δ/2)
Length of Curve (L) = Δ(100) = Δ2πR = ΔR
D 360° 57.2958
The deflection angle (Δ) is measured in decimal degrees. For English units D is measured in
decimal degrees and R is in feet.
In addition to simple curves, compound curves, reverse curves and vertical curves are also
used in highway work.
Compound curves are a combination of two or more simple curves and their use should be
avoided where a simple curve can be used. However, due either to right-of-way problems or
topographic considerations, a compound curve may occasionally be necessary. The curve
must not be compounded at a ratio greater than 2:1. If the adjacent curves differ by more than
2º-00', a transition curve should be used.
Illinois CURVES May 2001
A-2
A reverse curve is a combination of two simple curves of opposite curvature with a common
tangent. A tangent adequate in length to provide the superelevation transition required by the
Design Policies should be provided between the curves. If the reverse curves do not contain
any superelevations, a tangent between the curves is not required.
A “Broken-back” curve is a term used to denote two curves in the same direction separated by
a short tangent or by a flat curve whose radius is greater than twice the radius of either of the
two initial curves. This layout is particularly objectionable on highways and should be avoided
by using one simple curve or a compound curve if necessary.
The curves described above are illustrated in Figure A.2, page A-10.
In the past, there was a period when transition or spiral curves were extensively used in
connection with pavement widening at curves. As the curves became flatter and the pavement
wider, it became unnecessary to widen at curves in this manner. It remains, however, that
vehicular paths entering and leaving circular curves follow a spiral curve. For this reason or for
reasons of “fitting” an alignment into a problem area, transition curves may occasionally be
used.
Vertical curves are used to transfer the smoothly change from one slope to another along an
alignment to account for the change in terrain. See Figure A.3, page A-11 for a diagram of a
vertical curve.
CALCULATIONS FOR A HORIZONTAL CIRCULAR CURVE
Given:
P.I. Sta. 107+67.90, Δ = 11° 00' 00", D = 2° 30' 00"
Calculate the Radius
R = 5729.58/D
R = 5729.58/2° 30' 00"
R = 2291.83'
Tangent Distance
T = R(Tan Δ/2)
T = 2291.83(Tan 11° 00' 00"/2)
T = 220.68'
Length of Curve
L = 100 (Δ/D)
L = 100 (11° 00' 00"/ 2° 30' 00")
Illinois CURVES May 2001
A-3
L = 440.00'
External Distance
E = T(Tan Δ/4)
E = 220.68(Tan 11° 00' 00"/4)
E = 10.60'
CALCULATE P.C. AND P.T. STATIONS
P.C. Station
P.C. = P.I. Station – Tangent Distance
P.C. = 107+67.90 – 220.68
P.C. = 105+47.22
P.T. Station
P.T. = 105+47.22 + 440.00
P.T. = 109+87.22
CALCULATE DEFLECTION ANGLES
Deflection for 100' of Arc
100' Arc = D/2
100' Arc = 2° 30' 00"/2
100' Arc = 1° 15' 00"
Deflection for 50' of Arc
50' Arc = D/4
50' Arc =2° 30' 00"/4
50' Arc =0° 37' 30"
Deflection for 25' of Arc
25' Arc = D/8
25' Arc =2° 30' 00"/8
Illinois CURVES May 2001
A-4
25' Arc = 0° 18' 45"
Deflection for 1' of Arc
1' Arc = D/200
1' Arc = 2° 30' 00"/200
1' Arc = 0° 00' 45"
CALCULATE CHORD LENGTHS
Chord Length for 100' of Arc
100' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
100' Arc = (2)(2291.83)(Sin 1° 15' 00")
100' Arc = 99.99'
Chord Length for 50' of Arc
50' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
50' Arc = (2)(291.83)(Sin 0° 37' 30")
50' Arc = 50.00'
CALCULATE THE DEFLECTION FOR THE FIRST STATION FROM THE P.C. OR ANY ODD
STATION ALONG THE CURVE
1. Take the distance from the last point with a known deflection to the station you are
calculating.
2. Multiply this distance by the deflection of a 1' Arc (D/200), this will give you the
deflection between these two points.
Example: Find the deflection angle for Sta. 108+55.
(108+55 - 105+47.22) = 307.78'
307.78'(0° 00' 45") = (3° 50' 50") Note: Use decimal degrees for this calculation.
Illinois CURVES May 2001
A-5
EXAMPLE OF A FIELD BOOK SETUP FOR A HORIZONTAL CURVE
Deflections For Curve #1
Chord Deflection Total
Sta. Distance Distance Angle Deflection
105+00
+47.22 0 0 0 0 P.C.
+50 2.78' 2.78' 0° 02' 05" 0° 02' 05"
106+00 50.0 50.00 0° 37' 30" 0° 39' 35"
+50 50.0 50.00 0° 37' 30" 1° 17' 05"
107+00 50.0 50.00 0° 37' 30" 1° 54' 35"
+50 50.0 50.00 0° 37' 30" 2° 32' 05"
108+00 50.0 50.00 0° 37' 30" 3° 09' 35"
+50 50.0 50.00 0° 37' 30" 3° 47' 05"
109+00 50.0 50.00 0° 37' 30" 4° 24' 35"
+50 50.0 50.00 0° 37' 30" 5° 02' 05"
+87.22 37.22 37.22 0° 27' 55" 5° 30' 00" P.T.
Calculated by (Initials) (Date)
Checked by (Initials) (Date)
Δ = 11° 00'00" D = 2° 30' 00"
P.C. Marked by P.K. Nail
Deflection Angles Chord Lengths
100' of Arc = D/2 = 2° 30' 00" 100' of Arc = (2)(R)(Sin of Deflection
Angle)
= 1° 15' 00" = (2)(2291.83)(Sin 1° 15' 00")
50' of Arc = D/4 = 2° 30' 00"/4 = 99.99'
= 0° 37'30" 50' of Arc = (2)(R)(Sin of Deflection
Angle)
1’ of Arc = D/200 = 2° 30' 00"/200 = (2)(2291.83)(Sin 0° 37' 00")
= 0° 00' 45" = 50.00'
P.T. (Note: Total Deflection should equal Δ/2)
CALCULATIONS FOR VERTICAL CURVE
Please refer to Figure A.3 on page A-11 of this appendix for a graphic view of the
components of a vertical curve. Following are several definitions of the elements of a
vertical curve and a set of sample field notes that have been prepared for field use to stake
out a vertical curve.
Definitions:
PVC: “Point of vertical curve”. Station on centerline where the vertical curve starts.
PVI: “Point of vertical intersection”. Station at which the two tangent grade lines intersect.
Illinois CURVES May 2001
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PVT. “Point of vertical tangency”. Station on centerline where the vertical curve ends.
LVC. Length of vertical curve.
OFFSET: The vertical distance from the tangent grade line to the vertical curve.
e: a mathematical constant whose value is determined by the grades of the two intersecting
tangents and the length of the vertical curve.
e: Grade #2(%) – Grade#1 (%) x (LVC) (Stations)/8 (Constant) = (G2-G1)(LVC/100)/8
CL ELEV: = Tangent Elevation + offset.
Highpoint/low point locations.
Distance from PVC = G1 x (Stations)/G2 –G1
Sample set of field notes for a vertical curve.
Tangent Elevation
Station (X) Distance Elevation (Y)offset on Curve
45+50 429.34
46+00 50.00 428.84 0.03 428.87
47+00 150.00 427.84 0.26 428.10
48+00 250.00 426.84 0.71 427.55
49+00 350.00 425.84 1.40 427.24
49+50 400.00 425.34 1.83 427.17
50+00 450.00 424.84 2.31 427.15
50+50 500.00 424.34 2.86 427.20
51+00 550.00 423.84 3.46 427.30
52+00 650.00 422.84 4.83 427.67
52+50 700.00 422.34 5.60 427.94
53+00 650.00 423.44 4.83 428.27
54+00 550.00 425.64 3.46 429.10
55+00 450.00 427.84 2.31 430.15
56+00 350.00 430.04 1.40 431.44
57+00 250.00 432.24 0.71 432.95
58+00 150.00 434.44 0.26 434.70
59+00 50.00 436.64 0.03 436.67
59+50 437.74 437.74
e = (G2-G1)LVC/8 = (2.2-(-1.0))14/8 = 3.2(14)/8 = 5.60 Vertical offset at
VPI.
Y-Offset @ 46+00 & 59+00 = (50/700)2(5.60) = 0.03
47+00 & 58+00 = (150/700)2(5.60) = 0.26
48+00 & 57+00 = (250/700)2(5.60) = 0.71 Y = e(X/L)2
49+00 & 56+00 = (350/700)2(5.60) = 1.40
49+50 = (400/700)2(5.60) = 1.83
50+00 & 55+00 = (450/700)2(5.60) = 2.31
50+50 = (500/700)2(5.60) = 2.86
51+00 & 54+00 = (550/700)2(5.60) = 3.46
52+00 & 53+00 = (650/700)2(5.60) = 4.83
52+50 = (700/700)2(5.60) = 5.60
When using this method the elevation difference is calculated from both the tangent
gradients at specified distances from the PVC and PVT and then applied to the tangent
Illinois CURVES May 2001
A-7
elevations determined for the two gradients. The value of (e) is the offset at the VPI and
each individual station’s offset value is determined as a percentage of the external at the
VPI. The vertical offsets from a tangent to a parabola are proportional to the squares of the
distances from the point of tangency.
STATION AND ELEVATION OF LOW POINT OF VERTICAL CURVE
X (In Station from VPC) = G1L/(G2-G1) = 1.0(14)/(2.2-(-1.0)) = 1.0(14)/3.2 = 4.375 Sta. or
437.50'.
Station of Low Point = 45+50 + (4+37.5) = 49+87.50
X-Distance = 437.50 Tangent Elev. = 429.34 – 437.50(0.01) = 424.96'
Y-Offset (437.50/700)2(5.60) = 2.19'
Elevation of Low Point = 424.96 + 2.19 = 427.15'
ALTERNATE METHOD OF CALCULATING CURVE ELEVATIONS
An alternative method of calculating the elevations of a vertical curve is as follows: calculate
the value of (a) using the following formula: The tangent elevations are computed using the
elevation of the PVC and the slope of the forward tangent. An elevation is computed for
each station needed. Then compute the offsets from the forward tangent to the curve. The
offset equals aX2. Apply the offset values to the tangent elevation to obtain the curve
elevation.
a = 100(G2-G1)/2L = 100(2.2-(-1.0))/2*1400 = 0.114
2a = 0.228
Station Tangent Offsets from AV=aX2 Curve. Check
El. =G1X Elev. 1st Diff. 2nd Diff
45+50 429.34 429.34
46+00 428.84 -.114*0.502 =0.03 428.87
47+00 427.84 -.114*1.52 =0.256 428.10 -0.77
48+00 426.84 -.114*2.52 =0.71 427.55 -0.55 0.22
49+00 425.84 -.114*3.52 =1.40 427.24 -0.31 0.24
49+50 425.34 -.114*4.02 =1.83 427.17
50+00 424.84 -.114*4.52 =2.31 427.15 -0.09 0.22
50+50 424.34 -.114*5.02 =2.86 427.20
51+00 423.84 -.114*5.52 =3.46 427.30 +0.15 0.24
52+00 422.84 -.114*6.52 =4.83 427.67 +0.37 0.22
52+50 422.34 -.114*7.02 =5.60 427.94
53+00 421.84 -.114*7.52 =6.43 428.27 +0.60 0.23
54+00 420.84 -.114*8.52 =8.26 429.10 +0.83 0.23
55+00 419.84 -.114*9.52 =10.31 430.15 +1.05 0.23
56+00 418.84 -.114*10.52 =12.60 431.44 +1.29 0.24
57+00 417.84 -.114*11.52 =15.11 432.95 +1.51 0.22
58+00 416.84 -.114*12.52 =17.86 434.70 +1.75 0.24
59+00 415.84 -.114*13.502 =20.83 436.67 +1.97 0.22
59+50 415.34 -.114*14.002 =22.40 437.74
Illinois CURVES May 2001
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One method of checking your calculations is to calculate the first and second differences of
the curve elevation between the full stations. The second difference should be the same
and is equal to 2a, which is the percent of constant change in slope per station.
If other methods of calculating the station vertical offsets are desired, see a survey textbook
for the procedures.
A-9
Figure A.1
A-10
Figure A.2
A-11
SURVEY MANUAL
May 2001
Illinois CURVES May 2001
A-1
APPENDIX - A
CURVES
The radius can be used to define a curve in lieu of the degree of curve definition to describe
horizontal curves.
For projects designed or constructed in English units of measure the arc definition of degree of
curve is primarily used. The arc definition is defined as the change of direction of the central
angle per an arc length of 100 feet. The relationship between the degree of curve and the
radius in feet is stated by the following equation:
R = 5729.58
D
Some of the various definitions and relationships of a simple curve are illustrated in Figure A.1,
page A-9 and given as follows:
(Δ) = Deflection Angle
Tangent distance (T) = R(Tan Δ/2)
External distance (E) = R(Sec Δ/2 -1) = R(Exsec Δ/2) = T (Sin Δ/2) - M
Middle Ordinate (M) = R(1-Cos Δ/2) = R(Vers Δ/2) = R – R (Cos Δ/2)
Long Chord (L.C.) = 2R(Sin Δ/2) = 2T(Cos Δ/2)
Length of Curve (L) = Δ(100) = Δ2πR = ΔR
D 360° 57.2958
The deflection angle (Δ) is measured in decimal degrees. For English units D is measured in
decimal degrees and R is in feet.
In addition to simple curves, compound curves, reverse curves and vertical curves are also
used in highway work.
Compound curves are a combination of two or more simple curves and their use should be
avoided where a simple curve can be used. However, due either to right-of-way problems or
topographic considerations, a compound curve may occasionally be necessary. The curve
must not be compounded at a ratio greater than 2:1. If the adjacent curves differ by more than
2º-00', a transition curve should be used.
Illinois CURVES May 2001
A-2
A reverse curve is a combination of two simple curves of opposite curvature with a common
tangent. A tangent adequate in length to provide the superelevation transition required by the
Design Policies should be provided between the curves. If the reverse curves do not contain
any superelevations, a tangent between the curves is not required.
A “Broken-back” curve is a term used to denote two curves in the same direction separated by
a short tangent or by a flat curve whose radius is greater than twice the radius of either of the
two initial curves. This layout is particularly objectionable on highways and should be avoided
by using one simple curve or a compound curve if necessary.
The curves described above are illustrated in Figure A.2, page A-10.
In the past, there was a period when transition or spiral curves were extensively used in
connection with pavement widening at curves. As the curves became flatter and the pavement
wider, it became unnecessary to widen at curves in this manner. It remains, however, that
vehicular paths entering and leaving circular curves follow a spiral curve. For this reason or for
reasons of “fitting” an alignment into a problem area, transition curves may occasionally be
used.
Vertical curves are used to transfer the smoothly change from one slope to another along an
alignment to account for the change in terrain. See Figure A.3, page A-11 for a diagram of a
vertical curve.
CALCULATIONS FOR A HORIZONTAL CIRCULAR CURVE
Given:
P.I. Sta. 107+67.90, Δ = 11° 00' 00", D = 2° 30' 00"
Calculate the Radius
R = 5729.58/D
R = 5729.58/2° 30' 00"
R = 2291.83'
Tangent Distance
T = R(Tan Δ/2)
T = 2291.83(Tan 11° 00' 00"/2)
T = 220.68'
Length of Curve
L = 100 (Δ/D)
L = 100 (11° 00' 00"/ 2° 30' 00")
Illinois CURVES May 2001
A-3
L = 440.00'
External Distance
E = T(Tan Δ/4)
E = 220.68(Tan 11° 00' 00"/4)
E = 10.60'
CALCULATE P.C. AND P.T. STATIONS
P.C. Station
P.C. = P.I. Station – Tangent Distance
P.C. = 107+67.90 – 220.68
P.C. = 105+47.22
P.T. Station
P.T. = 105+47.22 + 440.00
P.T. = 109+87.22
CALCULATE DEFLECTION ANGLES
Deflection for 100' of Arc
100' Arc = D/2
100' Arc = 2° 30' 00"/2
100' Arc = 1° 15' 00"
Deflection for 50' of Arc
50' Arc = D/4
50' Arc =2° 30' 00"/4
50' Arc =0° 37' 30"
Deflection for 25' of Arc
25' Arc = D/8
25' Arc =2° 30' 00"/8
Illinois CURVES May 2001
A-4
25' Arc = 0° 18' 45"
Deflection for 1' of Arc
1' Arc = D/200
1' Arc = 2° 30' 00"/200
1' Arc = 0° 00' 45"
CALCULATE CHORD LENGTHS
Chord Length for 100' of Arc
100' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
100' Arc = (2)(2291.83)(Sin 1° 15' 00")
100' Arc = 99.99'
Chord Length for 50' of Arc
50' Arc = (2)(R)(Sin of Deflection Angle) Deflection Angle = D/2
50' Arc = (2)(291.83)(Sin 0° 37' 30")
50' Arc = 50.00'
CALCULATE THE DEFLECTION FOR THE FIRST STATION FROM THE P.C. OR ANY ODD
STATION ALONG THE CURVE
1. Take the distance from the last point with a known deflection to the station you are
calculating.
2. Multiply this distance by the deflection of a 1' Arc (D/200), this will give you the
deflection between these two points.
Example: Find the deflection angle for Sta. 108+55.
(108+55 - 105+47.22) = 307.78'
307.78'(0° 00' 45") = (3° 50' 50") Note: Use decimal degrees for this calculation.
Illinois CURVES May 2001
A-5
EXAMPLE OF A FIELD BOOK SETUP FOR A HORIZONTAL CURVE
Deflections For Curve #1
Chord Deflection Total
Sta. Distance Distance Angle Deflection
105+00
+47.22 0 0 0 0 P.C.
+50 2.78' 2.78' 0° 02' 05" 0° 02' 05"
106+00 50.0 50.00 0° 37' 30" 0° 39' 35"
+50 50.0 50.00 0° 37' 30" 1° 17' 05"
107+00 50.0 50.00 0° 37' 30" 1° 54' 35"
+50 50.0 50.00 0° 37' 30" 2° 32' 05"
108+00 50.0 50.00 0° 37' 30" 3° 09' 35"
+50 50.0 50.00 0° 37' 30" 3° 47' 05"
109+00 50.0 50.00 0° 37' 30" 4° 24' 35"
+50 50.0 50.00 0° 37' 30" 5° 02' 05"
+87.22 37.22 37.22 0° 27' 55" 5° 30' 00" P.T.
Calculated by (Initials) (Date)
Checked by (Initials) (Date)
Δ = 11° 00'00" D = 2° 30' 00"
P.C. Marked by P.K. Nail
Deflection Angles Chord Lengths
100' of Arc = D/2 = 2° 30' 00" 100' of Arc = (2)(R)(Sin of Deflection
Angle)
= 1° 15' 00" = (2)(2291.83)(Sin 1° 15' 00")
50' of Arc = D/4 = 2° 30' 00"/4 = 99.99'
= 0° 37'30" 50' of Arc = (2)(R)(Sin of Deflection
Angle)
1’ of Arc = D/200 = 2° 30' 00"/200 = (2)(2291.83)(Sin 0° 37' 00")
= 0° 00' 45" = 50.00'
P.T. (Note: Total Deflection should equal Δ/2)
CALCULATIONS FOR VERTICAL CURVE
Please refer to Figure A.3 on page A-11 of this appendix for a graphic view of the
components of a vertical curve. Following are several definitions of the elements of a
vertical curve and a set of sample field notes that have been prepared for field use to stake
out a vertical curve.
Definitions:
PVC: “Point of vertical curve”. Station on centerline where the vertical curve starts.
PVI: “Point of vertical intersection”. Station at which the two tangent grade lines intersect.
Illinois CURVES May 2001
A-6
PVT. “Point of vertical tangency”. Station on centerline where the vertical curve ends.
LVC. Length of vertical curve.
OFFSET: The vertical distance from the tangent grade line to the vertical curve.
e: a mathematical constant whose value is determined by the grades of the two intersecting
tangents and the length of the vertical curve.
e: Grade #2(%) – Grade#1 (%) x (LVC) (Stations)/8 (Constant) = (G2-G1)(LVC/100)/8
CL ELEV: = Tangent Elevation + offset.
Highpoint/low point locations.
Distance from PVC = G1 x (Stations)/G2 –G1
Sample set of field notes for a vertical curve.
Tangent Elevation
Station (X) Distance Elevation (Y)offset on Curve
45+50 429.34
46+00 50.00 428.84 0.03 428.87
47+00 150.00 427.84 0.26 428.10
48+00 250.00 426.84 0.71 427.55
49+00 350.00 425.84 1.40 427.24
49+50 400.00 425.34 1.83 427.17
50+00 450.00 424.84 2.31 427.15
50+50 500.00 424.34 2.86 427.20
51+00 550.00 423.84 3.46 427.30
52+00 650.00 422.84 4.83 427.67
52+50 700.00 422.34 5.60 427.94
53+00 650.00 423.44 4.83 428.27
54+00 550.00 425.64 3.46 429.10
55+00 450.00 427.84 2.31 430.15
56+00 350.00 430.04 1.40 431.44
57+00 250.00 432.24 0.71 432.95
58+00 150.00 434.44 0.26 434.70
59+00 50.00 436.64 0.03 436.67
59+50 437.74 437.74
e = (G2-G1)LVC/8 = (2.2-(-1.0))14/8 = 3.2(14)/8 = 5.60 Vertical offset at
VPI.
Y-Offset @ 46+00 & 59+00 = (50/700)2(5.60) = 0.03
47+00 & 58+00 = (150/700)2(5.60) = 0.26
48+00 & 57+00 = (250/700)2(5.60) = 0.71 Y = e(X/L)2
49+00 & 56+00 = (350/700)2(5.60) = 1.40
49+50 = (400/700)2(5.60) = 1.83
50+00 & 55+00 = (450/700)2(5.60) = 2.31
50+50 = (500/700)2(5.60) = 2.86
51+00 & 54+00 = (550/700)2(5.60) = 3.46
52+00 & 53+00 = (650/700)2(5.60) = 4.83
52+50 = (700/700)2(5.60) = 5.60
When using this method the elevation difference is calculated from both the tangent
gradients at specified distances from the PVC and PVT and then applied to the tangent
Illinois CURVES May 2001
A-7
elevations determined for the two gradients. The value of (e) is the offset at the VPI and
each individual station’s offset value is determined as a percentage of the external at the
VPI. The vertical offsets from a tangent to a parabola are proportional to the squares of the
distances from the point of tangency.
STATION AND ELEVATION OF LOW POINT OF VERTICAL CURVE
X (In Station from VPC) = G1L/(G2-G1) = 1.0(14)/(2.2-(-1.0)) = 1.0(14)/3.2 = 4.375 Sta. or
437.50'.
Station of Low Point = 45+50 + (4+37.5) = 49+87.50
X-Distance = 437.50 Tangent Elev. = 429.34 – 437.50(0.01) = 424.96'
Y-Offset (437.50/700)2(5.60) = 2.19'
Elevation of Low Point = 424.96 + 2.19 = 427.15'
ALTERNATE METHOD OF CALCULATING CURVE ELEVATIONS
An alternative method of calculating the elevations of a vertical curve is as follows: calculate
the value of (a) using the following formula: The tangent elevations are computed using the
elevation of the PVC and the slope of the forward tangent. An elevation is computed for
each station needed. Then compute the offsets from the forward tangent to the curve. The
offset equals aX2. Apply the offset values to the tangent elevation to obtain the curve
elevation.
a = 100(G2-G1)/2L = 100(2.2-(-1.0))/2*1400 = 0.114
2a = 0.228
Station Tangent Offsets from AV=aX2 Curve. Check
El. =G1X Elev. 1st Diff. 2nd Diff
45+50 429.34 429.34
46+00 428.84 -.114*0.502 =0.03 428.87
47+00 427.84 -.114*1.52 =0.256 428.10 -0.77
48+00 426.84 -.114*2.52 =0.71 427.55 -0.55 0.22
49+00 425.84 -.114*3.52 =1.40 427.24 -0.31 0.24
49+50 425.34 -.114*4.02 =1.83 427.17
50+00 424.84 -.114*4.52 =2.31 427.15 -0.09 0.22
50+50 424.34 -.114*5.02 =2.86 427.20
51+00 423.84 -.114*5.52 =3.46 427.30 +0.15 0.24
52+00 422.84 -.114*6.52 =4.83 427.67 +0.37 0.22
52+50 422.34 -.114*7.02 =5.60 427.94
53+00 421.84 -.114*7.52 =6.43 428.27 +0.60 0.23
54+00 420.84 -.114*8.52 =8.26 429.10 +0.83 0.23
55+00 419.84 -.114*9.52 =10.31 430.15 +1.05 0.23
56+00 418.84 -.114*10.52 =12.60 431.44 +1.29 0.24
57+00 417.84 -.114*11.52 =15.11 432.95 +1.51 0.22
58+00 416.84 -.114*12.52 =17.86 434.70 +1.75 0.24
59+00 415.84 -.114*13.502 =20.83 436.67 +1.97 0.22
59+50 415.34 -.114*14.002 =22.40 437.74
Illinois CURVES May 2001
A-8
One method of checking your calculations is to calculate the first and second differences of
the curve elevation between the full stations. The second difference should be the same
and is equal to 2a, which is the percent of constant change in slope per station.
If other methods of calculating the station vertical offsets are desired, see a survey textbook
for the procedures.
A-9
Figure A.1
A-10
Figure A.2
A-11
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