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3. Figure in slide 4, equal spiral curves join a circular curve
to the main tangents;
• T.S. tangent to spiral point.
• LS the length of the spiral curve.
• S.C. spiral to circular curve point.
• C.S. curve to spiral point.
• S.T. spiral to tangent point.
• TS tangent distance from the T.S. or the S.T to the P.I.
• R the radius of the circular curve.
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4. P.I.

TS
TS
Circular curve
T.S. S.T.
C.S.
S.C.
LS LS
R
R
1st spiral curve 2nd spiral
curve
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5. S.P.I. S.C.
Circular curve
T.S.
R
Enlargement of Spiral Curve
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6. Figure in slide 5, the spiral curve on the left side redrawn
to a much larger scale.
• S.P.I. the spiral point of intersection.
• S the spiral angle.
• S.T. and L.T. are the short and long tangents of
the spiral curve.
• c the degrees of curvature of the circular curve.
• XS the distance measured from the T.S. along the
main tangent to a point where a perpendicular line
to the tangent hits the S.C.
• YS the distance measured perpendicular from the
XS coordinate to the S.C.
• L.C. the long chord from the T.S. to the S.C.
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7. • P.C. the point at which the circular curve becomes
parallel to the spiral. The curves will be a distance p
apart.
• S the deflection angle from the T.S. to the S.C.
• CS the correction factor, negligible when   15.
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8. Spiral Curve Equations
5729.58
R

LS  C
S 
200
 S 2 S 4 
X S  L S 1 
  
 10 216 

 S S3 
YS  LS 
 3  42 

 
L.C.  X S  YS
2 2
YS
S .T . 
sin  S
L.T .  X S  S .T . cos  S
S
S 
3
p  YS  R (1  cos  S )

TS  X S  R sin  S  ( R  p ) tan
2
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9. Steps of Laying Out A Spiral Curve
• LS is selected considering: traffic design, speed, No. of
lanes, c and the length needed for super-elevation.
• The values for R, S, XS, YS, L.C., S.T., L.T., S, p and TS
are computed.
• The chord lengths are assumed and the deflection angle
L
is determined from   ( L ) 2
S
S
• The curve will be staked out in identical manner used for
circular curves.
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10. Approximate Solution for Spiral Problems
LS
S.C.
(2/3) S
T.S.
S = (1/3) S
Basic Assumption: LS  Long Chord, LS  L.C.
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11. Approximate Equations
Y  LS sin  S
X  LS  Y 2
2
1
q X
2
1
p Y
4
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12. Approximate Equations
Using the sine law, we obtain the following:
L.T . LS

2 sin (180   S )
sin (  S )
3
2 LS
L.T .  sin (  S ) 
3 sin  S
Using the sine law, we obtain:
S .T . LS

S sin  S
sin
3
S LS
S .T .  sin 
3 sin  S
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13. An Example of Compared Values
Referring to previous solved problem:
Precise Methods Approximate Methods
Parameters (ft) Parameters (ft)
Y 10.46 Y 10.47
X 299.67 X 299.82
q 149.94 q 149.91
p 2.61 p 2.62
L.T. 200.15 L.T. 200.20
S.T. 100.10 S.T. 100.16
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