Formulation of shift of a circular curve with unequal transition length

Formulation of shift of a circular curve with unequal transition length September 22, 2013 Formulation of co-ordinates for circular curve. Normally rail or road alignment moves through different curvatures. It may consists of straight lines, circular curves and compound curves. This paper presents the generation of co-ordinates or the alignment having straight lines and circular curves.

First of all the relations will be drawn in local co-ordinate system which can be very easily converted into WGS-84 or any other type of global co-ordinate system. Global co-ordinate does not mean only standard co-ordinates system like WGS-84 but a co-ordinate system adopted for a project.

Straight Line For given two number of points other points between them can easily be find out as the relationship between them will remain linear.

Circular Curve Let us consider a circular curve of radius ’R’ between the tangents T1 O and T2 O . The deviation angle between the tangents are δ. The center of the circular curve is c1 . The circular curve touches the left and right hand tangents at A and B respectively.

Take a case when the circular curve is shifted in such a way that the center of the curve assigns a new position c2 from its old position c1 . Absolute distance between c1 and c2 is s. c2 M and c2 N are perpendicular to the tangents from c2 . Refer figure 1 on page no. 4 Transition curve starts from T1 and T2 . Transition curves are widely accepted in the form of cubical parabola. Let us take a cubical parabola as

1

x = ay 3 + k,

(1)

where,a and k is a constant dx/dy = 3ay 2 d2 x/dy 2 = 6ay 1 R

=

d2 x dy 2 3

[1+( dx )2 ] 2 dy

Assuming as, dx/dy ≈ 0, within transition length This assumption fairly accurate if L R 2

= ddyx2 , at point P1 where transition curve touches circular curve, curvature of both curves will be same, 1 R

at P1 , y = L1 , x = 0 1 R

= 6ay = 6aL1

a=

1 6L1 R

at T1 , x = 0, y = 0, therefore, k = 0 Assuming again, scos(φ1 ) ,y 2

at M, x =

= L1 /2

Therefore, s cos(φ1 ) 2

= a( L21 )3

s cos(φ1 ) 2

=

1 ( L1 )3 6L1 R 2

s cos(φ1 ) 2

=

L21 48R

scos(φ1 ) =

L21 24R

(2)

This way the above mentioned assumption, basically serves as a boundary 2

condition to find out the value of s. Refer figure no. 2 and 3 Similarly, to the other side of the transition curve where the length of the transition curve is L2 , 6 (c1 c2 N ) = φ2 s cos(φ2 ) =

L22 24R

(3)

also, AM = s. sin(φ1 ), AM =

L21 . tan(φ1 ) 24R

similarly, L22 BN = 24R . tan(φ2 ) Say, S1 =

L21 24R

and S2 =

L22 24R

Hence, AM = S1 . tan(φ1 ) and BN = S2 . tan(φ2 ), where S1 =

L21 24R

and S2 =

L22 24R

Finding co-ordinate of different portions of a circular curve in a given co-ordinate system Let us take co-ordinate system (X, Y ) as shown in figure 1 on page no. 4 as our global co-ordinate system. Presented below the a typical way to find out the co-ordinates of different portions of a circular curve as represented in figure no. 1. Here the case has been represented as right hand curve.

1. First Transition Curve: The length of the transition curve is L1 , which extends from T1 to P1 . The chosen form of transition curve is a cubical parabola as explained above, x = ay 3 , where a =

1 , 6L1 R

and 0 ≤ y ≥ L1

2. Second Transition Curve: This transition curve which is of length L2 and extends between T2 and P2 as shown in figure 1. Let us first define the co-ordinates in local co-ordinate as shown in figure no 1 as X 0 , Y 0 x0 = ay 03 , where a =

1 , 6L2 R

and 0 ≤ y 0 ≤ L2

Now, the co-ordinates found out in local co-ordinate (X 0 , Y 0 ) could be transferred in global co-ordinate (X, Y ) by using transformation equations as ex3

Figure 1: Curve shown with shift plained below,

x = xo + x0 cos(θ) − y 0 sin(θ)

(4)

y = yo + x0 sin(θ) + y 0 cos(θ)

(5)

(xo , yo ) are the co-ordinate of origin of (X 0 , Y 0 ) with respect to (X, Y ) coordinate system. here in this case, θ = (180 − δ) deg, if δ is in deg and replace x0 by −x0 , 4

Figure 2: A zoom portion of curve so that the orientation of (X 0 , Y 0 ) be similar to that of (X, Y ) Now, consider triangle ∆T1 OT2 , where 6 O = 180 − δ, T1 O = 0.5L1 + S1 tan(φ1 ) + R tan(δ/2)

(6)

T2 O = 0.5L2 + S2 tan(φ2 ) + R tan(δ/2)

(7)

Also, xo = T2 O.sin(δ) and yo = T1 O + T2 O. cos(δ)

5

Figure 3: A zoom portion of curve

6

φ1 + φ2 = δ using equations 2 and 3 cos(φ1 ) cos(φ2 )

=

L21 L22

−1

φ1 = tan

[

L22 L21

− cos(δ) sin(δ)

]

(8)

Therefore, the equation 4 and 5 can be evaluated 3. Circular Curve: A curve of Radius R has to fit between the two transition curves. Refer figure no. 4. The circular curve AV B and its shifted portion cuts the bisector at point V and v. If V v = s1 then c1 v = R − s1 , c2 v = R and c1 c2 = s. From ∆(vc1 c2 ), 6 vc1 c2 = 180 − [φ1 − 2δ ] cos(6 vc1 c2 ) =

(vc1 )2 +(c1 c2 )2 −(vc2 )2 2×vc1 ×c1 c2

cos(180 − [φ1 − 2δ ]) =

(R−s0 )2 +(s)2 −(R)2 2×(R−s0 )×s

Simplifying above equation, if K = cos(φ1 − 2δ ) (s0 )2 − 2.s0 .(R + s.K) + s2 + 2.s.R.K There will be two solution for this quadratic equation, and for s0 < R

s0 = (R + s.K) −

q

R2 − s2 (1 − K 2 )

Therefore, intercept of V v on Y axis will be V vt = s0 × sin(δ/2) Co-ordinate of center c2 will be (R + s. cos(φ1 ), L21 c2 = (cx , cy ) cx = R + s. cos(φ1 ) and cy =

L1 2

Equation of circle will be, 7

(9)

Figure 4: Cuve showing shift

8

(x − cx )2 + (y − cy )2 = R2 , x = R + s. cos(φ1 ) − where, L1 ≤ y ≤

L1 2

q

R2 − (y −

L1 2 ) 2

+ s. sin(φ1 ) + R. sin(δ/2) − s0 . sin(δ/2)

Similarly, with respect to (x0 , y 0 ) also remaining portion of circular curve could be drawn suitably transformed to global co-ordinate system (x, y) by using transformation equations as explained in equations 4 and 5

Thanks Auther: Shailendra Kumar, Executive Engineer, Northern Railway, India Your valuable comments/suggesitions may be addressed to XENCUSBRL@GMAIL.COM or AMBASTHA.SEE@GMAIL.COM

9