The Banked Track Curve
Segments
Figure 2, below, illustrates the
curve segments of the banked
track. In all that follows it
will be assumed that the curves
at each end of the track are
equal, and that each curve is
symmetrical about the long axis
of the track. Point 0 at the
beginning of the curve is the
beginning of the transition to
the ascending segment of the
curve, point 1 marks the end of
the transition and the beginning
of the ascending segment, point
2 is the end of the ascending
segment and the beginning of the
transition to the constant
banking segment, and point 3 is
the end of that transition and
the beginning of the constant
banking segment.
Continuing, point 4 is the end
of the constant banking segment
and the beginning of the
transition to the descending
segment, point 5 is the end of
that transition and the
beginning of the descending
segment, point 6 is the end of
the descending segment and the
beginning of the transition to
the straightaway, and point 7 is
the end of that transition at
the end of the curve.
The Differential Element of
Arc Length
For a banked track, cylindrical
coordinates are the natural
coordinate system to use. In
Cartesian coordinates in three
dimensions, the differential
element of arc length, ds, is
given by
ds2 = dx2 + dy2 + dz2. (4)
In cylindrical coordinates the
differential element of arc
length is given by
ds2 = dr2 + (rdθ)2 + dz2 (5)
The differential element of arc
length in cylindrical
coordinates is illustrated in
Figure 3.
We can simplify equation (5) by
noting that r and z are both
functions of θ. We can then
write
dr =(∂r/∂θ)dθ (6a), dz = (∂z/∂θ)dθ
(6b)
and
ds = √((∂r/∂θ)2 + r2 +
(∂z/∂θ)2)dθ (7)
It is the differential in
equation (7) that we seek to
integrate in a piecewise fashion
around the seven segments. Note
that for a flat track = ∂r/∂θ =
∂z/∂θ = 0 and equation (7)
becomes simply ds = rdθ, which
when integrated around the
entire curve becomes the well
known result πr.
The Banked Track Cross
Section
Figure 4 illustrates a cross
section of the banked track for
lane i (1 ≤ i ≤ 6 typically) at
some angle θ around the turn. r0
is the radius to the track side
of the curb. di is the slant
distance from the curb to the
measure line of lane i. α is the
angle of banking at angle θ
(assumed constant across the
cross section) around the curve
(0 ≤ α ≤ α0, where α0 is the
maximum angle of banking in the
constant banking portion of the
curve, typically 8°  12°).
zi is the vertical distance
above the straightaway given by
zi = disinα (8a)
and
ri = r0 + dicosα (8b)
Constant Banking Segment Arc
Length
We will first consider the
easiest portion of the arc
length calculation, the constant
banking segment (from points 3
4 in Figure 2). Throughout this
segment, α = α0 and ∂r/∂θ =
∂z/∂θ = 0 and the problem
reduces to one dimension as
stated in Part I of this paper.
Thus, from (7) and (8b),
dsi = (r0 + dicosα0)dθ (9)
Since r0, di, and α0 are
constant, we obtain, upon
integration,
Sci = (r0 + dicosα0)(θ4 θ3)
(10)
with Sci the arc length of the
constant banking segment for
lane i, and θ3 and θ4 the angles
of the curve traversed (in
radians) between points 3 and 4.
Ascending and Descending
Segments Arc Length
Because of symmetry, the arc
length of the descending segment
is the same as the ascending
segment, so we will only
consider the ascending portion.
Because the builders of banked
tracks have not shared precise
construction details with the
author, certain assumptions must
be made. Two likely schemes come
to mind with regard to the
ascending segment. One where the
angle of banking increases
linearly with the angle
traversed, and one where the z
coordinate increases linearly.
In practice, because the angle
of banking is small, there will
be little difference in the two
approaches.
In what follows we assume the
angle of banking increases
linearly with angle traversed.
The angle of banking will
therefore be given by
α = α1 + (θ θ1)(α2  α1)/(θ2
θ1) (11)
with α1 and α2 the angles of
banking and θ1 and θ2 the angles
traversed at points 1 and 2,
respectively. By substitution
into (8a) and (8b), zi and ri
become
zi = disin[α1 + (θ θ1)( α2 
α1)/(θ2 θ1)] (12a)
ri = r0 + dicos[α1 + (θ θ1)(
α2  α1)/(θ2 θ1)] (12b)
∂zi/∂θ and ∂ri/∂θ are given by
∂zi/∂θ = dicos[α1 + (θ θ1)(α2
 α1)/(θ2 θ1)][(α2  α1)/(θ2
θ1)] (13a)
∂ri/∂θ =  disin[α1 + (θ
θ1)(α2  α1)/(θ2 θ1)][(α2 
α1)/(θ2 θ1)] (13b)
Substituting into (7) and
integrating (after
simplification), we obtain
Sai = ∫√(r02 + 2r0dicos[α1 + (θ
θ1)(α2  α1)/(θ2 θ1)] +
di2cos2[α1 + (θ θ1)(α2 
α1)/(θ2 θ1)] + di2[(α2 
α1)/(θ2 θ1)]2)dθ (14)
Where Sai is the arc length of
the ascending (and descending)
segment of the curve in lane i.
Equation (14) is not integrable
in closed form, but since the
integrand does not change
rapidly with θ it can be readily
integrated with high precision
using a standard numerical
method such as Simpsons Rule.
In Part III of this paper we
will calculate arc length for
the most difficult segments of
the curve, the transition
segments, and summarize the
three parts of the paper. 
