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Section 2 - Horizontal Alignment

This section will define the mathematical relationships and equations for parent curve segments used to define the geometry of horizontal alignments.

General

Parameter Equation
Start Curvature $$\kappa_{s} = \frac{1}{R_{s}}$$
End Curvature $$\kappa_{e} = \frac{1}{R_{e}}$$
Cumulative change in curvature $$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = L\left( \frac{1}{R_{e}} - \frac{1}{R_{s}} \right)$$
Angle of tangent line, RefDirection = $$\cos\left( \theta(s) \right),\sin\left( \theta(s) \right)$$ $$\theta(s) = \int_{}^{}{\kappa(s)ds}$$
X-ordinate as a function of curve length s $$x(s) = \int_{}^{}{\cos\left( \theta(s) \right)ds}$$
Y-ordinate as a function of curve length s $$y(s) = \int_{}^{}{\sin{\left( \theta(s) \right)\ ds}}$$

IfcCurveSegment Placement

The IfcCurveSegment placement is provided in the IFC. This placement is represented in matrix form as

$$M_{PC} = \begin{bmatrix} dx & - dy & 0 & x \\ dy & dx & 0 & y \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Column 1 represents the RefDirection (dx,dy,0)

Column 3 represents the Axis (0,0,1). Axis is always "up" normal to the 2D plane

Column 4 represents the Location (x,y,0). Z is always zero because the horizontal curve lies in the X-Y plane

Column 2 is the Y direction is the cross product of RefDirection and Axis.

Line

When IfcLine is used as a parent curve, it represents a tangent run of the alignment.

Source Model: FHWA Alignment Model

  1. Parent Curve Parametric Equations

The equation of a line is parameterized as

$$\theta(t) = \theta$$

$$\kappa(t) = 0$$

$$\lambda(u) = C + L\left( \int_{0}^{u = 1}{\cos\left( \theta(t) \right)}dt\ x,\ \int_{0}^{u = 1}{\sin\left( \theta(t) \right)}dt\ y \right)$$

$$\lambda(u) = C + \left( \int_{0}^{u = L}{\cos\left( \theta(t) \right)}dt\ x,\ \int_{0}^{u = L}{\sin\left( \theta(t) \right)}dt\ y \right)$$

⚠️ [Discuss that a line (and the other parent curves) can be parametetized differently and there isn't a standards therefore the rule that IfcCurveSegment.Start and Length IfcCurveMeasureSelect must be IfcLengthMeasure.] ⚠️

Semantic Definition to Geometry Mapping

Mapping of the semantic definition of the linear segment to the geometric definition is described with the following example.

Given a horizontal alignment segment is a line segment starting at point (500,2500), bearing in the direction 5.70829654085293 and has a length of 1956.785654. This segment is represented as

#23=IFCCARTESIANPOINT((500.,2500.));
#32=IFCALIGNMENTHORIZONTALSEGMENT($,$,#23,5.70829654085293,0.,0.,1956.785654,$,.LINE.);

The geometric representation is an IfcLine. The line can be defined in different ways and is generally not important, as will be shown in the example below. IfcLine is an infinitely long line that passes through a specified point at some direction in the X-Y plane. An easy way to define the IfcLine parent curve is to have in the X-axis direction passing through point (0,0).

Define the parent curve

#51=IFCCARTESIANPOINT((0.,0.));
#52=IFCDIRECTION((1.,0.));
#53=IFCVECTOR(#52,1.);
#54=IFCLINE(#51,#53);

IfcCurveSegment defines the portion of the line used and how it is placed and oriented in the X-Y plane.

IfcCurveSegment.SegmentStart defines where the parent curve trimming begins. It can be anywhere along the line. It's easiest to begin the trimming at the origin but could be anywhere. The SegmentStart attribute is 0.0 in this case.

IfcCurveSegment.SegmentLength defines where the parent curve trimming ends relative to SegmentStart. This is the length of the curve we want to trim. The SegmentLength attribute is 1956.785654 for this example.

The SegmentStart and SegmentLength attributes trims a portion of the IfcLine parent curve, which is oriented in the direction (1,0) with origin at (0,0). The trimmed portion of the curve needs to be placed and oriented into the horizontal alignment, but in this case, it is trivial and the IfcCurveSegment.Placement is sufficient to place and orient the trimmed line segment.

From IfcCurveSegment.Placement, the direction of the segment is

$$dy = \sin{(5.70829654085293)} = -0.54374144087698$$

$$dx = \cos{(5.70829654085293)} = 0.839252789970355$$

#49=IFCDIRECTION((0.839252789970355,-0.54374144087698));

The trimmed line segment is placed at (500,2500). The curve segment placement is

#50=IFCAXIS2PLACEMENT2D(#23,#49);

Finally, the IfcCurveSegment can be defined as

#55=IFCCURVESEGMENT(.CONTSAMEGRADIENT.,#50,IFCLENGTHMEASURE(0.),IFCLENGTHMEASURE(1956.785654),#54);

Compute Point on Curve

Compute the position matrix for a point 100 m from the start of the curve segment.

The IfcCurveSegment.Placement is sufficient to locate a point on the segment. The position and orientation of the parent curve are irrelevant for IfcLine.

From the curve segment placement

$$C = (500,2500)$$

$$dx\ = 0.839252789970355,\ dy\ = -0.54374144087698$$

$$\theta = \tan^{-1}{\left( \frac{-0.54374144087698}{0.839252789970355} \right) = -0.574888766\ }$$

The point on the curve segment can be computed as

$$\lambda(u) = C + \left( \int_{0}^{u = L}{\cos\left( \theta(t) \right)}dt\ x,\ \int_{0}^{u = L}{\sin\left( \theta(t) \right)}dt\ y \right)$$

$$x = 500 + \int_{0}^{100}{\cos{(-0.574888766)}} = 500 + ( - \sin{(-0.574888766))(100m - 0m) = 583.925249}$$

$$y = 2500 + \int_{0}^{100}{\sin{( -0.574888766)}} = 2500 + (\cos{(-0.574888766))(100m - 0m) = 2445.625886}$$

Though it is much easier to use the following calculation

$$x(u) = p_{x} + u(dx)$$

$$y(u) = p_{y} + u(dy)$$

$$x = 500 + 0.839252789970355(100) = 583.925249$$

$$y = 2500 - 0.54374144087698(100) = 2445.625886$$

The resulting position matrix is

$$M_{h} = \begin{bmatrix} 0.83925279 & 0.54374114 & 0 & 583.925249 \\ -0.54374114 & 0.83925279 & 0 & 2445.625886 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Circular Arc

When IfcCircle is used as a parent curve, it represents a circular curve of the alignment.

Source Model: GENERATED__HorizontalAlignment_CircularArc_100.0_300_1000_1_Meter.ifc

Parent Curve Parametric Equations

$$\theta(s) = \frac{s}{R}$$

$$\kappa(s) = \frac{1}{R}$$

$$x(s) = \int_{}^{}{\cos\left( \theta(s) \right)ds}$$

$$y(s) = \int_{}^{}{\sin{\left( \theta(s) \right)\ ds}}$$

Semantic Definition to Geometry Mapping

IfcCircle is defined by its center (IfcCircle.Position) and radius (IfcCircle.Radius). The position of the circle is irrelevant. The mapping from the semantic definition of radius to the geometric definition of radius is trivial.

Consider a horizontal curve segment that starts at (0,0), has a radius of 300 m, and an arc length of 100 m. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 300., 300., 100., $, .CIRCULARARC.);

The parent curve can be defined as follows:

#45 = IFCCIRCLE(#46, 300.);
#46 = IFCAXIS2PLACEMENT2D(#47, #48);
#47 = IFCCARTESIANPOINT((0., 300.));
#48 = IFCDIRECTION((0., -1.));

This is a circle centered at point (0,300) with the local X-axis aligned with the negative global Y-axis. This aligns the trimmed curve with its final placement.

The trimmed curve segment is placed such that the start point is at (0,0) and the tangent at the start point is in the direction (1,0).

#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));

The curve segment is defined by its placement at a segment trimmed from the parent curve starting at 0.0 for a length of 100.0 along the curve.

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(0.),IFCLENGTHMEASURE(100.), #45);

Compute Point on Curve

Compute the placement matrix for a point 50 m from the start of the curve segment.

Compute point on parent curve at $u = 50$

$$\theta(s) = \frac{s}{R}$$

$$x(s) = \int_{}^{}{\cos\left( \theta(s) \right)ds}$$

$$y(s) = \int_{}^{}{\sin{\left( \theta(s) \right)\ ds}}$$

$$x(50) = \int_{0}^{50}{\cos\left( \frac{s}{300} \right)ds} = 300\sin\frac{50}{300} - 300\sin\frac{0}{300} = 49.76883981$$

$$y(50) = \int_{0}^{50}{\sin{\left( \frac{s}{300} \right)\ ds}} = - 300\cos\frac{50}{300} - \left( - 300\cos\frac{0}{300} \right) = 4.1570305$$

This is the point on the parent curve relative to the start of the trim.

Position this point using the curve segment placement. Direction of the curve tangent (RefDirection) is (1,0). Apply the rotation

$$x = 49.76883981(1) + 4.1570305(0) = 49.76883981$$

$$y = - 49.76883981(0) + 4.1570305(1) = 4.1570305$$

The curve segment is placed at (0,0). Apply the translation

$$x = 49.76883981 + 0 = 49.76883981$$

$$y = 4.1570305 + 0 = 4.1570305$$

The resulting placement matrix is

$$M_{h} = \begin{bmatrix} 1 & 0 & 0 & 49.76883981 \\ 0 & 1 & 0 & 4.1570305 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Clothoid

Source Model: GENERATED__HorizontalAlignment_Clothoid_100.0_300_1000_1_Meter.ifc

Parent Curve Parametric Equations

$$\kappa(s) = \frac{A}{\left| A^{3} \right|}s$$

$$\theta(s) = \int_{}^{}{\kappa(s)ds} = \frac{\pi}{2}\frac{A}{|A|}s^{2}$$

$$x(u) = A\sqrt{\pi}\int_{0}^{u}{\cos{\left( \frac{\pi}{2}\frac{A}{|A|}t^{2} \right)\ }dt}$$

$$y(u) = A\sqrt{\pi}\int_{0}^{u}{\sin{\left( \frac{\pi}{2}\frac{A}{|A|}t^{2} \right)\ }dt}$$

A clothoid curve is parameterized by $$u = \frac{s}{\left| A\sqrt{\pi} \right|}\ $$ with $- \infty < u < \infty$. When $\theta = \pi/2$, $u = 1.0$.

Semantic Definition to Geometry Mapping

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

$$A = \frac{L}{\sqrt{|f|}}\frac{f}{|f|}$$

Consider a horizontal clothoid segment that starts at (0,0) with a start direction of (1.0,0.0). The radii at the start and end of the segment are 300 and 1000, respectively. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 300., 1000., 100., $, .CLOTHOID.);

From the semantic definition, compute the clothoid constant, $A$

$$R_{s} = 300,\ R_{e} = 1000,\ L = 100$$

$$f = \frac{100}{1000} - \frac{100}{300} = -0.23333$$

$$A = \frac{100}{\sqrt{| -0.23333|}}\frac{-0.23333}{| -0.23333|} = -207.0196678$$

The curve segment starts where the radius is 300. The distance from the origin where this radius occurs can be computed from the curvature equation. The curvature at the start is 1/300 and the clothoid constant is -207.0196678. Solve for distance from origin.

$$\kappa(s) = \frac{A}{\left| A^{3} \right|}s$$

$$\frac{1}{300} = \frac{- 207.0196678}{\left| - {207.0196678}^{3} \right|}s$$

$$s = - 142.8571429$$

Place the parent curve at (0,0) with a tangent of (1,0)

#45 = IFCCLOTHOID(#46, -207.019667802706);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Place the curve segment at (0,0) with the tangent in the direction (1,0).

#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));

Define the curve segment

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(-142.857142857143), IFCLENGTHMEASURE(100.), #45);

Compute Point on Curve

These example calculations will reference the steps in the algorithm presented in 1.5.3.

Step 1 -- Evaluate the parent curve at IfcCurveSegment.SegmentStart

Compute the placement matrix for a point 50 m from the start of the curve segment.

Start by computing the point and curve tangent at the start of the parent curve trim.

Compute point on parent curve at $u = \frac{-142.8571428}{\left| - 207.0196678\sqrt{\pi} \right|} = -0.3893278$

$$x(-0.3893278) = - 207.0196678\sqrt{\pi}\int_{0}^{-0.3893278}{\cos{\left( \frac{\pi}{2}\frac{-207.0196678}{| - 207.0196678|}t^{2} \right)\ }dt} = -142.04941746210602$$

$$y(-0.3893278) = -207.0196678\sqrt{\pi}\int_{0}^{-0.3893278}{\sin{\left( \frac{\pi}{2}\frac{-207.0196678}{| - 207.0196678|}t^{2} \right)\ }dt} = 11.292042785713347$$

$$\theta(-0.3893278) = \frac{\pi}{2}\frac{-207.0196678}{|-207.0196678|}(-0.3893278)^{2} = -0.238095237$$

$$dx = \cos(-0.238095237) = 0.971788979$$

$$dy = \sin{(-0.238095237) = -0.235852028}$$

In matrix form

$$M_{PCS} = \begin{bmatrix} 0.971788979 & 0.235852028 & 0 & -142.04941746210602 \\ -0.235852028 & 0.971788979 & 0 & 11.292042785713347 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 2 -- Create parent curve translation matrix

Remove the start point location so the point under consideration is relative to (0,0).

$$M_{T} = \begin{bmatrix} 1 & 0 & 0 & 142.04941746210602 \\ 0 & 1 & 0 & -11.292042785713347 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 3 -- Create parent curve rotation matrix

Remove the start point rotation so the point under consideration tangent at the start point in the direction (1,0).

$$M_{R} = \begin{bmatrix} 0.971788979 & -0.235852028 & 0 & 0 \\ 0.235852028 & 0.971788979 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 4 -- Create matrix for IfcCurveSegment.Placement

Represent IfcCurveSegment.Placement in matrix form. In this example, the placement is at (0,0) with RefDirection (1,0) which results in an identity matrix. This is not true in all cases.

$$M_{CSP} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 5 -- Compute position on IfcCurveSegment

Step 5a -- Evaluate parent curve at IfcCurveSegment.SegmentStart + IfcCurveSegment.SegmentLength

Compute point and curve tangent at 50 m from the start,

$$ -142.8571428 + 50 = -92.8571428$$

$$u = \frac{-92.8571428}{\left|-207.0196678\sqrt{\pi} \right|} = -0.25306307$$

$$x(-0.25306307) = -207.0196678\sqrt{\pi}\int_{0}^{-0.25306307}{\cos{\left( \frac{\pi}{2}\frac{-207.0196678}{|-207.0196678|}s^{2} \right)\ }ds} = - 92.763220844871512$$

$$y(-0.25306307) = -207.0196678\sqrt{\pi}\int_{0}^{-0.25306307}{\sin{\left( \frac{\pi}{2}\frac{-207.0196678}{|-207.0196678|}s^{2} \right)\ }ds} = 3.1114126254443812$$

$$\theta(-0.25306307) = \frac{\pi}{2}\frac{-207.0196678}{|-207.0196678|}(-0.25306307)^{2} = -0.100595238$$

$$dx = \cos(-0.100595238) = 0.994944564$$

$$dy = \sin{(-0.100595238) = -0.100425663}$$

In matrix form

$$M_{PC} = \begin{bmatrix} 0.994944564 & 0.100425663 & 0 & -92.763220844871512 \\ -0.100425663 & 0.994944564 & 0 & 3.1114126254443812 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 5b -- Map the parent curve point to the curve segment.

Apply the translation, rotation, and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_{R} M_{T} M_{PC} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.971788979 & -0.235852028 & 0 & 0 \\ 0.235852028 & 0.971788979 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 142.04941746210602 \\ 0 & 1 & 0 & -11.292042785713347 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.994944564 & 0.100425663 & 0 & -92.763220844871512 \\ -0.100425663 & 0.994944564 & 0 & 3.1114126254443812 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$M_{h} = \begin{bmatrix} 0.99056175921247780 & -0.13706714116038599 & 0 & 49.825200930152405 \\ 0.13706714116038599 & 0.99056175921247780 & 0 & 3.6744032279728032 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Cubic Transition Curve

Source Model: GENERATED__HorizontalAlignment_Cubic_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

The approach for calculating the coordinates of a horizontal cubic curve is unique compared to the other curve types. For a horizontal alignment, the coordinate point and curve tangent angle is generally defined by parametric equations where the parameter is the distance along the curve. A cubic curve is represented with the IfcPolynomialCurve parent curve type. When IfcPolynomialCurve is used as a cubic transition curve, the $x$ coefficients are [0,1] and the $y$ coefficients are [0,0,0, $A_{3}$] resulting in a function of the form $y(x) = A_{3}x^{3}$.

The distance along the curve is defined by

$$s = \int_{}^{}{\sqrt{\left( y'(x) \right)^{2} + 1}\ dx}$$

$$y'(x) = 3A_{3}x^{2}$$

$$s = \int_{}^{}{\sqrt{9A_{3}^{2}x^{4} + 1}\ dx}$$

This equation is solved for $x$ and then $y$ can be computed. This can be accomplished with numeric methods.

  1. For a distance $u$ along the curve, find $x$ for $s - u = 0$

  2. Compute $y(x) = A_{3}x^{3}$

  3. Compute curve tangent slope as $\frac{dy}{dx} = 3A_{3}x^{2}$

Semantic Definition to Geometry Mapping

$$A_{0} = A_{1} = A_{2} = 0$$

$$A_{3} = \frac{1}{6R_{e}L} - \frac{1}{6R_{s}L}$$

Consider a horizontal cubic transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., $,.CUBIC.);

Compute the polynomial curve constants

$$A_{3} = \frac{1}{6(300m)(100m)} - \frac{1}{6(\infty)(100m)} = 5.55555 \bullet 10^{- 6}\ m^{- 2}$$

⚠️

There is a flaw in the IFC Specification. IfcAlignmentHorizontalSegment indicates the cubic formula is $y=\frac{x^3}{6RL}$ which means the IfcPolynomialCurve.CoefficentY[3] attribute must have unit of $Length^{-2}$. This is a direct contradiction to IfcPolynomialCurve which clearly states the coefficent are real values (i.e. scalar values) with IfcReal.

Why is this a problem? The calculation of a point on the polynomal curve is different depending on how the coefficients are defined.

The parametric form of the polynomial curve is $Q(u) = ( x(u), y(u), z(u) )$. The polynomial curve represents real geometry so the resulting values, $x$, $y$, and $z$ must have units of $Length$.

When $u$ is parametrized as a Length measure, as required by IfcCurveSegment, the parametric terms of the polynomal equation are:

$x(u) = \sum\limits_{i=0}^{l} a_i u^i $

$y(u) = \sum\limits_{j=0}^{m} b_j u^j $

$z(u) = \sum\limits_{k=0}^{n} c_k u^k $

For $x$, $y$, and $z$ to have values in $Length$ measure, the implicit unit of measure of the coefficients must be:

$Length^{(1-i)}$ for $a_i$

$Length^{(1-j)}$ for $b_j$

$Length^{(1-k)}$ for $c_k$

When $u$ is parametrized as a scalar, as required by IfcPolynomialCurve, the parametric terms of the polynomial equation are:

$x(u) = L\sum\limits_{i=0}^{l} a_i u^i $

$y(u) = L\sum\limits_{j=0}^{m} b_j u^j $

$z(u) = L\sum\limits_{k=0}^{n} c_k u^k $

The parameter $u$ and the coefficients are scalar so they must be multipled with the curve length $L$ to result values of Length.

The equations for $x$, $y$, and $z$ differ depending on the parameterization.

Concept Template 4.2.2.1.5 Cubic Transition Segment and IfcAlignmentHorizontalSegment provide clear direction (not withstanding typographical errors) that IfcPolynomalCurve is to be evaluated as $y = CoefficientsY[3]x^3$ which dictates that $CoefficientsY[3]$ must be in units of $Length^{-2}$ and must be encoded in the IfcPolynomialCurve.CoefficientY attribute as an IfcReal.

While I've never been able to find an offically documented Implementer's Agreement, this interpetation is consistent with several different discussions on GitHub.

⚠️

The geometric representation is

#45 = IFCPOLYNOMIALCURVE(#46, (0., 1.), (0., 0., 0., 5.55555555555556E-6), $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Evaluate Point on Curve

Compute the curve coordinates at a distance along the curve, $u = 100$

X Coefficients = ($0 m^1$, $1 m^0$)

Y Coefficients = ($0 m^1$, $0 m^{0}$, $0 m^{-1}$, $5.55555E{-06} m^{-2}$)

$$y(x) = \left( 5.55555 \bullet 10^{-6} m^{-2}\right)x^{3}$$

Find $x$ such that

$$s - u = \int_{0}^{x}\sqrt{\left( y'(x) \right)^{2} + 1}dx - u = 0$$

$$y'(x) = 3\left( 5.55555 \bullet 10^{-6} m^{-2}\right)x^{2}$$

Solve numerically

$$x = 99.72593255 m$$

Check solution

$$s = \int_{0}^{99.72593255m}\sqrt{\left( 3 \bullet (5.55555 \bullet 10^{-6} m^{-2})(x m)^{2} \right)^{2} + 1}dx = 100m$$

Compute y

$$y(x) = (5.55555 \bullet 10^{-6} m^{-2}{)(99.72593255m)}^{3} = 5.5100m$$

The tangent vector at $u = 100m$ along the curve is

$$dy = y'(99.72593255m) = 3\left( 5.55555 \bullet 10^{-6} m^{-2}\right)(99.72593255m)^{2} = 0.165753$$

$$dx = 1.0$$

Normalizing the tangent vector (RefDirection)

$$dx = \frac{1}{\sqrt{1^{2} + {0.615753}^{2}}} = 0.986539$$

$$dy = \frac{0.165753}{\sqrt{1^{2} + {0.615753}^{2}}} = 0.1635219$$

The Y-direction vector is (-dy,dx)

Axis vector is (0,0,1)

The resulting matrix is

$$M_{PC} = \begin{bmatrix} 0.986539 & -0.1635219 & 0 & 99.72593255 \\ 0.1635219 & 0.986539 & 0 & 5.5100 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

For this example, $M_{CSP}$, $M_{R}$, and $M_{T}$ are identity matrices. The resulting point is

$$M_{h} = \begin{bmatrix} 0.986539 & -0.1635219 & 0 & 99.72593255 \\ 0.1635219 & 0.986539 & 0 & 5.5100 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Helmert Transition Curve

The Helmert curve is unique in that it is modeled in two halves. The business logic is represented geometrically by two IfcSecondOrderPolynomialSpiral objects.

Source Model: GENERATED__HorizontalAlignment_HelmertCurve_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

$$\theta(t) = \frac{t^{3}}{3A_{2}^{3}} + \frac{A_{1}}{2\left| A_{1}^{3} \right|}t^{2} + \frac{t}{A_{0}}$$

$$\kappa(t) = \frac{1}{A_{2}^{3}}t^{2} + \frac{A_{1}}{\left| A_{1}^{3} \right|}t + \frac{1}{A_{0}}$$

Semantic Definition to Geometry Mapping

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

First Half

$$a_{0} = \frac{L}{R_{s}},\ A_{0} = \frac{L}{\left| a_{0} \right|}\ \frac{a_{0}}{\left| a_{0} \right|}$$

$$a_{1} = 0,\ A_{1} = 0$$

$$a_{2} = 2f,\ A_{2} = \frac{L}{\sqrt[3]{\left| a_{2} \right|}}\ \frac{a_{2}}{\left| a_{2} \right|}$$

Second Half

$$a_{0} = -f + \frac{L}{R_{s}},\ A_{0} = \frac{L}{\left| a_{0} \right|}\ \frac{a_{0}}{\left| a_{0} \right|}$$

$$a_{1} = 4f,\ A_{1} = \frac{L}{\sqrt{\left| a_{1} \right|}}\ \frac{a_{1}}{\left| a_{1} \right|}$$

$$a_{2} = -2f,\ A_{2} = \frac{L}{\sqrt[3]{\left| a_{2} \right|}}\ \frac{a_{2}}{\left| a_{2} \right|}$$

Consider a horizontal Helmert transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., $, .HELMERTCURVE.);

Compute the curve parameters

$$L = 100\ m$$

$$R_{s} = \infty$$

$$R_{e} = 300\ m$$

$$f = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

First Half

$$a_{0} = \frac{100}{\infty} = 0,\ A_{0} = 0$$

$$A_{1} = 0$$

$$a_{2} = 2(0.33333) = 0.66667,\ A_{2} = \frac{100\ m}{\sqrt[3]{|0.66667|}}\frac{0.66667}{|0.66667|} = 114.4714255\ m$$

Second Half

$$a_{0} = -0.33333 + \frac{100}{\infty} = -0.33333,\ A_{0} = \frac{100\ m}{| -0.33333|}\frac{-0.33333}{|-0.33333|} = - 300\ m$$

$$a_{1} = 4(0.33333) = 1.33333,\ A_{1} = \frac{100\ m}{\sqrt{|1.33333|}}\frac{1.33333}{|1.33333|} = 86.602540378\ m$$

$$a_{2} = -2(0.33333) = -0.66667,\ A_{2} = \frac{100\ m}{\sqrt[3]{| -0.66667|}}\frac{-0.66667}{|-0.66667|} = -114.4714255\ m$$

End point and end slope of first half is the placement of the second half.

$$\theta(t) = \frac{t^{3}}{3A_{2}^{3}} + \frac{A_{1}}{2\left| A_{1}^{3} \right|}t^{2} + \frac{t}{A_{0}} = \frac{t^{3}}{3(114.4714255)^{3}}$$

$$x_{1} = \int_{0}^{L/2}{\cos{\theta(t)}}\ dt = 49.99724\ m$$

$$y_{1} = \int_{0}^{L/2}{\sin{\theta(t)}}\ dt = 0.347204\ m$$

$$\theta_{1}\left( \frac{L}{2} \right) = \frac{{(50m)}^{3}}{3(114.4714255m)^{3}} = 0.0277777777$$

$$dx = \cos{0.027777777} = 0.999614$$

$$dy = \sin{0.027777777} = 0.0277742$$

Start point of second parent curve

$$\theta(t) = \frac{t^{3}}{3A_{2}^{3}} + \frac{A_{1}}{2\left| A_{1}^{3} \right|}t^{2} + \frac{t}{A_{0}} = \frac{t^{3}}{3( - 114.4714255\ m)^{3}} + \frac{86.602540378\ m}{2\left| (86.602540378\ m)^{3} \right|}t^{2} + \frac{t}{- 300\ m}$$

$$x_{2} = \int_{0}^{L/2}{\cos{\theta(t)}}\ dt = 49.9664\ m\ $$

$$y_{2} = \int_{0}^{L/2}{\sin{\theta(t)}}\ dt = - 1.73556\ m$$

$$\theta_{2}\left( \frac{L}{2} = 50\ m \right) = \frac{{(50\ m)}^{3}}{3( - 114.4714255\ m)^{3}} + \frac{86.602540378\ m}{2\left| (86.602540378\ m)^{3} \right|}({50\ m)}^{2} + \frac{50\ m}{- 300\ m} = - 0.0277777$$

The end point of the first parent curve and the start point of the second parent curve are offset so a rotation and translation for the placement of the second parent curve is needed.

$$\theta_{p} = \theta_{1} - \theta_{2} = 0.0277777777 - (-0.0277777777) = 0.05555555$$

$$dx = \cos{0.05555555} = 0.998457$$

$$dy = \sin{0.05555555} = 0.055526$$

$$x_{p} = x_{1} - x_{2}\cos\theta_{p} + y_{2}\sin\theta_{p}$$

$$y_{p} = y_{1} - x_{2}\sin\theta_{p} - y_{2}\cos\theta_{p}$$

$$x_{p} = 49.9972 - 49.9664\cos{0.0555555} - 1.73556\sin{0.0555555 =}0.011503\ m$$

$$y_{p} = 0.347204 - 49.9664\sin{0.055555} + 1.73556\cos{0.0555555} = -0.6943693\ m$$

The geometric representations are

First half

#45 = IFCSECONDORDERPOLYNOMIALSPIRAL(#46, 114.471424255333, $, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Second half

#52 = IFCSECONDORDERPOLYNOMIALSPIRAL(#53, -114.471424255333, 86.6025403784439, -300.);
#53 = IFCAXIS2PLACEMENT2D(#54, #55);
#54 = IFCCARTESIANPOINT((0.0115725277555399, -0.694297741112199));
#55 = IFCDIRECTION((0.998457186998745, 0.0555269820047339));

The curve segments are

First half

#36 = IFCCURVESEGMENT(.CONTSAMEGRADIENTSAMECURVATURE., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(50.), #45);
#37 = IFCLOCALPLACEMENT($, #38);
#38 = IFCAXIS2PLACEMENT3D(#39, $, $);
#39 = IFCCARTESIANPOINT((0., 0., 0.));

Second half

⚠️ [Write this up better] ⚠️

The start placement of the second half is

$$x = x_s + x_1 cos(\theta_s) - y_1 sin(\theta_s)$$ $$y = y_s + x_1 sin(\theta_s) + y_2 cos(\theta_s)$$ $$dx = cos(\theta_s + \theta_1)$$ $$dy = sin(\theta_s + \theta_1)$$

where $x_s, y_s, \theta_s$ are the start point and angle at start of the first half

$$x = 0.0 + 49.99724 cos(0.0) - 0.347204 sin(0.0) = 49.99724$$ $$y = 0.0 + 49.99724 sin(0.0) + 0.347204 cos(0.0) = 0.347204$$ $$dx = cos(0. + 0.02777777) = 0.999614$$ $$dy = sin(0. + 0.02777777) = 0.027774$$

#48 = IFCCURVESEGMENT(.CONTINUOUS., #49, IFCLENGTHMEASURE(50.), IFCLENGTHMEASURE(50.), #52);
#49 = IFCAXIS2PLACEMENT2D(#50, #51);
#50 = IFCCARTESIANPOINT((49.9972443634885, 0.347204361427475));
#51 = IFCDIRECTION((0.999614222337484, 0.0277742056705088));
  1. Evaluate Point on Curve

See clothoid curve for general calculation procedure.

⚠️ [Add calculation] ⚠️

Bloss Transition Curve

Parent curve type: IfcThirdOrderPolynomialSpiral

Source Model: GENERATED__HorizontalAlignment_BlossCurve_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

$$\theta(t) = \frac{A_{3}}{4\left| A_{3}^{5} \right|}t^{4} + \frac{1}{3A_{2}^{3}}t^{3} + \frac{A_{1}}{2\left| A_{1}^{3} \right|}t^{2} + \frac{t}{A_{0}}$$

$$\kappa(t) = \frac{A_{3}}{\left| A_{3}^{5} \right|}t^{3} + \frac{t^{2}}{A_{2}^{3}} + \frac{A_{1}}{2\left| A_{1}^{3} \right|}t + \frac{1}{A_{0}}$$

Semantic Definition to Geometry Mapping

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

$$a_{0} = \frac{L}{R_{s}},\ A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|}$$

$$A_{1} = 0$$

$$a_{2} = 3f,\ A_{2} = \frac{L}{\sqrt[3]{\left| a_{2} \right|}}\frac{a_{2}}{\left| a_{2} \right|}$$

$$a_{3} = - 2f,\ A_{3} = \frac{L}{\sqrt[4]{\left| a_{3} \right|}}\frac{a_{3}}{\left| a_{3} \right|}$$

Consider a horizontal Bloss transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., $, .BLOSSCURVE.);

$$L = 100\ m$$

$$R_{s} = \infty$$

$$R_{e} = 300\ m$$

$$f = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

$$a_{0} = \frac{100}{\infty} = 0,\ A_{o} = 0$$

$$a_{2} = 3(0.33333) = 1,\ A_{2} = \frac{100}{\sqrt[3]{|1|}}\frac{1}{|1|} = 100\ m$$

$$a_{3} = - 2(0.33333) = -0.66667,\ A_{3} = \frac{100\ m}{\sqrt[4]{| -0.66667|}}\frac{-0.66667}{|-0.66667|} = - 110.668192\ m$$

#45 = IFCTHIRDORDERPOLYNOMIALSPIRAL(#46, -110.668191970032, 100., $, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Evaluate Point on Curve

Evaluate at the end of the curve $u = 100$

$$\theta(t) = \frac{-110.668192}{4\left|-{110.668192}^{5} \right|}t^{4} + \frac{1}{3(100)^{3}}t^{3}$$

$$x = \int_{0}^{100}{\cos{\theta(t)}}\ dt = 99.7486$$

$$y = \int_{0}^{100}{\sin{\theta(t)}}\ dt = 4.98981$$

$$\theta(100) = \frac{-110.668192}{4\left|-{110.668192}^{5} \right|}(100)^{4} + \frac{1}{3(100)^{3}}(100)^{3} = \frac{1}{6} = 0.1666667$$

$$dx = \cos\left( \frac{1}{6} \right) = 0.98614$$

$$dy = \sin{\left( \frac{1}{6} \right) = 0.16589}$$

The parent curve point is then positioned using the curve segment start point and orientation as illustrated for the clothoid curve example.

Cosine Transition Curve

Source Model: GENERATED__HorizontalAlignment_CosineCurve_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

$$\kappa(t) = \frac{1}{A_{0}} + \frac{1}{A_{1}}\cos\left( \frac{\pi}{L}t \right)$$

$$\theta(t) = \frac{1}{A_{0}}t + \frac{L}{\pi A_{1}}\sin\left( \frac{\pi}{L}t \right)$$

$$x = \int_{0}^{L}{\cos{\theta(t)}}\ dt$$

$$y = \int_{0}^{L}{\sin{\theta(t)}}\ dt$$

Semantic Definition to Geometry Mapping

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

Constant Term: $a_{0} = 0.5f + \frac{L}{R_{s}},\ A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|}$

Cosine Term: $a_{1} = \ -0.5f,\ A_{1} = \frac{L}{\left| a_{1} \right|}\frac{a_{1}}{\left| a_{1} \right|}$

Consider a horizontal cosine transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., $, .COSINECURVE.);

$$R_{s} = \infty,\ R_{e} = 300\ m,\ L = 100\ m$$

$$f = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

$$a_{0} = 0.5(0.33333) + \frac{100}{\infty} = 0.16667,\ A_{0} = \frac{100\ m}{|0.16667|}\ \frac{0.16667}{|0.16667|} = 600\ m$$

$$a_{1} = -0.5(0.33333) = -0.16667,\ A_{1} = \frac{100}{|-0.16667|}\ \frac{-0.16667}{|-0.16667|} = - 600\ m$$

#45 = IFCCOSINESPIRAL(#46, -600., 600.);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Evaluate Point on Curve

Evaluate at the end of the curve $u = 100$

$$\theta(t) = \frac{1}{600}t - \frac{100}{600\pi}\sin\left( \frac{\pi}{100}t \right)$$

$$x = \int_{0}^{100}{\cos{\theta(t)}}\ dt = 99.7485\ m$$

$$y = \int_{0}^{100}{\sin{\theta(t)}}\ dt = 4.9458\ m$$

$$\theta(100) = \frac{1}{600}100 - \frac{100}{600\pi}\sin\left( \frac{\pi}{100}100 \right) = \frac{1}{6} = 0.1666667$$

$$dx = \cos\left( \frac{1}{6} \right) = 0.98614$$

$$dy = \sin{\left( \frac{1}{6} \right) = 0.16589}$$

The parent curve point is then positioned using the curve segment start point and orientation as illustrated for the clothoid curve example.

Sine Transition Curve

Source Model: GENERATED__HorizontalAlignment_SineCurve_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

$$\kappa(t) = \frac{1}{A_{0}} + \frac{A_{1}}{\left| A_{1} \right|}\left( \frac{1}{A_{1}} \right)^{2}t + \frac{1}{A_{2}}\sin\left( \frac{2\pi}{L}t \right)\ \ $$

$$\theta(t) = \frac{1}{A_{0}}t + \frac{1}{2}\left( \frac{A_{1}}{\left| A_{1} \right|} \right)\left( \frac{t}{A_{1}} \right)^{2} - \frac{L}{2\pi A_{2}}\left( \cos\left( \frac{2\pi}{L}t \right) - 1 \right)$$

$$x = \int_{0}^{L}{\cos{\theta(t)}}\ dt$$

$$y = \int_{0}^{L}{\sin{\theta(t)}}\ dt$$

Semantic Definition to Geometry Mapping

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

Constant Term: $a_{0} = \frac{L}{R_{s}},\ A_{o} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|}$

Linear Term: $a_{1} = f,\ A_{1} = \frac{L}{\sqrt{\left| a_{1} \right|}}\frac{a_{1}}{\left| a_{1} \right|}$

Sine Term: $a_{2} = -\frac{1}{2\pi}f,\ A_{2} = \frac{L}{\left| a_{2} \right|}\frac{a_{2}}{\left| a_{2} \right|}$

Consider a horizontal sine transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length of the segment is 100. The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., $, .SINECURVE.);

$$R_{s} = \infty,\ R_{e} = 300\ m,\ L = 100\ m$$

$$f = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

$$a_{0} = \frac{100}{\infty} = 0,\ A_{0} = 0$$

$$a_{1} = 0.33333,\ A_{1} = \frac{100}{\sqrt{|0.33333|}}\frac{0.33333}{|0.33333|} = 173.2050808\ m$$

$$a_{2} = -\frac{1}{2\pi}(0.33333) = -0.053051647,\ A_{2} = \frac{100}{|-0.053051647|}\frac{-0.053051647}{|-0.053051647|} = -1884.955592\ m$$

#45 = IFCSINESPIRAL(#46, -1884.95559215388, 173.205080756888, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Evaluate Point on Curve

Evaluate at the end of the curve $u = 100$

$$\theta(t) = \frac{1}{2}\left( \frac{t}{173.2050808} \right)^{2} + \frac{100}{2\pi(1884.955592)}\left( \cos\left( \frac{2\pi}{100}t \right) - 1 \right)$$

$$x = \int_{0}^{100}{\cos{\theta(t)}}\ dt = 99.757\ m$$

$$y = \int_{0}^{100}{\sin{\theta(t)}}\ dt = 4.70132\ m$$

$$\theta(100) = \frac{1}{2}\left( \frac{100}{173.2050808} \right)^{2} + \frac{100}{2\pi(1884.955592)}\left( \cos\left( \frac{2\pi}{100}100 \right) - 1 \right) = \frac{1}{6} = 0.1666667$$

$$dx = \cos\left( \frac{1}{6} \right) = 0.98614$$

$$dy = \sin{\left( \frac{1}{6} \right) = 0.16589}$$

The parent curve point is then positioned using the curve segment start point and orientation as illustrated for the clothoid curve example.

Viennese Transition Curve

Parent curve type: IfcSeventhOrderPolynomialSpiral

Source Model: https://github.com/bSI-RailwayRoom/IFC-Rail-Unit-Test-Reference-Code/blob/master/alignment_testset/IFC-WithGeneratedGeometry/GENERATED__HorizontalAlignment_VienneseBend_100.0_inf_300_1_Meter.ifc

Parent Curve Parametric Equations

$$\theta(s) = \frac{A_{7}}{8\left| A_{7}^{9} \right|} s^{8} + \frac{1}{7 A_{6}^{7}} s^{7} + \frac{A_{5}}{6\left| A_{5}^{7} \right|} s^{6} + \frac{1}{5 A_{4}^{5}} s^{5} + \frac{A_{3}}{4\left| A_{3}^{5} \right|} s^{4} + \frac{1}{3 A_{2}^{3}} s^{3} + \frac{A_{1}}{2\left| A_{1}^{3} \right|} s^{2} + \frac{1}{A_{0}} s$$

$$\kappa(s) = \frac{A_{7}}{\left| A_{7}^{9} \right|}s^{7} + \frac{1}{A_{6}^{7}}s^{6} + \frac{A_{5}}{\left| A_{5}^{7} \right|}s^{5} + \frac{1}{A_{4}^{5}}s^{4} + \frac{A_{3}}{\left| A_{3}^{5} \right|}s^{3} + \frac{1}{A_{2}^{3}}s^{2} + \frac{A_{1}}{\left| A_{1}^{3} \right|}s + \frac{1}{A_{0}}$$

Semantic Definition to Geometry Mapping

$h_{cg}$ = Gravity Center Line Height = IfcAlignmentHorizontalSegment.GravityCenterLineHeight

$d_{rh}$ = Rail Head Distance = IfcAlignmentCant.RaliHeadDistance

$D_{sl}$ = IfcAlignmentCantSegment.StartCantLeft

$D_{el}$ = IfcAlignmentCantSegment.EndCantLeft

$D_{sr}$ = IfcAlignmentCantSegment.StartCantRight

$D_{er}$ = IfcAlignmentCantSegment.EndCantRight

$$\theta_s = \frac{(D_{sr} - D_{sl})}{d_{rh}}$$, Start Cant Angle

$$\theta_e = \frac{(D_{er} - D_{el})}{d_{rh}}$$, End Cant Angle

$$cf = -420.\left ( \frac{h_{cg}}{L} \right) \left( \theta_e - \theta_s \right)$$, Cant Factor

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}$$

Constant term $$a_{0} = \frac{L}{R_{s}}, A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|}$$

Linear term $$a_{1} = 0, A_{1} = 0$$

Quadratic term $$a_{2} = cf, A_{2} = \frac{L}{\sqrt[3]{\left| a_{2} \right|}}\frac{a_{2}}{\left| a_{2} \right|}$$

Cubic term $$a_{3} = -4cf, A_{3} = \frac{L}{\sqrt[4]{\left| a_{3} \right|}}\frac{a_{3}}{\left| a_{3} \right|}$$

Quartic term $$a_{4} = 5cf + 35f, A_{4} = \frac{L}{\sqrt[5]{\left| a_{4} \right|}}\frac{a_{4}}{\left| a_{4} \right|}$$

Quintic term $$a_{5} = -2cf - 84f, A_{5} = \frac{L}{\sqrt[6]{\left| a_{5} \right|}}\frac{a_{5}}{\left| a_{5} \right|}$$

Sextic term term $$a_{6} = 70f, A_{6} = \frac{L}{\sqrt[7]{\left| a_{6} \right|}}\frac{a_{6}}{\left| a_{6} \right|}$$

Septic term $$a_{7} = -20f, A_{7} = \frac{L}{\sqrt[8]{\left| a_{7} \right|}}\frac{a_{7}}{\left| a_{7} \right|}$$

Example

Consider a horizontal Viennese Bend transition curve segment that starts at (0,0) with a start direction of 0.0. The radius at the start is infinite and the radius at the end is 300. The arc length is 100. The Gravity Center Line Height is 1.8 (this optional parameter is required for the Viennese Bend). The semantic definition is

#28 = IFCCARTESIANPOINT((0., 0.));
#29 = IFCALIGNMENTHORIZONTALSEGMENT($, $, #28, 0., 0., 300., 100., 1.8, .VIENNESEBEND.);

Viennese Bend transition curves are unique in that the horizontal geometry of the curve depends on the cant. For the example, the cant segment horizontal length is 100, the start cant is 0.0 and the end cant is 0 at the left rail and 0.1 at the right rail. The semantic definition is

#64 = IFCALIGNMENTCANTSEGMENT($, $, 0., 100., 0., 0., 0., 1.E-1, .VIENNESEBEND.);

$$R_{s} = \infty,\ R_{e} = 300\ m,\ L = 100\ m$$

$$f = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

$$ \theta_s = \frac{(0. - 0.)}{1.5} = 0.0$$

$$ \theta_e = \frac{0.1 - 0.0}{1.5} = 0.066667$$

$$ cf = -420\left(\frac{1.8}{100}\right)(0.066667-0.)= -0.504$$

$$ a_{0} = \frac{100}{\infty} = 0, A_{0} = 0 $$

$$ a_{1} = 0, A_{1} = 0 $$

$$ a_{2} = -0.504, A_{2} = \frac{100}{\sqrt[3]{\left| -0.504 \right|}}\frac{-0.504}{\left| -0.504 \right|} = -125.6579069m$$

$$ a_{3} = -4(-0.504) = 2.016, A_{3} = \frac{100}{\sqrt[4]{\left| 2.016 \right|}}\frac{2.016}{\left| 2.016 \right|} = 83.92229813m$$

$$ a_{4} = 5(-0.504) + 35(0.33333) = 9.1466655, A_{4} = \frac{100}{\sqrt[5]{\left| 9.1466655 \right|}}\frac{9.1466655}{\left| 9.1466655 \right|} = 64.231406m$$

$$ a_{5} = -2(-0.504) - 84(0.33333) = -26.9919999, A_{5} = \frac{100}{\sqrt[6]{\left| -26.9919999 \right|}}\frac{-26.9919999}{\left| -26.9919999 \right|} = -57.7378785m$$

$$ a_{6} = 70(0.33333) = 23.33333333, A_{6} = \frac{100}{\sqrt[7]{\left| 23.33333333 \right|}}\frac{23.33333333}{\left| 23.33333333 \right|} = 63.76388134m$$

$$ a_{7} = -20(0.33333) = -6.66666666, A_{7} = \frac{100}{\sqrt[8]{\left| -6.66666666 \right|}}\frac{-6.66666666}{\left| -6.66666666 \right|} = -78.8880838m$$

#75 = IFCSEVENTHORDERPOLYNOMIALSPIRAL(#76, -78.8880838459446, 63.7638813456506, -57.7378785242934, 64.2314061308743, 83.922298125931, -125.657906854859, $, $);
#76 = IFCAXIS2PLACEMENT2D(#77, $);
#77 = IFCCARTESIANPOINT((0., 0.));

The parent curve point is then positioned using the curve segment start point and orientation as illustrated for the clothoid curve example.