2_Horizontal_Alignments.md

Chapter 2 — Horizontal Alignments

2.0 Introduction

The geometric representation of a horizontal alignment is accomplished with an IfcCompositeCurve. The composite curve consists of a sequence of IfcCurveSegment entities whose geometry is defined by a parent curve. This section defines the mathematical relationships and equations for each parent curve type and the algorithm for evaluating points on those curves.

Table 2.0-1 maps each IfcAlignmentHorizontalSegment.PredefinedType to its corresponding parent curve type.

Business Logic (IfcAlignmentHorizontalSegment.PredefinedType)	Geometric Representation (IfcCurveSegment.ParentCurve)
LINE	IfcLine
CIRCULARARC	IfcCircle
CLOTHOID	IfcClothoid
CUBIC	IfcPolynomialCurve
HELMERTCURVE	IfcSecondOrderPolynomialSpiral
BLOSSCURVE	IfcThirdOrderPolynomialSpiral
COSINECURVE	IfcCosineSpiral
SINECURVE	IfcSineSpiral
VIENNESEBEND	IfcSeventhOrderPolynomialSpiral

Table 2.0-1 — Mapping of business logic to geometric representation for horizontal alignment

This chapter covers:

• The three-step evaluation algorithm for computing position and tangent direction at any point on a horizontal curve segment.
• Parametric equations and geometry mapping examples for all nine curve types in Table 2.0-1.
• The zero-length closing segment required at the end of every IfcCompositeCurve.

2.1 Conventions

Each curve is parameterized by arc-length $s$, where $s = 0$ at the start of the parent curve. The start and end radii $R_s$ and $R_e$ are taken from IfcAlignmentHorizontalSegment.StartRadiusOfCurvature and EndRadiusOfCurvature; a value of zero indicates infinite radius (zero curvature, i.e. a straight line). Radii follow a positive-left sign convention: a positive value indicates curvature to the left of the direction of travel; a negative value indicates curvature to the right. The segment length $L$ is IfcAlignmentHorizontalSegment.SegmentLength. The tangent angle $\theta(s)$ is measured from the positive $x$-axis; its cosine and sine form the RefDirection of the curve at $s$.

The $x$ and $y$ coordinates of all horizontal parent curves are defined by the integrals:

$$x(s) = \int \cos\theta(s)\ ds \qquad y(s) = \int \sin\theta(s)\ ds$$

2.2 Curve Segment Evaluation Algorithm

This algorithm evaluates the 2D position and tangent direction of a point on an IfcCurveSegment in the alignment coordinate system. Let $s_0$ = SegmentStart and $s$ = the arc-length parameter of the point to evaluate, where $s_0 \leq s < s_0 + \mathtt{SegmentLength}$.

Step 1 — Form the curve segment placement matrix $M_{CSP}$

$M_{CSP}$ places the trimmed segment into the alignment coordinate system. It is constructed directly from IfcCurveSegment.Placement, where $(x_p, y_p)$ is the Location and $\theta_p$ is the bearing of the RefDirection.

$$M_{CSP} = \begin{bmatrix} \cos\theta_p & -\sin\theta_p & 0 & x_p \\ \sin\theta_p & \cos\theta_p & 0 & y_p \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Compute the position $(x_0, y_0)$ and tangent angle $\theta_0$ of the parent curve at $s_0$. This establishes the frame of the parent curve at the point where trimming begins.

$M_N$ simultaneously translates the trim-start point to the origin and rotates so that the tangent at $s_0$ aligns with the positive $x$-direction.

$$M_N = \begin{bmatrix} \cos\theta_0 & \sin\theta_0 & 0 & -x_0\cos\theta_0 - y_0\sin\theta_0 \\ -\sin\theta_0 & \cos\theta_0 & 0 & x_0\sin\theta_0 - y_0\cos\theta_0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 3 — Evaluate and map each point

For the point at arc-length $s$, compute the parent curve position $(x(s), y(s))$ and tangent angle $\theta(s)$ and form:

$$M_{PC} = \begin{bmatrix} \cos(\theta(s)) & -\sin(\theta(s)) & 0 & x(s) \\ \sin(\theta(s)) & \cos(\theta(s)) & 0 & y(s) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Apply the normalization and placement in sequence:

$$M_h = M_{CSP} M_N M_{PC}$$

The position of the point in the alignment coordinate system is the fourth column of $M_h$, and its tangent direction is the first column. Numerical examples of this algorithm are given in Sections 2.3 through 2.11.

2.3 Line

A tangent run is geometrically represented with a segment trimmed from an IfcLine parent curve.

2.3.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

$$\kappa(t) = 0$$

The point on line equation can be parameterized with a unit parameterization where the parameter $u$ ranges from $0$ to $1$ over the full extent of the curve, independent of its physical length.

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

It can also be parameterized with an arc length parameterization where the parameter $u$ equals the distance along the line in units of length, so $u = s$.

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

Other parent curve types can also be parameterized by unit value or arc length. However, the polynomial spirals do not have a well defined unit value parameterization. For this reason, the IFC specification mandates that all parameterization for alignment geometry be by arc length. For IfcCurveSegment, the SegmentStart and SegmentLength attributes must be of type IfcLengthMeasure. See Section 8.9.3.28 IfcCurveSegment, Informal Proposition 1.

This matters because IfcCurveSegment.SegmentStart and IfcCurveSegment.SegmentLength are of type IfcCurveMeasureSelect, which can be either IfcParameterValue (a dimensionless scalar based on unit value) or IfcLengthMeasure (a physical length).

2.3.2 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 S 57 E (5.70829654085293 radian) and has a length of 1956.785654.

Figure 2.3.2-1 Tangent segment

The semantic definition of this segment is:

#31=IFCCARTESIANPOINT((500.,2500.));
#32=IFCALIGNMENTHORIZONTALSEGMENT($,$,#31,5.70829654085293,0.,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. A valid, but unnecessarily complex parent curve is shown in Figure 2.3.2-2. The IfcLine passes through point (1000,-800) and is directed at 135° from the x-axis. The IfcCurveSegment trim starts 100 m from (1000,-800) and the trimmed segment has a length of 1956.796 m.

Figure 2.3.2-2 IfcLine at arbitrary placement

Figure 2.3.2-2 represents a valid parent curve definition and implementations must be prepared to properly interpret this parent curve geometry. This guide defines general algorithms to accommodate this geometry.

In practice, it is far easier to define the parent curve passing through point (0,0) and in the direction of the X-axis. The trimming can begin at the origin as shown in Figure 2.3.2-3.

Figure 2.3.2-3 IfcLine parent curve at origin

The parent curve definition is:

#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 is 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 trim 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. The tangent segment begins at (500,2500) and runs in the direction 5.70829654085293 radian.

From the segment direction,

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

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

The segment placement is

#31=IFCCARTESIANPOINT((500.,2500.));
#49=IFCDIRECTION((0.839252789970355,-0.54374144087698));
#50=IFCAXIS2PLACEMENT2D(#31,#49);

The IfcCurveSegment is defined as:

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

2.3.3 Compute Point on Curve

Compute the position matrix for a point 1500 m from the start of the curve segment.

Step 1 — Form the curve segment placement matrix $M_{CSP}$

From the IfcCurveSegment.Placement:

$$x = 500, y = 2500$$

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

$$M_{CSP} = \begin{bmatrix} 0.839252789970355 & 0.54374144087698 & 0 & 500 \\ -0.54374144087698 & 0.839252789970355 & 0 & 2500 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0 = 0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Evaluate the parent curve at $u = 1500$

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

$$\theta(1500) = \tan^{-1}\left(\frac{0}{1}\right) = 0$$

$$x = 0 + \cos(0)\int_{0}^{1500}{dt} = 0 + 1(1500\ m - 0\ m) = 1500\ m$$

$$y = 0 + \sin(0)\int_{0}^{1500}{dt} = 0 + 0(1500\ m - 0\ m) = 0\ m$$

Though it is much easier to use the following calculation

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

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

$$x = 0 + 1(1500) = 1500$$

$$y = 0 - 0(1500) = 0$$

$$M_{PC} = \begin{bmatrix} 1 & 0 & 0 & 1500 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

The resulting position matrix is

$$M_h = M_{CSP} M_N M_{PC}$$

$$M_h = \begin{bmatrix} 0.839252789970355 & 0.54374144087698 & 0 & 500 \\ -0.54374144087698 & 0.839252789970355 & 0 & 2500 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \mathbf{I} \begin{bmatrix} 1 & 0 & 0 & 1500\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$M_{h} = \begin{bmatrix} 0.83925279 & 0.54374114 & 0 & 1758.879185 \\ -0.54374114 & 0.83925279 & 0 & 1684.3878387 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.4 Circular Arc

Circular alignment curves are geometrically represented by an arc trimmed from an IfcCircle parent curve.

2.4.1 Parent Curve Parametric Equations

The curve tangent angle, curvature, and point on the curve are given by the following equations

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

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

$$x(s) = \int{\cos\left( \theta(s) \right)ds} = R\sin\left(\theta(s)\right)$$

$$y(s) = \int{\sin\left( \theta(s) \right)ds} = R\left(1 - \cos\left(\theta(s)\right)\right)$$

2.4.2 Semantic Definition to Geometry Mapping

IfcCircle is defined by its center (IfcCircle.Position) and radius (IfcCircle.Radius).

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 as shown in Figure 2.4.2-1.

Figure 2.4.2-1 — Curve segment trimmed from an IfcCircle parent curve

The parent curve is placed such that a trim starting at 0.0 is at (0,0) and the tangent is in the direction (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);
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));

2.4.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

The trim begins where the local x-axis of the circle intersects the circumference. The circle is centered at (0,300) with radius = 300 and the local x-axis is in the direction of the global y-axis. This puts the trim start point at $x_0 = 0, y_0 = 0$ and the tangent direction (1,0) with $\theta_0 = 0$

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Compute point on parent curve at $u = 50$

$$\theta(s) = \frac{s}{R} = \frac{50}{300} $$

$$d_x = \cos(\theta(s)) = \cos\left(\frac{50}{300}\right) = 0.98614323$$

$$d_y = \sin(\theta(s)) = \sin\left(\frac{50}{300}\right) = 0.16589613$$

$$x(s) = R\sin\left(\theta(s)\right) = 300\sin\left(\frac{50}{300}\right) = 49.76883981$$

$$y(s) = R\left(1 - \cos\left(\theta(s)\right)\right) = 300\left(1 - \cos\left(\frac{50}{300}\right)\right) = 4.1570305$$

$$M_{PC} = \begin{bmatrix} 0.98614323 & -0.16589613 & 0 & 49.76883981\\ 0.16589613 & 0.98614323 & 0 & 4.1570305\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} $$

The resulting position matrix is

$$M_h = M_{CSP} M_N M_{PC}$$

$$M_h = \mathbf{I} \mathbf{I} \begin{bmatrix} 0.98614323 & -0.16589613 & 0 & 49.76883981\\ 0.16589613 & 0.98614323 & 0 & 4.1570305\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$M_{h} = \begin{bmatrix} 0.98614323 & -0.16589613 & 0 & 49.76883981\\ 0.16589613 & 0.98614323 & 0 & 4.1570305\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.5 Clothoid

Spiral transition curves are geometrically represented by a segment trimmed from one of the many spiral types in IFC. IfcClothoid geometry is used for clothoid curves.

2.5.1 Parent Curve Parametric Equations

The curve tangent angle, curvature, and point on the curve are given by the following equations

$$\theta(s) = \frac{\pi}{2}\frac{A}{\left|A\right|}s^{2}$$

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

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

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

IFC provides a unit parametrization of clothoid curve.

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

Care must be taken when evaluating clothoids because IfcCurveSegment.SegmentStart and IfcCurveSegment.SegmentLength are defined with arc length. The details are illustrated in the example calculations.

2.5.2 Semantic Definition to Geometry Mapping

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, $A$

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

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}}= \frac{100}{1000} - \frac{100}{300} = -0.23333$$

$$A = \frac{L}{{\left|f\right|}^{\frac{1}{2}}}\frac{f}{\left|f\right|} = \frac{100}{{\left|-0.23333\right|}^{\frac{1}{2}}}\frac{-0.23333}{\left|-0.23333\right|} = -207.0196678$$

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

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

The curve segment starts where the radius is 300. A positive radius indicates a counter-clockwise curve which is a curve towards the left. 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.

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

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

$$s = -142.8571429$$

Figure 2.5.2-1 shows the parent curve clothoid with the segment having $\kappa_s = \frac{1}{300}$ and $\kappa_e = \frac{1}{1000}$ highlighted. This is the segment that IfcCurveSegment must trim from the parent curve. The negative clothoid constant results in the parent curve in quadrants 2 and 4. The curvature is positive in quadrant 2. For this reason, the SegmentStart attribute is negative. From this starting point, the trimming progresses to a larger radius, which is towards the origin. For this reason SegmentLength is a positive value. If the trimming were to progress to a smaller radius, SegmentLength would be a negative value. The SegmentStart and SegmentLength attributes can be positive and negative values in any combination. This is true for all parent curve types.

Figure 2.5.2-1 — IfcCurveSegment trimmed from an IfcClothoid parent curve

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);

2.5.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

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

Recall that the clothoid equations are in terms of a unit parameterization. Compute the parametric position on the curve.

$$u = \frac{-142.8571428}{\left| - 207.0196678\sqrt{\pi} \right|} = -0.3893278$$

Compute the point on curve tangent direction

$$\theta_0(-0.3893278) = \frac{\pi}{2}\frac{-207.0196678}{\left|-207.0196678\right|}(-0.3893278)^{2} = -0.238095237$$

and compute the point on the curve

$$x_0(-0.3893278) = - 207.0196678\sqrt{\pi}\int_{0}^{-0.3893278}{\cos{\left( \frac{\pi}{2}\frac{-207.0196678}{\left|-207.0196678\right|}t^{2} \right)\ }dt} = -142.04941746210602\ m$$

$$y_0(-0.3893278) = -207.0196678\sqrt{\pi}\int_{0}^{-0.3893278}{\sin{\left( \frac{\pi}{2}\frac{-207.0196678}{\left|-207.0196678\right|}t^{2} \right)\ }dt} = 11.292042785713347\ m$$

The terms of the normalization matrix are:

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

$$\sin(-0.238095237) = -0.235852028$$

$$-(-142.04941746210602\ m) \cos(-0.238095237) - (11.292042785713347\ m) \sin(-0.238095237) = 140.705309648\ m$$

$$(-142.04941746210602\ m) \sin(-0.238095237) - (11.292042785713347\ m) \cos(-0.238095237) = 22.529160405\ m$$

$$M_N = \begin{bmatrix} \cos\theta_0 & \sin\theta_0 & 0 & -x_0\cos\theta_0 - y_0\sin\theta_0 \\ -\sin\theta_0 & \cos\theta_0 & 0 & x_0\sin\theta_0 - y_0\cos\theta_0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0.971788979 & -0.235852028 & 0 & 140.705309648 \\ 0.235852028 & 0.971788979 & 0 & 22.529160405 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 3 — Evaluate and map each point

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

$$ s = -142.8571428 + 50 = -92.8571428$$

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

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

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

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

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

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

In matrix form

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

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_N M_{PC} = \mathbf{I} \begin{bmatrix} 0.971788979 & -0.235852028 & 0 & 140.705309648 \\ 0.235852028 & 0.971788979 & 0 & 22.529160405 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.994944564 & 0.100425663 & 0 & -92.763220844871512 \\ -0.100425663 & 0.994944564 & 0 & 3.1114126254443812 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

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

2.6 Cubic Transition Curve

IfcPolynomialCurve is the geometric type for cubic transition spirals.

2.6.1 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 $y(x) = A_{3}x^{3}$. See 4.2.2.1.5 Cubic Transition Segment.

The distance along a curve is defined by

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

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

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

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

1. For a distance $u$ along the curve, find $x$ for $d - u = 0$, within a tolerance consistent with the modeled elements. See Chapter 11 for a discussion on tolerances.

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

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

2.6.2 Semantic Definition to Geometry Mapping

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 coefficients

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

$$A_{3} = \frac{1}{6R_{e}L} - \frac{1}{6R_{s}L} = \frac{1}{6(300\ m)(100\ m)} - \frac{1}{6(\infty)(100\ m)} = 5.55555 \cdot 10^{-6}\ m^{-2}$$

The geometric representation is

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(100.), #45);
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));
#45 = IFCPOLYNOMIALCURVE(#46, (0., 1.), (0., 0., 0., 5.55555555555556E-6), $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Warning: 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 coefficients are real values (i.e. scalar values) with IfcReal.

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

The parametric form of the polynomial curve is $\lambda(u) = ( x(u), y(u), z(u) )$. 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 polynomial 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 multiplied with the curve length $L$ to result in 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 IfcPolynomialCurve 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.

An official Implementation Agreement does not appear to exist; however, this interpretation is consistent with discussions on GitHub [4].

2.6.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0=0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Compute point and curve tangent at 100 m from the start.

From the IfcPolynomialCurve definition,

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

Y Coefficients = ($0 m^1$, $0 m^{0}$, $0 m^{-1}$, $5.55555 \cdot 10^{-6} m^{-2}$)

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

Find $x$ such that

$$\left|d - u\right| = \left| \int_{0}^{x}\left(\sqrt{\left( y'(x) \right)^{2} + 1}\right) dx - u \right| < 10^{-4} \approx 0$$

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

Solve numerically

$$x = 99.72593255 m$$

Check solution

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

Compute y

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

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

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

Normalizing the tangent vector

$$dx = \frac{1}{\sqrt{1^{2} + {0.165753}^{2}}} = 0.986539$$

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

The resulting matrix is

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

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP}\ M_N\ M_{PC} = \mathbf{I} , \mathbf{I} \begin{bmatrix} 0.986539 & -0.1635219 & 0 & 99.72593255 \\ 0.1635219 & 0.986539 & 0 & 5.5100 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ $$M_{h} = \begin{bmatrix} 0.986539 & -0.1635219 & 0 & 99.72593255 \\ 0.1635219 & 0.986539 & 0 & 5.5100 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

2.7 Helmert Transition Curve

The Helmert curve is unique in that it is semantically defined as a single layout segment (IfcAlignmentHorizontalSegment) but geometrically defined with two curve segments (IfcCurveSegment), each representing half of the transition curve. The parent curve for a helmert transition curve is IfcSecondOrderPolynomialSpiral.

2.7.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

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

The polynomial coefficients carry a second subscript to indicate first half, $1$, and second half, $2$. For example, $A_{21}$ is coefficient $A_2$ for the first half and $A_{02}$ is coefficient $A_0$ for the second half.

First Half

$$a_{01} = \frac{L}{R_{s}},\ A_{01} = \frac{L}{\left| a_{0} \right|}\ \frac{a_{0}}{\left| a_{0} \right|}$$

$$a_{11} = 0,\ A_{11} = 0$$

$$a_{21} = 2f,\ A_{21} = \frac{L}{{\left| a_{2} \right|}^{\frac{1}{3}}}\ \frac{a_{2}}{\left| a_{2} \right|}$$

Second Half

$$a_{02} = -f + \frac{L}{R_{s}},\ A_{02} = \frac{L}{\left| a_{0} \right|}\ \frac{a_{0}}{\left| a_{0} \right|}$$

$$a_{12} = 4f,\ A_{12} = \frac{L}{{\left| a_{1} \right|}^{\frac{1}{2}}}\ \frac{a_{1}}{\left| a_{1} \right|}$$

$$a_{22} = -2f,\ A_{22} = \frac{L}{{\left| a_{2} \right|}^{\frac{1}{3}}}\ \frac{a_{2}}{\left| a_{2} \right|}$$

2.7.2 Semantic Definition to Geometry Mapping

Consider a horizontal Helmert transition curve segment that starts at (0,0) with a start direction of (1,0). 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

$$L = 100\ m$$

$$R_{s} = \infty$$

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

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

First Half

$$a_{01} = \frac{100}{\infty} = 0,\ A_{0} = 0$$

$$a_{11} = 0,\ A_{11} = 0$$

$$a_{21} = 2(0.33333) = 0.66667,\ A_{2} = \frac{100\ m}{{\left|0.66667\right|}^{\frac{1}{3}}}\frac{0.66667}{\left|0.66667\right|} = 114.4714255\ m$$

The geometric representation of the first half is

#36 = IFCCURVESEGMENT(.CONTSAMEGRADIENTSAMECURVATURE., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(50.), #45);
#37 = IFCLOCALPLACEMENT($, #38);
#38 = IFCAXIS2PLACEMENT3D(#39, $, $);
#39 = IFCCARTESIANPOINT((0., 0., 0.));
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));
#45 = IFCSECONDORDERPOLYNOMIALSPIRAL(#46, 114.471424255333, $, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

Note: when a polynomial coefficient is zero, the correspond term is unused and coded as $ in the IFC entity

Second Half

$$a_{02} = -0.33333 + \frac{100}{\infty} = -0.33333,\ A_{0} = \frac{100\ m}{\left|-0.33333\right|}\frac{-0.33333}{\left|-0.33333\right|} = - 300\ m$$

$$a_{12} = 4(0.33333) = 1.33333,\ A_{1} = \frac{100\ m}{{\left|1.33333\right|}^{\frac{1}{2}}}\frac{1.33333}{\left|1.33333\right|} = 86.602540378\ m$$

$$a_{22} = -2(0.33333) = -0.66667,\ A_{2} = \frac{100\ m}{{\left|-0.66667\right|}^{\frac{1}{3}}}\frac{-0.66667}{\left|-0.66667\right|} = -114.4714255\ m$$

Figure 2.7.2-1 shows the first and second half parent curves, without any adjustments.

Figure 2.7.2-1 First and second half of parent curves, without placement adjustments

Two-level placement for the second half

IfcCurveSegment evaluation normalizes the parent curve at SegmentStart, then applies the curve segment Placement. For the second half, SegmentStart = L/2, so the second half parent curve's value at $t = L/2$ is the normalization origin. The second half parent curve's own Position attribute is therefore set so that the point at $L/2$ exactly matches the endpoint of the first half.

With this alignment the normalization cancels the offset, and the curve segment Placement supplies only the world-space join position.

Level	Attribute	Purpose
IfcSecondOrderPolynomialSpiral.Position	$(x_p,\ y_p,\ \theta_p)$	Shift/rotate the raw spiral so its value at $t = L/2$ equals $(x_1, y_1, \theta_1)$
IfcCurveSegment.Placement	World-space join point	Place the normalized second-half curve at the correct position in the composite curve

Table 2.7.2-1 — Two-level placement for the second Helmert half

Finding the parent curve Position

Evaluate the raw second half spiral (with Position at origin) at $t = L/2$ to obtain $(x_2, y_2, \theta_2)$. The rigid-body transform that maps this to $(x_1, y_1, \theta_1)$ is

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

$$x_{p} = x_{1} - x_{2}\cos\theta_{p} + y_{2}\sin\theta_{p} = 49.9972 - 49.9664\cos(0.0555555) - 1.73556\sin(0.0555555) = 0.011503\ m$$

$$y_{p} = y_{1} - x_{2}\sin\theta_{p} - y_{2}\cos\theta_{p} = 0.347204 - 49.9664\sin(0.055555) + 1.73556\cos(0.0555555) = -0.6943693\ m$$

$$dx_p = \cos(\theta_p) = \cos(0.055555555) = 0.998457$$

$$dy_p = \sin(\theta_p) = \sin(0.055555555) = 0.055527$$

Figure 2.7.2-2 shows the same first and second half parent curves, with the origin of the second half curve translated $(x_p,y_p)$ and rotated so that the curve tangent is $(dx_p,dy_p)$

Figure 2.7.2-2 — First and second half parent curves with second half curve positioned

Finding the curve segment Placement

The Placement is the world-space position and direction at the join point, obtained by rotating $(x_1, y_1)$ by the overall segment start direction $\theta_s$ and translating by the start point $(x_s, y_s)$:

$$x_{join} = x_s + x_1\cos\theta_s - y_1\sin\theta_s$$

$$y_{join} = y_s + x_1\sin\theta_s + y_1\cos\theta_s$$

$$\theta_{join} = \theta_s + \theta_1$$

For the example with $\theta_s = 0$:

$$x_{join} = 0.0 + 49.99724\cos(0.0) - 0.347204\sin(0.0) = 49.99724\ \text{m}$$

$$y_{join} = 0.0 + 49.99724\sin(0.0) + 0.347204\cos(0.0) = 0.347204\ \text{m}$$

$$\theta_{join} = 0.0 + 0.02777777 = 0.02777777\ \text{rad}$$

The RefDirection of the curve segment Placement is $(\cos\theta_{join},, \sin\theta_{join})$:

$$\cos\theta_{join} = \cos(0.02777777) = 0.999614$$

$$\sin\theta_{join} = \sin(0.02777777) = 0.027774$$

The geometric representation of the second half

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

Figure 2.7.2-3 shows the trimmed and positioned first and second half parent curves resulting in the full Helmert transition curve.

Figure 2.7.2-3 — Final Helmert transition curve

2.7.3 Compute Point on Curve

Compute the curve coordinates at a distance along the curve, $u = 100$ (the end of the full Helmert segment).

The point falls in the second half ($50 &lt; u \leq 100$). The parent curve parameter is $t = \text{SegmentStart} + (u - 50) = 50 + 50 = 100$.

Step 1 — Form the curve segment placement matrix $M_{CSP}$

The curve segment Placement for the second half is at $(x_{join}, y_{join}) = (49.9972,\ 0.347204)$ with direction $\theta_{join} = 0.027778\ \text{rad}$:

$$M_{CSP} = \begin{bmatrix} 0.999614 & -0.027774 & 0 & 49.9972\\ 0.027774 & 0.999614 & 0 & 0.347204\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 2 — Evaluate the second half parent curve at SegmentStart and form the normalization matrix $M_N$

The second half parent curve has Position at $(x_p, y_p) = (0.011503,\ -0.694370)$ with direction $\theta_p = 0.055556\ \text{rad}$. This was chosen so that the parent curve at $t = L/2 = 50$ yields $(x_1, y_1, \theta_1) = (49.9972,\ 0.347204,\ 0.027778)$ — the endpoint of the first half.

$M_N$ maps $(x_1, y_1, \theta_1)$ to the origin with the $x$-axis tangent direction:

$$M_N = \begin{bmatrix} \cos\theta_1 & \sin\theta_1 & 0 & -x_1\cos\theta_1 - y_1\sin\theta_1\\ -\sin\theta_1 & \cos\theta_1 & 0 & x_1\sin\theta_1 - y_1\cos\theta_1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$M_N = \begin{bmatrix} 0.999614 & 0.027774 & 0 & -50.0\\ -0.027774 & 0.999614 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Step 3 — Evaluate and map each point

Evaluate the second half parent curve at $t = 100$. The raw spiral (before applying Position) uses $(A_{02},\ A_{12},\ A_{22}) = (-300,\ 86.6025,\ -114.4714)$:

$$\theta_{raw}(t) = \frac{t^3}{3(-114.4714)^3} + \frac{86.6025}{2\left|86.6025^3\right|}t^2 + \frac{t}{-300}$$

$$\theta_{raw}(100) = -0.22222 + 0.66666 - 0.33333 = 0.11111\ \text{rad}$$

$$x_{raw} = \int_0^{100}\cos\theta_{raw}(t),dt = 99.8942272\ \text{m}$$

$$y_{raw} = \int_0^{100}\sin\theta_{raw}(t),dt = -0.00146765266\ \text{m}$$

Apply the parent curve Position $(x_p, y_p, \theta_p)$:

$$x_{pos} = x_p + x_{raw}\cos\theta_p - y_{raw}\sin\theta_p = 0.011503 + 99.8942272(0.998457) - (-0.00146765266)(0.055527) = 99.7517\ \text{m}$$

$$y_{pos} = y_p + x_{raw}\sin\theta_p + y_{raw}\cos\theta_p = -0.694370 + 99.8942272(0.055527) + (-0.00146765266)(0.998457) = 4.8510\ \text{m}$$

$$\theta_{pos}(100) = \theta_p + \theta_{raw}(100) = 0.055556 + 0.11111 = 0.16\overline{6}\ \text{rad}$$

Note: $\theta_{pos}(100) = \frac{1}{6} = \frac{L}{2R_e} = \frac{100}{2 \times 300}$, confirming the total tangent angle change matches a Clothoid with the same endpoints.

Form $M_{PC}$ from the parent curve point at $t = 100$:

$$M_{PC} = \begin{bmatrix} \cos\theta_{pos} & -\sin\theta_{pos} & 0 & x_{pos}\\ \sin\theta_{pos} & \cos\theta_{pos} & 0 & y_{pos}\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0.98615 & -0.16590 & 0 & 99.7517\\ 0.16590 & 0.98615 & 0 & 4.8510\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Apply normalization and curve segment placement:

$$M_h = M_{CSP},M_N,M_{PC}$$

$$M_h = \begin{bmatrix} 0.999614 & -0.027774 & 0 & 49.9972\\ 0.027774 & 0.999614 & 0 & 0.347204\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.999614 & 0.027774 & 0 & -50.0\\ -0.027774 & 0.999614 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.98615 & -0.16590 & 0 & 99.7517\\ 0.16590 & 0.98615 & 0 & 4.8510\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$M_{h} = \begin{bmatrix} 0.98614323 & -0.16589613 & 0. & 99.75176295 \\ 0.16589613 & 0.98614323 & 0. & 4.85106157 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.8 Bloss Transition Curve

IfcThirdOrderPolynomialSpiral is the geometric type for Bloss transition spirals.

2.8.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

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

2.8.2 Semantic Definition to Geometry Mapping

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.);

Compute the polynomial spiral parameters

$$L = 100\ m$$

$$R_{s} = \infty$$

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

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

$$a_{0} = \frac{L}{R_{s}}= \frac{100}{\infty} = 0$$ $$A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|} = 0$$

$$A_{1} = 0$$

$$a_{2} = 3f = 3(0.33333) = 1$$ $$A_{2} = \frac{L}{{\left| a_{2} \right|}^{\frac{1}{3}}}\frac{a_{2}}{\left| a_{2} \right|} = \frac{100}{{\left|1\right|}^{\frac{1}{3}}}\frac{1}{\left|1\right|} = 100\ m$$

$$a_{3} = -2f = -2(0.33333) = -0.66667$$ $$A_{3} = \frac{L}{{\left| a_{3} \right|}^{\frac{1}{4}}}\frac{a_{3}}{\left| a_{3} \right|} = \frac{100\ m}{{\left|-0.66667\right|}^{\frac{1}{4}}}\frac{-0.66667}{\left|-0.66667\right|} = -110.668192\ m$$

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(100.), #45);
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));
#45 = IFCTHIRDORDERPOLYNOMIALSPIRAL(#46, -110.668191970032, 100., $, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

2.8.3 Compute Point on Curve

Compute the curve coordinates at a distance along the curve, $u = 50$

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0=0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

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

$$x = \int_{0}^{50}{\cos{\theta(s)}}\ ds = 49.9962109$$

$$y = \int_{0}^{50}{\sin{\theta(s)}}\ ds = 0.416638875$$

$$\theta(50) = \frac{-110.668192}{4\left|-{110.668192}^{5} \right|}(50)^{4} + \frac{1}{3(100)^{3}}(50)^{3} = 0.03125$$

$$dx = \cos\left( 0.03125 \right) = 0.999512$$

$$dy = \sin\left( 0.03125 \right) = 0.031245$$

In matrix form

$$M_{PC} = \begin{bmatrix} 0.999512 & -0.031245 & 0 & 49.9962109\\ 0.031245 & 0.999512 & 0 & 0.416638875\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_N M_{PC}= \mathbf{I} , \mathbf{I} \begin{bmatrix} 0.999512 & -0.031245 & 0 & 49.9962109\\ 0.031245 & 0.999512 & 0 & 0.416638875\\ 0 & 0 & 1 & 0\\ 0&0&0&1 \end{bmatrix}$$ $$M_{h} = \begin{bmatrix} 0.999512 & -0.031245 & 0 & 49.9962109\\ 0.031245 & 0.999512 & 0 & 0.416638875\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.9 Cosine Transition Curve

IfcCosineSpiral is the geometric type for cosine transition spirals.

2.9.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

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

2.9.2 Semantic Definition to Geometry Mapping

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.);

Compute the cosine spiral parameters

$$R_{s} = \infty,\ R_{e} = 300\ m,\ L = 100\ m$$ $$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

Constant Term:

$$a_{0} = 0.5f + \frac{L}{R_{s}} = 0.5(0.33333) + \frac{100}{\infty} = 0.16667$$ $$A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|}= \frac{100\ m}{\left|0.16667\right|}\ \frac{0.16667}{\left|0.16667\right|} = 600\ m$$

Cosine Term: $$a_{1} = -0.5f= -0.5(0.33333) = -0.16667$$ $$A_{1} = \frac{L}{\left| a_{1} \right|}\frac{a_{1}}{\left| a_{1} \right|}= \frac{100}{\left|-0.16667\right|}\ \frac{-0.16667}{\left|-0.16667\right|} = -600\ m$$

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(100.), #45);
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));
#45 = IFCCOSINESPIRAL(#46, -600., 600.);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

2.9.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0=0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Compute point and curve tangent at 100 m from the start.

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

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

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

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

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

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

In matrix form

$$M_{PC} = \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.7485\\ 0.16589 & 0.98614 & 0 & 4.9458\\ 0 & 0 & 1 & 0\\ 0&0&0&1 \end{bmatrix}$$

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_N M_{PC} = I\ I\\ \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.7485\\ 0.16589 & 0.98614 & 0 & 4.9458\\ 0 & 0 & 1 & 0\\ 0&0&0&1 \end{bmatrix}$$ $$M_{h} = \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.7485\\ 0.16589 & 0.98614 & 0 & 4.9458\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.10 Sine Transition Curve

IfcSineSpiral is the geometric type for sine transition spirals.

2.10.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

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

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

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

2.10.2 Semantic Definition to Geometry Mapping

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.);

Compute the sine spiral parameters

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

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

Constant Term: $$a_{0} = \frac{L}{R_{s}} = \frac{100}{\infty} = 0$$ $$A_{0} = \frac{L}{\left| a_{0} \right|}\frac{a_{0}}{\left| a_{0} \right|} = 0$$

Linear Term: $$a_{1} = f = 0.33333$$ $$A_{1} = \frac{L}{{\left| a_{1} \right|}^{\frac{1}{2}}}\frac{a_{1}}{\left| a_{1} \right|} = \frac{100}{{\left|0.33333\right|}^{\frac{1}{2}}}\frac{0.33333}{\left|0.33333\right|} = 173.2050808\ m$$

Sine Term: $$a_{2} = -\frac{1}{2\pi}f = -\frac{1}{2\pi}(0.33333) = -0.053051647$$ $$A_{2} = \frac{L}{\left| a_{2} \right|}\frac{a_{2}}{\left| a_{2} \right|} = \frac{100}{\left|-0.053051647\right|}\frac{-0.053051647}{\left|-0.053051647\right|} = -1884.955592\ m$$

#36 = IFCCURVESEGMENT(.CONTINUOUS., #42, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(100.), #45);
#42 = IFCAXIS2PLACEMENT2D(#43, #44);
#43 = IFCCARTESIANPOINT((0., 0.));
#44 = IFCDIRECTION((1., 0.));
#45 = IFCSINESPIRAL(#46, -1884.95559215388, 173.205080756888, $);
#46 = IFCAXIS2PLACEMENT2D(#47, $);
#47 = IFCCARTESIANPOINT((0., 0.));

2.10.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0=0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Compute point and curve tangent at 100 m from the start.

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

$$x = \int_{0}^{100}{\cos{\theta(t)}}\ dt = 99.75698\ m$$

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

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

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

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

In matrix form

$$M_{PC} = \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.75698\\ 0.16589 & 0.98614 & 0 & 4.70132\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_N M_{PC} = \mathbf{I} , \mathbf{I} \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.75698\\ 0.16589 & 0.98614 & 0 & 4.70132\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$ $$M_{h} = \begin{bmatrix} 0.98614 & -0.16589 & 0 & 99.75698\\ 0.16589 & 0.98614 & 0 & 4.70132\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.11 Viennese Transition Curve

The Viennese Bend is a transition spiral developed for high-speed railway applications. Unlike every other transition curve type in this guide — all of which define horizontal geometry purely from geometric parameters (start radius, end radius, and arc length) — the Viennese Bend incorporates vehicle dynamics: its curvature variation is shaped by the superelevation (cant) transition that accompanies it. The result is a curve that simultaneously satisfies continuity requirements for curvature, cant, and their first derivatives, producing lower lateral jerk experienced by passengers.

IfcAlignmentHorizontalSegment.GravityCenterLineHeight is required for VIENNESEBEND but is optional for all other horizontal segment types. The associated cant segment (IfcAlignmentCantSegment with PredefinedType = VIENNESEBEND) must span the same horizontal length.

The IFC geometric type is IfcSeventhOrderPolynomialSpiral.

2.11.1 Parent Curve Parametric Equations

The curve tangent angle and curvature are given by the following equations

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

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

2.11.2 Semantic Definition to Geometry Mapping

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., 0.1, .VIENNESEBEND.);

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

$D_{rh}$ = Rail Head Distance = IfcAlignmentCant.RailHeadDistance

$D_{sl}$ = IfcAlignmentCantSegment.StartCantLeft

$D_{el}$ = IfcAlignmentCantSegment.EndCantLeft

$D_{sr}$ = IfcAlignmentCantSegment.StartCantRight

$D_{er}$ = IfcAlignmentCantSegment.EndCantRight

$\beta_s = \frac{(D_{sr} - D_{sl})}{D_{rh}}$, Start Cross Slope

$\beta_e = \frac{(D_{er} - D_{el})}{D_{rh}}$, End Cross Slope

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

Compute the polynomial spiral parameters

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

$$f = \frac{L}{R_{e}} - \frac{L}{R_{s}} = \frac{100}{300} - \frac{100}{\infty} = 0.33333$$

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

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

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

Constant term

$$a_{0} = \frac{L}{R_{s}}= \frac{100}{\infty} = 0$$

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

Linear term

$$a_{1} = 0$$

$$A_{1} = 0$$

Quadratic term

$$a_{2} = cf= -0.504$$

$$A_{2} = \frac{L}{{\left| a_{2} \right|}^{\frac{1}{3}}}\frac{a_{2}}{\left| a_{2} \right|} = \frac{100}{{\left| -0.504 \right|}^{\frac{1}{3}}}\frac{-0.504}{\left| -0.504 \right|} = -125.6579069\ m$$

Cubic term

$$a_{3} = -4cf= -4(-0.504) = 2.016$$

$$A_{3} = \frac{L}{{\left| a_{3} \right|}^{\frac{1}{4}}}\frac{a_{3}}{\left| a_{3} \right|}= \frac{100}{{\left| 2.016 \right|}^{\frac{1}{4}}}\frac{2.016}{\left| 2.016 \right|} = 83.92229813\ m$$

Quartic term

$$a_{4} = 5cf + 35f = 5(-0.504) + 35(0.33333) = 9.1466655$$

$$A_{4} = \frac{L}{{\left| a_{4} \right|}^{\frac{1}{5}}}\frac{a_{4}}{\left| a_{4} \right|} = \frac{100}{{\left| 9.1466655 \right|}^{\frac{1}{5}}}\frac{9.1466655}{\left| 9.1466655 \right|} = 64.231406\ m$$

Quintic term

$$a_{5} = -2cf - 84f= -2(-0.504) - 84(0.33333) = -26.9919999$$

$$A_{5} = \frac{L}{{\left| a_{5} \right|}^{\frac{1}{6}}}\frac{a_{5}}{\left| a_{5} \right|} = \frac{100}{{\left| -26.9919999 \right|}^{\frac{1}{6}}}\frac{-26.9919999}{\left| -26.9919999 \right|} = -57.7378785\ m$$

Sextic term

$$a_{6} = 70f = 70(0.33333) = 23.33333333$$

$$A_{6} = \frac{L}{{\left| a_{6} \right|}^{\frac{1}{7}}}\frac{a_{6}}{\left| a_{6} \right|} = \frac{100}{{\left| 23.33333333 \right|}^{\frac{1}{7}}}\frac{23.33333333}{\left| 23.33333333 \right|} = 63.76388134\ m$$

Septic term

$$a_{7} = -20f = -20(0.33333) = -6.66666666$$

$$A_{7} = \frac{L}{{\left| a_{7} \right|}^{\frac{1}{8}}}\frac{a_{7}}{\left| a_{7} \right|} = \frac{100}{{\left| -6.66666666 \right|}^{\frac{1}{8}}}\frac{-6.66666666}{\left| -6.66666666 \right|} = -78.8880838\ m$$

#66 = IFCCURVESEGMENT(.CONTINUOUS., #72, IFCLENGTHMEASURE(0.), IFCLENGTHMEASURE(100.), #75);
#72 = IFCAXIS2PLACEMENT2D(#73, #74);
#73 = IFCCARTESIANPOINT((0., 0.));
#74 = IFCDIRECTION((1., 0.));
#75 = IFCSEVENTHORDERPOLYNOMIALSPIRAL(#76, -78.8880838459446, 63.7638813456506, -57.7378785242934, 64.2314061308743, 83.922298125931, -125.657906854859, $, $);
#76 = IFCAXIS2PLACEMENT2D(#77, $);
#77 = IFCCARTESIANPOINT((0., 0.));

2.11.3 Compute Point on Curve

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

Step 1 — Form the curve segment placement matrix $M_{CSP}$

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.

$$M_{CSP} = \mathbf{I}$$

Step 2 — Evaluate the parent curve at the trim start and form the normalization matrix $M_N$

Because the parent curve is located at (0,0) in the direction (1,0), $x_0 = 0, y_0 = 0, \theta_0=0$.

Since $x_0 = 0$, $y_0 = 0$, and $\theta_0 = 0$, $M_N = \mathbf{I}$.

Step 3 — Evaluate and map each point

Compute point and curve tangent at 100 m from the start.

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

$$x = \int_{0}^{100}{\cos{\theta(s)}}\ ds = 99.76319823871657 $$

$$y = \int_{0}^{100}{\sin{\theta(s)}}\ ds = 4.499916129045404 $$

$$\theta(100) = \frac{-78.8880838}{8\left| (-78.8880838)^{9} \right|} 100^{8} + \frac{1}{7 (63.76388134)^{7}} 100^{7} + \frac{−57.7378785}{6\left| (−57.7378785)^{7} \right|} 100^{6} + \frac{1}{5 (64.231406)^{5}} 100^{5} + \frac{83.92229813}{4\left| (83.92229813)^{5} \right|} 100^{4} + \frac{1}{3 (−125.6579069)^{3}} 100^{3} = \frac{1}{6} = 0.1666666667$$

$$dx = \cos\left( 0.1666666667\right) = 0.9861432315629198$$

$$dy = \sin\left( 0.1666666667\right) = 0.16589613269344608$$

In matrix form

$$M_{PC} = \begin{bmatrix} 0.9861432315629198 & -0.16589613269344608 & 0 & 99.76319823871657\\ 0.16589613269344608& 0.9861432315629198& 0 & 4.499916129045404\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Apply the normalization and curve segment placement to the parent curve point

$$M_{h} = M_{CSP} M_N M_{PC} = \mathbf{I} , \mathbf{I} \begin{bmatrix} 0.9861432315629198 & -0.16589613269344608 & 0 & 99.76319823871657\\ 0.16589613269344608 & 0.9861432315629198 & 0 & 4.499916129045404\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$ $$M_{h} = \begin{bmatrix} 0.9861432315629198 & -0.16589613269344608 & 0 & 99.76319823871657\\ 0.16589613269344608 & 0.9861432315629198 & 0 & 4.499916129045404\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

2.12 Zero Length Segment

IFC Concept Templates 4.1.4.4.1.1, 4.1.4.4.1.2, 4.1.7.1.1.1, 4.1.7.1.1.2, 4.1.7.1.1.3, and 4.1.7.1.1.4 require that the final IfcAlignmentSegment nested within IfcAlignmentHorizontal, IfcAlignmentVertical, and IfcAlignmentCant has a SegmentLength of zero. A corresponding zero-length IfcCurveSegment is required in each associated geometric representation (IfcCompositeCurve, IfcGradientCurve, or IfcSegmentedReferenceCurve).

The specification imposes no constraint on which curve type backs the zero-length geometric segment. As best practice, use .LINE. (IfcAlignmentHorizontalSegment), .CONSTANTGRADIENT. (IfcAlignmentVerticalSegment), or .CONSTANTCANT. (IfcAlignmentCantSegment) as the PredefinedType of the semantic closing segment, and an IfcLine as the parent curve of the corresponding IfcCurveSegment. The placement of both the semantic and geometric closing segments must match the endpoint and end direction of the preceding segment.

2.12.1 Semantic Closing Segment

The StartPoint and StartDirection of the semantic segment are taken from the end state of the preceding segment, and SegmentLength is set to zero. One implementation strategy is to map the preceding semantic segment definition to its geometric representation as described in this chapter, as well as in Chapters 3 and 4, and evaluate it at its end point.

// Last non-zero length segment
#88=IFCCARTESIANPOINT((7790.93212831259,4006.73076454875));
#89=IFCALIGNMENTHORIZONTALSEGMENT($,$,#88,-1.23849516741382,0.,0.,2112.2850844257,$,.LINE.);

// Final segment, zero length.
#91=IFCCARTESIANPOINT((8480.,2010.));
#92=IFCALIGNMENTHORIZONTALSEGMENT($,$,#91,-1.23849516741382,0.,0.,0.,$,.LINE.);

2.12.2 Geometric Closing Segment

The zero-length IfcCurveSegment is placed at the endpoint of the alignment with its axis aligned to the end tangent direction. The parent curve is an IfcLine whose point and direction vector match that placement. SegmentStart and SegmentLength are both zero. Because the segment has zero length, the choice of IfcLine here has no geometric effect — it represents a point at the end of the alignment.

// Last non-zero length segment
#169=IFCCARTESIANPOINT((0.,0.));
#170=IFCDIRECTION((1.,0.));
#171=IFCVECTOR(#170,1.);
#172=IFCLINE(#169,#171);
#173=IFCDIRECTION((0.326219162729524,-0.945294164727599));
#174=IFCAXIS2PLACEMENT2D(#88,#173);
#175=IFCCURVESEGMENT(.CONTSAMEGRADIENTSAMECURVATURE.,#174,IFCLENGTHMEASURE(0.),IFCLENGTHMEASURE(2112.2850844257),#172);

// Final segment, zero length
#176=IFCCARTESIANPOINT((0.,0.));
#177=IFCDIRECTION((1.,0.));
#178=IFCVECTOR(#177,1.);
#179=IFCLINE(#176,#178);
#180=IFCDIRECTION((0.326219162729524,-0.945294164727599));
#181=IFCAXIS2PLACEMENT2D(#91,#180);
#182=IFCCURVESEGMENT(.DISCONTINUOUS.,#181,IFCLENGTHMEASURE(0.),IFCLENGTHMEASURE(0.),#179);

2.13 Summary and Implementation Checklist

#	Item	Notes
1	Use IfcLengthMeasure for SegmentStart and SegmentLength on every IfcCurveSegment	Arc-length parameterization is mandatory per IFC schema Informal Proposition 1; IfcParameterValue (unit parameterization) is not valid for alignment geometry
2	Handle negative SegmentStart and SegmentLength	A negative SegmentStart locates the trim origin on the opposite side of the parent curve's reference point; a negative SegmentLength indicates trimming in the direction of decreasing parameter. Both can occur in any combination.
3	For clothoid curves, convert SegmentStart from arc-length to unit parameter before evaluating	Compute $u = s / \left
4	Expect one IfcAlignmentHorizontalSegment with PredefinedType = HELMERTCURVE to map to two IfcCurveSegment entities	The Helmert curve is the only horizontal type where one semantic segment produces two geometric segments; evaluate the correct half based on the query distance relative to $L/2$
5	Treat IfcPolynomialCurve coefficients as having implicit units of $\text{Length}^{(1-i)}$	For a cubic curve, CoefficientsY[3] has implicit units of $m^{-2}$; evaluate as $y = b_3 x^3$ where $x$ is a length. Do not multiply by curve length $L$ as you would for a unit-parameterized polynomial.
6	Viennese Bend horizontal geometry requires GravityCenterLineHeight and a paired cant segment	IfcAlignmentHorizontalSegment.GravityCenterLineHeight is mandatory for VIENNESEBEND; the associated IfcAlignmentCantSegment with PredefinedType = VIENNESEBEND must span the same horizontal length, and its cant values enter the curvature formula via the cant factor $cf$
7	Do not assume the parent curve is centered at the origin or aligned with the x-axis	The three-step algorithm ($M_{CSP}$, $M_N$, $M_{PC}$) handles arbitrary parent curve placement; SegmentStart locates where trimming begins regardless of the parent curve's own position and orientation
8	Apply the positive-left sign convention for radii	Positive StartRadiusOfCurvature / EndRadiusOfCurvature indicates a curve to the left of the direction of travel; negative indicates a curve to the right; zero indicates infinite radius (straight tangent)