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Chapter 2 · 6 hours

Geometric Modeling Fundamentals

Practice questions

Practice questions and answers

5 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Differentiate between geometry and topology in solid modelling. A rectangular block has a square hole drilled completely through it, from the top face to the bottom face. It has 16 vertices, 24 edges and 10 faces. Use the Euler-Poincare formula to find the number of rings (inner loops) and check that the solid is valid.

Answer

Geometry and topology

Geometry is the metric information of the model: actual coordinates of points, equations of curves and surfaces, lengths, angles, radii. Topology is the connectivity information: how vertices, edges and faces are joined, and which side of a face is material. It does not change when the shape is stretched.

PointGeometryTopology
DescribesSize and shapeConnectivity and adjacency
ExamplesCoordinates, line equation, radiusVertex-edge-face relations, loops, shells
Under scalingChangesUnchanged
Used forDisplay, measurementValidity checks, B-rep data structure

Euler-Poincare formula

For a valid solid with holes (genus) and inner loops:

V−E+F−R=2 (S−H)V - E + F - R = 2\,(S - H)

where VV = vertices, EE = edges, FF = faces, RR = rings (inner loops on faces), SS = shells, HH = through-holes (genus).

Numerical

Given: V=16V = 16, E=24E = 24, F=10F = 10, one shell S=1S=1, one through-hole H=1H=1.

16−24+10−R=2(1−1)=02−R=0R=2\begin{aligned} 16 - 24 + 10 - R &= 2(1-1) = 0 \\ 2 - R &= 0 \\ R &= 2 \end{aligned}

The two rings are the square hole boundaries on the top face and on the bottom face. Each of these faces has an outer loop and one inner loop, so R=2R = 2, which agrees with the physical block. The formula is satisfied, so the topology is valid.

Answer: R=2R = 2 inner loops (top and bottom faces); V−E+F−R=0=2(S−H)V-E+F-R = 0 = 2(S-H), so the solid is valid.

  • Practice · 6 marks

Explain the Cartesian, cylindrical and spherical coordinate systems used in geometric modelling. (a) Convert the Cartesian point (3, 4, 5) into cylindrical and spherical coordinates. (b) Convert the spherical point r = 10, theta = 60 deg (measured from the +Z axis), phi = 45 deg (azimuth from +X in the XY plane) into Cartesian coordinates.

Answer

Coordinate systems

  • Cartesian (x,y,z)(x, y, z): three mutually perpendicular axes; simplest for prismatic shapes and used internally by CAD.
  • Cylindrical (ρ,ϕ,z)(\rho, \phi, z): radial distance in the XY plane, azimuth angle from +X, and height; suited to shafts, holes and revolved parts.
  • Spherical (r,θ,ϕ)(r, \theta, \phi): distance from origin, polar angle from +Z, azimuth from +X; suited to spheres and radial scanning.

Relations:

ρ=x2+y2,ϕ=tan⁡−1yx,r=x2+y2+z2,θ=cos⁡−1zr\rho=\sqrt{x^2+y^2},\quad \phi=\tan^{-1}\frac{y}{x},\quad r=\sqrt{x^2+y^2+z^2},\quad \theta=\cos^{-1}\frac{z}{r} x=rsin⁡θcos⁡ϕ,y=rsin⁡θsin⁡ϕ,z=rcos⁡θx=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta

(a) Point (3, 4, 5)

ρ=32+42=5ϕ=tan⁡−1(4/3)=53.13∘r=9+16+25=7.071θ=cos⁡−1(5/7.071)=45∘\begin{aligned} \rho &= \sqrt{3^2+4^2} = 5 \\ \phi &= \tan^{-1}(4/3) = 53.13^\circ \\ r &= \sqrt{9+16+25} = 7.071 \\ \theta &= \cos^{-1}(5/7.071) = 45^\circ \end{aligned}

Cylindrical: (5, 53.13∘, 5)(5,\ 53.13^\circ,\ 5). Spherical: (7.071, 45∘, 53.13∘)(7.071,\ 45^\circ,\ 53.13^\circ).

(b) Spherical to Cartesian

x=10sin⁡60∘cos⁡45∘=10(0.8660)(0.7071)=6.124y=10sin⁡60∘sin⁡45∘=6.124z=10cos⁡60∘=5.000\begin{aligned} x &= 10\sin 60^\circ\cos 45^\circ = 10(0.8660)(0.7071) = 6.124 \\ y &= 10\sin 60^\circ\sin 45^\circ = 6.124 \\ z &= 10\cos 60^\circ = 5.000 \end{aligned}

Answer: (a) cylindrical (5, 53.13 deg, 5); spherical (7.071, 45 deg, 53.13 deg). (b) Cartesian (6.124, 6.124, 5.000) in the same length units.

  • Practice · 5 marks

Two vectors are A = 2i + 3j + k and B = 4i - j + 2k. Find (a) A.B, (b) A x B, (c) the angle between them, (d) the unit vector normal to the plane of A and B, and (e) the area of the parallelogram formed by A and B. State where each result is used in geometric modelling.

Answer

Given

A=(2,3,1)\mathbf A = (2, 3, 1), B=(4,−1,2)\mathbf B = (4, -1, 2).

(a) Dot product

A⋅B=2(4)+3(−1)+1(2)=8−3+2=7\mathbf A\cdot\mathbf B = 2(4)+3(-1)+1(2) = 8-3+2 = 7

(b) Cross product

A×B=∣ijk2314−12∣=(6+1)i−(4−4)j+(−2−12)k=7i+0j−14k\mathbf A\times\mathbf B= \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&1\\4&-1&2\end{vmatrix} = (6+1)\mathbf i-(4-4)\mathbf j+(-2-12)\mathbf k = 7\mathbf i+0\mathbf j-14\mathbf k

(c) Angle

∣A∣=14=3.742|\mathbf A|=\sqrt{14}=3.742, ∣B∣=21=4.583|\mathbf B|=\sqrt{21}=4.583.

cos⁡θ=73.742×4.583=0.4082,θ=65.9∘\cos\theta=\frac{7}{3.742\times4.583}=0.4082,\qquad \theta=65.9^\circ

(d) Unit normal

∣A×B∣=49+0+196=15.652|\mathbf A\times\mathbf B|=\sqrt{49+0+196}=15.652

n^=(7,0,−14)15.652=(0.447, 0, −0.894)\hat{\mathbf n}=\frac{(7,0,-14)}{15.652}=(0.447,\ 0,\ -0.894)

(e) Area of parallelogram

Area =∣A×B∣=15.65=|\mathbf A\times\mathbf B| = 15.65 square units.

Uses in modelling

  • Dot product: angle between edges, projection of a point on a line, checking perpendicularity (dot product = 0).
  • Cross product: surface normal of a face (used in shading, offsetting, STL facet normals), area of a face, and the sign of a turn.

Answer: A.B = 7; AxB = (7, 0, -14); angle = 65.9 deg; unit normal = (0.447, 0, -0.894); area = 15.65 square units.

  • Practice · 6 marks

Differentiate between parametric and non-parametric representation of curves. Why is parametric representation preferred in CAD? Represent a circle of radius 5 centred at the origin in both forms.

Answer

Non-parametric representation gives a direct relation between coordinates: explicit y=f(x)y=f(x) or implicit f(x,y)=0f(x,y)=0. Parametric representation expresses each coordinate as a function of an independent parameter uu: x=x(u)x=x(u), y=y(u)y=y(u), z=z(u)z=z(u), with uu in a range, usually 0≤u≤10\le u\le1.

PointNon-parametricParametric
Formy=f(x)y=f(x) or f(x,y)=0f(x,y)=0P(u)=[x(u),y(u),z(u)]\mathbf P(u)=[x(u),y(u),z(u)]
Multi-valued curves (circle)Explicit form fails (two yy values per xx)No problem
Vertical tangentSlope becomes infinitedy/dudy/du, dx/dudx/du remain finite
Axis dependenceDepends on chosen axesIndependent of axes
Point generationSolve equation, may need iterationSubstitute uu directly
Bounded segmentsNeeds extra conditionsLimits of uu define the segment
3D curvesNeeds two equationsEasy to extend
TransformationEquation must be re-derivedTransform control points / vector

Why parametric is preferred

  1. Curves can be closed, multi-valued or have vertical tangents.
  2. Easy to generate points at equal steps of uu for display and tool paths.
  3. Tangent vector is simply dP/dud\mathbf P/du.
  4. Segments are bounded and can be joined with continuity conditions.
  5. Same equation works in 2D and 3D, and with any axes.

Circle of radius 5

Non-parametric (implicit): x2+y2=25x^2+y^2=25; explicit form needs two branches y=±25−x2y=\pm\sqrt{25-x^2}.

Parametric:

x=5cos⁡θ,y=5sin⁡θ,0≤θ≤2πx = 5\cos\theta,\qquad y = 5\sin\theta,\qquad 0\le\theta\le 2\pi

For example θ=90∘\theta=90^\circ gives (0,5)(0,5) directly, where the explicit form has an infinite slope.

  • Practice · 8 marks

Explain the three types of geometric modelling: wireframe, surface and solid modelling. Compare them with their advantages, limitations and applications.

Answer

Geometric modelling is the computer representation of the shape of an object. Three types are used, in increasing order of completeness.

Wireframe modelling

The object is represented by its edges (lines, arcs, curves) and vertices only; there are no faces.

  • Advantages: small data, fast to create and display, little memory, good for simple layouts and tool-path skeletons.
  • Limitations: ambiguous (a cube can be read in several ways), no hidden-line removal, no surface or volume data, no mass properties, no shading.

Surface modelling

The object is described by its bounding surfaces (plane, ruled, revolved, Bezier, B-spline, Coons patches), stored with edges.

  • Advantages: can model free-form shapes, supports hidden-line removal and shading, gives surface normals for NC machining and for aerodynamic shapes.
  • Limitations: the inside/outside of the object is unknown; no volume or mass properties; surfaces may have gaps or overlaps.
  • Applications: car bodies, aircraft skins, ship hulls, consumer product casings, dies and moulds.

Solid modelling

The object is represented as a closed volume with a definite inside and outside. Two main schemes:

  • CSG (constructive solid geometry): primitives (block, cylinder, sphere, cone) combined with Boolean operations (union, intersection, difference); stored as a binary tree.
  • B-rep (boundary representation): faces, edges, vertices with topology.
  • Advantages: unambiguous and complete, volume, mass, centre of gravity and moment of inertia available, interference checking, FEA meshing, CAM links.
  • Limitations: more memory and computing; free-form shapes are harder.
FeatureWireframeSurfaceSolid
Data storedEdges, verticesSurfaces + edgesVolume + topology
Hidden line removalNoYesYes
Mass propertiesNoNoYes
AmbiguityHighLowNone
MemoryLeastMediumMost
Typical useLayout, conceptFree-form shapesEngineering parts

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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