UNIVERSITEIT
STELLENBOSCH
UNIVERSITY
DEPARTEMENT MEGANIES EN MEGATRONIESE INGENIEURSWESE
DEPARTMENT OF MECHANICAL AND MECHATRONICS ENGINEERING
Descriptive Geometry
for Engineering Students
AH Basson, DNJ Els, PJ Prior and K Schreve
Version 4.2
© Copyright 2016: University of Stellenbosch
Contents
1 Introduction
1
2 Descriptive geometry of points and lines
2.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Projection planes and co-ordinates . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Projectors and the projection of a point . . . . . . . . . . . . . . . . . . . .
2.2.3 Projection drawings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 New projection planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Conventions for new projection planes . . . . . . . . . . . . . . . . . . . .
2.3.2 Projecting to a new projection plane . . . . . . . . . . . . . . . . . . . . . .
2.4 Projections and trace points of a straight line . . . . . . . . . . . . . . . . . . . . .
2.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Construction of the trace points of a straight line . . . . . . . . . . . . . .
2.5 Condition for two lines to intersect . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 The true length and true angles of a line . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Constructions involving the trace points, true angles and true length of straight
lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.1 Projecting the θ triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.2 Projecting the φ triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.3 Determining the front and top views of a line, given the angles θ and φ
and true length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.4 Swing method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.5 Example: Constructing a triangle, given various true lengths and/or angles
2.8 Constructions using new projection planes and the properties of lines . . . . .
2.8.1 Shortest distance between two lines . . . . . . . . . . . . . . . . . . . . . .
2.8.2 True shape of a triangle by means of new projection planes . . . . . . . .
2.8.3 The line of intersection and angle between two triangles . . . . . . . . . .
3 Descriptive geometry of planes
3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Important characteristics of trace lines and lines in planes . . . . . . . . . . . . .
3.3 Constructions using the properties of trace lines . . . . . . . . . . . . . . . . . .
3.3.1 The line of intersection between two planes . . . . . . . . . . . . . . . . .
3.3.2 The true angle between two trace lines . . . . . . . . . . . . . . . . . . . .
3.3.3 Completion of the projections of a point on a plane . . . . . . . . . . . . .
3.3.4 Determination of a trace line from the projections of a point on the plane
3.3.5 Determination of a trace line given a point on a trace line, when the
point of intersection cannot be determined . . . . . . . . . . . . . . . . .
3.4 Converting an oblique plane into an inclined plane using a new projection plane
3.5 Constructions with an oblique plane converted to an inclined plane . . . . . . .
3.5.1 Completion of the projections of a point on a plane . . . . . . . . . . . . .
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CONTENTS
3.5.2 Point of intersection between a plane and a straight line . . . . . . . . . .
3.5.3 Shortest distance between a point and a plane . . . . . . . . . . . . . . . .
3.5.4 The true angle between a plane and a straight line . . . . . . . . . . . . .
3.6 The true angle between planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.2 θ and φ for an oblique plane . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.3 Determination of θ and φ using new projection planes . . . . . . . . . . .
3.6.4 Swing method for determination of θ and φ . . . . . . . . . . . . . . . . .
3.6.5 Determination of the trace lines of a plane if θ and φ are given . . . . . .
3.6.6 Construction of a trace line if the other trace line and one true angle is
given . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7 The true angle between two oblique planes . . . . . . . . . . . . . . . . . . . . . .
3.7.1 Stategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7.2 New projection plane method . . . . . . . . . . . . . . . . . . . . . . . . .
3.7.3 Folding flat method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Shadows
4.1 Introduction . . . . . . . . . . . . . . . . .
4.2 Shadow of a point on one plane . . . . .
4.3 Shadow of a line on one or more planes .
4.4 Shadow of an object on a plane . . . . . .
4.5 Shadow lines of an object on itself . . . .
4.6 The shadow of one object upon another
4.6.1 Shadow of a point on an object . .
4.6.2 Shadow of a line on an object . . .
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5 Interpenetrations
5.1 Interpenetration points . . . . . . . . . . . . . . . . . . . . . .
5.1.1 A straight line through a sphere . . . . . . . . . . . . . .
5.1.2 A straight line through a prism . . . . . . . . . . . . . .
5.1.3 A straight line through a cone . . . . . . . . . . . . . . .
5.2 Simple interpenetration curves . . . . . . . . . . . . . . . . . .
5.2.1 A rectangular bar through a cone . . . . . . . . . . . . .
5.2.2 One cylinder through another cylinder . . . . . . . . .
5.2.3 A cylinder through a cone . . . . . . . . . . . . . . . . .
5.2.4 A cylinder through a sphere . . . . . . . . . . . . . . . .
5.3 Interpenetration curves requiring the use of oblique planes .
5.3.1 An oblique cone interpenetrated by a line. . . . . . . .
5.3.2 A cone interpenetrated by a cylinder . . . . . . . . . .
5.3.3 The interpenetrating curves between two given cones
6 Developments
6.1 Introduction . . . . . . . . . . . . . .
6.2 Conventions . . . . . . . . . . . . . .
6.3 Prism examples . . . . . . . . . . . .
6.3.1 Rectangular prism . . . . . .
6.3.2 Skew prism . . . . . . . . . .
6.4 Cylinder examples . . . . . . . . . .
6.4.1 Cylinder with one square end
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CONTENTS
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A Miscellaneous
A.1 First and third angle projection . . . . . . . . . . . .
A.2 Isometric projection . . . . . . . . . . . . . . . . . .
A.2.1 Descriptive Geometry background . . . . . .
A.2.2 Isometric projection of a sphere . . . . . . .
A.2.3 Isometric drawings vs. isometric projections
A.2.4 Inclined lines and curves . . . . . . . . . . .
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B Stellenbosch University Drawing Standards
B.1 Introduction . . . . . . . . . . . . . . . . .
B.2 Drawing techniques . . . . . . . . . . . .
B.2.1 Line thicknesses . . . . . . . . . .
B.2.2 Script . . . . . . . . . . . . . . . . .
B.2.3 Projection symbol . . . . . . . . .
B.2.4 Title frame . . . . . . . . . . . . . .
B.3 Guidelines for dimensioning . . . . . . .
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6.5
6.6
6.7
6.8
6.4.2 Cylinder with circular ends .
6.4.3 Junction of two circular pipes
Rectangular pyramid example . . .
Cone examples . . . . . . . . . . . .
6.6.1 Upright cone . . . . . . . . .
6.6.2 Skew cone . . . . . . . . . . .
Sphere and torus examples . . . . .
Transition examples . . . . . . . . .
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C The Metric Vernier Caliper
103
D Geometrical Constructions and Tangency
105
E Sectioning and hatching
113
F Dimensioning
115
G True Lengths and Auxiliary Views
117
H Conic Sections and Interpenetrations of Solids
123
Oefeninge/Exercises
125
v
Chapter 1
Introduction
The study of Descriptive Geometry is a well established way to develop the three-dimensional
perception and drawing capabilities of engineering students. The ability to represent threedimensional objects on paper, and to correctly interpret the two-dimensional representations of three-dimensional objects, is essential for all engineers. Furthermore, the outputs of
designs must often be portrayed as engineering drawings. Engineering students must therefore be able to prepare such drawings, to interpret them, and to judge their correctness and
completeness and ultimately, as an engineer, approve them.
The ability to conceptualize and communicate three-dimensional objects is an essential
ability to all engineering disciplines. Typical examples where these principles are essential
are: calculating the radar cross-section of a ship, determining the best layout of a chemical
processing plant, determining the volume of material that must be moved where a road cuts
through a hill, and planning the flow of material in a packaging plant.
In Descriptive Geometry we study methods that can be used to represent three-dimensional
objects on paper (i.e. in two dimensions) and to obtain relationships between objects by
means of construction. We refer to objects as being “in space”. This does not mean that the
objects are necessarily in outer space. Space contains everything on, as well as under, the
surface of the earth!
Although Computer Aided Design (CAD) programs can do some of the operations that we
consider here, it is still important for all engineers to be able to visualize three-dimensional
concepts that can only be represented by two-dimensional drawings. To produce threedimensional drawings using CAD, requires the same insight as drawing them on paper!
The benefit of Descriptive Geometry is first and foremost to develop the ability to visualize objects as well as their manipulation three-dimensionally. The knowledge of the techniques for the manipulation of objects in space is an incidental bonus.
Although many prospective engineering students now take technical drawing courses in
high school, this book presumes no prior knowledge. However, the book is intended to augment the material presented in other books on Engineering Drawing. In particular, “Manual
of Engineering Drawing” by C Simmons and D Maguire, is taken as point of departure.
This book is the result of contributions of various staff members of the Department of
Mechanical Engineering, Stellenbosch University. The material has been taught in Descriptive Geometry and Engineering Drawing courses for decades and the written notes for the
courses have been written through the years. The most recent significant contributors are
Prof Anton Basson, Prof Gerrie Thiart, Dr Danie Els, Mr Peter Prior and Mr Thys Neethling.
When augmented with a suitable book on Engineering Drawing, the material presented
here is suitable for a single semester course at first-year engineering level.
1
Chapter 2
Descriptive geometry of points and lines
2.1
Background
In Descriptive Geometry we study methods that can be used to represent three-dimensional
objects on paper (i.e. in two dimensions). Some of the fundamental principles involved are
derived from working with only points and lines in three-dimensional space. In the next
chapter, the application of planes will be addressed.
Objects are often represented by means of a number of lines. These lines usually represent edges (changes of plane) on an object. A straight line can be defined by its two end
points in space. The basis for the representation of objects is therefore the projection of
points. If the idea of projecting a point feels too abstract, just imagine that the point that has
to be projected is on the corner of an object and that the representation of that corner on the
projection planes is being sought.
2.2
Conventions
2.2.1
Projection planes and co-ordinates
As you will see during this course, orthographic projection (“ortografiese projeksie”) is one of
the most powerful methods to represent three-dimensional objects in two dimensions. Because projection is actually a symbolic representation of a real object, there must be universal agreement about the symbols that are used. We must therefore adhere to the conventions
that are in general use in this subject field.
Orthographic projection is based on the use of three mutually perpendicular projection
planes, as shown in figure 2.1. The names of the three planes and their abbreviations are:
• The horizontal plane (HP)
• The vertical plane (VP)
• The side plane (SP)
(“horisontale vlak”)
(“vertikale vlak”)
(“syvlak”)
You are probably familiar with the practice of indicating the position of a point with three
co-ordinates (typically X , Y , and Z ). According to our convention:
•
•
•
•
The X -axis is located at the intersection of the HP and the VP, i.e. at Y = Z = 0.
The Y -axis is located at the intersection of the VP and the SP, i.e. at X = Z = 0.
The Z -axis is located at the intersection of the HP and the SP, i.e. at X = Y = 0.
O (NB not “zero”, but “oh”) is located at the intersection of the HP, VP, and SP, i.e. at
X = Y = Z = 0.
• The X , Y and Z -axes are arranged according to the right-hand rule (the fingers of the
right-hand representing the axes: thumb for X , forefinger for Y ,and middle finger for
Z ).
3
Chapter 2. Descriptive geometry of points and lines
Figure 2.1: Projection planes, projectors and projections
If a point with coordinates (20, 10, 15) is mentioned, then the meaning is that the point is
20 to the right of O (X -direction)
10 above the HP (Y -direction)
15 in front of the VP (Z -direction)
Similarly, the point (-12,-6,-3) is
12 to the left of O
6 below the HP
3 behind the VP
Note that we speak as if the observer stands on the HP and looks towards the VP with O on
his/her left hand side and X on his/her right hand side.
As you will see later, the HP and VP are usually sufficient for our constructions and we
rarely use the SP. The HP and VP divide the space into four quadrants. Note where the “first”,
“second”, “third” and “fourth” quadrants are indicated in figure 2.1 at the end of the X -axis.
By now you probably have already encountered the use of “first angle projection” (the object
is located fully in the first quadrant) and “third angle projection” (the object is located fully
in the third quadrant) in creating working drawings (see §A.1). In Descriptive Geometry, objects can be located anywhere in space. Objects can therefore intersect any of the projection
planes!
2.2.2
Projectors and the projection of a point
The image of a point is obtained by projecting (“projekteer”) it to the relevant projection
plane. A projector (“projektor”) runs from the point in space, e.g. A in figure 2.1, to a projection (“projeksie”) of the point, e.g. a0 , on one of the projection planes, i.e. the VP in the case
4
2.2. Conventions
of a0 . Examples of projectors are the lines Aa0 , Bb and Aa00 in figure 2.1. The projection of
point A in the HP is therefore a, the projection of point B in the VP is b0 , etc.
In orthographic projections, the projectors are perpendicular to the projection plane (or
picture plane). Therefore the projectors are mutually perpendicular or parallel. The case
where they are not parallel, but converge to one point, is called perspective projection. Orthographic projection is mostly used in general engineering drawing work, because it gives
an accurate representation of scale. In this book we shall work with the orthographic projections of a point, a straight line, planes and solid objects.
The following conventions are used in Descriptive Geometry:
1. True points in space are indicated by means of uppercase letters, e.g. A, B, etc.
2. The projection of a point in the HP is indicated by a lowercase letter, e.g. a, b, etc. This
image is called the top view (“bo-aansig”).
3. The projection of a point in the VP is indicated by a lowercase letter with a single prime,
e.g. a0 , b0 , etc. (We say “b prime”, or, in Afrikaans, “b aksent.” This image is called the
front view (“vooraansig”).
4. The projection of a point in the SP is indicated by a lowercase letter with a double
prime, e.g. a00 , b00 , etc. (We say “b double prime”, or, in Afrikaans, “b dubbel aksent.”
This image is called the side view (“syaansig”).
2.2.3
Projection drawings
In §2.2.2 it was explained how to obtain the projection of a point. The projection is, however,
still a point in three-dimensional space: the top view points are located in the HP and the
front view points in the VP. We actually want to show everything on a flat sheet of paper.
In order to achieve this, we draw the various projection planes on top of each other on the
paper. This is done in a very specific manner, in order to still represent the three-dimensional
information accurately.
The process that we use to get all the projection planes on the same sheet of paper is
illustrated in figures 2.2, 2.3 and 2.4. Note that the HP and the VP intersect at the OX line.
Note further that the points A, B, C and D are located in the four quadrants 1, 2, 3 and 4
respectively. The top views are given by a, b, c and d and the front views by a0 , b0 , c0 and d0 ,
respectively.
In order to now show these points on a flat sheet of drawing paper, the VP is folded flat as
indicated by the arrows. In this instance, the drawing paper is considered to correspond to
the HP, and the VP is hinged back around the OX line (i.e. the part of the VP that is above the
HP folds away from the first quadrant) to also fall in the plane of the paper. The procedure is
therefore: stand on the HP with O on your left hand side and X on your right hand side, and
fold the positive part of the VP away from you. It follows, of course, that the negative part of
the VP comes up to the HP from below. The end result is called a projection drawing.
The OX line on a projection drawing is very important, because it is the reference position from where all distances are measured. Therefore we draw the OX line as a thick (0,7 mm
thick) or dark (0,5 mm thick) line. It is good practice to draw the OX line first when a projection drawing is started. Remember to always add the letters O and X (above the OX line).
The HP and the VP are shown on top of each other in a projection drawing. When we look
at the top view (the projection on the HP — as if we are hanging high above the HP, looking
down), the VP is only “visible” as a line on top of the OX line. When we look at the front view
(the projection on the VP — as if we stand far in front of the VP, looking horizontally) the HP
is only “visible” as a line on top of the OX line. Note that the top view and front view of a
5
Chapter 2. Descriptive geometry of points and lines
Figure 2.2: Folding down of the VP
Figure 2.3: A projection drawing
6
2.3. New projection planes
Figure 2.4: In front of/behind/above/below in projection drawings
point are located on a common straight line that is perpendicular to the OX line. Note also
that, with O on your left hand side and X on your right hand side, the following hold:
• For the front view:
– The perpendicular distance that a front view point (e.g. a0 ) is above/below the
OX line in a projection drawing, is the height/depth of the point itself (e.g. A)
above/below the HP.
– Below (“onder”) the HP means that y < 0, therefore the front view projection appears below the OX line on a projection drawing.
– Above (“bo”) the HP means that y > 0, therefore the front view projection appears
above the OX line on a projection drawing.
• For the top view:
– The perpendicular distance that a top view point (e.g. a) is in front of/behind the
OX line in a projection drawing, is the distance of the point itself (e.g. A) in front
of/behind the VP.
– Behind (“agter”) the VP means that z < 0, therefore the top view projection appears above the OX line on a projection drawing.
– In front of (“voor”) the VP means that z > 0, therefore the top view projection
appears below the OX line on a projection drawing.
When a projection drawing is being interpreted, it must be remembered that the front
view and top view fall on top of each other (different from the case for Working Drawings). It
helps to ignore the top view points when you are looking at the front view, and vice versa.
The SP can be folded down in a similar manner. The SP is usually swung around the OY
line, but can also be folded flat around the OZ line.
2.3
New projection planes
Before we consider the projection of straight lines and methods to determine the trace points,
true angles and true length of a straight line, a very powerful technique is introduced, i.e. the
use of new projection planes. This technique is used extensively in Descriptive Geometry.
New projection planes are used when one wants to view an object in a direction which
is not perpendicular to the HP or VP. By adding a new projection plane perpendicular to the
direction from where you want to view the object, and projecting the object to that plane,
the required image can be obtained. We often refer to these new projection planes as “new
vertical planes”, although they may not necessarily be vertical.
7
Chapter 2. Descriptive geometry of points and lines
It is extremely important to note that a new projection plane must always be placed perpendicular to one of the previous projection planes (since we are using orthographic projections). The procedure to obtain a projection plane perpendicular to the required view
direction may entail using a sequence of new projection planes, e.g. first a new projection
plane is placed perpendicular to the top view of the view direction, and then a subsequent
new projection plane can be placed perpendicular to the view direction. This concept is
quite difficult to visualize at this stage, and we will come back to it later.
Hint: An auxiliary view in a Working Drawing is obtained through the use of a new projection plane. If you are familiar with auxiliary views, using those concepts will help you to
understand how new projection planes are applied.
2.3.1
Conventions for new projection planes
The conventions for new projection planes are as follows:
1. A new projection plane is always perpendicular to one of the existing projection planes
(i.e. the HP, VP, or a previous “new” projection plane). We call this plane its mother
plane.
2. Every new projection plane has a new OX line. The positions of new planes are indicated on the projection drawing by O1X1, O2X2, etc.
3. To go from the three-dimensional space to the projection drawing, a new projection
plane is folded down with respect to its mother plane according to the following rule:
stand on the mother plane with On on your left hand side and Xn on your right hand
side, and push the upper part of the new projection plane away from you. Remember
that the part of the new projection plane that was below its mother plane, will end up
on the opposite side of the OnXn line on the mother plane during the folding down
process. The direction in which a plane is folded (as indicated by the positioning of On
and Xn) is chosen as convenient. Usually the direction that will produce the clearest
construction is chosen.
4. The projections of points in the new projection plane are distinguished by the addition
of a subscript with the same number as that of the new OX line. Also, primes (0 ) are
added when the projections of the mother plane do not have primes, and vice versa.
For example: a10 refers to a projection in O1X1 that has been projected from a.
2.3.2
Projecting to a new projection plane
Projections to new planes are drawn just as for the HP and the VP: the projectors are perpendicular to OnXn. Remember that each OnXn “connects” two projection planes (the mother
plane and the new projection plane). The projections of a point in these two planes will
always be on a line perpendicular to the particular OnXn.
The “height” of projections of points in the new projection plane are determined according to the projections relative to the previous OX line (On−1Xn−1). To illustrate this, refer
to figure 2.5 and visualize a situation where the mother plane is horizontal and projection
of B in that plane is b2. This projection was obtained by projecting across O2X2 from b10 .
Now place a new projection plane at O3X3 perpendicular to the mother plane and project B
to that plane to obtain b30 . The important aspect to note is that the “height” of b10 “above”
O2X2 must be the same as the “height” of b30 “above” O3X3. Conversely, if b10 is below
O2X2, then b30 must be the same distance below O3X3.
To distinguish between what is above an OnXn and what is below in a projection drawing,
remember the procedure followed when folding the new projection plane flat (as described
8
2.4. Projections and trace points of a straight line
Figure 2.5: New projection plane
in §2.3.1. When viewing the projection drawing with On on your left, then all projections in
that plane that end up on the far side of OnXn are above the mother plane. Note that when
considering a projection from a previous plane, e.g. b10 with respect to O2X2, then Xn must
be on one’s left.
2.4
2.4.1
Projections and trace points of a straight line
Definitions
One of the characteristics of orthographic projection is that the projection of a straight line
on a flat plane is always straight. The views of a straight line are therefore simply the lines
connecting the projections of the end points of the line. In figures 2.6 and 2.7, ab is the top
view of the line AB, and a0 b0 is the front view. Note that the views of the line can be on any
side of the OX line. Construct the top and front views of the lines AB, AC, AD, BC, BD and CD
in figure 2.3 in order to illustrate this yourself.
The trace point of a straight line (“peilpunt”) is the intersection of the line, or its extension, with a plane. In other words, the VP trace point is located at the intersection of the line
or its extension with the vertical plane, and the HP trace point is located at the intersection
of the line or its extension with the horizontal plane. We use the abbreviations HPTP and
VPTP for these two points.
Figure 2.6 also illustrates the trace points of a line AB. The extension of AB intersects the
HP at the HPTP and the extension of BA intersects the VP at the VPTP. Figure 2.7 shows the
9
Chapter 2. Descriptive geometry of points and lines
Figure 2.6: Projections and trace points of a straight line
Figure 2.7: Projection drawing of a straight line and its trace points
10
2.4. Projections and trace points of a straight line
Figure 2.8: Lines with trace points in other quadrants
corresponding construction on a projection drawing and is explained in the next section.
Figure 2.8 shows two other cases where one of the trace points is not in the first quadrant.
Many other variations can be found.
2.4.2
Construction of the trace points of a straight line
The projection drawing of figure 2.7 illustrates the construction. In order to obtain the HPTP,
we have to look at the front view in order to see where the line intersects the HP. On the
construction, this is where the extension of a0 b0 intersects the OX line. The corresponding
point in the top view can be obtained by drawing the projector perpendicular to the OX line
through the point of intersection up to the point where it intersects the extension of ab.
The VPTP is obtained in a similar manner by looking, in the top view, where the extension
of the line intersects the VP, and then projecting the point of intersection to the extension of
the front view.
11
Chapter 2. Descriptive geometry of points and lines
Figure 2.9: Lines that intersect/do not intersect
Figure 2.10: Application of lines that intersect/do not intersect
2.5
Condition for two lines to intersect
If two lines really intersect, then the point of intersection must be located on the same projector in the top and front views (i.e. they must be located directly above/below each other,
on a line perpendicular to the OX line), as shown in figure 2.9.
Consider the following task, which is a typical example of how the above can be extended:
Task: Complete the front and top views of the rods AB and CD shown on the left hand side
of figure 2.10.
Solution: Two rods AB and CD are shown in front view (above the OX line in this case) and
top view (in front of the OX line in this case). Recall that the top view is what an observer
hanging high above the HP will see when looking down, and front view is what an observer
standing on the HP and looking towards the VP will see. In both cases the O of the OX line
must be on the observers left hand side.
We must therefore know which rod is above in the top view, and which is in front in the
front view. Let us start with the top view. Refer to figure 2.10. If we project the apparent point
of intersection in the top view (e, f) up (perpendicular to the OX line) to the front view, we
see that e0 (on rod CD) is higher than f0 (on rod AB), i.e. the line CD is higher than the line AB
at the apparent point of intersection (e, f) in the top view. Therefore we must draw the lines
12
2.6. The true length and true angles of a line
of rod CD through solidly in the top view. Similarly, if the apparent point of intersection in
the front view (g0 , h0 ) is projected down to the top view, we find that g on rod AB is in front
of point h on rod CD. In the front view, therefore, rod AB is in front of rod CD (further away
from the VP) opposite the apparent point of intersection (g0 , h0 ), hence the lines of rod AB
must be drawn through solidly in the front view.
The completed projection drawing is then as shown in figure 2.10. Note how the ends of
the rods are also completed.
Hint: Create a model of the problem: Fold a piece of paper to represent the HP and VP and
use two pencils to represent the rods.
2.6
The true length and true angles of a line
Note that the orthographic projection of a line on a flat plane can never be longer than the
line itself. Usually the projection will be shorter than the line. The projection of the line will
only be as long as the line itself when the line is parallel to one of the projection planes. We
must distinguish between the length of the projections (views) of a line and the true length
(“ware lengte”). When working with projection drawings, we see only the projections of the
lines on paper and not normally the true lengths. Constructions to determine the true length
of a line will be explained in §2.7.
The definition of the true angle between a line and a plane (“ware hoek tussen ’n lyn en
’n vlak”) is: the angle between the line itself and its projection in that particular plane. The
true angle is the smallest angle that can be found between the line and the plane. Two true
angles are used so often, that they have special symbols:
θ − is the true angle between a line and the HP, i.e. the angle between the line itself and its
top view
φ − is the true angle between a line and the VP, i.e. the angle between the line itself and its
front view
Figure 2.11 shows a line AB and its θ and φ. Consider the triangle A-a-HPTP, which
contains the line AB and θ, and view it face on (perpendicular) (as shown in the top right
hand corner of figure 2.11). Note the following characteristics: Firstly we see that the true
length of the line AB, the top view length ab and the height difference between A and B form
a right-angled triangle, with the angle between the true length and top view length equal to θ,
i.e. the angle between AB and the HP (therefore we call this triangle the θ-triangle). Note that
we also see the HPTP. Note finally that the height difference is perpendicular to the top view
length (because the height difference line comes from a projector which is perpendicular to
the projection plane, which contains the projection, i.e. the top view in this case). In order
to construct the right-angled triangle we require only two of the following four variables:
True length
θ
Height difference (y-difference)
Top view length
The remaining two variables can then be determined from the construction of the triangle.
For example, if we know the true length of AB as well as the true angle θ, then the triangle
can be constructed, and the height difference and top view length can be obtained.
Similarly, we can consider the triangle B-b0 -VPTP (as shown in the bottom right hand
corner of figure 2.11) and see that this is also a right-angled triangle, in this case containing
13
Chapter 2. Descriptive geometry of points and lines
Figure 2.11: True length and true angles θ en φ of a line
the true length of AB and φ (the φ-triangle). The four potentially available variables for the
right-angled triangle are:
True length
φ
z-difference
Front view length
If any two of these four variables are known, the triangle can be constructed and the other
two variables determined. See the example in §2.7.5 (figure 2.19 on page 21).
Note that the θ-triangle is perpendicular to the HP and that the φ-triangle is perpendicular to the VP.
Hint: make a cardboard model of the HP, VP, the line AB, its θ-triangle and its φ-triangle.
Mark up all the captions of figure 2.11 on it.
2.7
Constructions involving the trace points, true angles and true
length of straight lines
One way to construct the trace points was explained in §2.4. The following construction,
is an alternative which also yields the true length and true angles, θ and φ. The method
explained below uses the concept of a new projection plane, as introduced in §2.3, and is
aimed at finding projections that, respectively, show the two right-angled triangles on the
right hand side of figure 2.11.
2.7.1
Projecting the θ triangle
Consider the θ-triangle in figure 2.11. Note that it is perpendicular to the HP (this must be
so because Aa and Bb are perpendicular to the HP). We place a new projection plane parallel
14
2.7. Constructions involving the trace points, true angles and true length of straight lines
Figure 2.12: New projection plane to construct the θ-triangle
to the θ-triangle and perpendicular to the HP as shown in figure 2.12. The placement of the
new projection plane is indicated in the projection drawing (figure 2.13) by the O1X1 line,
which is parallel to ab. Then the points A and B are projected against it (the lines from a and
b perpendicular to the O1X1 line in the projection drawing). The heights of A and B in the
new plane are of course the same as in the vertical plane. Therefore the distance of a10 above
the O1X1 line is the same as the distance of a0 above the OX line, and the same applies to B.
The line AB of the triangle has been projected onto a1b1 in the new projection plane.
Since the new projection plane is parallel to the top view of AB, a1b1 is just as long as AB and
therefore yields the true length.
The angle θ (i.e. the angle between AB and the HP) can now also be seen in the new
projection plane, because in the latter plane, the HP is seen as a line on the O1X1. Note that
true shape of the triangle that contains the true length, φ, the front view length and the Zdifference has been obtained by means of the construction.
Remember: To see the true length of a line, place a new projection plane parallel to one of
the line’s projections. The projection in the new plane will show the true length.
Remember: To see the true shape of a triangle, find a projection plane that is exactly parallel
to the triangle. The projection in that plane will show the triangle’s true shape, i.e. the true
length of all its sides, the true angles between its sides and its true area.
Note that, if O1 and X1 were swopped around, then the new projection plane would have
folded down in the opposite direction. For the sake of clarity, the direction in which the
plane is folded down is chosen, wherever possible, so as to keep the various constructions
away from each other.
Note that if, e.g., B was below the HP, then b10 would have been on the opposite side of
the O1X1 line on the projection drawing.
Note also that the new projection plane is used to obtain a new projection of the line.
The advantage of using a new projection plane is that further projections of the line can be
15
Chapter 2. Descriptive geometry of points and lines
Figure 2.13: Projection drawing of new projection planes to construct the θ- and
φ-triangles
made, which is not the case when using the Swing Method (§2.7.4).
2.7.2
Projecting the φ triangle
Consider the φ-triangle in figure 2.11. Note that it is perpendicular to the VP (this must be
so because Aa0 and Bb0 are perpendicular to the VP). Similarly to the previous construction,
the φ-triangle can therefore be found by placing a new projection plane perpendicular to
the VP (since the φ-triangle is perpendicular to the VP) and parallel to the front view of AB as
shown in figure 2.14. When the points A and B are projected against it (the lines from a0 and
b0 perpendicular to the O2X2 line in the projection drawing), the respective distances from
a2 and b2 to O2X2 are the same as the distances from OX to a and b.
In this way the true length and φ can be determined (in the O2X2 projection plane, the
VP is seen as a line on the O2X2. Note that true shape of the triangle that contains the true
length, φ, the front view length and the Z-difference has been obtained by means of the
construction.
The true length must of course be identical in both the θ- and the φ-triangles.
Both the θ- and φ-triangles could alternatively have been found by projecting in the opposite directions. As an exercise, try to apply the constructions of figure 2.13 if the new projection planes are folded towards the OX line. An application of the swing method is to find
the projections of a line if the true angles and length of the line is given, as contained in the
following example:
2.7.3
Determining the front and top views of a line, given the angles θ and φ and true
length
Task: Determine the front and top views of line AB, for given angles θ and φ and true length
TLAB.
16
2.7. Constructions involving the trace points, true angles and true length of straight lines
Figure 2.14: New projection plane to construct the φ-triangle
Figure 2.15: Construction of a line with θ, φ and the true length given
Solution: The solution starts with the construction of the θ- and φ-triangles. Because no
positions are prescribed, we can decide for the sake of convenience that A is located in the
HP and B in the VP. Refer to figure 2.15 for the explanation of the construction.
Start by constructing the θ-triangle, using the angle θ and TLAB, as shown on the right
hand side of figure 2.15. The vertical side of the triangle is equal to the height difference
between A and B. Since we chose A to be in the HP, a0 will have to be on the OX. B can be
either above or below the HP. If it is above, then draw a line parallel to OX at the height
determined by the θ-triangle, and choose a position for b0 . Since we are using orthographic
17
Chapter 2. Descriptive geometry of points and lines
projection, b must be on a line through b0 , perpendicular to OX. We chose B to be in the VP.
Therefore b has to lie on the OX, and on the perpendicular line through b0 .
Having determined its front and top views, the position of B has been uniquely determined, using what was given in the Task and the choices we made. Note that we could have
chosen to place B below the HP and then b0 would have been on the opposite side of the OX.
Since the horizontal side of the θ triangle is the top view length of AB, we can draw an
arc, centred at b, with this length. We now know that a must be located somewhere on this
arc.
We have now fully exploited the information contained in the θ-triangle. Let’s turn our
attention to the φ-triangle and draw it using the given values of φ and TLAB. We can now
use the Z-difference: choosing A to be in front of the VP, and knowing that b has already
been located on the OX line, we can draw a line parallel to the OX with the Z-difference
obtained from the φ-triangle. We know therefore that a must be somewhere on this line.
The previously constructed arc must also pass through a, and therefore a must be on an
intersection of the parallel line and the arc. In general, this results in two possible positions
of a. Since both would satisfy all the requirements given in the task, we can choose any
one. In figure 2.15 we chose the intersection closest to O. Once we have chosen one of the
intersections as the position of a, we can find a0 by simply projecting to the OX, since we
chose A to be on the HP, and therefore its front view must be on the OX.
Due to the choices we made in the construction above, we have not used the knowledge
of the front view length that the φ-triangle has given us. We can use it now to check the
accuracy of our construction.
Hint: There are many alternative approaches to obtain the same or other correct solutions,
e.g. instead of using the Z-difference to find the position of a, the front view length could
have been used to find the position of a0 (given that we chose A to be in the HP). The position
of a could then be derived from that of a0 and the arc giving the top view length. Another
alternative is to have started with the φ-triangle instead of the θ-triangle. Try each of these
alternatives, as well as constructions with A and B at various distances from the HP and VP
respectively.
Think-a-bit: Is any combination of θ and φ physically possible, or is θ+φ limited to a certain
maximum value?
2.7.4
Swing method
The swing method can be used as a short cut in some constructions. However, it is not consistent with orthographic projection and should therefore be used with extreme caution.
Always remember that one cannot project from the points and lines produced by a swing
method construction!
The construction is based on the premise that the true length of a line, when projected,
can only be seen if the line is parallel to the relevant projection plane. The point of departure
of this method is therefore that the line is swung until it is parallel to one of the projection
planes, while distances to the other projection plane are kept constant.
Figure 2.16 illustrates the swing method to determine θ and the true length of the line.
With Aa as hinge, swing the θ-triangle, A-a-HVPP, parallel to the VP. The new position of B
is then called b1, but its height above the HP is still the same. The angle between the line AB
and the HP is therefore kept constant during the swinging process. Note further that there is
an arc with centre at a and radius ab, because the top view projection length stays the same.
When the θ-triangle is parallel to the VP and the front view of it is drawn, the true length
can be seen, as well as the true angle that the line makes with the HP, i.e. θ. In figure 2.16 b10
18
2.7. Constructions involving the trace points, true angles and true length of straight lines
Figure 2.16: Swing method to determine θ and the true length of a line
Figure 2.17: Swing method to determine φ and the true length of a line
19
Chapter 2. Descriptive geometry of points and lines
Figure 2.18: Construction for the swing method
and b0 are at the same distance from the OX line, i.e. the distance that B is above the HP. b10 a0
is the true length of the line and the angle that b10 a0 makes with the horizontal line is θ.
On a projection drawing the construction is as follows (as illustrated in figure 2.18): With
a as centre and ab as radius, swing b to b1 so that ab1 is parallel to the OX line. This is the top
view of the process to swing the θ-triangle parallel to the VP. From b1 project perpendicular
to the OX line, because b10 must be located right above it. In the front view, draw a line
through b0 parallel to the OX line to determine b10 (because b1 stays on the height of B when
the θ-triangle is swung). Line b10 a0 is therefore the true length of the line AB and the angle
between b10 a0 and the horizontal, is θ. Triangle a0 b10 c0 is therefore the familiar θ-triangle.
A similar process can be followed to determine the φ-triangle (figure 2.17 and figure 2.18):
By hinging the φ-triangle (which is represented in the front view by a0 b0 ) about Aa0 and
swinging it parallel to the HP, b20 can be determined from b. Project then perpendicular
to the OX line through b20 to find b2. When the φ-triangle was swung, it appeared in the
top view as if b moved parallel to the OX line . Draw therefore bb2 parallel to the OX line.
Line ab2 is then also the true length of the line and just as long as a0 b10 . The angle b1ab2 is
equal to φ, the angle that the line makes with the VP. Triangle ab2d is therefore the familiar
φ-triangle.
Note that when the one view of a line is parallel to the OX line, then the other view of the
line shows the true length.
2.7.5
Example: Constructing a triangle, given various true lengths and/or angles
Task: Draw the top and front views of the triangle ABC and write the coordinates of the
vertices in a table. Use scale 2:1. The following is known: A(70; 0; 15), B(90; > 0; 60), C(20; <
y(B); 35), TLAB = 60, θ(BC) = 30◦ .
Solution: The construction is as follows (see figure 2.19): The position of A is given fully,
therefore a0 and a can be drawn. The top view of B is also given, therefore b can also be
drawn. A line can then be drawn through b and perpendicular to the OX line, because b0
must be located somewhere along this line. The top view of C is also given, therefore c can
be drawn. Once again, a line through c and perpendicular to the OX line can be drawn,
20
2.8. Constructions using new projection planes and the properties of lines
Figure 2.19: Solution of triangle construction
because c0 must be located on it. The top view of triangle ABC is now complete.
The front view points must now be determined by means of further constructions. First
see what information is already available. Except for the data given in the task, the top view
lengths ab, bc and ac are also known. Also see what information is still required: only the
heights of B and C. Because the height of A is already known, the height differences between
A and the other two points can be used. Therefore: use the height difference from the θtriangle of AB to determine the height of B.
Use the top view length, ab, and the true length, TLAB, to construct the θ-triangle for AB,
as shown on the right hand side of figure 2.19. Add the height difference of AB, obtained from
the triangle, to the height of a0 , to give the height of b0 (the requirements given for the task
includes that the Y-coordinate of B is greater than 0; since A is on the HP (yAB=0), b0 must
be above the OX). Draw a line parallel to the OX at the height of b0 , knowing that b0 must be
somewhere on that line. The intersection of that line and the projector through b fixes the
position of b0 .
Now c0 must still be found. With θ(BC) given, and the top view length bc known from
the construction, the θ-triangle for BC can be constructed. Because the height of B is now
known, the height difference from the θ-triangle of BC can be utilized to determine the
height of C.
The requirements given for the task includes that the Y-coordinate of C is less than that
of B. Therefore construct a vertical line from b0 downwards, with length equal to the height
difference given by the θ-triangle for BC. A horizontal line through the end of this line defines
the possible locations of c0 . The intersection of this horizontal line and the projector through
c therefore gives the position of c0 . The front view is now complete. All that remains is to
measure the coordinates of the points and set up the table in a block as shown.
2.8
2.8.1
Constructions using new projection planes and the properties of
lines
Shortest distance between two lines
The shortest distance between two lines can be seen when you look precisely along one of
the two lines, i.e. by seeing one of the two lines as a point. In that view a perpendicular can
be drawn to the other line. The perpendicular will, at the same time, also be a true length
line. The construction uses two important characteristics of projections:
21
Chapter 2. Descriptive geometry of points and lines
Figure 2.20: Construction for the determination of the shortest distance between
two lines
• A new projection plane parallel to a view of a line in another view, shows the true
length of that line.
• A new projection plane perpendicular to a true length view of a line, shows the line as
a point.
NB: the same does not apply if the true length has not been obtained by means of
projection (e.g. by means of the folding or swing methods).
The next example illustrates this (with reference to figure 2.20):
Task: Given the front and top views of two lines AB and CE, determine the shortest distance
between them, as well as the front and top views of the shortest distance line.
Solution: Insert a new projection plane (O1X1) parallel to the top view of one of the lines,
e.g. ab. Now project AB and CE to the new plane (perpendicular to O1X1) from ab and ce,
respectively, to obtain a10 b10 and c10 e10 . Because the O1X1 line is parallel to ab (and therefore also to AB), a10 b10 shows the true length of AB (note that c10 e10 is not a true length).
Remember that both the original VP and the new projection plane (O1X1) are perpendicular
to the HP, and therefore that all the points a10 , b10 , c10 and e10 are at the same respective
distances above the O1X1 line as a0 , b0 , c0 and e0 are above the OX line (both views show the
heights of A, B, C and D).
Now insert a new projection plane (represented by O2X2) perpendicular to a10 b10 in order to see BA as a point1 . Project from a10 , b10 , c10 and e10 perpendicularly over the O2X2 line
1
This second new OX line may initially be hard to visualize. One way to visualize it is as follows: Because
we want to visualize the orientation in space, we must first ignore the folding down of the projection planes. If
an observer were standing on the O1X1 projection plane, it would have appeared to him/her that the original
22
2.8. Constructions using new projection planes and the properties of lines
to obtain a2, b2, c2 and e2. Note that the distance from a2 to the O2X2 line is the same as the
distance from a to the O1X1 line, and similarly for the other points. This is so because both
the HP and the O2X2 projection plane are perpendicular to the O1X1 projection plane, and
because both views show how far the points A, B, C and E are away from the O1X1 projection
plane.
The result of the construction is that a2 and b2 are located on top of each other, i.e. the
line AB is seen as a point because we are looking along the line. In figure 2.20 b2 is written
before a2, because b2 will be seen before a2 when an observer looks from the O1X1 plane
towards the O2X2 plane.
From the point b2a2, draw now a perpendicular on c2e2 or its extension, to meet at m2.
This line, with length |SD|, is a true length line and is, in fact, the shortest distance between
the two lines AB and CE. Project now m2 back to the previous view, to obtain m10 on c10 e10 .
The line m2n2 is a true length line and therefore m10 n10 must be parallel to the O2X2 line.
The intersection of a10 b10 with a line through m10 parallel to the O2X2 line therefore gives
n10 . The top and front views can be completed by projecting back from the other views to
obtain lines mn and m0 n0 .
Think-a-bit: If a10 b10 is a true length view with MN perpendicular to AB, m10 n10 is then
always perpendicular to a10 b10 even though m10 n10 is not a true length view?
Hint: The best way to visualize this construction is to make a cardboard model of it.
2.8.2
True shape of a triangle by means of new projection planes
This construction, shown in figure 2.21, is based thereon that the triangle must first be seen
as a single line, and that you then project to a plane parallel to this line in order to see the
true shape of the triangle.
Think-a-bit: an alternative construction would be to determine the true lengths of all three
sides and then to draw the triangle’s true shape.
Task: Given the front and top views of a triangle ABC, determine the true shape of the triangle.
Solution: In order to see the triangle as a line, we must look along any line that is located
in the plane of the triangle. As was explained in §2.8.1, we can look along a line by placing
a new projection plane perpendicular to a true length view. To obtain a true length view, a
projection plane must be placed parallel to a view. Let us therefore choose a line in the front
view that is parallel to the HP (a0 d0 in figure 2.21). The top view of this line (ad) will then be a
true length view.
Think-a-bit: why can we be sure that AD is located in the plane of the triangle?
If we now look in the direction of ad (i.e. place a new projection plane perpendicular to ad),
then AD will be seen as a point and the triangle as a line. The O1X1 line is the indication of
the new projection plane in figure 2.21, and c10 a10 b10 is the triangle shown as a line.
By now placing a new projection plane (represented by the O2X2 line) parallel to the
triangle c10 a10 b10 , we can look perpendicularly onto the triangle. The resultant view will be
the true shape of the triangle. Remember that both the HP and the O2X2 projection plane
HP was perpendicular thereon. The “heights” that would have been seen by the observer, correspond with the
distances from O1X1 line up to the top view points. The second new projection plane, represented by O2X2
line, is now also placed perpendicular to the O1X1 projection plane. The same “heights” which were just now
referred to, will now, in the new projection plane, correspond to the distances from O2X2 line up to a2, b2, etc.
23
Chapter 2. Descriptive geometry of points and lines
Figure 2.21: Construction for the true shape of a triangle by means of new projection planes
are now perpendicular to the O1X1 projection plane, and that both views show how far the
points A, B, C and D are away from the O1X1 plane. Therefore the distance from e.g. a to the
O1X1 line is the same as the distance from a2 to the O2X2 line.
a2c2b2 is the required true shape of the triangle.
2.8.3
The line of intersection and angle between two triangles
Task: Given the front and top views of the vertices of two oblique, opaque triangles ABC
and DEF, find the line of intersection between the triangles, show the hidden detail and
determine the angle between the triangles.
Hint: Build a cardboard model of two intersecting triangles if you have trouble visualizing
the following construction.
Solution: The overall strategy followed here is to start by finding any two points on the
intersection line or its extension. Since the triangles are flat, their intersection line must be a
straight line, therefore any two points on the line define its general position. Once the line’s
general position is known, its visible and hidden parts can be determined.
The strategy used to find a point on the intersection line is based on the approach that
we look for two intersecting lines (one in the plane of each triangle) that intersect. The intersection point of these lines then must be on the intersection line of the triangles, or their
extensions.
The important clue to be able to do this, is to notice that when two lines intersect, then
they are in a common flat plane. Conversely, when any two lines lie in a flat plane, then those
lines will intersect unless they are parallel. Therefore, if we know two lines are in a flat plane,
and we obtain an intersection point in one view (e.g. the front view), then we can project
that intersection point to another view (e.g. the top view).
The construction is shown in figure 2.22 and figure 2.23 where the front and top views of
the vertices of the two triangles are given. First consider figure 2.22. One point on the line
of intersection can be determined by placing a vertical plane on line ac (ac was chosen as
a matter of convenience; other choices are also possible) and then finding the front view of
24
2.8. Constructions using new projection planes and the properties of lines
Figure 2.22: Construction for line of intersection between two triangles
the line where this new vertical plane intersects the other triangle, DEF. The point of intersection of line DE with the new vertical plane (m) can clearly be seen in the top view and
can therefore be projected from the top view up to d0 e0 . Similarly, the point of intersection
between DF and the new vertical plane in the top view (n) can be projected up to the front
view line d0 f0 . The intersection line between the new vertical plane (on ac) and DEF, passes
through M and N since these two points are in both the new vertical plane and in DEF.
Now note that lines MN and AC are both in the new vertical plane. Therefore the intersection of m0 n0 with a0 c0 at g0 is the front view of the intersection of MN and AC. Note that
AC obviously lies in triangle ABC and that MN lies in triangle DEF (as stated before). G is
therefore also a point on the line of intersection between the two triangles.
In order to obtain a second point on the line of intersection between the triangles, a similar procedure is followed: A vertical plane is placed on the top view line bc and p and q (the
top views of the points of intersection of the vertical plane with lines DE and DF, respectively) are projected to the front view to determine the point of intersection h0 . In this case
BC and PQ are both located in the new vertical plane, and PQ is also in triangle DEF. H is
therefore another point on the line of intersection between the two triangles. Therefore g0 h0
is the front view of the line of intersection. Project g0 and h0 down to the top view to obtain
gh.
25
Chapter 2. Descriptive geometry of points and lines
In the preceding construction, other top view lines could have been chosen instead of
ac and bc. Vertical planes placed on some top view lines may at first not appear to give a
point of intersection in the front view. However, because the planes are flat, the lines can be
extended as necessary to obtain points on the extension of the line of intersection between
the triangles. For example, a vertical plane on line df will probably give a point of intersection
off the drawing paper. The true line of intersection can however not extend outside any one
of the two triangles.
To determine which parts of the triangles are visible and which parts are hidden in each
view, consider each apparent intersection of their edge lines in each view (e.g. the intersection of fd and cb), then use the other view to determine which one of the lines is closer to the
observer (continuing the last example, the front view shows that c0 b0 is higher than f0 d0 above
the q and therefore cb remains visible beyond q, while fd becomes hidden.
Refer now to figure 2.23, which is simply an extension of figure 2.22. By placing a new
projection plane at O1X1, parallel to gh, the true length of GH, i.e. g10 h10 , is obtained and
then O2X2 can be placed perpendicular to g10 h10 so that GH is seen as a single point, h2g2.
By projecting a single point on each of the two triangles, e.g. b and f, to the O2X2 plane via
the O1X1 plane, the true angle can be constructed and measured.
Remember: To see the true angle between two planes, look exactly along the intersection
line, i.e. find a projection that shows the intersection line as a point. Also project a point on
each plane, which is not on the intersection line, to the same projection plane.
26
2.8. Constructions using new projection planes and the properties of lines
Figure 2.23: Construction for the true angle between two triangles
27
Chapter 3
Descriptive geometry of planes
Descriptive Geometry using points and lines was introduced in the previous chapter. In this
chapter, these concepts are extended and applied to planes. Readers should ensure that they
have mastered the concepts in Ch. 2 before attempting this chapter.
3.1
Definitions
The three main projection planes, i.e. the vertical plane (VP), the horizontal plane (HP) and
the side plane (SP), were introduced in the previous chapter. These are the planes in the
familiar X Y Z coordinate system. The planes that are referred to most often, are the VP and
the HP. The concept of “new projection planes” was also introduced in §2.3. The side plane
can be considered as a particular new projection plane.
The planes that are introduced in this chapter are not necessarily projection planes, but
often are planes that form part of three-dimensional objects. In some instances the planes
are considered to be infinitely large, often for the sake of the construction. These planes can
be placed in any position relative to the main projection planes. Such a plane is represented
in a projection drawing by the lines of intersection of the plane with the projection planes, as
shown in figures 3.1 to 3.4. The line of intersection between a plane and the horizontal plane
(HP) is called the Horizontal Plane Trace Line, HPTL (“horisontale vlak peillyn, HVPL”). The
line of intersection between a plane and the vertical plane (VP) is called the Vertical Plane
Trace Line, VPTL (“vertikale vlak peillyn, VVPL”).
Planes can be subdivided into the following groups:
horizontal planes
vertical planes
inclined planes
oblique planes
(“horisontale vlakke”)
(“vertikale vlakke”)
(“skuinsvlakke”)
(“skewe vlakke”)
(Fig. 3.1)
(Fig. 3.2)
(Fig. 3.3)
(Fig. 3.4)
The first three groups are sometimes classified as perpendicular planes (“loodregte vlakke”)
because these planes are perpendicular to at least one projection plane.
29
Chapter 3. Descriptive geometry of planes
Figure 3.1: A horizontal plane
Figure 3.2: Vertical planes
30
3.1. Definitions
Figure 3.3: Inclined planes
31
Chapter 3. Descriptive geometry of planes
Figure 3.4: An oblique plane
3.2
Important characteristics of trace lines and lines in planes
1. The trace lines of a plane will intersect at one point on the OX line, except if these trace
lines are parallel.
2. If a plane is parallel to one of the projection planes, it will not have a trace line on that
plane.
3. If a plane is parallel to the OX line, its trace lines will be parallel to the OX line.
4. If a line is located in/on a plane, its trace points will be located in/on the trace lines of
the plane.
5. Horizontal lines that are located in/on a plane, are parallel to the the HPTL of the plane
and therefore the top view of such a line is parallel to the HPTL.
Think-a-bit: What does the front view of such a horizontal line look like?
6. Any line that is parallel to the VP and which is located in/on an oblique plane, is parallel
to the VPTL and therefore the front view of the line is parallel to the VPTL.
Think-a-bit: What does the top view of such a line look like?
7. All trace lines are infinitely long and can be extended to both sides.
8. When a plane is perpendicular to a projection plane (i.e. it is an inclined plane relative
to that projection plane), then the projection of any point in the plane will be on the
trace line on that projection plane.
Hint: Build a cardboard model of an oblique plane and use it to investigate each one of the
aforementioned characteristics.
3.3
3.3.1
Constructions using the properties of trace lines
The line of intersection between two planes
Figure 3.5 shows two planes that intersect and their line of intersection AB. The construction
to determine the line of intersection is demonstrated in the following example:
32
3.3. Constructions using the properties of trace lines
Figure 3.5: The line of intersection between two oblique planes
Figure 3.6: Projection drawing for the construction of the line of intersection
between two oblique planes
33
Chapter 3. Descriptive geometry of planes
Task: Given the VPTLs and HPTLs of two oblique planes, determine the views of the line of
intersection.
Solution: The determination of the line of intersection on a projection drawing is shown in
figure 3.6. B is the point of intersection between the two VPTLs, and is therefore located in
the VP. The front view of B (b0 ) is therefore on the point of intersection of the VPTLs. Because
B is located in the VP, its top view is located on the OX. Project therefore b0 down perpendicular to the OX line to find b. Similarly, A is the point of intersection between the two HPTLs
and is located in the HP. The top view of A (a) is located on the point of intersection of the
HPTLs. Because A is located in the HP, the front view of A is located on the OX line. Project a
up perpendicular to the OX line and find a0 . Connect a0 b0 and ab respectively, to obtain the
front view and top view of the line of intersection AB of the two planes.
3.3.2
The true angle between two trace lines
Figure 3.7 shows the true angle α (angle AKB) between the HPTL and the VPTL. The construction for the determination of α is illustrated by means of the following example:
Task: Given the HPTL and VPTL of an oblique plane, determine the true angle between
them.
Strategy: Fit a line AB into the plane that is perpendicular to the HPTL and which reaches
from the VPTL to the HPTL. Point A is located in the VP and B in the HP and triangle ABK is
a right-angled triangle with angle ABK = 90◦ . In the front view the true length of KA i.e. ka0 , is
shown, because it is located in the VP. The true length of KB, i.e. kb, is shown in the top view,
because it is located in the HP. The true shape of the triangle AKB, which is inclined in space,
can be seen in a projection drawing by swinging it about KB until it is located in the HP. The
true shape will then be visible in the top view. Angle bka1 gives the true angle α.
Construction: The construction is shown in figure 3.8: From any point on the OX line draw
ab perpendicular to the HPTL and extend it. From a draw aa0 perpendicular to the OX line
Figure 3.7: True angle between trace lines
34
3.3. Constructions using the properties of trace lines
Figure 3.8: Construction for determining the true angle between trace lines
to intersect the VPTL at a0 . With k as centre and ka0 as radius, draw a circular arc to intersect
the extension of ab at a1. Connect a1 with k; then the angle bka1 gives the true angle α.
Think-a-bit: where is the true angle of AB given?
3.3.3
Completion of the projections of a point on a plane
If one of the projections of the point is given, the other projection can be obtained in a number of ways. Three approaches are discussed here by means of an example:
Task: Given the front view p0 of point P and the trace lines of the plane on which P is located.
Find the top view p.
Strategy 1: Use a horizontal line in the plane, and which also contains the point. The construction is shown in figure 3.9: Line a0 p0 is such a horizontal line in the oblique plane (front
view parallel to the OX line) of which point A is located in the VP, i.e. a0 lies on the VPTL and
a lies on the OX line. The top view of such a horizontal line is parallel to the HPTL (see §3.2).
Draw ap parallel to the HPTL to intersect the projector p0 p (which is perpendicular to the OX
line) at p. The point of intersection p is the required top view of the point that is located in
the plane.
Strategy 2: Use a line that is parallel to the vertical plane, and which also contains the point.
The construction is shown in figure 3.9 with dashed lines.
Think-a-bit: which of the characteristics stated in §3.2 are used in this construction?
Strategy 3: Take an arbitrary line that is located in the plane, which extends through the
relevant point and which ends in the HP and VP respectively, as shown in figure 3.10.
The construction is shown in figure 3.11: Take any line a0 p0 b0 that passes through p0 , and
which extends from the OX line to the VPTL. Since B is located in the VPTL and therefore in
the VP, b will be located on the OX line. Project b0 down to the OX line in order to obtain b.
Since a0 is located on the OX line, A is located in the HP, but the point is also located in the
plane, i.e. A is located in the HPTL. Project a0 down, perpendicular to the OX line, to intersect
the HPTL at a. Line ab is the top view of the line AB that contains the point P and therefore
35
Chapter 3. Descriptive geometry of planes
Figure 3.9: Constructions with horizontal and vertical lines to determine a view
of a point in a plane
Figure 3.10: An arbitrary line used to determine a view of a point in a plane
the top view of P must also be located in ab. Project p0 down, perpendicular to the OX line,
to intersect ab at p. The required top view has therefore been found.
3.3.4
Determination of a trace line from the projections of a point on the plane
Similar strategies to that used in the previous paragraph can be followed here, as illustrated
by the following example:
Task: Given the views p and p0 of a point P on the plane, and the HPTL of the plane. Determine the VPTL.
Construction: The construction is shown in figure 3.12: Draw pa parallel to the HPTL and
aa0 perpendicular to the OX line to intersect line p0 a0 (parallel to the OX line) at a0 . Connect
ka0 , which is the required new VPTL.
36
3.3. Constructions using the properties of trace lines
Figure 3.11: Construction with an arbitrary line to determine a view of a point in
a plane
Figure 3.12: Construction for the determination of a trace line from the projections of a point on the plane
Think-a-bit: which strategy has been followed in figure 3.12? How can the other strategies
be applied?
3.3.5
Determination of a trace line given a point on a trace line, when the point of
intersection cannot be determined
If the trace lines of a plane are nearly parallel to the OX line, it is not always possible to use the
construction discussed in the previous paragraph. An alternative construction is illustrated
by means of an example:
Task: Given one trace line and a point (P) on the other trace line, but the point of intersection
of the trace lines on the OX line cannot be determined, find the second trace line of the plane.
Construction: The construction is shown in figure 3.13: Through p0 (the given point on the
VPTL) draw a perpendicular on the OX line to intersect the HPTL at e. Draw any line Hp0 and
connect He. Take any other point A on the OX line. Draw c0 d perpendicular to the OX line
37
Chapter 3. Descriptive geometry of planes
Figure 3.13: Construction for the determination of a trace line from a point on a
trace line if the point of intersection is not determinable
through A. Make Bd parallel to He and Bc0 parallel to Hp0 . The point of intersection between
Bc0 and the perpendicular c0 d, is another point on the VPTL. Connect c0 p0 ; this gives the
required VPTL.
Think-a-bit: Prove that the HPTL and VPTL intersect on the OX line by making use of congruent triangles).
3.4
Converting an oblique plane into an inclined plane using a new
projection plane
A new projection plane (NPP) can be used to convert an oblique plane to an inclined plane
(with respect to the NPP). The line of intersection between the NPP and the oblique plane
(see figure 3.14) is called the New Projection Plane Trace Line (NPPTL). The very useful result of this construction is that the entire inclined plane is seen as a single line in the NPP
(i.e. every point in the inclined plane projects onto the NPPTL in the NPP).
Since it is easier to be visualized, the new projection plane is usually a vertical plane. The
terms “New Vertical Plane” (NVP) and “New Vertical Plane Trace Line” (NVPTL) are therefore
often used.
The construction of such a NPPTL is illustrated by means of the following example:
Task: Given the HPTL and VPTL of an oblique plane, create a projection in which the plane
is seen as a line (i.e. convert it to an inclined plane) and determine the NPPTL.
Strategy: Place a new VP (which must still be perpendicular to the HP) perpendicular to the
oblique plane (i.e. perpendicular to the HPTL).
Construction: The construction is shown in figure 3.15: O1X1 shows the position of the
NPP, which is perpendicular to the HPTL (remember the conventions in connection with the
folding down of the NPP that were discussed in §2.3.1). Let the point where the O1X1 line
intersects the OX line be called G. Line g0 b0 is then the line of intersection between the VP
and the NPP. Note that B is located on the VPTL (therefore in the oblique plane), but also
on the VP and the NPP. It must therefore also be located on the NPPTL. By obtaining the
projection of B in the NPP (b10 ), we can obtain one point on the NPPTL. Since B is located in
38
3.4. Converting an oblique plane into an inclined plane using a new projection plane
Figure 3.14: A new projection plane and the NPPTL
Figure 3.15: Construction and use of the NPPTL
the NPP, we can set off the height of B (g0 b0 ) perpendicular to the O1X1 line to give g10 b10 (G
is on the OX line and on the O1X1 line, therefore g, g0 and g10 are on top of each other in the
projection drawing).
Note further that A is located on the NPP and the HPTL of the oblique plane. A is therefore located on both the oblique plane and the NPP and is therefore a point on the NPPTL.
Because A is located on the HP and O1X1, a and a10 are located on top of each other.
A line through b10 and a10 gives the NPPTL because these two points are located on the
NPP and the oblique plane.
Note that in principle the O1X1 line can be drawn at any position along the HPTL. The
accuracy of the construction is however improved by choosing the O1X1 line as far as is
practically possible away from the point of intersection of the HPTL and the OX line.
Beware: there are other construction methods used in technical drawings. They do not
always work. The method presented here is recommended since it is easily derived from
first principles.
39
Chapter 3. Descriptive geometry of planes
Figure 3.16: Construction to find the front view of a point on an oblique plane
by means of a NPPTL
Think-a-bit: If the VPTL, and therefore also b0 , were located under the HP opposite G, how
would the construction have progressed?
Think-a-bit: If O1 and X1 were exchanged, what would the NPPTL have looked like?
3.5
Constructions with an oblique plane converted to an inclined plane
3.5.1
Completion of the projections of a point on a plane
The construction which converts an oblique plane to an inclined plane can be used as a further alternative to complete the projections of a point on the oblique plane. Other strategies
were discussed in §3.3.3. The next example illustrates the concept:
Task: Given the HPTL and VPTL of a plane, and also the top view p of a point P on the plane,
find the front view p0 of P.
Construction: The construction is shown in figure 3.16. After the NPPTL has been found,
as described §3.4, project from p to the NPP (perpendicular on the O1X1 line) to find p10 .
Because P is located on the oblique plane, p10 must be located on the NPPTL (all points
in the oblique plane project to the NPPTL because the NPP is perpendicular to it). From
this it follows that cp10 is the height of the point P above the HP. By now projecting from p
perpendicular towards the OX line and to the VP, and by setting out the height of P from the
NPP from the OX line, p0 can be obtained. This is the required front view of the point P.
Think-a-bit: If p0 , HPTL and VPTL are given, how would you find p?
Think-a-bit: If p were located to the left of the HPTL, how would it have influenced the
construction? Remember: a trace line is infinitely long and can be extended as necessary!
40
3.5. Constructions with an oblique plane converted to an inclined plane
Figure 3.17: Construction to determine the point of intersection between a
straight line and an oblique plane
3.5.2
Point of intersection between a plane and a straight line
The next example illustrates how the NPPTL can be used to determine where a straight line
penetrates an oblique plane:
Task: Given the HPTL and VPTL of an oblique plane, and also the front and top views of line
AB. Find the point where the line penetrates the plane.
Strategy: Look perpendicularly onto the oblique plane (i.e. the plane must be seen as a line)
to see where the line penetrates the plane.
Construction: The construction is shown in figure 3.17. Convert the given plane into an
inclined plane as described in §3.4. Obtain also the projection of the line AB in the NPP, i.e.
project a and b perpendicular over the O1X1 line and obtain a10 and b10 by transferring their
respective heights above the HP from the front view (marked “x” and “/” in figure 3.17). The
plane is now seen as a line and the point of intersection of the new projection of the line
and the NPPTL therefore gives the point of intersection c10 . Project c10 back to ab obtain c.
Project perpendicular on the OX line to a0 b0 to obtain c0 . Points c and c0 are the top and front
views of the point of intersection of the line AB with the plane.
Hint: If the projectors of C are not nearly perpendicular to ab or a0 b0 , it may be more accurate to transfer the height of c10 above O1X1 to find c0 , rather than projecting from c.
3.5.3
Shortest distance between a point and a plane
The shortest distance between a point and a plane can be determined by drawing a perpendicular from the point to the plane, as described in the next example:
Task: Given the HPTL and VPTL of an oblique plane, and also the front and top views of
a point A that is not in the plane. Find the shortest distance from A to the plane and the
projections of the shortest distance line, i.e. the perpendicular onto the plane through A.
41
Chapter 3. Descriptive geometry of planes
Figure 3.18: Construction to determine the shortest distance between a point
and a plane
Strategy: Look perpendicularly onto the oblique plane (i.e. the plane must be seen as a line)
to the point to see where the shortest distance line will be.
Construction: The construction is shown in figure 3.18: Convert the given plane to an inclined plane as described in §3.4. Obtain also the projection of the point A in the NPP, i.e.
project a perpendicular over the O1X1 line and obtain a10 by transferring its height above
the HP from the front view. Draw now a perpendicular from a10 to the NPPTL at d10 . The line
a10 d10 is the true length of AD. The top view of AD must be parallel to the O1X1 line (since
a10 d10 is a true length line. In order to find d, project from d10 perpendicular over the O1X1
line and draw a line from a parallel to O1X1 to intersect the projector at d. In order to find d0 ,
project from d perpendicular over the OX line and transfer the height of D from the NPP (the
distance of d10 above the O1X1 line).
Think-a-bit: is the top view of a perpendicular on an oblique plane necessarily perpendicular to the HPTL? And what about the front view of the perpendicular and the VPTL?
3.5.4
The true angle between a plane and a straight line
The true angle between a plane and a straight line can be determined by drawing a perpendicular from a point on the line to the plane, and then using the resultant right-angled
triangle to determine the angle. This concept is illustrated in the next example:
Task: Given the HPTL and VPTL of an oblique plane, and also the front and top views of a
line AB. Determine the true angle Q between AB and the plane.
Strategy: The true angle between a line and a plane has been defined as follows in §2.6:
the angle between the line itself and its projection in the relevant plane. The line itself is
given here, but its projection must be found. The orthographic projection of the line on the
oblique plane can be obtained by using one of the end points of the projection of the line and
the point of intersection between the line and the plane (other strategies are also possible,
42
3.5. Constructions with an oblique plane converted to an inclined plane
Figure 3.19: Construction to determine the true angle between a line and a plane
but this one is the most convenient). The projection of e.g. A on the oblique plane is given
by the perpendicular from A to the oblique plane (therefore the projector of A on the oblique
plane). After the projection has been found, the angle between the projection and the line
must be found, just as before with θ and φ for a straight line. The strategy is therefore:
1. Find the point of intersection and the perpendicular from A to the plane, and thus the
projection of AB.
2. Find the angle between the projection and the line.
Construction: Refer to figure 3.19: The construction of the point of intersection c10 and a
perpendicular a10 d10 are discussed in previous paragraphs. Because AD is perpendicular on
the plane and therefore on CD, the triangle ACD is a right-angled triangle. Two pieces of
information regarding the triangle are already available: the 90◦ angle and the length of AD.
One further piece of information is necessary and the most convenient is the true length of
AC. The line AC’s θ triangle can be constructed by using the top view length ac and taking the
height difference from the front view. The resulting triangle is shown on the left in figure 3.19.
The true length of AC can be taken from this triangle.
Now the right-angled triangle containing the true angle between AB and the oblique
plane can be constructed using the true length of AC and the perpendicular distance from A
to the plane (the latter is constructed by drawing a line parallel to the NPPTL through a10 ).
The angle Q, i.e. the angle between the line and its projection in the plane, can then be
measured.
43
Chapter 3. Descriptive geometry of planes
Hint: build a 3-dimensional cardboard model of figure 3.19, comprising the HP, the VP, the
new projection plane, the oblique plane, the line AB and the triangle containing the true
angle between AB and the oblique plane. Indicate on the model the axes, the trace lines
(including the NPPTL), the O1X1 and the projections of A and B on each projection plane.
3.6
The true angle between planes
3.6.1
Definitions
Here, as in the case with lines, θ and φ have specific meanings:
θ− the true angle between a plane and the HP
φ− the true angle between a plane and the VP
The angles θ and φ are special cases of true angles between planes. In general there are at
least three equivalent ways to define a true angle between two planes:
1. The true angle is the smallest angle between the two planes.
2. The true angle is the angle that can be seen when one looks along the trace line. To use
this, consider the true angle between planes 1 and 2. We must find a projection plane
(NPP) in which the trace line where 1 and 2 intersect is seen as a point. Then the true
angle can be measured between the respective trace lines that planes 1 and 2 make in
the NPP.
3. The true angle is the angle between two lines which have the following characteristics:
• There is one line in each of the two planes
• The lines are perpendicular to the line of intersection between the two planes
(trace line)
• The lines meet on the trace line
In figure 3.20, the angle between FH and FK is θ because
• CE is the trace line of the plane with the horizontal projection plane
• FK is a line in the oblique plane and perpendicular to the trace line CE
• FH is a line in the horizontal projection plane and perpendicular to the trace line
CE
• FK and FH meet at F on the trace line
The angle φ is, correspondingly, the angle between a line in the oblique plane and
perpendicular to the VPTL, and a line in the VP that is also perpendicular to the VPTL.
Think-a-bit: Where will the VPTL of the plane be located?
3.6.2 θ and φ for an oblique plane
If one looks at the inclined planes in figure 3.3, one will see that they make easily identifiable
angles θ and φ with the HP and VP, respectively. For the oblique planes the definitions of the
true angles are somewhat more complicated.
Think-a-bit: What is the value of θ and φ for the parallel planes?
Consider the oblique plane shown in figure 3.21. According to the definition of true
angles, θ is given by the angle between the lines AG and AB, because
• AG and AB are located in the different planes
• AG and AB are both perpendicular to the HPTL
• AG and AB meet on the HPTL
44
3.6. The true angle between planes
Figure 3.20: True angle between an oblique plane and the HP
A characteristic of AB and AG that is possibly not noted immediately, is that BG is perpendicular to the horizontal plane. In fact, ABG is a right-angled triangle that is perpendicular
to both the HP and the oblique plane.
Similarly, φ is given by the angle between lines CG and CD. In this case GD is perpendicular to the VP and the right-angled triangle CGD is perpendicular to both the VP and the
oblique plane.
For the sake of convenience, the two triangles ABG and CDG were chosen such that the
point G is common to both of them. This is not necessary to obtain θ and φ, but is necessary
for a construction that will be shown later. Figure 3.22 shows the two triangles hatched.
3.6.3
Determination of θ and φ using new projection planes
To determine θ, the right-angled triangle AGB must be projected to a plane parallel to it. The
construction described in §3.4 to create a NPP which shows the oblique plane as a line, does
exactly that. The angle between O1X1 and the NPPTL in figure 3.15 is θ because O1X1 is
perpendicular to the HPTL and in the HP, while the NPPTL is perpendicular to the HPTL and
in the oblique plane.
A similar construction, with a NPP perpendicular to the VP and to the VPTL, will clearly
show φ.
3.6.4
Swing method for determination of θ and φ
This approach is often used as a short cut since it does not require the use of new projection planes. However, the construction is not based purely on orthographic projection and
should therefore be used with considerable care. In particular, the results of the construction
may not be used in further projections.
To determine θ, the right-angled triangle AGB is made “visible” on paper by “swinging”
it parallel to one of the main-projection planes and then projected against it. There are two
possibilities:
45
Chapter 3. Descriptive geometry of planes
Figure 3.21: θ and φ for an oblique plane
Figure 3.22: Perspective view of θ and φ for an oblique plane
46
3.6. The true angle between planes
Figure 3.23: Swing method for the determination of θ and φ of an oblique plane
1. “Swing” triangle AGB with a hinge at BG until point A (or line GA) is located on the OX
line and the true shape of AGB (and θ) is visible in the front view.
2. “Swing” triangle AGB with a hinge at AG so that triangle AGB is located on the HP and
the true shape of AGB (and θ) is visible in the top view.
To determine φ, the right-angled triangle CGD is relevant. In order to “see” CGD, one of
the following approaches can be used:
1. “Swing” triangle CGD about GD until the triangle is located flat on the HP and φ is
visible in the top view.
2. “Swing” triangle CGD about CG to be located flat on the VP so that φ becomes visible
in the front view.
The constructions are shown in figure 3.23. Note now in figure 3.22 that GA is located in
the HP and that it is perpendicular to the HPTL. Therefore ga in figure 3.23 is also perpendicular to the HPTL. Also, GB in figure 3.22 is located in the VP, perpendicular to the OX line,
and extends to the VPTL. This gives gb0 in figure 3.23. We also know that triangle AGB is a
right-angled triangle, i.e. GA is perpendicular to GB. By drawing therefore in figure 3.23 ga1
perpendicular to gb0 with ga1 = ga, the angle ga1b0 will be the true angle θ.
In a similar way φ is determined by drawing c1g and gd perpendicular to each other in
figure 3.23.
An alternative construction, where the triangles are folded down into the HP (θ) and in
the VP (φ), respectively, is shown in figure 3.24.
3.6.5
Determination of the trace lines of a plane if θ and φ are given
Figure 3.22 gives the following important clues required to find the trace lines from θ and φ:
• G is located in both the triangles AGB and CGD.
• R is located in both the triangles AGB and CGD, because R is the point of intersection
of lines AB and CD, both of which are located in the plane;
• GR is therefore the line of intersection between triangles AGB and CGD;
• triangles AGB and CGD are both perpendicular to the oblique plane;
• therefore GR must be perpendicular to the oblique plane;
47
Chapter 3. Descriptive geometry of planes
Figure 3.24: Folding method for the determination of θ and φ of an oblique plane
• GR must also be perpendicular to AB and CD (the inclined sides of the two rightangled triangles);
• the length of GR depends on the position of G relative to the point of intersection of
the HPTL and the VPTL.
By using these facts, an easy construction can be done to determine the trace lines of
the plane, as shown in figure 3.25: Since GR must be common to and perpendicular to the
inclined sides of the triangles, a common circle with radius GR can be drawn. The radius of
the circle can be chosen arbitrarily (just as G in figure 3.22 was chosen arbitrarily). Choose
therefore a point g and draw a circle with centre g and a convenient radius. Draw a tangent
a1b0 to the circle so that it makes an angle θ with the OX line. Complete the AGB triangle
by drawing line gb0 perpendicular to OX. We know from figure 3.22 that B was chosen to be
on the VPTL, and therefore the VPTL passes through b0 . Draw an arc with radius a1g, with
centre at g. We know that the HPTL must be tangent to this line if AG is perpendicular to it,
as shown in from figure 3.22.
To construct the φ-triangle (CGD, draw line c1d tangent to the same circle with radius GR
used before, so that it makes an angle φ with the OX line. Complete the triangle by drawing
line gd perpendicular to the OX. From figure 3.22 we know that D was chosen to be on the
HPTL, and therefore the HPTL passes through d.
We previously determined that the HPTL must be tangent to the arc with radius a1g. We
can therefore now draw the HPTL through d to be tangent to that arc. The HPTL and VPTL
must intersect on the OX (here labled E). Therefore the VPTL can be drawn from e through
b0 .
As a check, a circular arc with g as centre and with radius gc1 can be drawn, which should
then be tangent to eb0 .
Note that in figure 3.25 the φ-triangle is drawn on the opposite side of the OX as the θtriangle purely as a matter of convenience. If only θ and φ were given, the θ- and φ-triangles
can both be drawn on any side of the OX. Since the trace lines are infinitely long, the projection drawing can show them on either side of the OX line.
The construction is in many ways the inverse of the swing method for the determination
of θ and φ.
48
3.6. The true angle between planes
Figure 3.25: Construction of trace lines if θ and φ are given
Figure 3.26: Construction of the HPTL if the VPTL and θ are given
Think-a-bit: how many possible solutions are there if only θ and φ are given? What will the
construction look like if e is located to the right of g or if A is behind the VP?
Think-a-bit: Are there physically impossible combinations of values for θ and φ? What
would it mean if point d ended up inside the arc with radius a1g?
Think-a-bit: How will you determine the trace lines of an inclined plane? Remember, it is
much easier!
3.6.6
Construction of a trace line if the other trace line and one true angle is given
Two variations are illustrated here by means of examples:
Task: Given the VPTL and θ. Find the HPTL.
Construction: The construction is shown in figure 3.26 and starts with constructing the θtriangle: Choose a point b0 anywhere on the VPTL (at a comfortable distance from e). Draw
line gb0 perpendicular to the OX line. Draw also a1b0 so that it makes an angle θ with the
OX line. This completes the θ-triangle, b0 ga1. With g as centre and ga1 as radius, draw
a circular arc because the HPTL must be this distance from g. Knowing that the HPTL must
pass through e and be tangent to the arc, draw es and extend as required to obtain the HPTL.
49
Chapter 3. Descriptive geometry of planes
Figure 3.27: Construction of the HPTL if the VPTL and φ are given
Task: Given the VPTL and φ. Find the HPTL.
Construction: The construction is shown in figure 3.27 and starts with constructing the φtriangle. Choose any point c0 on the VPTL (at a comfortable distance from e). Draw line c0 g
perpendicular to the VPTL. Draw a line through g perpendicular to the OX line and extend it
below the OX. Mark the length of gc0 off on the OX line to get gc1. From c1 draw line c1d at an
angle φ with the OX line, to intersect the line through g at d. This completes the φ-triangle,
i.e. c1gd. From the properties of the φ triangle, we know that d is on the HPTL. Therefore
connect ed and extend as required to give the HPTL.
3.7
3.7.1
The true angle between two oblique planes
Stategies
Figure 3.28 shows two oblique planes that intersect on a line AB. Three equivalent ways of
defining the true angle between the planes were given in §3.6.1. Two methods of construction to determine the angle are given in the following paragraphs. The NPP method (§3.7.2)
is based on the definition that the true angle can be seen when one looks precisely along the
line of intersection. The folding method (§3.7.3) utilizes the definition that the true angle is
the angle between two lines perpendicular to the line of intersection (CF and FD).
3.7.2
New projection plane method
The strategy that is followed here, is to project the line of intersection of the two planes so
that it is seen as a point. Then one looks precisely along the line of intersection and thus
one can see the true angle. In order to project the line of intersection as a point, a true length
projection must first be obtained, i.e. a projection parallel to a view of the line of intersection.
The construction is shown in figure 3.29: Place a new projection plane parallel to the top
view of the line of intersection, ab, and project the new front view over the O1X1 line in order
to see the true length of the line of intersection AB, i.e. a10 b10 . Place another new projection
plane, O2X2, perpendicular to the previous one and perpendicular to a10 b10 in order to see
the line of intersection as a point at b2a2. Project any point on each of the planes (e.g. C and
D), via the O1X1 line over the O2X2 line (e.g. to give c2 and d2). The angle c2f2d2 is the true
angle between the two planes, therefore angle α.
Points c, d, e and f in figure 3.29 are for a variation of the folding method and are discussed
in the following section.
50
3.7. The true angle between two oblique planes
Figure 3.28: True angle between two oblique planes
Figure 3.29: NPP method for the determination of the true angle between two
planes
51
Chapter 3. Descriptive geometry of planes
Figure 3.30: Folding flat method for the determination of the true angle between
two planes
3.7.3
Folding flat method
This method is a short-cut, but uses a construction that is not consistent with orthographic
projection, and should therefore be used with caution.
The strategy that is followed here, is to obtain a line in each plane that is perpendicular
to the line of intersection (CF and FD) and then to determine the angle between them. If
figure 3.28 is studied carefully, it will be found that line CD is perpendicular to the top view
of the line of intersection, ab. This is a key to the construction, which is shown in figure 3.30:
Draw the top view of the line of intersection. Draw any line cd perpendicular to the top view
of the line of intersection, ab, so that the HPTLs of the two planes are intersected at c and d
respectively, and ab itself at e. Find the true length of the line of intersection (AB) by swinging
it into the VP. Swing e also into the VP and connect e10 perpendicular to TL.AB in order to
obtain e10 f10 , the shortest distance between line AB and line ced. The true angle between the
planes is the angle CFD, or its complimentary angle. The triangle CFD is perpendicular to
the line of intersection and both planes. By setting off the distance e10 f10 from e on ab, find
f2. In this way triangle CDF is folded flat about cd, and the true angle between the planes,
angle α, is determined.
An alternative construction is shown in figure 3.29, in which the O1X1 line is used to
determine the true length of FD, as required in the folding method: Draw a line ced perpendicular over ab to intersect both HPTLs, in c and d respectively. In the NPP they are therefore
located behind each other on the HP. Draw line f10 d10 from d10 perpendicular to a10 b10 in
order to again see the triangle that contains the true angle between the planes, as a line. i.e.
Set off distance (∥) f10 d10 from e in the top view. Join cf and df to see the angle.
52
Chapter 4
Shadows
4.1
Introduction
We cover only those shadows that originate from a parallel system of light rays, e.g. rays from
the sun. If such light rays impinge on an opaque object, one part of it will be illuminated
and another part will be in shadow. The line separating light and dark areas is called the
separation line, and it is the trajectory of the light rays that just shave the object. This line
also describes the border of the shadow cast by one object on any other object.
The following properties of shadows are important for the construction thereof:
1. The shadow cast by any point on a plane is the trace point of the light ray through the
point on the plane.
2. The shadow cast by a straight line on a plane is a straight line connecting the shadows
cast by the end points of the line.
3. If a plane object is parallel to one of the projection planes, then the shadow cast by it
on the parallel plane will be of the same shape and size as the object. For example, the
shadow cast on the VP by a circular disc which is parallel to the VP will be circular and
of the same size as the disc.
4.2
Shadow of a point on one plane
In figure 4.1 it is shown how the shadows of a given point due to light rays of given direction
on respectively the HP and VP are determined. Note that, if both the HP and the VP are
opaque, then the trace point of the light ray through the point in the first quadrant will be
the shadow cast by the point.
Figure 4.1: Constructions of the shadow of a point on a plane
53
Chapter 4. Shadows
Figure 4.2: Constructions of the shadow of a line on one or more planes
4.3
Shadow of a line on one or more planes
Figure 4.2 shows how to determine the shadow of a given line on one or more planes due to
light rays of given direction.
It is clear from figure 4.2 that the shadow of a line is the connecting line between the
shadows of the end points of the line. Note however that if the shadows of the end points are
not located in the same plane, the end points cannot be connected willy-nilly. Determine
then first the shadows of both end points on the same plane (consider the other plane as
transparent), and obtain in this manner the part of the shadow that is located on that plane.
4.4
Shadow of an object on a plane
Considering that any object is built up from lines and neighbouring points, the determination of the shadow of an object is in essence the same as the determination of the shadows
of a number of neighbouring points.
Consider for example a horizontal disc. Figure 4.3 shows the way the shadow of the disc
on the HP due to light rays of given direction is determined: since the disc is parallel to the
HP, the shadow on the HP will be circular.
Figure 4.4 shows the shadow of the disc when the disc is not horizontal, but for e.g. perpendicular to the direction of the light rays.
Figure 4.5 shows the shadow of the disc when the disc is horizontal, but where part of
the shadow is cast on the vertical plane. Since the disc is parallel to the HP, the portion of
the shadow cast on the HP will be circular. Furthermore, it is clear from figure 4.5 that the
54
4.4. Shadow of an object on a plane
Figure 4.3: Construction of the shadow of a horizontal disc on the HP
Figure 4.4: Construction of the shadow of a non-horizontal disc on the HP
55
Chapter 4. Shadows
Figure 4.5: Construction of the shadow of a horizontal disc on both the HP and
VP
part adcb of the disc casts a shadow against the VP. Take therefore points such as c0 , and
determine the VP shadow points.
4.5
Shadow lines of an object on itself
It is clear that certain areas of an object will be illuminated and other areas will be in shadow.
There must therefore be separation lines between light and darkness on the object. These
separation lines are located where the light just shaves the object. The shadows of these
lines constitute the outlines of the shadow on the HP and the VP.
As an example of such a case, the construction of the shadow of a cone on itself and on
the HP and VP, is shown in figure 4.6.
4.6
4.6.1
The shadow of one object upon another
Shadow of a point on an object
Figure 4.7 shows how to determine the shadow of a point on a vertical cylinder due to light
rays of a given direction. E is the shadow on the cylinder.
The construction in figure 4.7 is based thereon that we are looking for the interpenetration point of the light ray line that contains the point, and the point of the object (the
cylinder).
A more complicated example is shown in figure 4.8, i.e. the shadow cast by a point on
a vertical cone. We can consider the problem as follows: The point must be able to cast a
shadow on the object (the cone) as well as on the HP. The shadow of the point is cast on a
point on the cone, which must in turn be able to cast the same shadow on the HP as the original point. Therefore, by determining the shadow cast by the point, in this case point B, on
56
4.6. The shadow of one object upon another
Figure 4.6: Construction of the shadow of a vertical cone on itself and on the HP
and VP
Figure 4.7: Construction of the shadow of a point on a vertical cylinder
57
Chapter 4. Shadows
Figure 4.8: Construction of the shadow of a point on a vertical cone
the HP, the shadow cast on the HP by the point (on the cone) that we are looking for, is determined. We can therefore find the point on the object by doing an “inverse” construction.
Because this point must be located on the surface of the cone, it must be located on
a generator line. Determine therefore the shadow of the generator line that contains the
shadow of point B on the HP. Determine then the top view and front view of this generator
line. Line dtHV is the shadow of the generator line td. The shadow of B is therefore cast on td
and t0 d0 at projections c and c0 of point C.
4.6.2
Shadow of a line on an object
Figure 4.9 shows how to determine the shadow of a line on a horizontal disc due to light
rays of a given direction. The method is similar to that for the shadow of a point on a cone
discussed in §4.6.1, i.e., we look for the line on the disc that casts the same shadow as line AB
on the HP.
A more complicated example is shown in figure 4.10, i.e. the shadow of a line due to light
rays of given direction on a sphere.
By sectioning the sphere using vertical planes containing the light rays (i.e. planes that
are parallel to the direction of the light rays), the shadow point of any part of the line AB on
the relevant section through the sphere can be seen. Thus, by drawing an O1X1-line parallel
to the top view direction of the light rays, the true length view of the light rays is obtained, and
arbitrary sections can be drawn and then projected back (as shown for point A, for example).
The light ray through point B misses the sphere. The exact location of c2, which is the “last”
shadow point on the sphere, can be obtained by means of an auxiliary view perpendicular to
the true length view of the light rays (the view over the O2X2 line). This point can then also
be projected back.
58
4.6. The shadow of one object upon another
Figure 4.9: Construction of the shadow of a line on a horizontal disc
59
Chapter 4. Shadows
Figure 4.10: Construction of the shadow of a line on a sphere
60
Chapter 5
Interpenetrations
5.1
Interpenetration points
The problem here is to obtain the projections of a point where a body is penetrated by a
line, i.e. the point where the line just touches the body. A number of specific cases are considered here in order to highlight the general principles for determining such interpenetration points.
5.1.1
A straight line through a sphere
A sphere, the line AB and their relative positions are given, as shown in figure 5.1.
The construction is as follows: Take a vertical plane that contains the line AB. Draw the
section of this vertical plane through the sphere (a circle, of course) and the projection of the
line AB in the plane. Thus, the section through the sphere and the line are both located in the
same plane, and therefore the points c10 en d10 where the line a10 b10 touches the section are
the interpenetration points. Project back to obtain the top views c en d and the front views
c0 and d0 .
5.1.2
A straight line through a prism
A prism, the line AB and their relative positions are given, as shown in figure 5.2.
Figure 5.1: Interpenetration of a sphere by a straight line
61
Chapter 5. Interpenetrations
Figure 5.2: Interpenetration of a prism by a straight line
Figure 5.3: A point on the outer surface of a cone
The construction is as follows: take, just as in the previous case, a vertical plane that
contains the line AB and draw the front view section of this plane with the prism. Note
that here it is not necessary to draw a further auxilliary view, because the front view of the
section can easily be determined. The front views of the interpenetration points (c0 and d0 )
are located where the line a10 b10 intersects the outer edge of the front view section. Project
down to find the top views (c and d).
5.1.3
A straight line through a cone
Before we consider this case, it is necessary to first discuss a few constructions relevant to
cones:
62
5.1. Interpenetration points
Generator lines: Suppose that a cone is given and also the front view p0 of a point P located on the outer surface of the cone (see figure 5.3). The top view of the point P can be
obtained by using the generator line running through point P: draw this generator line on
the front view and then obtain its top view as shown in the construction on the left hand side
of figure 5.3. Project p0 down to obtain the top view p on the top view of the generator line.
Horizontal sections: The foregoing problem can also be solved by means of a horizontal
section: take such a horizontal section containing p0 through the cone, as shown in the construction on the right hand side of figure 5.3. Obtain the top view of this section (in this case
it will be circular). Project p0 down to intersect the top view of the section at p.
Vertical sections: One method to determine the true shape of a vertical section through
a cone, is shown in figure 5.4: Take generator lines, such as AV, lying in the vicinity of the
section. The point t0 , which is the intersection of a0 v0 and the section, yields the height of the
section, and tt, in the top view, yields the width of the section. The figure shows clearly how
the construction progresses further.
A second method is shown in figure 5.5: Take horizontal sections through the cone and
the sectioning plane. Determine the top views of the sections. The widths of the sectioning
plane between the plane and the cone are then determined in the top view.
We can now proceed to determine the interpenetration points of a line AB with a cone,
figure 5.6. The method is quite similar to that for a sphere, discussed in §5.1.1: a vertical
section containing the line is taken. Draw the section of the plane through the cone; the
interpenetration points are located where the view of the line touches the section (any one of
the two methods to determine the section can be used - in figure 5.6 the method of horizontal
sections was used). C and D are the interpenetration points.
63
Chapter 5. Interpenetrations
Figure 5.4: Generator line method for determining a vertical section through a
cone
Figure 5.5: Method of horizontal sections for determining a vertical section
through a cone
64
5.1. Interpenetration points
Figure 5.6: Interpenetration of a cone by a straight line
65
Chapter 5. Interpenetrations
5.2
Simple interpenetration curves
An interpenetration curve is the intersection line between the two surfaces of two objects
that interpenetrate each other. This boundary line is therefore located on both surfaces (objects). The most efficient method to determine an interpenetration curve is to take sections.
Care must be taken that the section surfaces are formed by uniform lines such as straight
lines or circles. Choose therefore the sectioning planes such that it forms circular, triangular
or rectangular sections with the object.
Here also a number of specific cases are considered in order to highlight the general principles for determining such interpenetration curves.
5.2.1
A rectangular bar through a cone
The construction is shown in figure 5.7. In order to determine the interpenetration points,
we take horizontal sections so that the cone is sectioned into circles and the bar into rectangles. Take for example the rectangular section 1; it yields the circular and rectangular
sections as shown. The top views of the interpenetration points are located at the points of
intersection between the two sections, i.e. at a, b, c and d. Project up to the front view to
obtain the front views a0 , b0 , c0 and d0 . Repeat with other sections in order to obtain the interpenetration curve. (Alternatively, the generator line method can also be used.) Note: g and
h are obvious section points as well.
Figure 5.7: Interpenetration of a cone by a rectangular bar
66
5.2. Simple interpenetration curves
5.2.2
One cylinder through another cylinder
The construction is shown in figure 5.8. Take horizontal sections. Then the vertical cylinder
yields circular sections and the horizontal cylinder rectangular sections. The sections 1, 2, 3,
etc. are shown, and the widths of the rectangular sections can be obtained from circle a. Set
these widths off on the top view. The interpenetration points are located at the intersections
of the rectangular and the circular sections. In figure 5.9 the shape of a typical horizontal
section is shown. P and Q are two interpenetration points that are determined by means of
this particular section.
Figure 5.8: Interpenetration of a cylinder by another cylinder
Figure 5.9: A horizontal section through two interpenetrating cylinders
67
Chapter 5. Interpenetrations
5.2.3
A cylinder through a cone
The construction is shown in figure 5.10. In order to determine the interpenetration points,
we take horizontal sections so that circular sections are obtained on the cone and rectangular
sections on the cylinder. Take a horizontal section such as the one at height 6 and draw the
top view of the section through the cone and the cylinder. This gives a circular section with
cone radius 6 and the rectangular section L6. The top views of the interpenetration points,
i.e. the 6’s, are obtained where the two sections intersect. Project up to the front view to
obtain the 60 ’s. All the other sections, 3 to 8, can be obtained in a similar manner.
It is clear from figure 5.10 that the cylinder divides the cone into two separate pieces.
This problem can of course also be solved in other ways, for example by drawing an auxiliary view of the cone and cylinder on a vertical plane perpendicular to the centre line of the
cylinder, figure 5.11. The front view of the interpenetration curve will now be a circle, i.e. the
front view of the cylinder. The top view and then the front view of the interpenetration curve
can be determined by using generator lines, as shown in figure 5.11, or by taking horizontal
sections.
Figure 5.10: Interpenetration of a cone by a cylinder (first method)
68
5.2. Simple interpenetration curves
Figure 5.11: Interpenetration of a cone by a cylinder (second method)
69
Chapter 5. Interpenetrations
Figure 5.12: Interpenetration of a sphere by a cylinder
5.2.4
A cylinder through a sphere
The interpenetration curves between a cylinder and a sphere is shown in figure 5.12. Here
horizontal sections are taken, i.e. circular sections through the sphere and rectangular sections through the cylinder.
70
5.3. Interpenetration curves requiring the use of oblique planes
5.3
5.3.1
Interpenetration curves requiring the use of oblique planes
An oblique cone interpenetrated by a line.
Determine the intersecting points of the line through the oblique cone.
Given: Line AB and the oblique cone with a circular shaped base and top point C, as shown
in figure 5.13.
Method: Draw a plane that incorporates the top point of the cone and the line AB. The
section of the plane through the cone will be triangular with the line AB contained in it. The
cone stands on the horizontal plane, it is necessary to determine the HPTL of this plane that
contains the top point C. This will enable the HPTL of the plane to cut the base circle in the
lower points of the section triangle.
Because the plane contains the line AB, the HPTL will contain the HPTP of the line AB
(similar to shadows). Further, because the plane contains the top point C of the cone and
the line AB, it will also contain the line CD, where D is an arbitrary point on the line AB.
Determine thus the HPTP of the line CD, joining these two trace points to produce the HPTL.
The cutting points of the HPTL with the base circle, together with the top point C, produces the triangular section and because the line AB is in this section, E and F will be the
interpenetration points.
Alternatively
Given: An oblique cone with the top point C and a line AB, as shown in figure 5.14.
Method: This method is similar to the previous method.
An oblique plane that contains the line AB and a line parallel to AB through the top point
of the cone is used.
Determine the HPTP of the line AB and the HPTP of the line parallel to AB through C
(points x and y respectively). Join points x and y and lengthen this line to cut the base circle
of the cone. This line is the HPTL of the chosen oblique plane.
From here on the method is identical to the previous method.
5.3.2
A cone interpenetrated by a cylinder
In figure 5.15 it is clear that the use of horizontal and vertical sections is not possible. However, planes that utilise the top point of the cone and a line that is parallel to the centre line
of the cylinder can be used. The cone’s section will be triangular and the cylinder’s section
will be rectangular. Study figure 5.15.
A and B are the intersecting points of a triangular section and a parallel section determined in the top view of section 2. The plane utilises the top point of the cone to form a
triangular section and forms a section that is parallel to the centre line of the cylinder in the
front view.
a0 and b0 are projected into the front view, the generator lines are then drawn parallel to
the cylinder centre line. When c is projected to c0 and a generator line connected to the top of
the cone, the two sections will intersect to produce the points on the interpenetration curve.
5.3.3
The interpenetrating curves between two given cones
In figure 5.16, it is clear that horizontal and vertical sections will not help. A circular section
can be formed on one cone; a hyperbola will be created on the other cone. The general
71
Chapter 5. Interpenetrations
Figure 5.13: An oblique cone interpenetrated by a line
Figure 5.14: An oblique cone interpenetrated by a line, alternative method
72
5.3. Interpenetration curves requiring the use of oblique planes
Figure 5.15: A cone interpenetrated by a cylinder
73
Chapter 5. Interpenetrations
Figure 5.16: The interpenetrating curves between two given cones
method is to use a plane that utilises the top points of both cones as a hinge. This plane will
cut both cones as a triangle. Study figure 5.16.
Extend the line joining the two top points to meet the OX and O1X1 planes. The trace
point HPTP of this line will be found in the top view and the trace point SPTP in the auxiliary/side view. Remember that this side plane contains the base of the horizontal cone;
project the base circle of the horizontal cone to the side view. Project the height a0 into the
side view to obtain a10 , this is also the side plane trace point SPTP. From the SPTP point
draw a line tangent to the base circle and then project (extend) it to the O1X1 plane. Joining
this point to HPTP determines the outermost HPTL1 in the top view. The cutting points of
this line and the base of the vertical cone together with the top points of the cones, forms the
triangular section of the vertical cone and the side view triangular section of the horizontal
cone. It is now possible to determine the front view of these two triangles by projection.
Where the generator lines of these triangular sections intersect one another, produces points
on the interpenetration curve. Draw more planes i.e. section 2 and determine other points
in the same manner, to complete the interpenetration curves in the front and top views.
74
Chapter 6
Developments
6.1
Introduction
Developments (Afrikaans: “oopspalkings of ontvouings”) are used to find the shape of a piece
of flat sheet metal that can be folded or rolled to obtain a required three-dimensional hollow object. Even though many computer aids are currently available to do developments,
engineers must understand the principles behind the creation of developments so that the
software can be used with good judgement and that situations where the software fails can
still be handled from first principles.
Only true developments, i.e. where the sheet is not stretched, are considered here. The
sheet metal is assumed to deform symmetrically about the centre of the sheet’s thickness, i.e.
if the one surface is compressed, then the opposite surface experiences the corresponding
extension. The constructions presented below must therefore always be considered to be
midway between the two surfaces of the sheet metal.
The generation of true developments rely primarily on two notions: the tessellation of
surfaces that are not flat, and the use of true lengths. In some of the tessellations used here,
curved surfaces are approximated by triangles since a triangle is always flat when the sides
are straight. Quadrilaterals can be used for tessellation if they are parallelograms, i.e. guaranteed to be flat with opposite sides exactly parallel. Once a three-dimensional sheet metal
shape has been tessellated, the true lengths of all the triangles’ or parallelograms’ sides are
determined. The triangles or parallelograms are then drawn in the development, with adjoining triangles or parallelograms in the object, kept adjoining in the development wherever
possible. The following section outlines the drawing conventions for developments.
This is followed by a series of examples of developments that illustrate the various approaches to creating them.
6.2
Conventions
The following conventions are followed:
1. All nodes are numbered to explain the development. The node numbers on the orthographic projections follow the normal conventions with primes indicating front view
points. The corresponding points on the development are marked with an asterisk *.
2. The development is always drawn to show the outside of the object.
3. When it is necessary to indicate in which direction folds have to be made, a fold line
that represents a fold towards the inside of the object is indicated with an O, and a fold
towards the outside is indicated with an X.
Figure 6.1 illustrates the basic principles of developments and the conventions. In both
examples in figure 6.1, the three-dimensional objects’ front and top views are given on the
left. The true lengths naturally visible in these views are transferred to the developments on
the right.
75
Chapter 6. Developments
Figure 6.1: Two rudimentary developments
76
6.3. Prism examples
6.3
Prism examples
6.3.1
Rectangular prism
Task: The front and top views of a five-sided rectangular prism with a round hole all the way
through, and the projections of points R and S, are given in figure 6.2. The three-dimensional
view in figure 6.2 shows a pipe passing through the prism. Develop the sides of the prism and
find the front view of a taut string running from R to S along the outside of the prism.
Figure 6.2: Development of a rectangular prism
Solution: The development of the prism’s sides can be constructed using the true lengths
visible in the front and top views: the front view shows the true heights and the top view
shows the true length of the top and bottom edges. The development is drawn next to the
front view so that the heights of the points can conveniently be “projected” between the
development and the front view.
Since the centre line of the hole in the rear surface is perpendicular to the surface, the
hole’s development is a circle. The height of the circle’s centre is visible in the front view, and
the distance from the centre to a vertical edge of facet 4-5 can be measured in the top view.
The front hole can be found in the development by considering a number of points such as
P, one after another (due to symmetry, one quadrant can be constructed by taking, say, 4
points, and the other quadrants found by mirroring). For each point: select its position in
the front view and project to the top view. The position of the point in the development is
given by the height from the front view and the distance from a vertical edge of facet 2-3 in
the top view.
The positions of points R and S in the development can be found in similar fashion. The
taut string can then be drawn in the development, and the intersection between the string
and edges 2 and 3 transferred to the front view. The front view of the string can now be
completed.
77
Chapter 6. Developments
6.3.2
Skew prism
Task: The front and top views of a five-sided skew prism and the positions of points R and S
are given in figure 6.3. Develop the sides of the prism and find the front view of a taut string
running from R to S along the outside of the prism.
Figure 6.3: Development of a skew prism
Solution: Instead of using triangulation (i.e. dividing each parallelogram into two triangles),
a “short cut” construction can be used in this case. Note that all the skew edges are parallel
to each other and that the sides of the prism are therefore perfectly flat parallelograms. Visualise the following: Cut the prism along edge 1 and swing facet 1-6 while keeping edge 1 in
the same position. Point 6 will move in a plane perpendicular to skew edge 1. In the construction of the development therefore draw a line through point 60 perpendicular to edge 1
in the front view and a similar construction line through point 10 . Choose point 1* on the line
just drawn through 10 , for clarity slightly away from the front view. Measure the true length
of line 1-6 in the top view (since line 1-6 is horizontal). Using this true length, the position of
point 6* can now be marked off on the construction line previously drawn through point 60 .
The remaining points along the prism base’s development are found in a similar way
to point 6*. The skew edges in the development are easily found using the true length of
the skew edges seen in the front view and noting that the skew edges remain parallel in the
development.
The positions of R and S in the development are also found by constructing lines through
their front views perpendicular to skew edge 1 and using imaginary skew “edges” through
each point. The remainder of the construction is similar to the previous example.
78
6.4. Cylinder examples
6.4
6.4.1
Cylinder examples
Cylinder with one square end
In many cylinder developments, the cylinder is approximated by a number of rectangular
facets. When this loss in accuracy is not acceptable, the circumference of the cylinder can be
constructed as shown in figure 6.4. Figure 6.4 also shows how the circumference can easily
be divided into a number of equal length parts.
Task: The front and top views of a cylinder truncated by a skew plane are given in figure 6.4.
Construct the development using the true circumference.
Figure 6.4: Development of a cylinder with one square end
Solution: To construct the circumference of a circle, draw a vertical line (4-10) through the
circle and a line 30◦ off vertical through the centre (line through 11). Then draw a horizontal
line through 11 and find s, the intersection with the vertical line. Now draw line 4e with a
length of three diameters. Line es then gives the circumference to a high accuracy.
Line AC in the development can now be drawn next to the front view at the same height
as the top of the cylinder. This position for the development is chosen, as in the previous
examples, so that the lengths of vertical generators1 can easily be transferred from the front
view to the development. To construct twelve equal length segments on AC, draw AB at any
convenient angle and mark off 12 equal length segments (any convenient length) along it.
1
A cylinder can be considered to be composed of an infinite number of straight lines parallel to the centre
line. These lines are the cylinder’s generators or generator lines (Afrikaans: “mantellyne”).
79
Chapter 6. Developments
Then draw a line through 1 on AB and C, and draw the other lines that intersect AB and AC,
parallel to 1C. The properties of similar triangles guarantee that the points along AC will be
equally spaced.
After constructing the circumference of the top edge and the twelve equally spaced points,
the corresponding points, 1 to 12, can be constructed in the top view and projected to the
front view to give 10 to 120 . Since the generator lines are vertical, their front views are true
lengths. The true lengths of the generator lines through 10 to 120 can therefore be “projected”
from the front view to the development giving 1* to 12*. The development is completed by
drawing the curve through 1* to 12*.
6.4.2
Cylinder with circular ends
Task: The front view and an auxiliary view of an elliptical cylinder truncated by skew planes
are given in figure 6.5. Note that the ends of the cylinder form circles in the auxiliary view.
Construct the development by tessellating the cylinder into parallelograms.
Figure 6.5: Development of an elliptical cylinder with circular ends
Solution: The tessellation is accomplished as follows: Divide the base circle in the auxiliary
view into 12 chords, giving points 1 to 12; then find the front view of the corresponding
generator lines after projecting these points to the front view. Since the generator lines are
vertical, their true lengths are visible in the front view. These lengths are transferred to the
development by drawing horizontal construction lines from the front view.
Point 1* is chosen a convenient distance away from the front view, on the horizontal line
through 10 (note that the labels of the front view points have been omitted in figure 6.5 for
clarity). The height of 12* is given by the horizontal construction line through 120 and the
true distance from 1* to 12* is given in the auxiliary view (marked =). Since the circle was
divided into equal facets, the length “=” is also the true distance between all neighbouring
points. The positions of points 1* to 12* in the development are therefore all found in similar
fashion.
80
6.5. Rectangular pyramid example
The corresponding points along the top of the cylinder can be found in a similar way to
the base points, but it is simpler to just draw the vertical generator lines through 1* to 12* to
intersect with horizontal lines drawn from the front view of the cylinder’s top.
6.4.3
Junction of two circular pipes
Task: The front and side views of an off-centre right-angle junction of two circular pipes are
given in figure 6.6. Construct the development of the upper pipe.
Figure 6.6: Off-centre right-angle junction of two circular pipes
Solution: The construction of the interpenetration curve is not shown in figure 6.6, but can
be obtained as follows: The half-circle construction above the top view gives the distance
from each point, 1 to 12, to the pipe’s centreline in the side view. The side view then gives
the height where each corresponding generator line penetrates the lower pipe, which can be
projected from the side view to the front view. Once the interpenetration curve has been determined, the development can be constructed using a procedure similar to the first cylinder
example given above.
Remember: The generator lines are straight in the original object and in the development.
When seeking a specific point in the development, it is often useful to first find the generator
line containing that point in the orthographic views and the development.
6.5
Rectangular pyramid example
Task: The front and top views of a pyramid with a rectangular base are given, as well as the
vertical trace, VT, of a skew plane perpendicular to the vertical plane. Develop the sides of
the pyramid and find the interpenetration of the skew plane and the pyramid in all views.
Solution: Start by constructing the true length of A1, using the top view length (marked
=) and the height difference from the front view. Due to the symmetry of the pyramid, all
the oblique edges have the same true length. An arc, on which all the base points must be
located, can therefore be drawn around A*. The true lengths of 1-4, 1-2, 2-3 and 3-4 are
visible in the top view (since these lines are horizontal) and these true lengths can simply be
marked off along the arc constructed through 1*.
81
Chapter 6. Developments
Figure 6.7: Development of a rectangular pyramid
The positions where the skew plane intersects the edges, such as A1, are found by using proportions given in the front view, even though the front view does not show the true
lengths. The construction used to determine the true length of A1, was placed with this in
mind and now provides a convenient way to construct the respective true lengths from B, C
and D to A. Note that the similar triangles in the true length construction AE1 can be used to
find b01 , c01 , and d01 even when the development is constructed in a different position. The true
lengths from B, C and D to A can alternatively be found by first finding b, c, and d through
projecting down from the front view, and then constructing the true lengths using height differences and top view lengths. Once the true lengths from B, C and D to A have been found,
points B*, C* and D* can be placed in the development.
6.6
6.6.1
Cone examples
Upright cone
Task: The front and top views of an upright cone, as well as the vertical trace of a skew plane
perpendicular to the vertical plane, are given in figure 6.8. Find the development of the cone
82
6.6. Cone examples
Figure 6.8: Development of an upright cone
and the penetration curve. Also find the front view of the straight line B*C*, where B and C
are the matching points where the base circle is split for the development.
Solution: The development of the cone is obtained through a combination of the techniques
used for cylinders and the rectangular pyramid: The front view shows the true length of A1,
since it is parallel to the vertical plane. All the generator lines of this cone have the same
length since it is upright. The cone can therefore be tessellated into a number of equilateral triangles, and in the corresponding construction the circular base is approximated by
a twelve-sided polygon. The development is found by constructing arc B*C* with a radius
equal to the true length of the generator lines. Then the true length of, say, 1-12 (which is
visible in the top view since the line is horizontal), is used to mark off points 12*, 11*, ..., 1*
in the development. The development of the cone is completed by drawing a line from 1* to
A*.
The development of the interpenetration curve is illustrated by taking points P and Q on
the generator lines through 5 and 9, respectively. The particular generator lines are easily
drawn in the development from A* to 5* and 9*, respectively. The true distance from A* to P*
83
Chapter 6. Developments
and Q* is obtained by drawing a horizontal line through p0 q0 to a0 10 , giving p01 in figure 6.8.
The true length of AP is given by a0 p01 . The intersections of the arc through p01 and the generator lines through 5* and 9* give P* and Q* in the development. Points similar to P and Q can
be taken on each generator line, to obtain the interpenetration curve in the development.
The front view of line B*C* is found by reversing the procedure just followed for finding
the penetration curve. Points R and S in figure 6.8 illustrates this.
Remember: The key to finding points like P or R is often to identify the generator line
that they fall on. Since generator lines are straight in three-dimensional space, they remain
straight in the development and it is relatively easy to find specific point on a generator line
in the different views.
6.6.2
Skew cone
Task: The front and top views of a skew cone are given in figure 6.9. Construct the development.
Solution: As in the previous case, the cone is tessellated into twelve triangles, where two
sides of each triangle are formed by generator lines and the third by a side of a twelve-sided
polygon approximating the circular base of the cone. Since the base is horizontal, the true
lengths of the polygon’s sides are visible in the top view. However, except for the generator
lines through 1 and 7, the true lengths of the generator lines are not visible in either of the
original views. The true lengths therefore have to be constructed and in figure 6.9 this is done
by combining the top view length (e.g. the one marked with a ◦◦ in figure 6.9) and the height
difference.
The procedure for the construction is therefore as follows: Select A* at the same height
as a0 , and a convenient distance away. Draw a vertical line through A*. Now consider the
triangles in sequence, for example A-1-12. For convenience, the position of 1* is selected
where the true length of A-1 is constructed. To obtain the position of 12*, first measure the
top view length of A-12 and mark this length on the horizontal construction line in the development (the length marked ◦◦ in figure 6.9 is the corresponding construction for generator
A-10). A right angle triangle with the top view length and the height difference of A-12 has
therefore been constructed, giving |A-12| as the hypotenuse (the hypotenuse is not shown
in figure 6.9 to reduce the number of construction lines). An arc, centred at A*, with radius
equal to |A-12| is drawn and, finally, the distance |1-12|, taken from the top view, is marked off
as 1*-12* (the corresponding length for 3-4, marked =, is shown in figure 6.9). This completes
the construction of the true shape of triangle A-1-12.
The procedure is repeated for all the remaining facets.
Figure 6.10 shows an alternative construction of the true length of a generator, using the
swing method.
6.7
Sphere and torus examples
Task: Construct the development of a sphere by dividing the sphere into twelve "orange
peel" segments.
Solution: Draw a front and a side view of the sphere. Draw the topmost segment in the side
view (on the left in figure 6.11). Note that in the side view, the edges of he peels appear as
straight lines, thus ensuring that the peels will match up. The true length of the centreline of
the segment is visible in the front view. It is divided into six equal parts, giving points 1 to 7.
84
6.8. Transition examples
Using the length marked = to approximate the true length of one part, the development of
the centreline can be drawn.
Points 1 to 7 are then projected to the side view. Arcs, such as ab and cd can now be
constructed and the lengths transferred to the development to be able to complete it.
Task: Construct the development of a torus by dividing it into twelve segments.
Solution: The construction, shown in figure 6.12, is similar to that for the development of a
sphere.
6.8
Transition examples
Task: The front view of the centre line of an air duct transition is given. It connects a given
vertical channel at B to a given horizontal one at A. Each segment of the transition starts
and ends in a square shape. The channel width must change linearly along the centre line.
Construct the development of the channel.
Solution: Subdivide the centre line into a convenient number of equal segments, giving
points b0 , d0 , e0 , f0 , g0 and a0 in figure 6.13. Construct a development of the centre line, B*
to A*. Mark off the half-width of the flow channel at B* and A*, and join these points by a
straight line, thus giving a linear change in channel width. Draw vertical lines through D*,
E*, etc. to give M*, O*, etc. Now construct perpendicular lines through d0 , e0 , etc. in the front
view and mark off the positions of m0 , n0 , o0 , p0 , etc. along these perpendiculars using the
channel widths obtained from the development of the centreline. This completes the front
view.
Although lines such as MN and OP will be parallel, the lines MO and NP will not be. MOPN
therefore is not flat and has to be divided into two triangles by either joining MP or NO. In
figure 6.13, MP was joined. The sides of the transition are tessellated into triangles in this
way.
To construct the development, the true lengths of all the sides of the triangles must be
found. Lines such as m0 n0 are true lengths, but not lines such as n0 p0 and m0 p0 . The true
lengths of these lines can be constructed in the usual fashion once a top view has been completed. As a short cut, the top view lengths can also be taken from the construction of the
development of the centreline. The details of the construction are left as an exercise.
Task: The front and top views of a transition from square to round are given. Construct the
development.
Solution: In this type of transition, some triangles are naturally formed, i.e. GBC, ABF and
ECD. The remaining surfaces are partial skew cones and must be tessellated into triangles.
The development of, for example, BFG follows that of a skew cone with apex B given in a
preceding section. As an example, figure 6.14 shows the construction of the true length of
generator B4 using the swing method.
The further details of the construction are left as an exercise.
85
Chapter 6. Developments
Figure 6.9: Development of a skew cone
86
6.8. Transition examples
Figure 6.10: Alternative construction of the development of a skew cone
87
Chapter 6. Developments
Figure 6.11: Development of a sphere
Figure 6.12: Development of a torus
88
6.8. Transition examples
Figure 6.13: Development of an air duct transition
89
Chapter 6. Developments
Figure 6.14: Development of a square-to-round transition
90
Appendix A
Miscellaneous
A.1
First and third angle projection
A drawing depicts an object that is three dimensional (i.e. that has height, width and depth)
on the flat plane of the drawing paper. Different views of the object, e.g. the front view, side
view and top view, are systematically arranged on the drawing paper in order to convey the
necessary information. This type of drawing is known as orthographic projection.
Two kinds of orthographic projection are used in practice: first angle and third angle
projection. The difference between first and third angle projection is illustrated in figure A.1.
Note that, in third angle projection, all views are projected on planes that are located between
the observer and the object.
The principles of first angle projection are illustrated in figure A.2. All the views are projected on the planes behind the object, i.e. the object is located between the observer and
the plane on which a specific view is projected. There are six main directions from which an
object can be viewed: from above, below, the front, the rear, left and right. In figure A.3 it is
shown how these six views are rotated and positioned such that they can be displayed on a
single flat surface.
An example of the first and third angle projection drawings of a simple object is shown in
figure A.4. In working drawings, the ISO projection symbol must always be given to indicate
whether the drawing has been completed in first angle or third angle projection (see §B.2.3).
Figure A.1: Comparison between first angle and third angle projection
91
Appendix A. Miscellaneous
Figure A.2: First angle projection
Figure A.3: Unfolding of projection planes
92
A.1. First and third angle projection
Figure A.4: Orthographical projections of a simple object
93
Appendix A. Miscellaneous
Figure A.5: Two isometric projections of a cube: descriptive geometry construction
A.2
A.2.1
Isometric projection
Descriptive Geometry background
In order to investigate the characteristics of isometric projections, a cube is used. According
to definition the isometric projection of a cube will be such that all three the major axes of
the cube will make identical angles with the new projection plane, in other words, the new
projection plane is perpendicular to one of the diagonals of the cube. For isometric projections it is therefore essential to specify which diagonal must be perpendicular to the new
projection plane. Shown in figure A.5 are the front and top views of the isometric projection
along diagonal AG, seen in the direction from A to G (over the O2X2 line) and of another
isometric projection along diagonal BH, seen in the direction from B to H (over the O3X3
line).
Going back to the original front and top views of the cube and looking along the abovementioned diagonals, it means that if the cube is viewed from bottom left (along AG), then
the bottom surface AEHD will be completely visible; in the second case the cube is viewed
from top left and the top surface BFGC will be completely visible. This is confirmed by the
isometric projections shown in figure A.5.
The isometric projections of figure A.5 are not located conveniently with respect to the
original OX line, and therefore the coordinate axes for the projection is chosen such that
94
A.2. Isometric projection
Figure A.6: Practical presentation of isometric projections
Figure A.7: True shape vs. isometrically projected shape of a square surface
vertical lines remain vertical in the isometric projection drawing, as shown in figure A.6. The
axes will then make an angle of 30◦ with the horizontal.
It must also be noted that the sides of the cube in the isometric projection are shorter
than the true side length and that the diagonals FC and ED retain their original lengths. The
reason for this is that FC and ED are parallel to the new projection plane.
If the true shape of the plane BFGC is superimposed on its isometric projection, the result is as shown in figure A.7. The amount of shortening that the true lengths of the major axes
of the object undergo when the object is projected isometrically, can be obtained by means
of the scale diagram shown in figure A.8.
95
Appendix A. Miscellaneous
Figure A.8: Scale diagram for isometric projections
Figure A.9: Isometric projection of a sphere
A.2.2
Isometric projection of a sphere
Care must be taken with the construction of the isometric projections of objects with curved
surfaces. The simplest example of such an object is a sphere. The construction of the isometric projection is shown in figure A.9, and is carried out as follows:
1. Construct a cube around the sphere so that the sphere touches all the sides of the cube.
The radius of the sphere is r, therefore the side length of the cube is 2r.
2. Draw the isometric projection of all the sides of the cube. The true lengths of the front
and top views are scaled down.
3. The intersection of the diagonals of the cube, L, indicates the centre of the sphere.
4. The true radius of the sphere is shown parallel to the picture plane in the top view, i.e.
the radius dimension on the isometric projection stays the same as the true dimension
and is therefore not scaled down.
5. With centre L and radius r, draw the sphere on the isometric projection.
A.2.3
Isometric drawings vs. isometric projections
If the picture is scaled down, a true picture is obtained and is called an isometric projection.
If the true lengths in the directions of the axes are used to draw the picture, then the object
will appear bigger, and the picture is called an isometric drawing. An isometric drawing is
quicker to make than an isometric projection, because true lengths in the directions of the
96
A.2. Isometric projection
Figure A.10: Isometric drawing of a “flat” object as seen from the front, left, and
above
axes are used. The only case where a length does change, is that of a sphere, in which case
the diameter must be scaled up.
A.2.4
Inclined lines and curves
The true direction of an inclined line of an object that is to be represented on an isometric
drawing, is obtained by first finding the coordinates of its end points and then connecting
these end points with a line. This procedure is called “Drawing by co-ordinates.” An example
is shown in figure A.10; the construction is carried out as follows:
1.
2.
3.
4.
Draw the co-ordinate system.
Measure AF and AB along the correct axes.
H is located at a distance 20 from A and at a distance 40 above the horizontal plane.
On the isometric drawing find c at a distance 20 from A, on the AF-axis and distance
10 in the AB-axis direction and distance c vertically upwards.
5. Find G, D and E in a similar manner.
6. Complete the picture.
Drawing by co-ordinates is the only method that can be used for curved lines on the
front and top views to obtain the isometric drawing. It must also be used for circles (there
are simplifications, but these are not allowed in this course). Two such drawings are shown
in figure A.11 and figure A.12.
97
Appendix A. Miscellaneous
Figure A.11: Isometric drawing of an “oblique” object as seen from the front, left,
and above
Figure A.12: Isometric drawing of an object with curves, as seen from the front,
right, and below
98
Appendix B
Stellenbosch University Drawing Standards
B.1
Introduction
Engineering drawings are a very important means of communication. To be an effective
means of communication, a drawing has to be understood to have identical meanings by
the persons who created it and the persons who use it.
A good engineering drawing should speak for itself.
Drawing standards are essential elements in ensuring effective communication through
drawings. The standards’ main purpose is to prevent ambiguity. Since engineering drawings
are exchanged internationally, the need for international standards is clear. Many countries
have their own national standards, which are typically derived from the international standards (e.g. SABS 0111 Part 1 and Part 2).
Each engineering organization has its own set of standards that augment those given by
standards organization. The national and international standard represent a consensus of
the best practices, but allow much latitude to ensure their general acceptance. Organizationspecific standards define more precisely what a particular organization requires. In cases
where there are contradictions, the international or national standard should always prevail.
The particular standards required by Stellenbosch University are outlined here.
B.2
B.2.1
Drawing techniques
Line thicknesses
Usually only two line thicknesses are used on any drawing. These two thicknesses depend on
the size of the paper on which the drawing is made. In the drawing and mechanical design
courses at Stellenbosch University most drawings are done on A3 size paper, for which the
recommended line thicknesses are 0,7 mm and 0,3 mm. The 0,7 mm thickness is used to
show clearly those features that the drawing is all about, while the 0,3 mm thickness is used
for construction lines, centre lines, dimension lines, etc. If different line thicknesses are used,
they must have the same “darkness” or density. In practice drawings are made with black ink,
and therefore thin lines are just as black as thick lines.
In order to avoid the cost of special pencils in the Engineering Drawing and Machine
Drawing courses, the Stellenbosch University drawing standard allows students to deviate
from the use of different line thicknesses, by using 0,5 mm light and dark pencil lines. Note
that light lines must still be clearly visible.
An 0,7 mm or dark line is called a full line and is used for visible planes and changes of
plane. In Descriptive Geometry it is also used for the OX-line, the data and the answer.
An 0,3 mm or light line is called a thin line and is used for dimensioning lines, projection
lines and construction lines in Descriptive Geometry. Construction lines are never erased in
Descriptive Geometry, but always in Working Drawings. Thin lines are also used in various
other forms in working drawings (refer to SABS 0111 page 11).
99
Appendix B. Stellenbosch University Drawing Standards
Figure B.1: Shape and size of the ISO projection symbol
B.2.2
Script
A good drawing can easily be spoilt by poor script. The proper forms of lettering are shown in
SABS 0111 pages 18 and 19 (upper case letters). We always write using characters of at least
3,5 mm height. For captions, however, 5 mm high characters must be used. In the beginning
you might find it difficult to maintain the correct height — then you may write between two
very thin lines. You must always write on a line in order to prevent that your writing becomes
skew. You may use a stencil. We always use capitals (not italics) in working drawings. Results
and answers must always be placed in a block, e.g.:
|AB| = ...
B.2.3
Projection symbol
Unless otherwise instructed, all drawings must be made in first angle projection.
The ISO projection symbol must always be given in the top right hand corner of the drawing paper, to indicate whether the drawing has been completed in first angle or third angle
projection. The shape and approximate size of the projection symbol are shown in figure B.1.
NB: The sharp end of the cone must always point towards the inside of the drawing paper.
B.2.4
Title frame
All drawing work must be done on paper on which the title frame is printed. The following parts must always be filled in: STUDENTE-NOMMER (student number), TEKENAAR
(draughtsman), SKAAL (scale), MATE IN (measurements in), DATUM (date), TITEL (title) and
NOMMER (number).
In the Engineering Drawing course, the number that must be filled in consists of the
drawing hall number where you do your Engineering Drawing practicals, followed by your
desk number. The title is the same as the caption on the question paper, unless specified
otherwise. An example of a completed title frame and projection symbol is shown in figure B.2.
100
B.3. Guidelines for dimensioning
Figure B.2: Completed title frame and projection symbol
B.3
Guidelines for dimensioning
1. Projection lines
•
•
•
•
0,3 mm or thin lines
should not touch object (2 mm gap)
should project 2 mm past dimension line
avoid long projection lines and crossing projection lines
2. Dimension lines
• 0,3 mm or thin lines
• equal spacing between dimension lines for multiple dimensions
3. Arrowheads
• should be as shown in figure B.3 (length ≈ 3 × breadth)
• always filled in
• typical length is 3 to 5 mm
4. Dimensioning text
•
•
•
•
•
0,7 mm or thick lines, 3,5 mm high
always placed outside the object
legible from bottom right hand corner of paper
always 1 to 2 mm above the dimension line (not on it )
approximately in the middle of dimension line
5. Diameters
• always indicated by φ (e.g. φ100), except for pitch circles
• pitch circle diameters indicated by PCD (e.g. PCD100)
101
Appendix B. Stellenbosch University Drawing Standards
Figure B.3: Shape of arrowheads
• give φ rather than R wherever the diameter can be measured
• only two (2) crossing dimensions on a specific circular view
6. Radii
•
•
•
•
always indicated by R (e.g. R10)
give only one R dimension for many arcs with the same radius
do not dimension centre point position of fillet radii
the leader line should aim at or pass through the centre of the arc
7. Centre lines
• never used as projection lines (extend with projection lines where necessary)
• they always form part of drawing object =⇒ include in all views
• dimension to centres to show position of holes, only where hole is visible
8. Planning
• leave enough room between views for dimensions
• views do not need to be equal distances apart (different from Technical Drawing
taught at school)
• complete all views before starting dimensioning
• give any dimension only once on a drawing
9. Hidden detail
• never dimension to hidden detail
• do not dimension to centre position when aspect is hidden
• use partial sections to reveal detail, and then dimension
10. General
•
•
•
•
•
use symmetry to reduce the number of dimensions
use reference faces to prevent tolerance build-up
give dimensions that are either functionally important or easy to measure
for the sake of clarity, use e.g. “3 HOLES φ15” instead of “φ15 (3X)”
use only one language
11. Tolerances
• All dimensions have tolerances applied to them.
• The tolerance table is always shown on all DETAIL drawings for manufacture.
• Only those dimensions that need closer tolerances than the given tolerance table,
should be shown in detail.
• Large dimension on top; both dimensions shown above the dimension line.
102
Appendix C
The Metric Vernier Caliper
Figure C.1: Vernier caliper
Vernier Calipers (see figure C.1)are made with metric readings and may have both metric
and inch graduations on the same instrument. The main scale is graduated in millimetres
and every main division is numbered. Each numbered division has a value of 10 mm; for
example, #1 represents 10 mm, #2 represents 20 mm, etc. There are 50 graduations on the
sliding or vernier scale, with every fifth one numbered. These 50 graduations occupy the
same space as the 49 graduations on the main scale (49 mm). Therefore,
1 vernier division =
49 mm
50
= 0, 98 mm
The difference between 1 main scale division and 1 vernier scale division is
1 mm − 0, 98 mm = 0, 02 mm
To read a vernier caliper
1. The last numbered division on the left of the vernier scale represents the number of
millimetres multiplied by 10.
2. Note how many full graduations are showing between this numbered division and the
zero on the vernier scale. Multiply this number by 1 mm.
3. Find the line on the vernier scale that coincides with a line on the bar. Multiply this
number by 0,02 mm.
103
Appendix C. The Metric Vernier Caliper
Figure C.2: Vernier calliper example reading
Example:
Consider the vernier calliper in figure C.2:
• The large #4 graduation on the bar
• Three full lines past the #4 graduation
• The 9th line on the vernier scale
coincides with a line on the bar
4 × 10 mm = 40 mm
3 × 1 mm = 3 mm
9 × 0,02 mm = 0,18 mm
Total reading = 43,18 mm
104
Appendix D
Geometrical Constructions and Tangency
The following chapter is taken from:
Simmons, C.H. and Maguire, D.E., Manual of Engineering Drawing,
2nd edition, Elsevier Newnes, Oxford, 2005.
105
The following chapter is taken from:
Simmons, C and Maguire, D "Manual of Engineering Drawing",
2nd Edition, Elsevier Newnes, Oxford, 2004.
106
107
108
109
110
111
112
Appendix E
Sectioning and hatching
113
114
Appendix F
Dimensioning
115
116
Appendix G
True Lengths and Auxiliary Views
The following chapter is taken from:
Simmons, C.H. and Maguire, D.E., Manual of Engineering Drawing,
2nd edition, Elsevier Newnes, Oxford, 2005.
117
118
119
120
121
122
Appendix H
Conic Sections and Interpenetrations of Solids
The following chapter is taken from:
Simmons, C.H. and Maguire, D.E., Manual of Engineering Drawing,
2nd edition, Elsevier Newnes, Oxford, 2005.
123
124
Oefeninge
Exercises
1 Letters en Basiese Konstruksies
Lettering and Basic Constructions
1.1 Met verwysing na “Engineering drawing” handboek
§1.18 en bylae D bladsy 107 notas. Op die voorkant van
’n A3 vel (links bo), skryf onderstaande paragraaf oor in
HOOFLETTERS 3.5 mm hoog soos voorgeskryf. Voor u
begin skryf, teken twee ligte riglyne 4 mm apart. ’n 2 mm
gaping tussen die stel lyne. Slegs mm is in klein letters.
Refer to Engineering drawing text book §1.18 and to
notes Appendix D page 107. On the front of an A3 sheet
(top left), rewrite the paragraph below in CAPITAL letters 3.5 mm high as prescribed. Before you start printing,
draw two light guidelines lines 4 mm apart. A 2 mm space
between sets of lines. Only mm is in small letters.
“In Engineering drawing 123, I should write in capital letters not smaller than 3,5 mm high. Numbers
are written 0123456789. I will find the format of the
letters and numbers in the notes page 107 that are to
SABS 0111 standard. All titles and captions are in capital letters 5mm high. Engineering drawing consists of
practise, practise and more practise! When I write a
test or exam I read the question thoroughly.”
“In Ingenieurstekeninge 123 behoort ek nooit kleiner
as 3.5 mm hoog te skryf nie. Die formaat van die letters en getalle kry ek op bladsy 107 wat volgens SABS
0111 standaard is. Alle titels en opskrifte in hoofletters 5 mm hoog. 0123456789 is hoe syfers geskryf
moet word. Ingenieurstudentelewe bestaan uit swôt
afgewissel met werk! Wanneer ek toetse en eksamens skryf, LEES ek die vraag altyd deeglik.”
1.2 Teken die papierpakstuk getoon in figuur 1 skaal 1:1. Buitelyne moet deurlopend, dik wees. Begin met die hartlyne van die drie gate. Geen afmetings nodig nie. Verwys
na bladsy 115 in die notas.
Redraw the gasket in figure 1. No dimensions necessary.
Draw scale 1:1. Outside lines must be continuous thick.
Begin with the centre lines of the three holes. Refer page
115 notes.
1.3 Teken die pakstuk in figuur 2 oor. Geen afmetings nodig
nie. Skaal 1:1. Buitelyne moet deurlopend, dik wees. Begin met die hartlyne van die drie gate. Verwys na bladsy
115 in die notas.
Redraw the gasket in figure 2. No dimensions necessary. Draw scale 1:1. Outside lines must be continuous
thick. Begin with the centre lines of the three holes. Refer
page 115 notes.
1.4 Konstrueer ’n perfekte ellips met hoofasse van 100 mm
en 70 mm onderskeidelik wat loodreg op mekaar is (12
punte insluitend die as-punte), volgens slegs die“twocircle method”soos getoon op bladsy 116 notas. Verbind
die 12 punte van die ellips vryhand.
Construct a perfect ellipse with the major axis 100 mm
and the minor axis 70 mm that are perpendicular to
each other (Use 12 construction points including the axis
points). Follow the two-circle method only. Connect the
12 points of the ellipse freehand.
Figuur 1: Pakstuk
Figuur 2: Pakstuk
Figure 1: Gasket
Figure 2: Gasket
125
Oefeninge/Exercises
1.5 Konstrueer ’n reëlmatige veelhoek (poligoon met sewe
sye) lengte 40 mm.
Construct a regular seven sided polygon given the length
of sides (40 mm).
1.6 Teken al die konstruksies van Engineering drawing text
book 2.3.9 en 2.3.11 en van die notas bladsy 115.
Draw all the constructions from Engineering drawing
text book §2.3.9 and §2.3.11 and page 115 from the notes.
1.7 Teken op die agterkant van die A3 vel die konstruksie van
’n involute met sirkel diameter van 120 mm en ’n skaal
van 1:2.
Draw on the back of the A3 sheet the construction of the
involute of a circle 120 mm in diameter and at a scale of
1:2.
2 Ortografiese Projeksies
Orthographic Projection
2.1 Die tekening in figuur 3 toon die voor en bo-aansigte van
’n voorwerp. Met gebruik van tekening instrumente: projekteer en voltooi slegs die syaansig (gesien van links)
van hierdie voorbeeld. Gebruik eerstehoekse projeksie.
The drawing in figure 3 shows the front and top views of
an object. Using drawing instruments, copy these two
views, then: - Project and complete the side view only
(seen from the left) of this object using first angle projection.
2.2 Op die voorkant van ’n A4 vel, teken met instrumente
drie aansigte van die Steun in figuur 4 d.w.s. die vooraansig gegee, ’n bo-aansig en ’n linker-syaansig volgens
eerstehoekse projeksie en op skaal 1:1. Meet direk van
die onvolledige vooraansig en isometriese tekening gegee. Geen afmetings nodig nie. Geen konstruksielyne of
versteekte detail moet getoon word nie.
On the front of a A4 sheet, draw with instruments three
views i.e. the front view given, a top view and a left hand
side view, in first angle projection, scale 1:1, of the Support in figure 4. Measure direct from the incomplete front
view (on the left in figure 4) and isometric drawing (on
the right of figure 4) given. No dimensions necessary. No
construction lines or hidden detail must be shown.
2.3 Die tekeninge in figuur 5 toon telkens twee aansigte, van
vier verskillende voorwerpe. Voltooi vryhand die ISO
simbool op elke tekening, en teken vryhand die ontbrekende aansig met alle versteekte detail in die beskikbare
ruimte.
The drawings in figure 5 show two views of four different
components. Complete the ISO symbol for each drawing, and draw freehand the missing view with all hidden
detail in the space provided.
Figuur 3: Blok
Figure 3: Block
126
Oefeninge/Exercises
Figuur 4: Steunstuk
Figure 4: Support
2.4 Voltooi vryhand die drie primêre aansigte van die voorwerpe in figuur 5, in 3de - Hoekse projeksie. Die vooraansig is dieselfde as in die vorige vraag. Geen versteekte
detail nodig nie.
Complete freehand the three views of the components
in figure 5 in 3rd Angle projection. The front view is the
same as the previous question. No hidden detail necessary.
2.5 Op ’n A4 vel maak ’n vryhandtekening “EERSTEHOEKSE
PROJEKSIE”van die item in figuur 6. Teken die vooraansig in die rigting van die pyl A, ’n bo-aansig en ook ’n syaansig gesien van links. Skaal ±1 : 1. Geen afmetings nodig nie. Skryf u naam en studente nommer (Netjies). Voltooi die ISO-projeksie simbool.
On an A4 sheet make a freehand drawing “FIRST Angle
Projection” of the item in figure 6. Draw a front view in
the direction of the arrow A, a top view and also a side
view seen from the left. Scale ±1 : 1. No dimensions necessary. Write your name and student number (Neatly).
Complete the ISO-projection symbol.
Figuur 5: Ortografiese Aansigte
Figure 5: Orthographic Views
127
Oefeninge/Exercises
2.6 Op die voorkant van ’n A3 vel, teken met instrumente drie
aansigte van die voorwerp in figuur 6, d.w.s. vooraansig vanuit kykrigting soos aangetoon deur die pylpunt, ’n
bo-aansig en ’n syaansig gesien van links volgens 1steHoekse projeksie en op skaal 2:1. Geen afmetings nodig
nie. Geen versteekte detail nodig nie. Skryf u naam en
studente nommer en titel in die plek verskaf. (Netjies)
Onthou om die ISO-projeksiesimbool te voltooi.
On the front of an A3 sheet, draw with instruments three
views of the object in figure 6 i.e. a front view in the
direction of the arrow shown; a top view, and a side
view seen from the left, in 1st Angle projection, scale 2:1.
No dimensions necessary. No hidden detail necessary.
Write your name and student number and title in the
space provided (Neatly). Remember to complete the ISOprojection symbol.
A
Figuur 6: Steun
Figuur 7: Stut
Figure 6: Support
Figure 7: Support
2.7 Teken op ’n A4 tekenvel, volgens eerstehoekse projeksie,
’n vooraansig in die rigting van die pyl van die soliede
simmetriese voorwerp in figuur 7. Skaal 1:1. Teken ook
’n volsnit sy-aansig. Al die gate is regdeur. Vul die titelblok in. Teken die toleransietabel. Geen versteekte detail
moet getoon word nie. Toon alle nodige afmetings om
die voorwerp te vervaardig. Die twee groot gate moet ’n
toleransie van ±0.1 hê.
Draw, using first angle projection, on an A4 sheet a front
view of the symmetrical solid object in figure 7 in the direction of the arrow. Scale 1:1. Draw also a fully sectioned
side view. All the holes are through. Fill in the title block.
Draw the tolerance table. No hidden detail should be
shown. Show all the necessary dimensions of the object
for manufacture. The two large holes must have a tolerance of ±0.1.
2.8 Teken eerstehoekse projeksie die drie primêre aansigte
van die voorwerp in figuur 8. Die vooraansig is in die
rigting van die pyl. Die hoek wat die V-groef maak is
75◦ . Geen versteekte detail of konstruksielyne nodig nie.
Geen afmetings nodig nie.
Draw first angle projection the three primary views of the
object in figure 8. The front view is in the direction of the
arrow. The angle of the V-groove is 75◦ . No hidden detail
or construction lines necessary. No dimensions necessary.
2.9 Teken in eerstehoekse projeksie die drie primêre aansigte
van die simmetriese voorwerp in figuur 9, skaal 1:1. Die
vooraansig moet in die rigting van die pyl wees. Geen versteekte detail nie. Los slegs die konstruksielyne vir die
projeksie van die ;15 mm gat in die middel van die voorwerp. Gee slegs die afmetings nodig vir die vervaardiging
van die gate. Die dimensies vir die deursnee van die gate
moet ’n toleransie van ±0.1 mm hê. Gee ’n toleransietabel met ’n algemene toleransie van ±0.2 mm. Beplan die
uitleg van u tekeninge versigtig.
Draw in first angle projection the three primary views of
the symmetrical object in figure 9, scale 1:1. The front
view must be in the direction of the arrow. No hidden
detail. Leave only the construction lines for the projection of the ;15 mm hole in the centre of the object. Give
only the dimensions necessary for the manufacture of
the holes. The dimensions for the diameters must have a
tolerance of ±0.1 mm. Give a tolerance table with a general tolerance of ±0.2 mm. Plan the layout of your drawing carefully.
128
Oefeninge/Exercises
60
Figuur 8: V-Stut
Figuur 9: Pons Basis
Figure 8: V-Support
Figure 9: Punch Base
2.10 Die voor- en bo-aansig van ’n soliede voorwerp word
in figuur 10 getoon. Teken ’n volsit linkeraansig van
die voorwerp (gesny deur die middel van die voorwerp).
Skaal 1:1. Geen afmetings of versteekte detail nodig nie.
The front and top views of a solid object is shown in figure 10. Draw a fully sectioned left view of the object (sectioned through the middle of the object). Scale 1:1. No
dimensions or hidden detail necessary.
5
10
20
5
10
20
40
20
10
Figuur 10: Soliede Voorwerp
Figure 10: Solid Object
2.11 Dui die regte en verkeerde dimensie standaarde op die
voorwerpe in figuur 11. Omkring die foute met ’n sirkel.
Indicate the correct and incorrect dimensioning standards on the objects in figure 11. Mark the erros with a
circle.
2.12 Dui die regte en verkeerde dimensie standaarde op die
voorwerpe in figuur 12. Omkring die foute met ’n sirkel.
Indicate the correct and incorrect dimensioning standards on the objects in figure 12. Mark the errors with
a circle.
129
Oefeninge/Exercises
Ø15
30
30
30
30
15
30
15
30
30
15
15
30
30
15
15
30
15
30
Ø15
30
Ø15
R7.5
Figuur 11: Dimensionering
Figuur 12: Dimensionering
Figure 11: Dimensioning
Figure 12: Dimensioning
3 Isometriese Tekeninge
Isometric Drawing
3.1 Teken ’n isometriese projeksie van die voorwerp in figuur 13 soos gesien van voor, regs en bo. Skaal 1:1. Geen
versteekte detail is nodig nie.
Make an isometric projection of the object in figure 13 as
seen from front, right and above. Scale 1:1. No hidden
detail necessary.
3.2 Teken ’n isometriese tekening van die soliede voorwerp
in figuur 14 soos gesien van voor, links en bo. Skaal 1:1.
Geen versteekte detail is nodig nie.
Make an isometric drawing of the solid object shown in
figure 14 as seen from front, left and above. Scale 1:1. No
hidden detail necessary.
3.3 Teken ’n isometriese tekening, skaal 1:1, van die voorwerp in figuur 15 soos gesien van voor, links en bo. Geen
versteekte detail nodig nie.
Make an isometric drawing, scale 1:1, of the object shown
in figure 15 as seen from front, left and above. No hidden
detail necessary.
3.4 Teken die voorwerp in figuur 16 (eerstehoekse projeksie)
vryhand oor as ’n isometriese tekening gesien van voor,
links en bo. Die reghoekige gat is reg deur. Geen versteekte detail nodig nie. Skaal ±1:1.
Make a freehand isometric drawing seen from front, left
and above, of the object in figure 16 first angle projection.
Rectangular hole is right through. No hidden detail necessary. Scale ±1:1.
40
7
40
32
15
5
7
Ø25
50
35
10
13
60°
R15
Figuur 13: Blok
Figuur 14: Blok
Figure 13: Block
Figure 14: Block
130
12
10
38
15
50
7
Oefeninge/Exercises
40
Ø40
16
15
30
8
12
Figuur 15: Blok
Figuur 16: Blok
Figure 15: Block
Figure 16: Block
a’
20 x 45 ~
d’b’
15
e’
c’
15
h’f’
10
25
60
g’
7
30
45
f
ag
e
35
b
h
c
d
60
Figuur 17: Blok
Figuur 18: Blok
Figure 17: Block
Figure 18: Block
3.5 Teken vryhand die voorwerp in figuur 17 (eerstehoekse
projeksie) oor as ’n isometriese tekening soos gesien van
voor, links en bo. Geen versteekte detail nodig nie. Skaal
±1:1.
Make a freehand isometric drawing seen from front, left
and above, of the object in figure 17 first angle projection.
No hidden detail necessary. Scale ±1:1.
3.6 Teken ’n isometriese tekening van die voorwerp in figuur 18 soos gesien van voor, regs en bo. Toon alle versteekte detail en letters. Skaal 2:1. Meet direk vanaf die
tekening.
Make an isometric drawing of the object in figure 18 as
seen from front, right and above. Show all hidden detail
and letters. Scale 2:1. Measure directly from the drawing.
3.7 Twee aansigte van ’n voorwerp word in figuur 19 getoon.
Maak ’n isometriese tekening van voor, links en bo. Toon
slegs die konstruksielyne vir die gat. Geen versteekte detail nie.
Two views of an object are given in figure 19. Make an
isometric drawing from front, left and above. Show construction lines for the hole only. No hidden detail.
131
Oefeninge/Exercises
n20
15
78
41
20
35
30
20
10
15
8
50
34
44
8
60
Ø18 THRU
5
°
40
Figuur 19: Stut
Figuur 20: Blok
Figure 19: Support
Figure 20: Block
80
60
30
°
20
15
70
40
20
34
55
32
35
60
10
18
40
36
R25
22
Figuur 21: Blok
Figuur 22: Blok
Figure 21: Block
Figure 22: Block
132
Oefeninge/Exercises
3.8 Twee aansigte van ’n voorwerp word in figuur 20 getoon.
Maak ’n isometriese tekening van voor, regs en bo. Toon
alle konstruksielyne. Geen versteekte detail nodig nie.
Two views of an object are given in figure 20. Make an
isometric drawing from front, right and above. Show all
construction lines. No hidden detail necessary.
3.9 Drie aansigte van ’n voorwerp word in figuur 21 getoon.
Maak ’n isometriese tekening van voor, links en bo. Toon
alle konstruksielyne. Geen versteekte detail nodig nie.
Teken skaal 1:2.
Three views of an object are given in figure 21. Make an
isometric drawing from front, left and above. Show all
construction lines. No hidden detail necessary. Draw
scale 1:2.
3.10 Twee aansigte van ’n voorwerp word in figuur 22 getoon.
Maak ’n isometriese tekening van die voorwerp soos gesien van voor, regs en bo, skaal 1:1. Toon alle versteekte
detail. Gebruik die metode van projeksie om die ellips te
teken. Los alle konstruksie lyne.
Two views of an object are given in figure 22. Make an
isometric drawing of the object shown as seen from front,
right and above, scale 1:1. Show all hidden detail. Use
the method of projection to draw the ellipse. Leave all
construction lines.
4 Snitte en Dimensies
4.1 Voltooi die snit aansigte in figuur 23.
Sectioning and Dimensioning
Complete the section views in figure 23.
Figuur 23: Snit Voorbeelde
Figure 23: Sectioning Examples
133
Oefeninge/Exercises
A
A
Figuur 24: Blok
Figuur 25: Blok
Figure 24: Block
Figure 25: Block
4.2 Teken in eerstehoekseprojeksie die volgende twee aansigte van die voorwerp in figuur 24: Die vooraansig in die
rigting van die pyl A en ’n volsnit bo-aansig. Skaal 1:1.
Toon alle dimensies benodig. Geen versteekte detail of
konstruksielyne nodig nie. Die toleransie van die ;12 gat
moet ±0.1 wees.
Draw first angle projection, the following two views of the
object in figure 24: The front view in the direction of the
arrow A and a fully sectioned top view. Scale 1:1. Show all
dimensions necessary. No hidden detail or construction
lines. The tolerance of the ;12 hole must be ±0.1.
4.3 Teken die voor, bo en sy aansig van die soliede voorwerp
getoon in figuur 25 as die vooraansig in die rigting van
pyl A is. Skaal 1:1. Teken die projeksie simbool in die korrekte posisie. Projekteer ’n volsnit vooraansig. Tool alle
benodigde dimensies. Moet geen konstruksielyne of versteekte detail toon nie.
Draw the front, top and side view of the solid object
shown in figure 25 if the front view is in the direction of
the arrow A. Scale 1:1. Draw the projection symbol in the
correct position. Project the front view in full section.
Show all necessary dimensions. Do not show construction lines or hidden lines.
4.4 Teken op ’n A3 vel drie aansigte (Eerstehoekse Projeksie),
skaal 1:1, van die soliede voorwerp in figuur 26 in die rigting van die pyle soos getoon. Die vooraansig moet geplaas word sodat die bo-aansig in die normale posisie geteken kan word. Teken ’n volsnit sy-aansig deur die middel van die regterkantse vertikale gleuf. Toon slegs die dimensies wat nodig is om die gleuwe te vervaardig. Vee
alle konstruksielyne uit. Geen versteekte detail nodig nie.
Geen sakrekenaars toegelaat. Voltooi die titelblok.
On an A3 sheet, draw three views (First Angle Projection),
scale 1:1, of the solid object in figure 26 in the direction
of the arrows as shown. The front view must be placed so
that the top view is drawn in the normal position. Draw
a full section side view through the middle of the vertical
slot right side. Show only the dimensions necessary to
manufacture the slots. No construction lines. No hidden
detail. No calculators allowed. Complete the title block.
4.5 Teken die drie primêre aansigte van die voorwerp in figuur 27, skaal 1:1. Die vooraansig moet in die rigting van
die pyl wees. Die vooraansig moet in volsnit wees. Voltooi
al die afmetings nodig vir vervaardiging. Geen versteekte
detail of konstruksielyne moet getoon word nie. Die vulstrale is R5 en die afskuinings in 45◦ × 3. Alle afmetings in
mm.
Draw the three primary views of the object in figure 27,
scale 1:1. The front view must be in the direction of the
arrow. The front view must be in full section. Complete
all the dimensions necessary for manufacture. No hidden detail or construction lines must be shown. The fillet
radii are R5 and the chamfers are 45◦ × 3. All dimensions
in mm.
4.6 Dui die regte en verkeerde arseringstandaard in figuur 28
aan.
Indicate the correct and incorrect hatching standard in
figure 28.
134
Oefeninge/Exercises
VOORAANSIG
(FRONT VIEW)
SNIT
SECTION
B0-AANSIG
(TOP VIEW)
ALLE VULSTRALE R5
ALL FILLET RADII R5
Figuur 26: V-Blok
Figuur 27: Blok
Figure 26: V-Block
Figure 27: Block
Figuur 28: Arsering
Figure 28: Hatching
5 Deurdringing van Liggame
Interpenetration Curves
5.1 Teken die twee aansigte van die konus en silinder in figuur 29 oor en voltooi die vooraansig. Projekteer die
deurdringingskromme deur mantellyne te gebruik. Toon
vyf punte op die deurdringingskromme. Geen versteekte
detail nodig nie.
Copy the two views and complete the front view of the
solid cone and cylinder in figure 29. Project the intersecting curve using generator lines, showing five points. No
hidden detail necessary.
5.2 Vind die deurdringingskrommes (met versteekte detail)
van die twee soliede silinders soos getoon in figuur 30.
Toon alle konstruksielyne. Teken eers die twee aansigte
oor op ’n A4 tekenvel.
Find the interpenetration curves (with hidden detail) of
the two solid cylinders shown in figure 30. Show all construction lines. First copy the existing views onto an A4
sheet of paper.
5.3 Teken die twee aansigte van die blok en konus in figuur 31
oor. Voltooi die sy-aansig en teken die deurdringingskromme in die voor- en bo-aansig van die afgesnyde konus wat deurdring word tot die middel met die vierkantige blokkie. Skaal 1:1.
Copy the two views of the block and cone shown in figure 31. Complete the side view and draw the intersecting curve in the front and top views of the truncated
cone that is penetrated to the middle by the square block.
Scale 1:1.
5.4 Teken die twee aansigte van die solide blok in figuur 32
oor. Voltooi die vooraansig en maak ’n isometriese tekening van die voorwerp soos gesien van voor, links en bo.
Toon alle versteekte detail. Skaal 1:1.
Copy the two views of the solid block in figure 32. Complete the front view and make and isometric drawing of
the object as seen from front, left and above. Show all
hidden detail. Scale 1:1.
135
Oefeninge/Exercises
55
°
Ø6
24
120
45
60
0
20
Ø80
4
Ø70
90
Ø24
Figuur 29: Konus en Silinder
Figuur 30: Twee Silinders
Figure 29: Cone and Cylinder
Figure 30: Two Cylinders
60
40
10
13
Ø20
50
40
60
50
R30
Ø65
80
40
Figuur 31: Blok en Konus
Figuur 32: Blok
Figure 31: Block and Cone
Figure 32: Block
136
Oefeninge/Exercises
5.5 Teken die deurdringingskromme van die piramiede en silinder soos getoon in figuur 33. Toon die versteekte detail. Skaal 1:1.
(a) Gebruik snitte.
(b) Gebruik mantellyne. (Wenk: Teken ’n hulpaansig
waar die silinder as ’n sirkel gesien sal word.)
Draw the interpenetration curves of the pyramid and
cylinder shown in figure 33. Show the hidden detail.
Scale 1:1.
(a) Use sections
(b) Use generator lines (Hint: draw an auxiliary view
where the cylinder is seen as a circle.)
5.6 Teken die sy-aansig en voltooi die deurdringingskromme
van die konus en silinder in figuur 34. Geen versteekte
detail nodig nie. Toon alle konstruksielyne vir 5 punte op
die kromme.
Draw the side view and complete the intersecting curve
of the cone and cylinder in figure 34. No hidden detail
necessary. Show all construction lines for 5 points on the
curve.
5.7 Teken die twee aansigte in figuur 35 oor en voltooi die
bo-aansig. Geen versteekte detail of afmetings nodig nie.
Toon alle konstruksielyne vir 12 punte op die deurdringingskromme.
Draw the two views in figure 35 and complete the top
view. No hidden detail or dimensions necessary. Show
all construction lines for 12 points on the interpenetration curve.
8
n
39
30°
60
Ø3
0
Ø24
49
25
60
45°
51
Figuur 33: Piramiede en Silinder
Figuur 34: Konus en Silinder
Figure 33: Pyramid and Cylinder
Figure 34: Cone and Cylinder
10
30
30°
90°
Ø40
90
Figuur 35: Deurdringing van Silinder
Figure 35: Interpenetration of Cylinder
137
Oefeninge/Exercises
6 Hulpaansigte
Auxiliary Views
Copy the two views in figure 36 and draw only the auxiliary view of the oblique plane and the hole of the cylinder in the direction of the arrow A. Show the construction lines of point P only. Draw all dimensions necessary.
Scale 1:1.
6.2 Teken die voor-, linker- en bo-aansigte van die soliede
blok in figuur 37. Skaal 1:1. Teken dan die hulpaansig
wat nodig is om die ware dimensies van die ;12 ± 0.1 gat
te sien. Gee al die dimensies wat nodig is om die blok te
vervaardig. Gee die toleransie vir die gat asook ’n toleransietabel. Geen konstruksielyne of versteekte detail nodig
nie.
Draw the front, left and top view of the solid block in figure 37. Scale 1:1. Then draw the auxiliary view that is
necessary to see the true dimensions of the ;12±0.1 hole.
Give all the dimensions necessary for manufacture. Specify the hole tolerance and provide a tolerance table. No
construction lines or hidden detail necessary.
40°
6.1 Teken die twee aansigte in figuur 36 oor en teken slegs die
hulpaansig van die skewe vlak en die gat van die silider in
die rigting van die pyl A. Toon slegs die konstruksielyne
van die punt P. Teken alle nodige dimensies. Skaal 1:1.
60
A
20
P’
Ø60
Ø15
P
12
Figuur 36: Soliede Silinder met Gat
Figuur 37: Blok
Figure 36: Solid Cylinder with Hole
Figure 37: Block
6.3 Teken die twee aansigte van die blok in figuur 38 oor. Teken ’n hulpaansig, geprojekteer vanaf die vooraansig, in
die rigting van pyl A, in eerstehoekse projeksie. Toon alle
konstruksielyne en versteekte detail. Geen dimensies nodig nie.
Copy the two views of the block shown in figure 38. Draw
an auxiliary view, projected from the top view, in the direction of the arrow A, in first angle projection. Show all
construction lines and hidden detail. No dimensions are
necessary. Scale 1:1.
6.4 Teken die twee aansigte van die blok in figuur 39 oor. Teken ’n hulpaansig, geprojekteer vanaf die bo-aansig, in
die rigting van pyl A wat loodreg is met die vlak, in eerstehoekse projeksie. Toon alle konstruksielyne en versteekte
detail. Geen dimensies nodig nie. Skaal 1:1.
Copy the two views of the block in figure 39. Draw an auxiliary view, projected from the top view, in the direction of
the arrow A that is perpendicular to the face, in first angle
projection. Show all construction lines and hidden detail.
No dimensions necessary. Scale 1:1.
6.5 Teken ’n voor- en linkeraansig van die stut in figuur 40 in
eerstehoekseprojeksie, skaal 1:1. Die vooraansig moet in
die rigting van die pyl wees. Projekteer vanaf die linkeraansig ’n hulpaansig van die volledige onderdeel wat die
gat in ware vorm toon. Toon alle konstruksielyne. Geen
versteekte detail nodig nie.
Draw the front and left views of the support in figure 40
in first angle projection, scale 1:1. The front view must be
in the direction of the arrow. Project an auxiliary view of
the complete part from the left view that shows the hole
in its true shape. Show all construction lines. No hidden
detail necessary.
138
Oefeninge/Exercises
10 20
5
10 5
5
22
10
12
60
4 10
14
90
A
44
Figuur 38: Blok
Figuur 39: Blok
Figure 38: Block
Figure 39: Block
30
15
80
8
60
40
10
30 °
A
55
°
60
7
14
φ5
45°
14
20
75
15
Figuur 40: Stut
Figuur 41: Hulpaansig
Figure 40: Support
Figure 41: Auxiliary View
139
15
Oefeninge/Exercises
6.6 Teken die aansigte in figuur 41 en projekteer die hulpaansig in die rigting van die pyl. Skaal 1:1. Geen versteekte
detail of konstruksielyne is nodig nie.
Copy the two views in figure 41 and draw the auxiliary
view in the direction of the arrow. Scale 1:1. No hidden
detail or construction lines necessary.
6.7 Teken die aansigte in figuur 42 en projekteer die hulpaansig in die rigting van die pyl. Die V-groef maak ’n 90◦
hoek. Skaal 1:1. Geen versteekte detail of konstruksielyne
is nodig nie.
Copy the two views in figure 42 and draw the auxiliary
view in the direction of the arrow. The V-groove makes
a 90◦ angle. Scale 1:1. No hidden detail or construction
lines necessary.
6.8 Teken die aansigte in figuur 43 en projekteer die hulpaansig in die rigting van die pyl. Die V-groef maak ’n 90◦
hoek. Skaal 1:1. Geen versteekte detail of konstruksielyne
is nodig nie.
Copy the two views in figure 43 and draw the auxiliary
view in the direction of the arrow. The V-groove makes
a 90◦ angle. Scale 1:1. No hidden detail or construction
lines necessary.
60
8
°
5
60
10
30
5
30 °
17
15
20
22
0°
15
25 °
20
10
60
12
10
59
80
Figuur 42: Hulpaansig
Figuur 43: Hulpaansig
Figure 42: Auxiliary View
Figure 43: Auxiliary View
6.9 Teken die aansigte in figuur 44 en projekteer die hulpaansig in die rigting van die pyl. Skaal 1:1. Geen versteekte
detail of konstruksielyne is nodig nie.
Copy the two views in figure 44 and draw the auxiliary
view in the direction of the arrow. Scale 1:1. No hidden
detail or construction lines necessary.
6.10 Teken die aansigte van die voowerp in figuur 45 en projekteer die hulpaansig in die rigting van die pyl. Skaal 1:1.
Toon alle versteekte detail. Geen konstruksielyne is nodig
nie.
Copy the two views of the object in figure 45 and draw
the auxiliary view in the direction of the arrow. Scale 1:1.
Show all hidden detail. No construction lines necessary.
6.11 Teken die hulpaansig van die voorwerp in figuur 46 soos
gesien in die rigting van die pyl A. (Pyl A is loodreg op lyn
CD.) Tool alle versteekte detail. Skaal 1:1.
Draw the auxiliary view of the object in figure 46 in the direction of arrow A. (Arrow A is perpendicular to line CD.)
Show all hidden detail. Scale 1:1.
140
50
12,5
10
30
60
25
6
15
20
6
14
Oefeninge/Exercises
38
8
15
13
50
45°
50
Figuur 44: Hulpaansig
Figuur 45: Hulpaansig
Figure 44: Auxiliary View
Figure 45: Auxiliary View
15
50
C
D
50
50
A
Figuur 46: Hulpaansig
Figure 46: Auxiliary View
7 Punte en Lyne in die Ruimte
Points and Lines in Space
7.1 Teken ’n OX lyn. Toon die voor-, bo- en sy-aansigte van
die punt A hier onder. In watter kwadrant is die punt?
Punt B is dieselfde afstand as A van die vertikale vlak,
maar dit is 3 maal hoër bo die horisontale vlak en die
helfte so ver van die syvlak. Teken punt B. Verbind A en B
as ’n lyn en toon al drie projeksies van die lyn. Wat kan u
aflei van die tekening?
Draw an OX line. Show the front, top and side views of
the point A below. In which quadrant is this point? Point
B is the same distance from the vertical plane as A is, although it is 3 times the height above the horizontal plane
and half as far from the side plane. Draw B as well. Connect A to B as a line and show all three projections for this
line. What do we establish from this drawing?
A(30; 10; -20)
141
Oefeninge/Exercises
7.2 Op ’n A3 vel trek ’n OX lyn, en merk die nulpunt by O.
Teken dan die voor- en boaansigte van punte A, B, C en
D waarvan die koördinate hieronder gegee word. Skryf
elke projeksie se benaming by, bv. a en a’. Teken ook die
vooraansigte van lyne AB en BD en die boaansigte van
lyne AC en CD. Skaal 2:1.
On an A3 sheet, draw an OX line, and mark the zero point
at O. Now draw the front and top views of points A, B, C
and D of which the co-ordinates are given below. Write in
the name of each projection, e.g. a and a’. Draw also the
front views of lines AB and BD and the top views of lines
AC and CD. Scale 2:1.
A(5; 15; 25), B(15; -20; 25), C(25; 20; -25), D(35; -15; -20)
7.3
Teken die projeksies van die volgende punte op die HV
en die VV. Teken ’n linkeraansig van die punte. In watter
kwadrant is elke punt?
Draw the projections of the following points on the HP
and the VP. Draw a left hand view of the points. In what
quadrant is each point?
A(10; 30; 50); B(30; -30; -50); C(50; 30; -50); D(70; -30; 50)
Vir die punte in vraag 7.3, sit ’n nuwe projeksie vlak
(NPV) in. Die NPV is loodreg met die HV en maak ’n hoek
van 30◦ met die VV. Die NPV sny die OX by die oorsprong.
Teken die punte in die NPV.
For the points in question 7.3, put in a new projection
plane (NPP). The NPP is perpendicular to the HP and
makes an angle of 30◦ with the VP. The NPP intersects the
OX at the origin. Draw the points in the NPP.
7.5 Herhaal vraag 7.4, maar nou is die NPV loodreg met die
VV en maak ’n hoek van 30◦ met die HV.
Repeat question 7.4, but now the NNP is perpendicular
to the VP and makes an angle of 30◦ with the HP.
7.6 Punte A en B is die endpunte van ’n lyn. Die projektore
van A en B is 90 mm uit mekaar. A is in die derde kwadrant en lê 30 mm van die HV en 20 mm van die VV. B is
70 mm bo die HV en 45 mm agter die VV. Teken die lyn.
Neem aan dat die x waarde van A 0 is en dat B regs van
A is. In watter kwadrant is B? As die lyn die HV sny, tabuleer die koördinate van daardie punt. As die lyn die VV
sny, tabuleer die koördinate daardie punt. Teken die linkeraansig van die lyn AB.
Point A and B are the end points of a line. The projectors of A and B is 90 mm apart. A is in the third quadrant
and lies 30 mm from the HP and 20 mm from the VP. B is
70 mm above the HP and 45 mm behind the VP. Draw the
line, assuming that the x value of A is 0 and that B is to the
right of A. In what quadrant is B? If the line intersects the
HP, tabulate the coordinates of the intersection point. If
the line intersects the VP, tabulate the coordinates of the
intersection point. Draw the left side view of AB.
O1
X1
7.4
O
X
Figuur 47: Punte en Lyne
Figure 47: Points and Lines
7.7 Op die A3 vel, trek ’n OX-lyn en ’n O1X1 lyn soos getoon
in figuur 47. Die projeksievlak O1X1 is loodreg op die VV.
Teken die drie aansigte van punte E, F en G waarvan die
koördinate hieronder gegee word. Skryf elke projeksie
se benaming by. Voltooi die projeksies van die driehoek
EFG. Skaal 2:1.
On the A3 sheet, draw the OX line and an O1X1 line as
shown in figure 47. The projection plane O1X1 is perpendicular to the VP. Draw the three views of points E, F and
G of which the co-ordinates are given below. Write in the
name of each projection. Complete the projections of the
triangle EFG. Scale 2:1
E(15; 20; 25), F(25; 30; -20), G(-10; -20; 20)
142
Oefeninge/Exercises
7.8 Teken die projeksies van die volgende punte op die HV
en VV. Teken ’n linker sy-aansig van die punte. In watter
kwadrant is elke punt?
Draw the projections of the following points on the HP
and the VP. Draw a left hand side view of the points. In
what quadrant is each point?
A (-10;40;30) B (30;-40;-30) C (60;55;-45) D (90;-10;25)
7.9 Teken figuur 48 oor op ’n A4 tekenvel. Vind die ontbrekende projeksies van A, B, C, D en E in die HV, VV en O1X1
vlak (⊥ op die HV) en die O2X2 vlak (⊥ op die VV). Tabuleer die koördinate van die punte.
Copy figure 48 onto an A4 sheet. Find the missing projections of the points A, B, C, D and E in the HP, VP, O1X1
plane (⊥ on HP) and O2X2 plane (⊥ on VP). Tabulate the
co-ordinates of the points.
X2
e2
c’
a’
b’
O
X
O1
O2
c1’
e
b
X1
d
d1’
a
Figuur 48: Punte en Lyne
Figure 48: Points and Lines
143
Oefeninge/Exercises
7.10 Teken figuur 49 oor op ’n A4 tekenvel. Die voor- en boaansigte van A, B, C en D word gegee. Vind die projeksies
van die punte in die nuwe projeksievlakke aangedui deur
O1X1, O2X2, O3X3 en O4X4. Die O1X1 vlak is ⊥ op die
HV. Die O2X2 vlak is ⊥ op die O1X1 vlak. Die O3X3 vlak is
⊥ op die VV. Die O4X4 vlak is ⊥ op die O3X3 vlak.
Copy figure 49 onto an A4 sheet. The front and top views
of A, B, C and D are given. Find the projections of the
points in the new projection planes given as O1X1, O2X2,
O3X3 and O4X4. The O1X1 plane is ⊥ to the HP. The O2X2
plane is ⊥ to the O1X1 plane. The O3X3 plane is ⊥ to the
VP. The O4X4 is ⊥ to the O3X3 plane.
O3
X4
b’
O4
c’
X3
a’
O
X
d
d’
c
b
a
O1
O2
X1
X2
Figuur 49: Punte en Lyne
Figure 49: Points and Lines
7.11
Die voor- en bo-aansig van die lyne AB en CD word in
figuur 50 getoon. Die O1X1 vlak is loodreg op die HV. Die
O1X1 lyn maak ’n hoek van 80◦ met die OX lyn. Die koördinate van die punte word hieronder getoon. Teken die
aansig van die lyne in die O1X1 vlak. Bepaal die peilpunte
van die lyn AB op die HV, VV, en die O1X1 vlak.
The front and top views of the lines AB and CD are given
in figure 50. The O1X1 plane is perpendicular to the HP.
The O1X1 lines makes an angle of 80◦ with the OX line.
The coordinates of the points are shown below. Complete the view of the lines in the O1X1 plane. Determine
the tracepoints of the line AB on the HP, VP and the O1X1
plane.
A(10; 52; -7), B(75; 27; -20), C(20; 24; 16), D(62; 33; 25)
7.12 Herhaal vraag 7.11, maar laat die O1X1 lyn nou ’n hoek
van 70◦ met die OX lyn maak en laat dit die OX lyn by x=2 sny. Bepaal hierdie keer die pylpunte van die lyn CD op
die HV, VV en O1X1 vlak.
Repeat question 7.11, but let the O1X1 line make an angle
of 70◦ with the OX line and let it intersect the OX line at
x=-2. Determine the trace points of the line CD on the HP,
VP and the O1X1 plane.
144
X1
Oefeninge/Exercises
a’
d’
b’
c’
b
a
O
X
O1
c
d
Figuur 50: Punte en Lyne
Figure 50: Points and Lines
7.13 Teken die voor- en bo-aansigte van lyne AB en CD. Vind
die peilpunte van die twee lyne en tabuleer die koördinate daarvan in ’n blok.
Draw the front and top views of lines AB and CD. Find
the trace points of the two lines and tabulate their coordinates in a block.
A(17; 8; 16), B(60; 35; 28,5), C(0; 35; 35), D(75; 12,5; 12,5)
7.14 Punt A en B is die endpunte van ’n lyn. Die projektore van
A en B is 100 mm van mekaar. A is in die 3de kwadrant en
lê 10 mm van die HV en 30 mm van die VV. B is 60 mm bo
die HV en 70 mm agter die VV. Teken die lyn. Neem aan
die x-waarde van A is 0 en dat B regs van A lê. In watter
kwadrant is B? As die lyn die HV sny, tabuleer die koördinate van die snypunt. As die lyn die VV sny, tabuleer die
koördinate van die snypunt. Teken die linkeraansig van
die lyn AB.
Point A and B are the endpoints of a line. The projectors
of A and B are 100 mm apart. A is in the 3rd quadrant
and lies 10 mm from the HP and 30 mm from the VP. B is
60 mm above the HP and 70 mm behind the VP. Draw the
line, assuming that the x value of A is 0 and that B lies to
the right of A. In what quadrant is B? If the line intersects
the HP, Tabulate the coordinates of the intersection point.
Draw the left side view of the line AB.
7.15 Teken die volgende 3 punte. Sit ’n NPV in wat die 3 punte
bevat. Teken die punte in die NPV.
Draw the following 3 points. Put in a NPP that contains
these three points. Draw the points in the NPP.
A (0;0;10) B (40;50;50) C (80;-20;90)
7.16 Teken die volgende 3 punte. Sit ’n NPV in wat die 3 punte
bevat. Teken die punte in die NPV.
Draw the following 3 points. Put in a NNP that contains
these 3 points. Draw the points in the NPP.
A(0; 0; 10); B(30; 10; 50); C(60; 20; 40)
7.17 Herhaal vraag 7.16 met die onderstaande punte.
Repeat question 7.16 with the points below.
A(0; -10; -20); B(40; 0; 60); C(60; 5; -40)
145
Oefeninge/Exercises
7.18 Teken die punte hieronder. Sit ’n NPV loodreg met die
HV, wat ’n hoek van 50◦ met die VV maak en deur die oorsprong gaan, in. Teken die projeksies van die punte in
hierdie vlak. Sit nou ’n NPV loodreg met die eerste NPV
in. Hierdie vlak maak ’n hoek van 20◦ met die HV en bevat
die punt C. Teken die projeksies van die punte in hierdie
vlak.
Draw the points below. Put in a NPP perpendicular to
the HP, making and angle of 50◦ with the VP and passing
through the origin. Draw this projection. Now, put in a
NPP perpendicular to the first NPP. This plane makes an
angle of 20◦ with the HP and contains point C. Draw this
projection.
A(10; 30; 40); B(20; -40; 60); C(80; -20; -40); D(100; 70; -50)
7.19 Herhaal vraag 7.18, maar nou bevat die tweede NPV die
punte A en B.
Repeat question 7.18, but now the second NPP contains
points A and B.
7.20 Teken die punte A,B,C en D soos hieronder gegee en sit ’n
nuwe projeksie vlak (NPV) in. Die NPV is loodreg op die
HV en maak ’n hoek van 60◦ met die VV. Die NPV sny die
OX by die oorsprong. Teken die punte in die NVP.
Draw the points A,B,C and D given below. Put in a new
projection plane (NPP). The NPP is perpendicular to the
HP and makes an angle of 60◦ with the VP. The NPP intersects the OX at the origin. Draw the points in the NPP.
A (-10;40;30) B (30;-40;-30) C (60;55;-45) D (90;-10;25)
7.21 Herhaal die vorige vraag, maar nou is die NPV loodreg op
die VV en maak ’n hoek van 60◦ met die HV.
Repeat the previous question, but now the NPP is perpendicular to the VP and makes an angle of 60◦ with the
HP.
7.22 Teken die voor- en bo-aansigte van die lyne AB en BC,
waar die koördinate van A, B en C hieronder gegee word.
Bepaal dan:
(a) die ware lengte van elke lyn;
(b) die hoek wat AB met die VV maak;
(c) die hoek wat BC met die HV maak;
(d) die hoek ABC. (Wenk: bepaal die ware lengte van
|AC|)
Draw the front and top views of the lines AB and BC,
where the coordinates of A, B and C are given below. Then
determine:
(a) the true length of each line;
(b) the angle that AB makes with the VP;
(c) the angle that BC makes with the HP;
(d) the angle ABC. (Hint: determine the true length of
|AC|)
A (0; 17; 25), B (60; 55; 10), C (115; 25; 55).
7.23 Teken die voor- en bo-aansigte van die hartlyne AB, CD
en EF. Hierdie hartlyne verteenwoordig die drie pype in
’n chemise aanleg. Die pype het ’n deursnee van 8 mm
en moenie mekaar kruis nie. Voltooi dan ’n bo-, voor- en
linkerhand-syaansig van die pype. Die eindpunte van die
pype mag vryhand geteken word. Geen versteekte detail
nie.
Draw the front and top views of the centre lines AB, CD
and EF. These centre lines represent the lengths of three
8 mm diameter (chemical plant) pipes that must not intersect one another. Complete the top, front and lefthand side views of the pipes. The endpoints of the pipes
may be drawn freehand. No hidden detail.
A(0; 61; 23), B(67; 61; 23), C(0; 103; 46), D(67; 37; 32), E(49; 110; 7), F(0; 23; 60)
7.24 Met die inligting soos hieronder gegee, teken die voor- en
bo-aansigte van die driehoek ABC en bepaal θ AC .
Wenk: Teken eers die aansigte van B; bepaal dan die ware
lengte van BC (|BC|); gebruik nou (|BC|) om die aansigte
van C te teken; bepaal dan θ AC .
|AB|=90;
A(70; 35; 10);
φ AB = 30◦ ;
B(0; ?; > z A );
For the information given below, draw the top and front
views of the triangle ABC and determine θ AC .
Hint: First determine the views of B; then the true length
of BC (|BC|); now use |BC| to determine the views of C;
then finally determine θ AC .
φBC = 25◦ ;
C(?; 0; > z B );
146
θBC = 50◦ ;
B is higher than both A and C
B is hoër as beide A en C
Oefeninge/Exercises
7.25 ’n Maspaal van ’n radio antenne word met drie kabels AC,
AD en AE geanker. Die mas is 20 m hoog en van bo gesien
is die hoeke tussen die kabels almal 120◦ . Kabel AC maak
’n hoek van 45◦ met die grond sowel as die vertikale vlak.
D is 5 m verder van die paal as C en E is 3 m nader aan die
paal as C. Bepaal die ware lengte van elke kabel sowel as
die hoeke wat kabels AD en AE met die grond maak.
A radio antenna AB is anchored with three cables AC, AD,
and AE. The mast is 20 m high, and when seen from the
top the angles between the three cables are all 120◦ . Cable AC makes and angle of 45◦ with the ground and also
with the vertical plane. D is 5 m further from the mast
than C, and E is 3 m nearer to the mast than C. Determine
the true length of each cable as well as the angle that cables AD and AE make with the ground.
7.26 ’n Waarnemer is op ’n boot wat in die rigting N100◦ (d.i.
teen ’n hoek van 100◦ kloksgewys van Noord, op die horisontale vlak). Die waarnemer sien ’n vliegtuig die rigting N30◦ , 10 km van die boot (gemeet langs die siglyn)
en ’n inklinasiehoek van 30◦ . Nadat die boot 2 km afgelê
het, sien die waarnemer die vliegtuig in die rigting N300◦ ,
5 km weg en teen ’n inklinasie hoek van 60◦ . Bepaal die
ware afstand wat die vliegtuig afgelê het asook die vlugrigting en die hoogte verskil. Skaal 1:100 000.
Wenk: Neem Noord as die rigting van die negatiewe Z-as,
dan is Suid in die rigting van die positiewe Z-as, Oos is
in die rigting van die positiewe X-as, ens. Begin met die
boot by ’n posisie van Z=+7 km, wanneer op ’n A4 tekenvel gewerk word.
An observer is on a boat that is sailing in a direction
N100◦ (i.e. at an angle of 100◦ clockwise from North, on
the horizontal plane). The observer sees an aeroplane at
an angle of N30◦ , 10 km from the boat (measured along
the line of sight) and at an inclination angle of 30◦ . After the boat has covered 2 km, the observer sees the aeroplane at an angle of N300◦ , 5 km away at an inclination
angle of 60◦ . Determine the true distance that the aeroplane has covered, as well as the flight direction and the
height difference. Scale 1:100 000.
Hint: Take North in the direction of the negative Z-axis,
then South is in the direction of the positive Z-axis, East
is in the direction of the positive X-axis, etc. Start with
the boat at a position for which Z=+7 km, when drawing
on an A4 sheet.
7.27 Gegee die inligting hieronder en dat B regs van A is. C is
nader aan die vertikale vlak as A en laer as B. Teken die
voor- en bo-aansigte van ABC en vind θ AC .
Given the information below and that B is to the right of
A. C is nearer to the vertical plane than A and lower than
B. Draw the front and top views of ABC and find θ AC .
|AB| =90
A(0;0;z)
φ AB = 30◦
|BC|=80
B(x;y;0)
θBC = 25◦
|AC|=50
θ AB = 45◦
7.28 Punt B is 60 mm verder as A langs die X-as; A is 20 mm
bo die HV en 30 mm voor die VV; B is 40 mm bo die HV
en 40 mm voor die VV. C lê onder die HV en die lyne |AC|
en |BC| is onderskeidelik 60 mm and 70 mm lank. C lê
voor die VV. Die ware hoek wat BC met die horisontale
vlak maak is 45◦ . Bepaal die projeksies van die driehoek.
Tabuleer die koördinate van C. Toon alle konstruksies.
Point B is 60 mm further along the X-axis than point A; A
is 20 mm above the HP and 30 mm in front of the VP; B
is 40 mm above the HP and 40 mm in front of the VP. C
lies below the HP and the lines |AC| and |BC| are 60 mm
and 70 mm long respectively. C lies in front of the VP. The
true angle that BC makes with the horizontal plane is 45◦ .
Determine the projections of the triangle. Tabulate the
coordinates of C. Show all constructions.
7.29 Punt B is 60 mm verder as A langs die X-as; A is 20 mm bo
die HV en 30 mm voor die VV; B is 40 mm bo die HV en
40 mm voor die VV. C lê in die HV en die lyne |AC| en |BC|
is onderskeidelik 60 mm and 70 mm lank. C lê voor die
VV. Bepaal die projeksies van die driehoek. Tabuleer die
koördinate van C. Toon alle konstruksies.
Point B is 60 mm further along the X-axis than point A; A
is 20 mm above the HP and 30 mm in front of the VP; B
is 40 mm above the HP and 40 mm in front of the VP. C
lies in the HP and the lines |AC| and |BC| are 60 mm and
70 mm long respectively. C lies in front of the VP. Determine the projections of the triangle. Tabulate the coordinates of C. Show all constructions.
7.30 Gegee die inligting hieronder en dat C verder van die vertikale vlak is as A en laer is as B. Toon alle konstruksielyne.
(a) Vind die koördinate van B en C.
(b) Vind die HVPL en VVPL van die valk van driehoek
ABC.
(c) Bepaal die ware hoek tussen die twee pyllyne.
Given the information below and that C is further from
the vertical plane than A and is lower than B. Show all
construction lines.
(a) Find the coordinates of B and C.
(b) Find the HPTL and VPTL of the plane of the triangle
ABC.
(c) Determine the true angle between the two trace
lines.
A(0; 15; 26); B(30;+ ; 0); C(75;+ ;+ ); |AB|=60; |BC| =58; θBC = 12◦ .
147
Oefeninge/Exercises
7.31 Teken die voor- en bo-aansig van die driehoek ABC. Bepaal die posisie van ’n punt X wat op die skewe vlak ABC
lê en wat dieselfde afstand van al drie die sye van die driehoek is. Hoe ver bo die HV is punt X? Toon al die projeksies van punt X.
Draw the front and top view of the triangle ABC. Determine the location of a point X on the oblique plane
ABC which will be equidistant from the three edges of the
plane. How far above the HP is point X? Show point X in
all views.
A(0; 45; 40); B(25; 5; 0); C(50; 20; 25)
7.32 Teken die voor- en bo-aansig van die driehoek ABC. Wat
is die helling van die vlak wat ABC bevat? Bepaal die deursnee van die grootste sirkel wat op die vlak geteken kan
word en wat beperk word deur ABC.
Draw the front and top view of the triangle ABC. What
is the slope of the plane containing ABC? Determine the
diameter of the largest circle which could be drawn on
the plane as limited by ABC.
A(0; 10; 10); B(40; 60; 40); C(80; 30; 30)
7.33 Twee rioleringspype AB en CB kom bymekaar by ’n mangat by punt B. Punt A is 35 m noord, 10 m oos van B en
30 m bo B. Punt C is 20 m noord, 60 m wes van B en 15 m
bo B. Gebruik ’n geskikte skaal. Teken die twee rioleringspype. ’n Nuwe rioleringspyp moet in die vlak ABC lê en
moet begin by punt D wat 30 m reg wes van A lê. Gebruik
slegs twee aansigte en bepaal punt D in die vooraansig.
Gebruik soveel aansigte as nodig en bepaal die lengte van
elke rioleringspyp wat by punt B aansluit. Wat is die helling van die vlak?
Two sewer lines AB and CB converge at a manhole B.
Point A is 35 m north, 10 m east of B and 30 m above B.
Point C is located 20 m north, 60 m west of B and 15 m
above B. Use a suitable scale. Draw the two sewer lines. A
new sewer line is to be located in the plane ABC and beginning at the point D which is located 30 m due west of
A. Using two views only, locate point D in the front view.
Using as many views as necessary, determine the length
of each sewer line that converges at B. What is the slope
of the plane?
7.34 Twee pype word voorgestel deur die lyne AB en CD. Bepaal die warelengte, helling en rigting van die korste afstand tussen die twee pype.
Two pipelines are represented by lines AB and CD. Determine the true length, slope and bearing of the shortest
distance between the two pipelines.
A(0; 30; 60); B(90; 10; 45); C(10; 10; 10); D(100; 30; 35)
7.35 Punt A, 1550 m bo seevlak, is die ingang van ’n myntonnel wat N60◦ O loop teen ’n afwaartse helling van 30%. ’n
Tweede tonnel ingang by punt B lê 50 m suid, 110 m oos
van A en 50 m laer as A. Die tweede tonnel se rigting is
N45◦ O teen ’n opwaartse helling van 35%. Gebruik ’n geskikte skaal. Bepaal die ware lengte, rigting en gradiënt
van die kortste verbindingstonnel. Wat is die hoogte bo
seevlak by beide ente van die verbindingstonnel? Toon
die voorgestelde verbindingstonnel in al die aansigte.
Point A, at elevation 1550 m, is the portal of a mine tunnel
which bears N60◦ E on a downward grade of 30%. A second tunnel entrance is located at point B which is 50 m
south, 110 m east of A and 50 m below A. This second tunnel bears N45◦ E on an upward grade of 35%. Use a suitable scale. Determine the true length, bearing, and grade
of the shortest connecting tunnel. What would be the elevation at both ends of the connecting tunnel? Show this
proposed connecting tunnel in all views.
7.36 AB en CD is die hartlyne van twee 75 mm deursnee pyplyne. Punt B is 2 m oos, 3 m noord van A en 2 m onder
A. Punt C is 1.5 m oos, 1.5 m noord van A en op dieselfde
hoogte as A. Punt D is 1 m wes, 3 m noord van A en 2 m
onder A. Wat is die spasie, indien enige, tussen die twee
pype?
AB and CD are the centerlines of two 75 mm diameter
pipe lines. Point B is located 2 m east, 3 m north of A and
2 m below A. Point C is located 1.5 m east, 1.5 m north of
A and at the same elevation as A. Point D is located 1 m
west, 3 m north of A and 2 m below A. How much clearance, if any, is there between the two pipes?
7.37 Bepaal die snylyn tussen die twee vlakke ABC en DEF.
Toon versteekte detail. Wat is die rigting van die snylyn?
Determine the intersecting line between the two planes,
ABC and DEF. Show hidden lines. What is the bearing of
the intersecting line?
A(25; 25; 160); B(50; 85; 110); C(75; 50; 135); D(25; 45; 130); E(45; 75; 160); F(75; 45; 110)
148
Oefeninge/Exercises
7.38 ’n Gebuigde plaat, ABCD, is as volg geplaas: Punt B is
10 cm oos, 15 cm noord van A en 25 cm laer as A. Punt C is
40 cm oos, 10 cm noord van A en 15 cm laer as A. Punt D is
35 cm oos, 10 cm suid van A en 5 cm bo A. Die “buig lyn”is
AC. ’n Kabel, XY, moet deur die gebuigde plaat gaan. Punt
X is 10 cm noord, 8 cm wes van A en 10 cm laer as A. Punt
Y is 40 cm oos, 5 cm suid van A en 5 cm laer as A. Gebruik
’n geskikte skaal. Deur slegs twee aansigte te gebruik,
bepaal die punt, of punte, waar die kabel deur die plaat
gaan.
A bent plate, ABCD, is located as follows: Point B is 10 cm
east, 15 cm north of A and 25 cm below A. Point C is 40 cm
east, 10 cm north of A and 15 cm below A. Point D is
35 cm east, 10 cm south of A and 5 cm above A. The “bend
line”is AC. A cable, XY, must pass through the bent plate.
Point X is 10 cm north, 8 cm west of A and 10 cm below
A. Point Y is 40 cm east, 5 cm south of A and 5 cm below
A. Use a suitable scale. Using two view only, determine
the point, or points, where the cable will pass through the
bent plate.
7.39 Twee vlakke ABC en DEF sny mekaar. Punt B is 6 cm
noord, 4 cm oos van A en 6 cm laer as A. Punt C is 4 cm
suid, 12 cm oos van A en 4 cm laer as A. Punt D is 2 cm
suid, 4 cm oos van A en 6 cm laer as A. Punt E is 6 cm
noord, 2 cm oos van A en 2 cm laer as A. Punt F is 1 cm
noord, 10 cm oos van A en 5 cm laer as A. Bepaal die snylyn. Toon versteekte detail. Wat is die rigting van die snylyn?
Two planes ABC and DEF intersect each other. Point B is
6 cm north, 4 cm east of A and 6 cm below A. Point C is
4 cm south, 12 cm east of A and 4 cm below A. Point D is
2 cm south, 4 cm east of A and 6 cm below A. Point E is
6 cm north, 2 cm east of A and 2 cm below A. Point F is
1 cm north, 10 cm east of A and 5 cm below A. Determine
the line of intersection. Show hidden lines. What is the
bearing of the intersection line?
7.40 Twee driehoeke ABC en PQR sny mekaar op die lyn MN.
Bepaal die koördinate van die punte M en N, asook die
ware hoek tussen die vlakke waarin die driehoeke lê.
Wenk: gebruik nuwe projeksievlakke nadat die snylyn bepaal is.
Two triangles ABC and PQR intersect each other along
the line MN. Determine the coordinates of points M and
N, as well as the true angle between the two planes in
which the triangles are located. Hint: Use new projection
planes after the intersection line has been found.
A(30; 60; 50); B(10; 30; 10); C(70; 5; 5); P(0; 10; 45); Q(10; 65; 15); R(80; 30; 15)
7.41 Vind die snylyn KL tussen die vlakke PQRS en TUVW, en
bepaal ook die kleinste hoek tussen die vlakke.
Find the line of intersection KL between the planes PQRS
and TUVW and also the smallest angle between the
planes.
P(5; 21; 65); Q(0; 44; 46); R(35; 27; 6); S(41; 5; 25);
T(46; 0; 53); U(7; 14; 35); V(20; 49; 8); W(59; 34; 21);
7.42 Die driehoek ABC het ’n gat DEF daarin. Voltooi die boaansig van die gat.
(a) sonder om van ’n nuwe projeksievlak gebruik te
maak;
(b) deur van ’n nuwe projeksievlak gebruik te maak.
Triangle ABC has a hole DEF in it. Complete the top view
of the hole.
(a) without using a new projection plane.
(b) by using a new projection plane.
A(10; 24; 46); B(83; 87; 12); C(99; 11; 85); D(62;21; ?); E(57; 57; ?); F(87; 46; ?)
7.43 ’n Ertslaag word deur die driehoek PQR op die voor- en
bo-aansigte aangedui. AB stel ’n tonnel voor wat na die
ertslaag toe gegrawe moet word. Wat is die hoek (α) waarmee die tonnel die ertslaag sal tref? Gestel die punt R op
die ertslaag lê 1000 m onder die oppervlakte, op watter
diepte sal die tonnel die erstlaag ontmoet? Hoe ver moet
die tonnel nog verleng word? Skaal: 1 mm = 100 m.
An ore deposit layer is represented by the triangle PQR in
the front and top views. AB represents a tunnel that has
to be dug towards the ore deposit layer. What is the angle
(α) that the tunnel makes with the ore layer? Assuming
that point R on the deposit layer lies 1000 m under the
surface, at what depth will the tunnel meet the ore deposit layer? What is the distance that the tunnel must be
lengthened by to reach the ore deposit layer? Scale: 1 mm
= 100 m.
P(0; 13; 16); Q(45; 3; 27); R(21; 43; 52); A(51; 38; 7); B(35; 27.5; 17.5)
149
Oefeninge/Exercises
7.44 Vind die voor- en bo-aansigte van die lyn wat deur punt
P gaan en beide lyn AB en CD sny.
(a) Deur CD as ’n punt te projekteer en vind die korste
afstand tussen AB en CD.
(b) sonder hulpprojeksies deur gebruik te maak van
die punt waar AB deur die driehoek PCD sny.
(Wenk: gebruik die benadering wat toegepas word
om die snylyn tussen twee driehoeke te vind.)
Find the front and top views of the line that goes through
point P and intersects both lines AB and CD.
(a) By projecting CD as a point, and find the shortest
distance between the lines AB and CD.
(b) Without auxiliary projections, by using the point
where AB penetrates the triangle PCD. (Hint: use
the approach applied for finding the intersection
between two triangles.)
A(57; 13; 15); B(0; 48; 55); C(36; 5; 48); D(10; 35.5; 8); P(80; 46; 45)
7.45 Die lyn AM is stel ’n vertikale paal 8 m voor. Dit word in
plek gehou met drie anker drade wat as volg geplaas is:
AC is 10 m lank, het ’n 45◦ helling en ’n rigting van N60◦ W.
Anker tou AD is 8 m lank, maak ’n hoek van 30◦ met die
paal en het ’n rigting van N20◦ E. Die draad van E is 2 m
van die bokant van die paal daaraan vasgemaak, dit wys
reg suid en maak ’n hoek van 30◦ met die paal. As punt
E 0.5 m hoër is as B, wat is die ware lengte van die draad
vanaf punt E? Gebruik ’n geskikte skaal. Teken die anker
drade in die voor- en bo-aansig.
Line AM is a vertical pole 8 m high. It is held in position by
three guy wires which are located as follows: AC is 10 m
long, has a 45◦ slope and a bearing of N60◦ W. Guy wire
AD is 8 m long, makes an angle of 30◦ with the pole and
has a bearing of N20◦ E. The wire from E is fastened to the
pole 2 m from the top, is due south and makes an angle
of 30◦ with the pole. If point E is at an elevation of 0.5 m
above B, what is the true length of the wire from E? Use
a suitable scale. Locate the three wires in the front and
plan views.
7.46 Die twee gelykbenige pote van ’n landmeter se driepoot
is as volg geplaas relatief tot die skietlood: Been A staan
N30◦ W en het ’n helling van 30◦ ; Been B staan 1 m reg
oos van die loodlyn en teen dieselfde helling as die beginpunt. Die skietloodgewig raak die beginpunt by ’n vertikale afstand van 1.22 m onder die bokant van die lyn.
Been C staan S45◦ W en het ’n helling van 45◦ . Gebruik
’n geskikte skaal. Bepaal die warelengte van bene A, B en
C. Watter hoek maak been B met die skietlood? Toon die
bene in beide die bo- en vooraansig.
The 2 equal legs of a surveyor’s tripod are located in their
relationship to the plumb line as follows: Leg A bear
N30◦ W and has a slope of 30◦ ; Leg B is extended 1 m due
east of the plumb line and at the same elevation as the
bench mark. The plumb bob touches the benchmark at a
vertical distance of 1.22 m below the top of the line. Leg C
bears S45◦ W and has a slope of 45◦ . Use a suitable scale.
Determine the true length of legs A, B and C. What angle
does leg B make with the plumb line? Show the legs in
both the plan and front views.
7.47 ’n Produksielyn voerbelt seksie begin op die tweede vloer
van ’n fabriek. Die begin van die voerband is 7.32 m wes,
2.44 m en 4.88 m bo die ander end van die voerband. As
die skuins afstand van die tweede vloer tot die vals plafon
van die eerste vloer 3.43 m is, en die voerband 0.915 m bo
die tweede vloer begin, wat is die ware lengte, helling en
rigting van die voerband? Hoeveel van die voerband sal
uitsteek bo die tweede vloer en ook onder die vals plafon
van die eerste vloer?
An assembly line conveyor section begins on the second
floor of a manufacturing plant. The start of the conveyor
is located 7.32 m west, 2.44 m north and 4.88 m above the
other end of the conveyor. If the sloping distance from
the elevation of the second floor to the false ceiling of the
first floor is 3.43 m, and the conveyor starts 0.915 m above
the elevation of the second floor, what are the true length,
slope and bearing of the conveyor? How much of the conveyor will protrude above the second floor and below the
first floor ceiling?
8 Vlakke
Planes
8.1 Die pyllyne van skewevlakke word in figuur 51 gegee.
Vind, vir elke geval, die hoeke wat die vlak met die verwysingsvlakke maak, nl θ en φ.
The trace lines of oblique planes are given in figure 51.
Find, for each case, the angles that the plane makes with
the reference planes, i.e. θ and φ.
8.2 ’n Lyn deurdring vlakke A en B. Die lyn is parallel aan
beide die vertikale en die horisontale vlakke. Bepaal die
voor- en bo-aansigte van punte P en Q waar die lyn deur
vlakke A en B, onderskeidelik, dring. Sien figuur 52.
A line penetrates planes A and B. The line is parallel to
the vertical and horizontal planes. Determine the front
and top views of points P and Q where the line penetrates
planes A and B, respectively. See figure 52.
8.3 Bepaal die (ware lengte) loodregte afstand van die punt P
na die gegewe vlak asook die koördinate van die punt Q
wat in die vlak geleë is. Sien figuur 53.
Determine the (true length) perpendicular distance from
the point P to the given plane and also the co-ordinates
of the point Q that is located in the plane. See figure 53.
P(50; 55; 39); Q(60; ?; 25)
150
Oefeninge/Exercises
TL
HP
O
O
X
HP
TL
Q
X
T
VP
VP
T
X
L
L
P
B
A
O
Figuur 51: Vlakke
Figuur 52: Vlakke
Figure 51: Planes
Figure 52: Planes
45°
0
60
°
VP
TL
1
HP
TL
2
TL
VP
p’
X
15°
HPT
50
°
O
X
30°
°
60
O
L2
HP
TL
1
90
q
TL
VP
p
Figuur 53: Vlakke
Figuur 54: Vlakke
Figure 53: Planes
Figure 54: Planes
151
Oefeninge/Exercises
8.4 Punt P lê 40 mm weg van en bokant vlak 1. Bepaal sy
afstand vanaf vlak 2 met gebruik van ’n NVVPL. Sien figuur 54.
A point P lies 40 mm away from and above plane 1. Determine its distance from plane 2 using a NVPTL. See figure 54.
P(26; ?; 15)
8.5 Voltooi die projeksies van punt P op die vlak wat voorgestel word deur die pyllyne in figuur 55.
Complete the projection of point P on the plane represented by the trace lines in figure 55.
p'
TL
VP
TL
VP
p
X
O
X
O
HP
TL
HP
TL
p'
TL
HP
X
O
VP
T
L
Figuur 55: Projeksies van ’n Punt in ’n Vlak
Figure 55: Projections of a Point in a Plane
8.6 Voltooi die pyllyne in figuur 56. Gegee: Die punt P lê in
die vlak.
Complete the trace lines in figure 56. Given: Point P lines
in the plane.
8.7 Teken die snylyn tussen die vlakke wat voorgestel word
deur hul pyllyne in figuur 57.
Draw the intersection lines of the planes represented by
their trace lines in figure 57.
8.8 Vind α, die ware hoek tussen lyn CD en die vlak getoon
(figuur 58), asook die ware lengte van die deel van die lyn
CD bokant die vlak.
Find α, the true angle between the line CD and the plane
shown (figure 58), and also the true length of the part of
the line CD above the plane.
G(100; 0; 0); C(40; -40; -25); D(90; 30; -50)
8.9 Bepaal die ware hoek tussen die twee vlakke, wat deur hul
onderskeie peillyne beskryf word in figuur 59.
Determine the true angle between the two planes that are
given by their individual trace lines in figure 59.
8.10 Verbind AB en BC en beskou dit as twee reguit lyne wat
mekaar by punt B sny.
(a) AB en BC lê in dieselfde vlak. Vind die hoek θ van
die vlak en ook die hoek wat die HVPL en die VVPL
met die OX maak.
(b) Vind die hoek ABC.
Connect AB and BC and show them as two straight lines
that cross at point B.
(a) AB and BC lie in the same plane. Find the angle θ
to the plane and also the angle that the HPTL and
VPTL make with the OX.
(b) Find the angle ABC.
A(50; 12; 25); B(75; 50; 12); C(125; 20; 50)
152
Oefeninge/Exercises
p'
p
X
O
X
O
HP
TL
HP
TL
p
p'
p
X
O
VP
T
L
p'
Figuur 56: Bepaal Pyllyne vanaf die Projeksies van ’n Punt
Figure 56: Determine Trace Lines from the Projections of a
Point
8.11 Gegee in figuur 60: Die voor- en bo-aansigte van ’n vliegtuigkajuit se vensters. Vraag: Bepaal die hoek tussen die
kajuitvenster vlakke K en L en ook die hoek tussen L en
M. Tabuleer die antwoorde.
Given in figure 60: The front and top views of an aircraft’s
cabin windows. Question: Determine the angle between
the cabin window planes K and L and also the angle between L and M. Tabulate your results.
8.12 Bepaal waar die lyln AB in figuur 61 deur die vlak gaan en
gee die koordinate van die punt. Bepaal die hoek wat die
lyn AB met die vlak maak.
Determine where the line AB in figure 61 goes through
the plane and give the co-ordinates of that point. Determine the angle that line AB makes with the plane.
A(7, -21, 0); B(32, -43, 14); G(68, 0, 0)
8.13 ’n Vlak word voorgestel deur twee pyllyne. Die vertikale
vlak pyllyn is gegee deur lyn AB and die horisontale vlak
pyllyn is gegee deur lyn AC. ’n Tweede vlak word gegee
deur die vertikale vlak pyllyn deur lyn DE en die horisontale vlak pyllyn deur lyn DF. Vir elk van die volgende gevalle, teken die pyllyne en vind die snylyn tussen die twee
vlakke.
A plane is represented by two trace lines. The vertical
plane trace line is given by AB and the horizontal plane
trace line by AC. A second plane is given by the vertical
plane trace line DE and the horizontal plane trace line
DF. For each of the following cases, draw the trace lines
and find the intersection between the two planes.
(a) A(0; 0; 0); B(50; 40; 0); C(30; 0; 60); D(100; 0; 0); E(50; 40; 0); F(30; 0; 60)
(b) A(0; 0; 0); B(50; -10; 0); C(30; 0; 60); D(100; 0; 0); E(50; -20; 0); F(30; 0; 50)
(c) A(0; 0; 0); B(50; 40; 0); C(30; 0; -10); D(100; 0; 0); E(150; -20; 0); F(130; 0; 40)
8.14 Vind die ware hoek tussen die pyllyne in vraag 8.13.
Find the true angle between the trace lines in question
8.13.
8.15 Twee lyne AB en CD en punt P word gegee in figuur 62.
AB en punt P lê in die skuins vlak. Bepaal met gebruik
van ’n NPVPL watter lyn deur P sal ook albei gegewe lyne
sny. Gee die koördinate van die snypunte.
Two lines AB and CD and point P are given in figure 62.
AB and P are in the oblique plane. Determine using
a NPPTL what line goes through point P and also both
lines. Give the coordinates of the cutting points.
A(47; 22; 20); B(96; 96; 5); C(91; 72; 39); D(121; 34; 29); P(72; 42; 24)
153
Oefeninge/Exercises
1
TL
2
HP
TL
1
1
1
H PT L
HPT
TL
2
L2
O
X
VPTL2
O
HP
TL
1
X
X
O
HP
T
L2
VP
TL
1
Figuur 57: Snylyn tussen Twee Vlakke voorgestel deur hul
Pyllyne
Figure 57: Intersection Line between Two Planes Repre-
sented by their Trace Lines
154
L2
HPTL1
1
TL
HP
T
HP
O
1
2
TL
VP
TL
VP
X
L2
VP
HP
T
TL
VP
X
VPT
L1
HP
T
L2
HP
TL
1
VPTL VPTL2
1
O
L2
X
O
HP
T
VP
VP
T L2
TL
VP
Oefeninge/Exercises
8.16 ’n Televisie antenne steek bo die dak van ’n huis uit. Die
vlak ABCD beskryf die dak en die bokant van die antenne,
punt X, is soos hieronder geplaas. As ’n ankerdraad vanaf
punt X loodreg tot die dak is, hoe lank moet die ankerdraad wees? Hoe lank is die vertikale antenne? Toon die
ankerdraad in al die aansigte.
A television antenna extends above the roof of a house.
The roof plane ABCD and the top of the antenna, point X,
are located as shown below. If a supporting brace extending to point X is perpendicular to the roof, how long must
the brace be? How long is the vertical antenna? Show the
brace in all views.
A(75; 87; 162); B(150; 87; 187); C(162; 37; 150); D(87; 37; 125); X(112; 106; 144)
8.17 Die vlak ABC en punt X is gegee. Die vlak en punt X se
koördinate is hieronder gegee. Bepaal die warelengte en
rigting van die kortste lyn van X na die vlak wat terselfde
tyd ’n helling (hoek met die horisontale vlak) van 45◦ het.
Bepaal eers die pyllyne van die vlak ABC en gebruik ’n
NVVPL om die konstruksie te doen.
Plane ABC and point X are given. The plane and point X
are located as shown below. Determine the true length
and bearing of the shortest line from X to the plane having a slope (angle with horizontal plane) of 45◦ . First determine the trace lines of the plane ABC and use a NPPTL
to do the construction.
A(40; 30; 45); B(50; 10; 55); C(60; 25; 40); X(55; 30; 50)
8.18 Twee vlakke word deur hul pyllyne beskryf in figuur 63.
Die vlakke is parallel. Bepaal die kortste afstand tussen
hulle.
30°
O
X
55°
O
45° 6
0°
L
G
X
T
HP
H
PT
L2
30°
15
°
1
TL
P
V
2
TL
HPT
VP
VP
TL
Two planes are given by their trace lines in figure 63.
The planes are parallel. Determine the distance they are
apart.
L1
92
Figuur 58: Vlakke
Figuur 59: Vlakke
Figure 58: Planes
Figure 59: Planes
f’d’
0
O a
b’
d
b
c
G
a’
K
a
g
h
e
M
b’
b
L
f
Figuur 60: Vliegtuig Vensters
Figuur 61: Vlakke
Figure 60: Aircraft Windows
Figure 61: Planes
155
VP
TL
30°
45°
h’
a’
PT
L
e’c’
H
g’
X
Oefeninge/Exercises
VP
TL
AB
P
b’
c’
p’
d’
a’
47°
X
37°
O
HP
TL
AB
b
a
P
p
d
c
Figuur 62: Vlakke
T
HP
VP
TL
1
Figure 62: Planes
L2
55°
45°
O
X
20
55°
T
HP
VP
TL
2
45°
L1
Figuur 63: Vlakke
Figure 63: Planes
156
Oefeninge/Exercises
8.19 Die vlak ABC en punt Y se koördinate is gedeeltelik hieronder gegee. Teken die vlak ABC wat loodreg op die lyn
YA is. Tabuleer die koördinate van B en C.
Plane ABC and point Y are located as shown below. Draw
the plane ABC perpendicular to line YA. Tabulate the coordinates of B and C.
A(30; 20; 80); B(35; x; 65); C(40; 35; x); Y(10; 30; 65)
8.20 Vlak ABC en punt X word asvolg beskryf: Punt B is 100
oos, 80 noord van A en 120 laer as A. Punt C is 160 oos, 20
suid van A en 50 laer as A. Punt X is 60 oos, 20 suid van
A en 140 laer as A. Bepaal die kortste horisontale afstand
tussen punt X en die vlak ABC. Teken die pyllyne van ABC
en gebruik ’n NVVPL. Vind ook die kortste lyn van X na
die vlak wat ’n helle van 35% het. Bepaal die rigting van
die lyn vanaf X.
Plane ABC and point X are given as follows. Point B is 100
east, 80 north of A and 120 below A. Point C is 160 east,
20 south of A and 50 below A. Point X is 60 east, 20 south
of A and 140 below A. Determine the shortest horizontal
distance from point X to the plane ABC. Construct trace
lines of ABC and use a NPPTL. Also find the shortest line
from X to the plane having a grade of 35%. Determine the
bearing of each line from X.
8.21 Figuur 64 toon ’n oorgangstuk waar aan ’n kabel vasgemaak moet word wat deur punt X gaan. Bepaal die korste
moontlike afstand vanaf punt X na die naaste vlak van die
oorgangstuk. Toon die verbindingslyn in al die aansigte.
Figure 64 shows a transition piece which is to be connected by a cable passing through point X. Determine the
shortest possible distance from the point X to the nearest face of the transition. Show the connecting line in all
views.
x’
x
Figuur 64: Vlakke
Figure 64: Planes
8.22 In die volgende vrae, bepaal die ware lengte of die korste
afstand van punt X na die vlak ABC. Gebruik die metode
van pyllyne, bepaal dus eers die pyllyne van ABC en gebruik dan ’n NVVPL.
(a)
(b)
(c)
(d)
In the following questions, determine the true length of
the shortest distance from the point X to the plane ABC.
Use the method of trace lines, i.e. first determine the
trace lines of ABC and then use a NPPTL.
A(20; 30; 50); B(30; 10; 65); C(35; 25; 40); X(25; 5; 45)
A(10; -30; 60); B(25; 20; -60); C(40; 40; 55); X(30; -45; 70)
A(10; 15; 55); B(30; 5; -40); C(45; -30; 60); X(20; 5; 65)
A(15; 30; 40); B(20; 30; 50); C(45; 35; 60); X(35; 30; 75)
157
Oefeninge/Exercises
9 Developments
Developments
9.1 Teken op ’n A3 tekenvel die ontwikkeling van die konus
in figuur 65. Die boonste deel van die konus is afgesny.
Skaal 1:1.
On an A3 sheet, draw the development of the cone in figure 65 of which the top part is removed. Scale 1:1.
9.2 Maak ’n ontwikkeling van die helfte van die skewe piramiede wat getoon word in figuur 66. Let daarop dat die
bopunt van die piramiede afgesny is deur ’n skewe vlak
(d.i. ’n vlak wat ’n hoek maak moet die horisontale vlak
en loodreg is met die vertikale vlak). Neem TA en TE as
die snylyn. Skaal 1:1 op ’n A4 tekenvel.
Make a development of the one half of the oblique pyramid in figure 66 that is shown. Note that the top of the
pyramid is cut by an oblique plane (i.e. a plane that has
an angle with the horizontal plane and is perpendicular
to the vertical plane). Make TA and TE the cut-line. Scale
1:1 on an A4 sheet.
t’
55
75
90
29
φ80
c’g’
d’f’ e’
75
°
°
50
a’ b’h’
g
h
23
f
e
a
b
t
d
c
80
Figuur 65: Ontwikkeling van ’n Afgesnyde Konus
Figuur 66: Skewe Piramiede
Figure 65: Development of a Truncated Cone
Figure 66: Oblique Pyramid
The drawing in figure 67 shows the transition piece between two channels that look like flattened pipes. Develop one quarter of the object from AO to E4 only. Scale
1:1 on an A4 sheet.
9.3 Die tekening in figuur 67 toon ’n oorgangstuk tussen twee
kanale wat soos platgedrukte pype lyk. Ontwikkel slegs
een kwart van die voorwerp van AO tot E4. Skaal 1:1 op ’n
A4 tekenvel.
e’
45
a’
4’
o’
φ45
e
4
o
a
Figuur 67: Oorgangstuk
Figure 67: Transition Piece
158