# SSS 2 (BASIC 11) Technical Drawing FIRST TERM

SSS 2 (BASIC 11)
Technical Drawing

FIRST TERM e – LEARNING NOTES

SCHEME OF WORK

WEEK TOPICS

THEME: GEOMETRICAL CONSTRUCTIONS

Revision of last term’s work.

Special Curves:
Locus definition, applications and construction of special curves. E.g. Parabola, hyperbola, involutes, epicycloids, etc.

Special Curves: (b) Construction of special curves using different methods.

(b) Plotting the loci of points on link mechanism.

THEME: DEVELOPMENT OF GEOMETRICAL SOLIDS

True Shapes:
(a) True shapes of truncated cone, prisms, pyramids and cylinders.
(b) Development of models.

Intersection of Solids:
Line and curve of intersection
(i) two intersecting cylinders.
(ii) a cylinder intersecting a cone.
(iii) two intersecting prisms.
(iv) two intersecting pyramids.
(v) Surface development of intersecting solids.

MID – TERM: ASSIGNMENT

THEME: PICTORIL DRAWING

Perspective Drawing:
Uses and types of perspective drawings
(i) one point perspective
(ii) two points perspective.

Perspective Drawing:
(b) Identification and location of main terms in perspective drawing e.g. horizon, station point, vanishing points and planes.

Revision.

Examination.

WEEK 1

REVISION OF LAST TERM’S WORK.

WEEKS 2 & 3

TOPIC:SPECIAL CURVES: Locus: Definition, Application and Construction of special curves. E.g. Parabola, Hyperbola, Involute, and Epicycloid.
CONENT:
Locus definition
Construction and Application of Parabola and Hyperbola
Construction and Application of Involute and Epicycloid.
Sub-Topic 1: Locus definition and construction of Parabola and Hyperbola.
What is Locus- Locus is the path traced by the point when moving in accordance with a definite rule.
Construction of Parabola.
Parabola is the locus of a point which moves so that its distance from a fixed point called Focus and a fixed straight line called Directrix are always equal.
A
X P

H Q axis
Focus
Directrix
Vertex A
The ratio PF/PX= 1 is called eccentricity.
It means = (distance from focus)/(perpendicular distance from directrix)
Therefore, the eccentricity of a parabola is unity or 1.
To draw the curve by the locus method, first position the focus F and the directrix. The vertex V may be found by bisecting FH, thus satisfying the law for the curve. Draw lines such as AA, parallel to the directrix, and by drawing an arc centred at F, of radius HQ, to cut AA, fix two points on the parabola. Other points may be found in the same way and the parabola drawn with a French curve. Note that at V the parabola crosses the axis at 900. The ratio PF:XP, which is equal to unity, is the eccentricity of the curve.

Exercise 1: To Construct Parabola using Circumscribing Rectangular method.

Procedures:
Use the Span and Rise to lay out a rectangle
Divide it in two by the axis of symmetry
Divide the sides of the two rectangles so obtained into the same number of equal divisions.
Draw lines parallel to the axis through one set of divisions and join each of the other sets to O
Where the parallel through 2 cuts the line O to 2 is a point on the parabola, and so on.
Exercise: Construct a Parabola using rectangular method, given the Span and Rise 70mm and 45mm respectively.
Procedures: Follow the guide lines stated above.
The circumscribing rectangle method may also be used to draw an ellipse. In this case the sides of the rectangle are equal in length to the axes of the ellipse. The divisions along the span in the parabola above are laid out along the major axis, and the divisions on the rise along half the minor axis. The parallel lines above now radiate from the lower end of the minor axis and the points are plotted as before.
The parabola has the property that if a source of light, sound or heat is placed at the focus, rays reflected from the parabola form a parallel beam. This property is made use of in hand torches, car headlamps and some electric fires. Conversely, rays falling on the parabola from outside will be reflected to the focus. This explains the shape of some radar and radio dishes which have the form of a paraboloid.
Hyperbola.
The hyperbola is the locus of a point which moves so that the ratio of its distances from the focus and directrix is constant and greater than 1. The eccentricity is thus greater than 1.

To draw the curve for a given ratio, say 3/2, first position the focus and directrix as shown above. The vertex is found by dividing AF into five equal parts, when V will be two divisions from A. to fix further points, draw any line such as AA, parallel to the directrix, and with centre F and radius AH times the eccentricity, draw arcs to cut AA above and below the axis.
Lines which are tangents to the hyperbola at infinity are called Asymptotes. When the asymptotes are at right angles, the hyperbola is called rectangular or equilateral.
The ellipse, parabola and hyperbola are conic sections. i.e., they may be produced by passing planes through a right cone.

Exercise 2: To draw a hyperbola when given the Ordinate, the Vertex and the Transverse.

Procedures:
Indicate O and mark the ordinate each side of it OA and OB
At O draw a perpendicular to AB and mark the Vertex V and the Transverse axis VV1
Divide OA OB and the two perpendiculars at A and B into the same number of equal parts. Radiate lines as shown. The intersection of the lines gives the points for the hyperbolic curves.
Exercise 3: To draw a hyperbola when given the Foci and the Transverse Axis.

Procedures:
Draw the given foci F and F1, and transverse axis VV1.
Take any convenient unit and mark it off from F to give points A, B, C, D,E.
Use radius VA from one focal point and radius V1A from the other focal point to obtain the four points on the hyperbolic curve. The other points are obtained by using radii B, etc.
NB:A tangent at a given point P on the curve is obtained by joining P to the focal points F and F1 and then bisecting the angle FPF1. The bisector is the required tangent.

EVALUATION
Construction a hyperbola given the span, height and the transverse axis e.g. span 90mm, height= 70mm, transverse axis= 70mm.
Construct a parabola given the span and the height e.g. span=80mm, height =50mm.
What is hyperbola?
What is parabola?
List any two method of constructing parabola.
Sub- Topic: Construction of Involute.
Introduction: What is an Involute?
If a straight line is rolled round a polygon without slipping, points on the line will trace out involutes.
Construction of Involutes of a square

When the line rolls round the square it will pivot on successive corners, so the locus consists of a series of circular arcs whose radius increases as each corner becomes the pivot.

Construction of the Involute of a circle.

Procedures:
The circle may be thought of as a polygon with an infinite number of sides.
First draw the base circle and in some convenient position a line equal in length to its circumference.
Divide both into the same number of equal parts (say twelve).
From the points on the circle draw tangents as shown, to represent successive positions of the generating line. When this line is in the position of the tangent at point 1 it will have rolled one-twelfth of the circumference.
Take one-twelfth from the straight line along which the base circle circumference was set out.
Mark it off along the tangent at point 1, thus obtaining a point on the curve.
Repeat the procedure stepping off two-twelfths, three-twelfths of the circumference on succeeding tangents.

CYCLOID
The cycloid, epicycloid, hypocycloid and involute have an important application in the profiles of gear teeth. In the cycloidal system the profile consists of parts of an epicycloid and hypocycloid, above and below the base circle, whilst the rack tooth profile is part of a cycloid.
Construction of Cycloid
A cycloid is the path of a point on the circumference of a circle rolling along a straight line.

Procedures:
Draw horizontal lines from 0,1,2,3,- – – – – – .
Divide the line from 0 into the same 12 units as the circumference of the circle and erect perpendiculars.
With the intersection of the perpendiculars 1, 2, 3, 4, 5 – – – – – – – and the horizontal circle centre line as centres, draw the same radius as the first circle as the first circle.
Circle with centre perpendicular 1 cuts horizontal line 1 and circle with centre perpendicular 2 cuts horizontal line 2, etc. a curve drawn through these points gives the required cycloid.
EPICYCLOID
An epicycloid is the locus of a point on the circumference of a circle as it rolls without slipping around the outside of a larger circle.

Procedures:
Find φ in order to centralize the figure using φ= r/R X 360.
Where r= radius of the small circle (30mm) R= radius of the larger circle (90mm).
Draw the larger arc, also draw the vertical radius and share the φ half ways to indicate radius, 6, and 12.
Draw the small circle and divide it into 12 equal parts. Also step off same interval on the larger arc.
Draw the centre arc and also draw arcs 1, 2, 3, 4, – – – – – 12.
With centre point C1 and radius r (small circle), cut across arc 1 repeat the same for C2, C3 – – – – – C12. A curve drawn through these points gives the required Epicycloid.
EVALUATION:
What is a cycloid?
Define an Epicycloid.
GENERAL EVALUATION:
Which of the following is the true shapes of the section S-S below if the object is spherical.

(a) circle (b) ellipse (c)triangular (d)sphere (e) parabola

A conic section whose eccentricity is 4/3 is a/ an ——–. (a) Hyperbola (b) Parabola (c) Sphere (d) Circle (e) Ellipse.
The head lamp of a car is constructed by applying the principle of ———— (a) Hyperbola (b) Parabola (c) Involute (d) Cycloid (e) Ellipse
The profile of a screw thread is constructed using the principle of a/ an ——-
(a) Helix (b) Hyperbola (c) Involute (d) Cycloid
Essay
What is Epicycloid?
List at least 2 method of constructing parabola.
What is cycloid?
WEEKEND ASSIGNMENT:
Draw a cycloid when the radius of a circle is 40mm
Draw an Involute when the radius of a circle is 25mm.
REFERENCE BOOK:
Engineering Drawing Book 1 3rd Edition By Parker M.A and Pickup F. Pp 38-42.
Technical Drawing For School Certificate and GCE by J.N. Green. Pp 135-143
Element of Technical Drawing, By Akanno. Pp 75-80.

WEEK 4
CONTENT:
Plotting the loci of points on link mechanism
Mechanism is the combination of various links so paired that the motion is completely constrained. Mechanisms are used to transform the motion, e.g. Rotary to reciprocating, reciprocating to rotary, rotary to oscillating, etc. In other words, by available mechanisms motion is converted into useful motion.
Examples 1: OBA is a simple slider crank chain. OB is a crank of 30 mm length. BA is a connecting rod of 90 mm length. Slider A is sliding on a straight path passing through point O. Draw the locus of the mid-point of the connecting rod AB for one complete revolution of the crank OB.

Example 2: Fig. 4.13 shows a mechanism in which OB is a crank of 30 mm length revolving in clockwise direction. BC is a rod connected to the crank at the point B by turning pair and rod BC is constrained to pass through the guide at O1 called trunnion. Draw the loci of point P and C for one revolution of the crank. The point P is 30 mm from B on the rod BC. Length of BC is 150 mm. Point O1 is 80 mm on the right and 15 mm below the point O.

CLASS WORK:
Shown is a link mechanism. A and C are fixed points. Crank AB rotates in an anti-clockwise direction. Link BDEF is pin jointed at B and D. Link CD oscillates about point C.
(i) Using a line diagram to represent the mechanism, plot the locus of point F for one revolution of the crank AB.

Procedures:
Draw the given crank (circle) and divide it into 8 or 12 equal parts.
With the compass, open 100mm (length of link) and position at each of the division draw arc across the horizontal centre line, also draw line for each.
For each line drawn, determine the mid-point O. the curve drawn round the points is the locus at point O
Example1: In the figure the crank AB rotates clockwise about A and the crank CD rotates anticlockwise about C. the cranks are joined by the link BDF. Plot to a scale of 1mm to 1cm the loci of points E and F for one revolution of AB. AB and DC are 45cm, BD is 120cm and BE and DF are 30cm.

B
120mm

NB: The curves drawn in red are the paths traced by the points specified on the cranks.
GENERAL EVALUATION
If the crank OK in the diagram below moves anticlockwise the link KP with (A) rotate at Q (B) Slide toward K (C) Slide toward P (D) Remain stationary (E) Rotate at P.

ESSAY.
In the crank mechanism shown below AB oscillates as the crank CD rotates about D. Construct the locus of point P for a complete revolution of crank CD. DC=40mm, AB=60mm, BC=100mm, BP=60mm.

WEEKEND ASSIGNMENT:
Draw the locus of a point O. when the crank BC undergoes one complete revolution. The point A is made to reciprocate horizontally.

In the crank mechanism shown below, OA rotates about centre O, while CB is pivoted at C and oscillate as shown. Plot the locus of point P for all the possible of CB.

REFERENCE BOOK:
Engineering Drawing Book 1 3rd Edition By Parker M.A and Pickup F. Pp 38-42.
Technical Drawing for School Certificate and GCE by J.N. Green. Pp 135-143
Element of Technical Drawing, By Akanno. Pp 82-85.

WEEK 5
TOPIC:True Shapes
CONTENT: (a) True shapes of truncated cone, prisms, pyramids and cylinders.
(b) Development of models.
Sub-topic 1: True shapes of Sections
When an object, a geometric figure for instance, is cut with an incline plane, the object does not present its true shape of the cut surface. The true shape can only be seen if the observer stays at angle of 900 to the cut surface or the rotated to be at right angles to the observer. On this basis, can true shape be worked out in geometric constructions?
RECTANGULAR PRISM
To find the true shape of a rectangular prism cut at an angle with x-x as the cutting plane.

Procedures:
Project lines from all points and at angle 900 to the cutting plane x-x as shown below.
At any convenient point, draw a line DA parallel to the cutting plane line xx.
From A as centre, radius AB cut the projected lines to get the true shape. Note that AB is in its true length.
TRIANGULAR PRISM

Procedures:
Draw the given views of the prism.
Project lines perpendicular to the cutting plane line x-x from all points.
At any convenient point, locate the centre line Ay.
Transfer By and Cy to get the true shape.

HEXAGONAL PRISM

CYLINDER

Procedures:
Draw the given views.
Project lines perpendicular to the cutting line x-x from all points.
At any points, draw lines 1-7 parallel to x-x.
Transfer all the points to get the true shapes.
SQUARE PYRAMID

HEXAGONAL PYRAMID

CONE
The procedures are similar to those of cylinders and prisms.

WEEK 6

TOPIC: INTERSECTION OF SOLID
CONTENT: (a) Line and curve of intersection
(b) Two intersecting cylinders
(c) A cylinder intersecting a cone.
(d) Two intersecting prisms.
(e) Two intersecting pyramids.

Interpenetration

First of all we are going to break Interpenetration down into its most basic element, where we find the point where a line penetrates a flat surface or plane. Below is the dimensioned diagram you need to draw in order to practice this exercise.

Generally it is important in Technical Drawing to label points of importance, but in Interpenetration it is doubly important. This will become more noticeable as the drawings become more complicated but it is good practice to start the way you mean to continue.

From the Plan and Elevation you should be able to see that you have a plane surface at an angle to the Vertical Plane (VP). You are told a line L, 50mm above the Horizontal Plane (HP) and 45mm out from the VP penetrates the plane surface. You are asked to find where this penetration occurs.

You cannot see where the line penetrates the plane in the Elevation, but in the Plan you can see where the penetration occurs. This leads us to the solution of the problem.

Remember that when you are setting up this question draw all of your lines very lightly and label the points. Below you can see how to complete the question.

Set up your drawing according to the given dimensions, remembering to label your points and draw lightly.

In the Plan you can see where the line intersects with the plane. Project this point onto the line in the Elevation.

You should be able to see from the Elevation and Plan where the line is behind the plane from the direction that you are looking at it. You do not need to draw in hidden detail unless you are asked to do so in your question, but we do it here for clarity.
First of all we are going to break Interpenetration down into its most basic element, where we find the point where a line penetrates a flat surface or plane. Below is the dimensioned diagram you need to draw in order to practice this exercise.

Generally it is important in Technical Drawing to label points of importance, but in Interpenetration it is doubly important. This will become more noticeable as the drawings become more complicated but it is good practice to start the way you mean to continue.

Technical Drawing – Interpenetration – Solid Penetration
In this section we will look at how we draw the penetration of a plane surface by a solid square based prism. Below you can see the problem that you should lay out one your page if you wish to work along or attempt this question first. The labelling has been left out of the question. You should try and do this yourself before looking at the first step of the solution.

You should be able to see from the orthograhic projection that we have a solid square based prism parallel to the HP penetrating a plane surface which is at an angel to the VP. Below are the stages in the completion of this question.

You may not have labeled the points of importance the same but do not worry about this so long as you have the labeling format correct.

In the plan view it is easy to see where the square based prisms edges 1, 2, 3 and 4 intersect with the plane surface. Project these points into the Elevation as shown above.

We now show the lines which are hidden, (again you do not have to do this unless you have to). Part of the plane surfaces edge ac is behind the square based prism and the square based prism goes behind the plane surface before it reaches the edge bd.

Your completed drawing should look like this.

Your completed drawing without construction or hidden lines would look like the above drawing.
Technical Drawing – Interpenetration – Square into Square
In this the first section where we are looking at the Interpenetration of actual solids we will examine the simple exercise of finding the lines of Interpenetration between two square based prisms. Below you can see the setup drawing for this problem.

You can draw this yourself if you wish but you will have to come up with your own dimensions. To the left in grey there is a section view of the smaller square based prism which is penetrating the larger square based prism. We need this to help us rotate the Plan view into the Elevation view and maintain the correct dimensions. (You could just use measurements in the Elevation but it is good practice to do it as above as you will need this skill later where measurements will not work so easily.)

You can see the labelling that we have used here in this drawing and why we have used the section of the smaller square based prism to provide our Elevation drawing. Remember to draw the lines lightly at this stage. As you can see there is no need to label every corner of the prisms, just so long as we can recognise the edges.

It is obvious where the edges 1 and 3 of the small prism intersect with edge a of the larger prism and these are points we can use to start with. In the Plan you can see where edge 2 of the small prism intersects with the surface ab of the large prism. Project this point into the Elevation to find the point of intersection there. You will also notice that edge 4 of the small prism intersects the surface ad of the large prism along the same construction line. You can now draw in the lines of Interpenetration in the Elevation. Draw the lines following the section view, edge 1 joins to edge 2, joins to edge 3, joins to edge 4, and back to edge 1 again. You can only see the lines of Interpenetration joining edges 1, 2 and 3.

Line out your drawing and it should look like the drawing above.
Technical Drawing – Interpenetration – Square into Square Offset
Here we are going to look at the construction of the lines of Interpenetration between two square based prisms one of which is offset. Below you can see the problem which you can create yourself with your own dimensions.

You can see from the drawing that the horizontal square based prism (HSBP) is offset towards the back of the vertical square based prism (VSBP). Continue below to see how to do this problem.

In order to setup the question you need an End View as well as the Elevation and Plan. You will also need this End View later in order to find some of the points of Interpenetration. Do not forget to label the edges of the two solids.

For clarity we have removed the construction lines necessary to setup the question. You should not rub these out when you are doing problems.

From the Plan it is easy to see where the edges 1, 2, 3 and 4 of the HSBP intersect with the surfaces of the VSBP. Project these points into the Elevation to locate four of the points of Interpenetration. At this stage you should be beginning to see the value of labeling.

From the Plan we can see that the edge of the VSBP intersects the HSBP, but we cannot see where this intersection occurs. We need the End View for this. In the End View we can see that the edge a of the VSBP intersects the edge 1,2 of the HSBP at point 5, and the edge 2,3 of the HSBP at point 6. Project these points across to the Elevation.

Now you are ready to draw in the lines of Interpenetration in the Elevation. The simplest way to do this is to follow the order of the points in the End View. Draw these lines lightly first. 1 to 5 to 2 to 6 to 3 to 4 to 1. Now you need to determine which of these lines are hidden and which are visible from your point of view. If this is not clear the lines 5 to 1, 1 to 4, 4 to 3 and 3 to 6 are all behind the VSBP and so are hidden.

Finally line out your drawing remembering that the part of the edge a on the VSBP between points 5 and 6 no longer exists and so should not be lined in.
Technical Drawing – Interpenetration – Flat Surface Problems
Here are three problems that you should attempt in order to reinforce your learning so far. Click on the problem to see the solution.
Question 1

Determine the lines of Interpenetration between the hexagonal based prism and the square based prism.
Question 2

Show the lines of Interpenetration between these two square based prisms in the End Elevation to the left. The solution to this problem is shown without any of the question setup construction lines.
Question 3

Determine the lines of Interpenetration for the two square based prisms shown.
Example 1: A hexagonal pyramid, of base edge 50 mm and height 110 mm, rest on its base in H.P such that one pair of opposite base edges is parallel to V.P. Draw three views of the combination, showing the lines of intersection.

WEEK 7
MID- TERM BREAK
ASSIGNMENT

Q1. Two unequal pipes diameter 45 and 60 respectively intersect at 300 as shown below. Draw, full size, the following:
complete front elevation
plan;
development of part q with XX as the seam.

Q2. Draw the curve of interpenetration.

Q3.Shown is a link mechanism. A and C are fixed points. Crank AB rotates in an anti-clockwise direction. Link BDEF is pin jointed at B and D. Link CD oscillates about point C.
(i) Using a line diagram to represent the mechanism, plot the locus of point F for one revolution of the crankAB.

Q4. Project a new elevation of the solid from the plan, which will show the true shape of the surface A.

WEEK 8
TOPIC:Perspective Drawing

CONTENT:(a) Uses and types of perspective drawings
One point perspective
Two points perspective.

Sub-topic 1: Uses and types of perspective drawings
Naturally, distant objects appear smaller than their real sizes e.g. the stars, the sun, the moon etc. For instance, the farther we look down a long corridor in a building, the closer it appears to come. Perspective drawing therefore shows the pictorial representation of the apparent reduction in size of a distant object. Since architecture deals with large objects like buildings, perspective drawing becomes more useful in making architectural drawings realistic than either isometric or oblique drawing. Distant parts of a building are shown in perspective as tapering to a point, just as they appear to an observer of the actual building. Perspective projection can be in one point or two point perspective.

REFERENCE TEXTS
Elements of Technical Drawing for Senior Schools and Colleges by Osuji,U.S.APh.d and Akano,E.O. M.sc
Technical Drawing manual with solved past questions School curricular 1 & 2
PAST EXAMS QUESTIONS.
The foci of an ellipse are 110 mm apart. The minor axis is 70 mm long. Determine the length of the major axis and draw half ellipse by rectangle method and other half by concentric circles method. [ B.T.E.M.S. April 2000 (Civil)]
A circle of 50 mm diameter rolls along a straight line without slipping. Trace the path of a point on the circumference of the rolling circle for one complete revolution: Name the curve. [B.T.E.M.S April/May 1998]
A circle of 50 mm diameter rolls along the circumference of another circle of 150 mm diameter from outside. Trace the path of a point P on the circumference of the rolling circle for one complete revolution. Name the curve. [ B.T.E.M.S November 1999 (Electronics)]
The figure below shows three cylinder pipes (J,K,L) of same diameter 50 and of negligible thickness joined together.
Draw full size, the:
given view;
development of the pipe K using the shorter side as the seam.

Two views of a line are shown above.
Draw full size, the given views.
Determine the:
horizontal and vertical traces;
true angles of inclination to the horizontal and vertical planes; true length.