ENGINEERING DRAWING.

THE PLUMBING DRAWING HANDBOOK

AS OF THE UBTEB SYLLABUS.
1ST EDITION.

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LUBEGA ASHRAF
(VTD)
0752363485
DEDICATION.
I dedicate this book to you; the reader.

ACKNOWLEDGEMENTS.
I would like to thank the Almighty Allah, & the following people for the help they rendered and continue to render to me in the
development of this, and many other books of a similar kind.
1. Mr. & Mrs. Lubega Murshid (My dearest Parents)
2. Mr. Nkwanga David (Lugogo VTI)
3. Mr. Yonnah James Tubugwiha. (Ntinda VTI)
4. And you.

1 Lubega_Ashraf
Table of Contents
DEDICATION. ............................................................................................................................................................................... 1
ACKNOWLEDGEMENTS. ........................................................................................................................................................... 1
Introduction to Plumbing drawing. ................................................................................................................................................. 4
Drawing tools, care and use of drawing instruments. ................................................................................................................. 4
Scales and their application. ....................................................................................................................................................... 5
Drawing paper setting and lettering. ........................................................................................................................................... 6
Drawing of symbols for pipe networks and plumbing fixtures. ................................................................................................. 7
Lines............................................................................................................................................................................................ 9
Scales and their application. ..................................................................................................................................................... 11
Geometrical construction. ............................................................................................................................................................. 12
Construction of angles. ............................................................................................................................................................. 15
Triangles. .................................................................................................................................................................................. 20
Quadrilaterals. ........................................................................................................................................................................... 27
Development of polygons. ............................................................................................................................................................ 33
Inscribing & circumscribing triangles and other polygons. ...................................................................................................... 33
Enlargement and transformation of figures. ............................................................................................................................. 38
Circle. ........................................................................................................................................................................................ 48
Ellipse ....................................................................................................................................................................................... 51
Tangents and blending of arcs onto circles. .............................................................................................................................. 55
Development of shapes. ................................................................................................................................................................ 66
Parallel line development. ......................................................................................................................................................... 67
Radial line development. .......................................................................................................................................................... 73

2 Lubega_Ashraf
 Development of a funnel. .......................................................................................................................................................... 77
Interpenetration of solid figures. ................................................................................................................................................... 79
The interpenetration between two cylinders of the same diameter........................................................................................... 80
The interpenetration between two cylinders of the same diameter connected at angle 45°. .................................................... 81
Interpenetration between two cylinders of different diameters. ............................................................................................... 83
Interpenetration between two cylinders of different diameters and one is positioned at 45°. .................................................. 85
Principles of isometric projection. ................................................................................................................................................ 90
Isometric rectangle. ................................................................................................................................................................... 91
Isometric of a cuboid. ............................................................................................................................................................... 92
Isometric of a cylinder. ............................................................................................................................................................. 93
Non isometric lines. .................................................................................................................................................................. 94
Orthographic projection. ............................................................................................................................................................... 97
Orthographic views to Isometric projection. .............................................................................................................................. 104
Oblique projections. .................................................................................................................................................................... 107
3D sketches and views ................................................................................................................................................................ 110

3 Lubega_Ashraf
Introduction to Plumbing drawing.
Plumbing drawing is a branch of technical drawing, where we only handle drawings that are only applicable in the plumbing
world. Drawing is the accepted means of communication between the building team. All people on this team either make or
read sketches or drawings. Usually, ideas start as rough sketches, rough sketches are then refined and eventually produced and
we get a finished drawing that can be put to ground. This is typically what we are to discuss.
Drawing tools, care and use of drawing instruments.
There are a number of tools that can be used to produce drawings, but the most basic tools include;
 Geometrical set containing 30°/60° and 45° set squares, an aim ruler, a pair of compasses and dividers, a protractor,
sharpener, rubber, and technical drawing pencils i.e., a 2H for outlines and 5H for construction lines.
It is advisable you get bigger set squares and not just use the traditional small set squares from a mathematical set.

 Drawing board normally the board can accommodate an A2 paper, i.e., it should be a little bigger that the A2 paper.
 Tee square; it is advisable to get one with a transparent edge, it is used to draw horizontal lines and its on which other
drawing instruments for example setsquares are placed to draw lines at angles.
 Drawing paper; the size depends on the task at hand. There are various paper sizes.

ISO CODE PAPER SIZES (mm)

A0 841x1189
A1 594x841
A2 420x594
A3 297x420
A4 210x297

Care of drawing instruments.

a. Instruments should be cleaned prior to drawing to help produce clean drawing and also keep the drawing paper clean.
b. Care should be taken not to drop the instruments as they tend to break on impact with hard surfaces.
c. Compasses and dividers should be regularly tightened to improve their efficiency in producing accurate drawings.

4 Lubega_Ashraf
Scales and their application.
Scaling is a drawing method used to enlarge or reduce a drawing in size while keeping the proportions of the drawing the
same. Scales are generally expressed as ratios and the most common scales used are 1:1, 1:2, 1:5, and 1:10 for reducing and
possibly 2:1 for enlarging.

Scaling is used to either:

 reduce the drawing in size so that it will fit onto the page, or

 enlarge the drawing in size so that all required details are clearly visible.

The fraction of the ratio always dictates whether the scale reduces or enlarges the figure, i.e., if the fraction of the ratio is
greater than 1, then it’s an enlarging scale and the vice versa is true.

Drawings can be scaled up or down using either a calculator or a scale rule.

To scale a drawing using a calculator:

 divide the measurement by the scale if you want to reduce the drawing in size, or
 multiply the measurement by the scale if you want to increase it in size.

Example 1: Scaling down

 A 50mm line is to be drawn at a scale of 1:5 (i.e., 5 times less than its original size). The
measurement 50mm is divided by 5 to give 10mm. A 10mm line is drawn.

Example 2: Scaling up

 A 50mm line is to be drawn at a scale of 5:1 (i.e., 5 times more than its original size). The
measurement 50mm is multiplied by 5 to give 250mm. A 250mm line is drawn.

5 Lubega_Ashraf
Drawing paper setting and lettering.
Before any drawing work is done, the paper has to be set out with a boundary & title block as shown below.

Lettering/printing.
paper boundary
This is the presentation of
cell tape/clip to hold informational data on a drawing. It is
paper in place. drawing board an art of writing alphabetic letters and
numbers onto a drawing.
guidelines

ENGINEERS ALWAYS PRINT/LETTER BY HELP OF

GUIDELINES Lettering is an important part of a
30 70 drawing used to write explanatory
notes, dimensions, letters and other
NAME LUBEGA ASHRAF necessary information required for the
complete interpretation and
implementation of a drawing.
Take off some time and develop the
lettering skill. Remember to always
use guidelines when printing.
Requirements for good lettering.
a. The letters should be written in guidelines.
b. The letters should touch the guidelines.
c. The spacing between the letters should be constant.

6 Lubega_Ashraf
Drawing of symbols for pipe networks and plumbing fixtures.
Symbols for some plumbing fixtures.
The meaning of plumbing fixture is a part (such as a sink, toilet, faucet, etc.) that is attached to a system of pipes that carry
water through a building.
The following are some of the symbols for the common fixtures used in plumbing.

Double compartment
Bathtub Bidet Handicap lavatory
sink

Handicap Oval lavatory

Single bowl sink Urinal type 1
lavatory symbol - type 2

Wall hung
Urinal Water closet (toilet) Water closet (toilet)
water closet
type 2 symbol - type 1 symbol - type 2
(toilet)

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Shower
Hot water tank Drinking fountain Column shower symbol
head

Corner
Shower stall Eye wash symbol - type
Shower Eye wash symbol
symbol 2
symbol

Wash Wash fountain

fountain semi-circle
symbol symbol

8 Lubega_Ashraf
Lines.
 Visible Outlines, Visible. Edges: Type 01.2 (Continuous wide lines)
These are drawn to represent the visible outlines/ visible edges / surface boundary lines of objects and these should be
outstanding in appearance.

 Dimension Lines: Type 01.1 (Continuous

narrow Lines)
20
Dimension Lines are drawn to mark
dimension.

 Extension Lines: Type 01.1 (Continuous DIMENSION LINE

narrow Lines)
These are extended slightly beyond the
EXTENSION LINE.

10
respective dimension lines.
 Guide Lines: Type 01.1 (Continuous
Narrow Lines) GUIDELINE
Guide Lines are drawn for lettering and
should not be erased after lettering.

 Construction Lines: Type 01.1 (Continuous narrow Lines)

Construction Lines are thin and faint, drawn for constructing drawings and should not be erased after completion of the
drawing.

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 Hatching / Section Lines: Type 01.1 (Continuous Narrow Lines)
Hatching Lines are drawn for the sectioned portion of an object. These are drawn
inclined at an angle of 45° to the axis or to the main outline of the section.

HATCHING

 Break Lines: Type 01.1 (Continuous Narrow Lines with Zigzags)

Straight continuous narrow line with zigzags is used to represent break of an object.

 Dashed Narrow Lines: Type 02.1 (Dashed Narrow Lines). Hidden edges / Hidden outlines of objects are shown by
dashed lines of short dashes of equal lengths of about 3 mm, spaced at equal distances of about 1 mm.

 Center Lines: Type 04.1 (Long-Dashed Dotted Narrow Lines)

Center Lines are drawn at the center of the drawings symmetrical about an axis or both the axes. These are extended by
a short distance beyond the outline of the drawing.

centre-line

10 Lubega_Ashraf
Scales and their application.
Scaling is a drawing method used to enlarge or reduce a drawing in size while keeping the proportions of the drawing the
same. Scales are generally expressed as ratios and the most common scales used are 1:1, 1:2, 1:5, and 1:10 for reducing and
possibly 2:1 for enlarging.

Scaling is used to either:

 reduce the drawing in size so that it will fit onto the page, or

 enlarge the drawing in size so that all required details are clearly visible.

The fraction of the ratio always dictates whether the scale reduces or enlarges the figure, i.e., if the fraction of the ratio is
greater than 1, then it’s an enlarging scale and the vice versa is true.

Drawings can be scaled up or down using either a calculator or a scale rule.

To scale a drawing using a calculator:

 divide the measurement by the scale if you want to reduce the drawing in size, or
 multiply the measurement by the scale if you want to increase it in size.

Example 1: Scaling down

 A 50mm line is to be drawn at a scale of 1:5 (i.e., 5 times less than its original size). The
measurement 50mm is divided by 5 to give 10mm. A 10mm line is drawn.

Example 2: Scaling up

 A 50mm line is to be drawn at a scale of 5:1 (i.e., 5 times more than its original size). The
measurement 50mm is multiplied by 5 to give 250mm. A 250mm line is drawn.

11 Lubega_Ashraf
Geometrical construction.
Construction of a perpendicular line to another line.
1. Draw the line AB
2. At a convenient point c, draw a semi-circle as shown, and using a
convenient radius r, draw arcs using endpoints of the semi-circle as
centres, (as shown) that intersect at d. d
3. Join points c and d, hence

r
perpendicular line. r

A c B A c B
2
1

Construction of a perpendicular line to another line from a point P.

1. Draw the line AB and locate the point p.
2. Using p as centre, draw an arc, with convenient radius R that touches line AB twice as shown at points x and y
3. Using x and y as centres and any convenient radius r, draw arcs that meet at m & n as shown.
4. Join points p, m & n, hence perpendicular to a line from point p
p
p
m
R

A x y B
r
A B
2
1
n

12 Lubega_Ashraf
Constructing a line parallel to another at a distance d.
1. Draw a line AB as shown.
2. At any convenient points m & n, and radius d, draw arcs as shown
3. Draw a line tangential to the arcs as shown, hence parallel lines.
d

d
d

d
A m n B A m n B
1 2

Bisecting a line.
1. Draw line AB, and using points A & B as centres and any convenient radius r, draw arcs that meet at points m & n.
2. Join points m & n, hence the line AB has been bisected.

m m

r r r r

A B A B

1 2

n n

13 Lubega_Ashraf
Division of a line into parts.
i. Draw line XY with the required length.
ii. At any angle, draw another line below line XY, using your compass, divide the drawn line into the number of portions
line XY is to be divided into. In this case, the line is to be divided into 10 portions.
iii. Join point 10 to point Y.
Then use your instruments to slide line Y10 to the other points as they touch line XY
X Y
X Y

1
1
2
3 2
3
4
4
5
6
1
5
6
2
7
7
8
8
9
9
10
10

To divide a line in the ratio 2:3:5.

i. Draw the given line XY
ii. Draw XZ at any convenient angle and divide it into 10 equal parts (2+3+5)
iii. Join 10 to Y, draw line at points 5 and 2 that are parallel to line 10Y
iv. Line is divided in the given ratio.

X Y X Y

1 1
2 2
3 3
4 4
5 5
6 6
7
7
8
1 8 2
9
Z 9
10 Z
10

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Construction of angles.
An angle is the space between two convergent lines.
Construct geometrically angle 60° & 120°
 draw a horizontal line, select any convenient point a, using
a as centre and any convenient radius r, draw arc at b.
 with the same radius r, and b as center, draw another arc to
r intersect the first drawn arc.
 draw a line through the
60° intersection point of the
r
arcs and a, hence required
a b a b angle produced.
1 2 120°
Note: the other side of the angle gives us angle 120° as shown.
a b

Construct geometrically angle 30° & 150°. (Bisecting an angle)

30° is obtained by dividing angle 60° by 2, geometrically, this is done by bisecting angle 60°. This is illustrated below.
 construct angle 60°
 at any convenient radius R, draw arcs as shown.
R
 Draw line as shown to obtain the required angle.
150° is the other side of 30°.

R
150° 30°

a b a b

15 Lubega_Ashraf
Construct geometrically angle 15° & 165°.
 Construct angle 30° and bisect it to obtain 15°, 165° is the remaining angle.
Construct geometrically angle 90°.

 Draw horizontal line, with any convenient

radius r, mark off segment as shown.
 With convenient radius R, draw arcs to
R R intersect as shown.
 Draw line through intersection point of arcs
and centre of segment, = angle 90°
r
r

Construct geometrically angle 45° & 135°.

 This is constructed by bisecting angle 90°.

135°
45°

16 Lubega_Ashraf
Construct geometrically angle 105° & 75°.
 To construct 75°, construct 60° and then 15°(bisect the shown 30°), the sum of these gives us 75°
 The remaining angle is 105°.

30°

60°
105°
75°

1 2

17 Lubega_Ashraf
Chord of scales.
a. Draw the baseline AB of any convenient length. Erect a perpendicular at A.
b. With A as center and AB as radius, draw an arc to intersect the perpendicular at C
c. Use B as the center and BC as radius, draw an arc to cut AB produced at D.
d. Divide BC into 9 equal parts (first trisect the arc, then divide each chord into 3 equal parts as shown in the figure
below) with B as center, drop arcs (chord lengths) from these points onto AB extended. Number the division from 0° to
90° as shown.
C
C

30°
30°

A B C
D A B
1
2

30°

D A B
3

18 Lubega_Ashraf
 The chord can help us obtain angles such as 70°, 85°, etc., that have no
C 90 80
70 direct method of construction.
60

50

40
To construct an angle let’s say 70°.

30
1. Draw a horizontal line and at any point x, draw an arc with radius r
R
(0°-60°) from the chord of scales.
20 2. With compass at y and radius r (0°-70°), draw an arc that intersects
the first arc at z. join z to x. angle zxy = 70°. Use same procedure to
30°
10 construct any angle using the chord of scales.

90 80 70 60 50 40 30 20 10 0

r(0°-70°)

r(0°-60°)

70°

X Y

X Y

19 Lubega_Ashraf
Triangles.
A triangle is a plane figure bounded by three straight sides.
Forms of triangles.
a. Equilateral triangle; this has all sides equal.
b. Right angled triangle; this has one right angle and the side opposite to the right angle is called the hypotenuse.
c. Scalene triangle; this has all sides and angles unequal to each other.
d. Isosceles triangle; has two opposite sides and angles equal.
Let us look at the construction of some triangles. Ready? let’s go.
To construct an equilateral triangle when given the length of one side.
1) Draw line AB, equal to the given length
2) Adjust your compass to the length of the given side, use A and B as centre and make arcs respectively. The arcs will
intersect at C.
3) Join A to C and C to B, hence the required triangle.

C C

r r

A B A B

1 2

20 Lubega_Ashraf
To construct an isosceles triangle when given the perimeter and altitude.
1) Construct half the given perimeter FO
2) Erect a perpendicular at either O or F and mark off the altitude OR. (in this case F)
3) Join R to F and bisect it. The bisector will cut FO at X.
4) Produce FO so that OY is equal to XO
5) Join X to R and Y to R, RXY is the required triangle.

R R

F O F X O
HALF GIVEN PERIMETER
1 2

F X O Y

21 Lubega_Ashraf
To construct a triangle, when given the perimeter and the ratio of sides. (2:4:4)
1) Draw line AB equal to the given perimeter
2) Divide line AB into the required ratio (2:4:4).
3) With R as center and (A-R) as radius, construct an arc, likewise using Y as your center and (Y-B) as your radius,
construct an arc to intersect in Q
 The middle 4 is the base of the required triangle and QRY is the required triangle.

Q
AB = GIVEN PERIMETER

2 R 4 Y 3 2 R 4 Y 3
A B A B

1 1
2 2
3
3
4
4
5
5
6
7 6
8 7
9 1 8 2
10 9
10

To construct a triangle when given the base length, one base angle and the perimeter.
1) Construct the base length AS
2) Construct the given base angle at either A or S. in this cas A
3) Use A as your center, (perimeter-base length) as your radius and make an arc along the line that makes the base angle
with the given base length. The arc will intersect the line at G
4) Join G to S and bisect it, the bisector cuts A-G in F

22 Lubega_Ashraf
 5) Join F to S, AFS is the required triangle.

G G G

gth
en
el
as
-b
ter

F F
me
eri
np
ve
Gi

base angle

A A S A S
S
1 2 3

To construct a triangle when given two base angles (45º and 60 º) and perimeter.
1) Draw a horizontal line and mark off the given perimeter AB
2) Construct the base angles at end A and end B
3) Extend the lines that make the respective base angles to intersect at G
4) Bisect the base angles and extend the bisectors to intersect at H
5) Draw a parallel line to AG from H and project it to intersect the line AB at X.
6) Draw a parallel line to BG from H and project it to intersect the line AB at P.
7) XPH is the required triangle.

23 Lubega_Ashraf
 G G

°
60
45°

°
60
45°
A GIVEN PERIMETER B
A B

1 G 2

°
60
45°

X P 3
A B

To construct a triangle given the base length, altitude and vertical angle.
1) Draw a line AB equal to the base length
2) At either end A or B, construct the vertical angle 60º (angle GAB)
3) Erect a perpendicular at A on line AG to intersect with the bisector of AB at X
4) Use X as centre, AX/BX as radius and draw a circle.

24 Lubega_Ashraf
5) Construct a line parallel to AB at a height equal to the altitude, to intersect the circle at D and F. (D and F are
the vertices that can be joined to AB to produce the required triangle)
In this case D has been joined to A and B to produce the required triangle.

A B A B A B
°
60

1 2 3
G G
G

D F

A B

25 Lubega_Ashraf
To construct a triangle, when given two base angles (60° and 45°) and altitude.
1) Draw a horizontal line BF, at any convenient point (O) on the line construct a perpendicular.
2) Construct a line SD that is parallel to BF at a height that is equal to the altitude.
3) Construct the base angles at point G using SD as your reference line. Angle SGT (60°) and angle VGD (45°). (These
are alternate angles)
4) The line that makes 60° will intersect line BF at T and the
line that makes. G
S D

B F B O F
1 2

S G D

60

45°
°

B V O K F
3

26 Lubega_Ashraf
Quadrilaterals.
These are plane figures that are bounded by four straight lines.
Quadrilaterals any of the following forms;
 Square, this has all sides equal and all angles are right angles.
 Rectangle; has its opposite sides equal and all angles are right angles.
 Rhombus; has its sides and opposite angles equal, but contains no right angles.
 Parallelogram; has its opposite pairs of sides and opposite angles equal, but contains no right angles.
 Trapezium; has two sides parallel.
 Trapezoid; has four unequal sides.
 Kite; has its adjacent pairs of sides equal in length.

Let us look at some construction methods of quadrilaterals.

To construct a square when given the length of one of the sides.
1) Construct a horizontal line and mark off the given length AB
2) Erect a perpendicular at A, using A as center and AB as radius, draw an arc to intersect the perpendicular at D.
3) With B as centre and AB as radius, draw an arc, with D as centre and AB as radius, draw an arc to intersect the arc
from B at C.
4) Draw lines from D to C and C to B, ABCD is the required square.
D C D C
D

A B A B A B
1 2 3

27 Lubega_Ashraf
To construct a rectangle, when given the diagonal distance and the length of one of the sides.
1) Construct a horizontal line and mark off the diagonal length AC.
2) Bisect AC and use O as centre and AO as radius and construct a circle.
3) Using A as centre and the given length as radius, construct an arc to intersect the circle in B. do the at point C.
4) Join B to C, C to D, D to A and A to B. ABCD is the required rectangle.

B B
r

r
A O C A O C A O C

D D
1 2 3
To construct a trapezium given the lengths of the parallel sides, the perpendicular distance between them and one angle
(60°).
1) Draw a horizontal line and mark off the length of one of the parallel sides AB.
2) Construct the given angle at A to intersect the parallel line at C. the parallel line is at height equal to the perpendicular
distance.
3) Using C as centre and the remaining length of the parallel side as radius, draw an arc to intersect the parallel line at D.
4) Join D to B, ABDC is the trapezium.
C
C

A B A B A B

1 2 3

28 Lubega_Ashraf
To construct a parallelogram when given two sides and an angle (60°).
1) Draw a horizontal line and mark off AD equal to the length of the given side
2) Construct the given angle at A.
3) Using A as center and radius equal to the length of the other side, draw an arc to intersect the line that makes
the given angle at B, using D as center and AB as radius, draw an arc to intersect the previous one at C.
4) Join B to C and C to D, ABCD is the required parallelogram.
B
C
B B
C

A D A D A D
1 2 3
To construct a rhombus when given the diagonal distance and the length of the sides.
1) Draw a horizontal line and mark off the given diagonal distance AC.
2) Using A as center and the length of the given side as radius, draw an arc above and below the diagonal line,
do the same at center C, the arcs should intersect at B and E.
3) Join A to B, B to C, C to E, and E to A. ABCE is the required rhombus.
B
B

r r
r

C A C
A
A C

E E

1 2 3

29 Lubega_Ashraf
Regular Polygons.
Regular polygons have all sides equal.
Some common regular polygons include;
o Square – 4-sided figure.
o Pentagon – 5-sided figure.
o Hexagon – 6-sided figure.
o Heptagon – 7-sided figure.
o Octagon – 8-sided figure.
o Nonagon – 9-sided figure.
o Decagon – 10-sided figure.
Make research and add to this list.

To draw such regular polygons, there is a standard formula demonstrated below.

1. Draw a horizontal line and mark off the length of one side AB.
2. Draw a perpendicular bisector that meets AB at O. using O as center and OA as radius, draw the arc A4.
3. Using B as center and AB as radius, draw the arc A6.
4. Points 4 & 6 are centers for the 4 and 6 sided regular polygons.
5. Using these centers, draw circles as shown and mark off the length of the side. Join these to obtain the required
polygons.
Note; to draw a pentagon, divide the distance between points 4 & 6 to obtain point 5 as shown. Enjoy.

30 Lubega_Ashraf
 HEXAGON

PENTAGON

4 5

A O B
A O B

1 2

EXERCISE.
Draw a regular heptagon and nonagon of side 30mm.

31 Lubega_Ashraf
EXERCISE ON TRIANGLES.
1) Construct an equilateral triangle when given the length of one side as 45mm.
2) Construct a triangle when given;
a) The base length as 50mm, vertical angle as 50º and the altitude as 35mm.
b) The two base angles as 45º and 70º and the altitude as 40mm.
c) Perimeter as 125mm, base length as 50mm and base angle as 50º.
d) The base length as 50mm, vertical angle as 50º and the length of one side as 35mm.

EXERCISE ON QUADRILATERALS.
1. Construct a trapezium given that the parallel sides are 40mm and 72mm long and 25mm apart and one base
angle as 60º
2. Construct a square whose diagonal is 47mm.
3. Construct a rectangle given the diagonal as 50mm and the length of one side as 30mm.
4. Construct a parallelogram given two sides as 40mm and 90mm long and an angle of 60º between them.
5. Construct a rhombus given the given the diagonal as 56mm long and the length of one side as 35mm.

32 Lubega_Ashraf
Development of polygons.
Inscribing & circumscribing triangles and other polygons.

To inscribe a circle into a triangle. (Or any regular polygon).

1. Construct the given triangle ABC.
2. Bisect any two interior angles, i.e., angle BAC and ABC.
3. The bisectors will intersect in D.
4. Drop a perpendicular from D to line AB, in order to obtain the radius of the inscribed circle.
5. Using D as center and DF as radius, draw the circle.
C C C

1 2 3

D D

F F
B B B
A A A

To inscribe three equal circles in an equilateral triangle; each circle to touch one side and two other circles.
1. Construct the given equilateral triangle ABC.
2. Bisect any two interior angles so as to establish the center T of the triangle.
3. Draw a line from C passing through point T; these forms three equal triangles TAC, TCB, and TAB.
4. Choose one of the three small triangles; bisect one of the base angles to intersect the line from C at Z.
5. Point Z is the center of one of the required circles; draw a pitch circle (center T and radius ZT) to establish the other
two centers X and Y.
6. With center Z, X and Y and radius ZH, draw the required circles.

33 Lubega_Ashraf
 C
C C

1 2 3
X Y X D Y
D D

Z Z

A B A B A B

To inscribe three equal circles in an equilateral triangle; each circle to touch two sides and two other circles.
1. Construct the given equilateral triangle ABC.
2. Bisect any two interior angles so as to establish the center T of the triangle.
3. Draw a line from B passing through point T to G.
4. Choose one of the three small triangles; bisect angle CGT to intersect the line from CN at Y.
5. Y is the center of one of the required circles; draw a pitch circle (center T and radius TY) to establish the remaining
two centers X and Z.
6. With center Z, X and Y and radius TF, draw the required circles.
B B B

1
X 2 X
3

T T T

Z Y Z Y

A G C A G C C
A G

34 Lubega_Ashraf
To inscribe four equal circles in a square; each circle to touch one side and two other circles.
1. Construct the given square.
2. Draw diagonals XZ and WY. The diagonals will create four equal triangles; CXY, CYZ, CZW and CWX.
3. Inscribe the triangles. The pitch circle (centre CA) is used to establish other centers.
X Y X Y X Y

E E

C D C B D C B

A A

W Z W Z W Z

1 2 3

To inscribe four equal circles in a square; each circle to touch two sides and two other circles.
1. Construct the given square XYZW
2. Draw diagonals XZ and WY. Draw perpendiculars (AE & DB) passing through the intersection of the diagonals, this
forms four equal squares.
3. Choose one of the squares (EZBC) and draw another diagonal BE to intersect diagonal XZ at O.
4. Point O is one of the centers for the required circles. Draw a pitch circle (center C and radius CO) to obtain other
centers.
5. Draw the required circles using established centers (P, Q, O and S) and radius OT.

35 Lubega_Ashraf
 X Y X Y X Y

Q S Q S

C C C

O O
P P

W Z W Z W Z

To circumscribe a circle onto a triangle (or any regular polygon).

1. Construct the given triangle ABC.
2. Bisect at least any two sides AC and BC.
3. The bisectors will intersect in D (D is the center of the circle)
4. Using D as center and DC as radius, draw the circle.
C
C

D
D

B
A
B
A
B
A

1 2 3

36 Lubega_Ashraf
To escribe a circle to a given triangle (or any regular polygon).
1. Construct the given triangle ABC.
2. Choose any side of the triangle onto which the escribed circle is going to be constructed, in this case BC.
3. Extend or produce side AB to form an external angle, likewise side AC and bisect them.
4. The bisectors will intersect at O, which is the center of the required escribed circle.
5. Drop a perpendicular onto the extended side AC to get point H, using O as center and OH as radius, draw the circle.

C C

1
2

A B
A B

C C
O O

A B H A B H

3 4

37 Lubega_Ashraf
Enlargement and transformation of figures.

To construct a similar figure to the figure ABCDEF, but bearing a different base length AB’.
1. Draw the given polygon ABCDEF,
2. Extend the base AB, and mark off the new dimension AB’ and draw radial lines from A passing through point C, D, E
& F.
3. Draw a line from B’ parallel to BC to get C’, draw line from C’ parallel to DC to obtain D’. do the same for all other
sides as shown.

E' D'

E E
E D D
D
F'

C'
C F C
F C F

1 A B B' 3
A B A B B' 2

To construct a similar figure to the given figure ABCDEF, with the sides in the ratio of 2:5.
Let’s begin with understanding whether we are enlarging or reducing the given figure in the ratio.
First convert the given ration into a fraction, & if the fraction is greater than 1 then we are to enlarge but if less
than 1, then we are to reduce.
For the question at hand, you realize this is reduction.
Solution.

38 Lubega_Ashraf
 1. Draw the given polygon ABCDEF.
2. At A, draw a line away at any convenient angle as shown and mark off 5 equal arcs.
3. Join point 5 to B.
4. Slide line 5B to point 2 as shown to intersect base at B’. draw radial lines from A passing through points C, D, E & F.
5. Draw a line from B’ parallel to BC to get C’, draw line from C’ parallel to DC to obtain D’. do the same for all other
sides as shown.
E
D

F C C
F E' D'

F' C'

A B
1
2
3
A 1
2
B'
B E'
4
5 3 D'
4
5

2
1 E
D E
D
F'
If the ratio was, say 6:4, then the solution C'
would be an enlargement. (Because ratio as C C
F F
fraction is greater than 1)

A 1 B A 1 B B'
2 2
3 3
4 4
5 5
6
6

1 2

39 Lubega_Ashraf
To construct a similar figure to the given figure ABCDEF, with the area in the ratio of 5:3.
Ratio as a fraction is greater than 1, so its enlargement.
1. Draw the given polygon ABCDEF.
2. Draw a perpendicular line from B downwards, produce line AB from point B and with a convenient compass distance,
step off 5 equal arcs.
3. Draw radial lines from B through D, E and F.
4. Bisect A3 and draw a semi-circle to intersect the perpendicular from B at 3’.
5. Bisect A5 and draw a semi-circle to intersect the perpendicular from B at 5’.
6. Draw a line from 3’ to point A, slide this line to point 5’ to intersect the base AB at A’.
7. Draw a line at A’ parallel to AF to get F’. do the same for all other points.

E' D'

E E
D D
If the ratio was, say 3:5, solution
F' would be a reduction.
C' Try it out.
F C F C

A B1 2 3 4 5
A' A B
1 2 3 4 5

3'
3'
5'
5'

1
2

40 Lubega_Ashraf
To construct a similar figure to the given figure ABCDEF, with the area in the ratio of 3:5.
Ratio as a fraction is less than 1, so its reduction. (Borrow knowledge from previous solution)

E D
E D
D'
E'

C F C
F F'
C'

A B1 2 3 4 5
A A' B
1 2 3 4 5

3' 3'
5' 1 5' 2

To construct a square equal in area to the given regular hexagon ABCDEF.

1. Draw the given polygon ABCDEF.
2. Produce AB on both sides & step off the perimeter of the hexagon on it, i.e., the length of one side of the hexagon is
stepped off six times on the extended baseline.
3. Join 0 to G and G to 6, G06 is a triangle equal in area to the given regular hexagon. (Transform this triangle into a
square)
4. Construct a perpendicular at 0 & 6, the bisector of the altitude of the triangle GH will intersect the perpendicular at S &
J, giving us a rectangle.
5. Using 6 as centre and J6 as radius., draw an arc to intersect the extended base at Q. draw a semi-circle at points 0 & Q
as shown.
6. Extend J6 to meet the semi-circle at P as shown.
7. 6P is one of the sides of the required square. PMK6 is the required square.

41 Lubega_Ashraf
 E
D

G C
F

J
S

4 5 6 Q
0 1 2 3 H

E
D
P M

G C
F

J
S

K
4 5 6 Q
0 1 2 3 H

42 Lubega_Ashraf
To construct a square equal in area to a given irregular pentagon.
1. Draw the given polygon ABCDE.
2. First transform the irregular pentagon into a triangle of equal area. (Follow diagram to understand procedure.)
3. Transform the triangle into a rectangle of equal area (draw perpendiculars at the edges of the triangle and the
perpendicular bisector of the triangle’s altitude forms a rectangle.)
4. Transform the rectangle into a square as shown. (This is discussed in previous example)

D
D
D

C
C
E E
C
E

A B A B A B

1 D 2 3

C
E

A B

43 Lubega_Ashraf
 To construct a square equal in area to a given plane figure.
1. Draw the given polygon ABCDE.
2. Join A to D and draw a parallel line to DA from point E to intersect the base AB at F. join D to F.
3. Join B to D and draw a parallel line to DB from point C to intersect the base AB at H. join D to H.
4. DFH is a triangle equal in area to the given plane figure. Transom this triangle into a square of equal area. (This has
been discussed before)

C C
C
D D
D

E E E

A B A B A B
1 2 3
C

A B

44 Lubega_Ashraf
To divide any triangle into a given number of equal areas by lines drawn parallel to one side. (Consider 4 equal areas).
1. Construct the triangle XYZ and construct a semi-circle on one side of the triangle as shown
2. Divide XY into 4 equal parts
3. Draw perpendiculars on XY from the four divisions to intersect the semi-circle at K, H, and P.
4. Use X as center, XK, XH, and XP as radius and draw arcs to intersect XY at q, e and n respectively.
5. Draw lines from q, e and n parallel to line ZY. Hence the triangle has been divided equally into 4 portions.

K
H

X X 1 2 3 4
P

Z Y Z Y 2
1
K
K
H H

X 1 2
X 1 2
3 3 4
4
P P

Z Y Z Y 4
3

45 Lubega_Ashraf
To trisect a triangle from a specific point (P) on one side of the triangle.
1. Divide the sie bearing the point (P) into three equal parts.
2. Join P to B and draw lines from 1 and 2 parallel to PB. The parallel lines will intersect AB at G and CB at H.
3. Draw lines from H and G to the point P
4. Hence 3 equal triangles.
1 A
1 A 2
2 3
3
P P

C B C B

1 2

1 A
2
3
P

C B

46 Lubega_Ashraf
To construct a triangle equal in area to another with an altitude reduced by half (having a different altitude).
1. Construct the triangle XYZ.
2. Drop a perpendicular from the apex X onto the base ZY, the perpendicular meets the base at H.
3. Bisect the altitude XH, the bisector intersects the altitude at F.
4. Draw a parallel line to FY from X to intersect the extended base at T
5. Draw a parallel line to FZ from X to intersect the extended base at W
6. Join W and T to F, WFT is the required triangle. (Equal int area but with half the original altitude)
X
X

H H X
Z Y
Z Y

1 2
F

H
W Z Y T

47 Lubega_Ashraf
Circle.
A circle is the locus of a point, which moves so that its distance from a fixed point remains constant.
Definitions of parts of a circle.
a. Diameter; this is a straight line that runs through the centre, meeting the circumference at both ends.
b. Radius; this is a straight line that runs from the centre to the circumference.
c. Segment; this is a part of a circle enclosed by a chord and an arc.
d. Chord; this is a straight line running from one end of the circumference to another without passing through the centre.
e. Quadrant; this is a part of a circle enclosed by two radii meeting at right angles and an arc.
f. Sector; this is a part of a circle bounded by two radii and an arc.
g. Arc; this is any part of the circumference.
h. Tangent; is a straight line, which touches the circumference of a circle or an arc at a point of tangency. It is always at
right angles to the radius.
TANGENT AB -Diameter
QUADRANT OC -Radius
C CD – Chord
DB – Arc
SEGMENT
D Enough with the introduction, lets construct some circles
to satisfy given conditions, ready?
SECTOR
A B Let’s go.
O

SEMI CIRCLE

48 Lubega_Ashraf
To construct a circle to pass through given points. ABC.
1. Locate the three given points and join them together.
2. Bisect line AB and BC, the bisectors will intersect at O. (Line AB and BC are chords)
3. Using O as center and either AO or BO or CO as radius, draw a circle passing through all the points ABC.
B
B
B
B
A
O
A C

A O
A C

1 C C 4
2 3
To inscribe a circle into a triangle. (Or any regular polygon).
6. Construct the given triangle ABC.
7. Bisect any two interior angles, i.e., angle BAC and ABC.
8. The bisectors will intersect in D.
9. Drop a perpendicular from D to line AB, in order to obtain the radius of the inscribed circle.
10. Using D as center and DF as radius, draw the circle.
C C C

D D

F F
B B B
A A A

1 2 3

49 Lubega_Ashraf
To construct three circles; touching each other when given the positions of their centers.
1. Construct the given figure.
2. Join the centers together to form a triangle ABC,
3. Bisect angle BAC ABC, produce the bisectors to intersect at O. drop a perpendicular on any side of the triangle from O.
4. Use A as center, AQ as radius and draw the first circle. Use B as center, BQ as radius and draw the second circle, and
finally use C as center, DC as radius and draw the third circle.
B B

A A
Q
1 2
C C
B
B

O
O
A
Q
A
Q C
C

3 4

NOTE: Other kinds of circles have been discussed in an earlier topic about inscribing and circumscribing. Borrow
knowledge to complement this.

50 Lubega_Ashraf
Ellipse
To construct an ellipse using the rectangle method when given the major (AB) and minor (CD) axes.
1. Construct the rectangle; the length equals to the major axis while the width equals the minor axis.
2. Divide the width and the major axis into the same even number of equal parts (in this case 8 equal parts)
3. Draw lines from C to 1, 2 and 3 i.e., on both sides of the rectangle. Do the same from D.
4. Radiate lines from D through 1, 2, and 3 on major axis AB to intersect lines running from C to 1, 2 and 3. This gives us
points for the ellipse, join these to get the ellipse.
C C
4 4

3 3
2 2
1 1
A B A 0 1 2 3 4 3 2 1 0 B
1 1
2 2
3 3
4 4
D D
1 2
C C
4 4 4 4

3 3 3 3
2 2 2 2
1 1 1 1
A 0 1 2 3 4 3 2 1 0 B A 0 1 2 3 4 3 2 1 0 B
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
D D
3 4

51 Lubega_Ashraf
To construct an ellipse by using the concentric circle method when given the major and minor axis.
1. Construct two concentric circles, using half the major and minor axis as radii.
2. Divide the circles into 12 parts.
3. Where the sector lines intersect the circumference of the smaller circle, draw horizontal lines towards the
circumference of the larger circle.
4. Where the sector lines intersect the circumference of the larger circle, draw vertical lines to intersect the horizontal
lines from the smaller circle. Draw a neat curve through the intersection points.

C C

MINOR AXIS
A B A B

MAJOR AXIS

D D
2
1

C C

A B A B

D D
4

52 Lubega_Ashraf
To construct an ellipse using the foci method (intersecting arc method), when given the major and minor axes.
1. Draw the major and minor axes.
2. Obtain the two foci (F1 and F2); by using C as center, OB (half the major axis) as radius and swing an arc to intersect
the major axis at F1 and F2, divide the distance between the foci (F1 and F2); into any equal even parts (in this case 8).
3. With center F1, and radius A1, swing an arc above and below the major axis.
4. With center F2, and radius B1, swing an arc to intersect the previous arc at E and H.
5. Repeat procedure 3 and 4 for the remaining divisions.

C C

OB
A A 1 2 3 4 5 6 7 8 B
B
F1 O F2 F1 O F2

D D

1 2

C C

A 1 2 3 4 5 6 7 8 B A 1 2 3 4 5 6 7 8 B
F1 O F2 F1 O F2

D D

3 4

53 Lubega_Ashraf
 To construct an ellipse when given the relative positions of the focus (F) and the directrix (D) 50mm and eccentricity
ratio as ⅔. (Ellipse as a locus of a point).

1. Construct the directrix (D) and locate the focus (F1). Obtain the vertex V1 by using the given eccentricity 2:3. (this is
done by dividing the line F1-D into a ratio of 2:3 as shown).
2. Obtain the other vertex V2 by using the formula below.
F1V2 = Eccentricity x (DF1 + F1V2), where DF1 = 20 & eccentricity = ⅔.
Solving gives F1V2 = 100.
3. At any convenient point z, draw a horizontal line perpendicular to the directrix. Draw a vertical line through V1 to meet
the previously drawn horizontal line at y. mark off length V1F1 from y as shown to obtain point x.
4. Draw a diagonal line through z and x.
5. Choose convenient points between the vertices V1 & V2 and draw lines parallel to the directrix line through those
points to intersect the horizontal line zy.
6. Pick distance x (length between diagonal line and horizontal line) and standing at F1, draw arcs on the vertical line
(same line from which the length x was picked) with radius x.
7. Repeat procedure for y and for all the other vertical lines.
8. Join the points obtained to obtain the ellipse.

y
y
100 100 100

x
x

3 2
D V1 F1 F2 V2 V1 V2 V1 V2
1 D F1 F2 D F1 F2
1 1
2 2 2

x
x

3 3 3
4 4 4
5 5 5

DV1 x DV1

y
y
V1F1

V1F1

x
x

z
y
1 2 3

54 Lubega_Ashraf
Tangents and blending of arcs onto circles.
To construct an arc/ circle touching a straight line at a point C.
1. Draw line AB and mark off the point C.
2. At point C, draw a perpendicular as shown as mark off the radius r of the arc or the circle, to obtain the centre D.
3. Using D as centre and r as radius, draw the arc or the circle.

D D

r
A C B A C B A C B
1 2 3

To construct an arc/ circle touching another arc/circle at a point P.

 Construct the arc/circle and locate point p, draw a line through the centre c and the point p, at p, mark off the radius r of
the arc/circle, draw the arc/ circle as shown.

r r r r

p p p p

c c c c

arc to arc circle to arc arc to circle circle to circle

55 Lubega_Ashraf
To construct a common tangent onto two equal circles.
1) Draw the two equal circles with separate centers A and B and join them.
2) Construct perpendiculars at the centers, A and B.
3) The point of intersection between the circles and the perpendiculars gives us the points of tangency; Q and R
4) Draw a line passing through point Q and R hence the required tangent.

A B A B

1 2
To construct a common internal tangent between two equal circles.
1) Construct the two equal circles with separate centers A and B.
2) Bisect the distance between the centers, A and B. further bisect the distance between Q and B.
3) Construct a semi-circle using E as your center and QE/EB as radius. The semi-circle will intersect the circle at P (point
of tangency).
4) Draw line PB and later draw a parallel line to PB from A (point of tangency).
5) Draw a line passing through point P and T hence giving us the required internal tangent.

P P

Q E Q E
A B
A B

1 2

56 Lubega_Ashraf
To construct a tangent onto a circle from a given point (A) outside it.
1) Construct the given figure i.e., the circle and given point A.
2) Join point A to the center O of the given circle, bisect line OA and construct a semi-circle.
3) The semi-circle enables us to get the point of tangency T.
4) Draw a line from the given point A through the point of tangency, hence the required tangent and the line running from
the center O through the point of tangency T is called the normal.
T

O
O F
A
A

T 2
1

O
F
A

To construct the common internal tangent between two unequal circles.

57 Lubega_Ashraf
 1) Construct the given circles.
2) Bisect distance between the centers A and B and draw a semi-circle.
3) Use A as center, (Ra + Rb) as radius and draw an arc to intersect the semi-circle at Q.
4) Draw a line from A to Q; this line will intersect the circle at P (point of tangency).
5) Draw another line from B that is parallel to AQ; this line will intersect the circle at point D (point of tangency)
6) Draw a line passing through the points of tangency P and D, hence the required internal tangents.

Ra Ra P
Rb Rb
Rb
+
Ra

A B B
A

D
1 2
To construct the common external tangent between two unequal circles.
1) Construct the given circles
2) Bisect distance between the centers A and B and draw a semi-circle.
3) Use A as center, (Ra - Rb) as radius and draw a circle to intersect the semi-circle at Z.
4) Draw a line from A through Z; this line will intersect the circle at Q (point of tangency)
5) Draw another line from B that is parallel to AQ; this line will intersect the circle at point D (point of tangency)
6) Draw a line passing through the points of tangency Q and D, hence the required external tangents.

58 Lubega_Ashraf
 Q
Q
Ra
B Rb D
B

Ra - Rb

Ra - Rb
A B A B

1 2

To find the center of an arc/circle of radius R, which blends with two circles. (Internal arc)
1) Construct the given circles.
2) Use A as center, the radius to be used will be got by adding the radius (R) of the circle/arc to be constructed and the
radius (ra) of the circle whose center is A i.e. (R + ra) and swing an arc.
3) Use B as center, the radius to be used will be got by adding the radius (R) of the circle/arc to be constructed and the
radius (rb) of the circle whose center is B ie (R + rb) and swing an arc to intersect the previous one at H.
4) Draw lines from H to center A and B in order to establish the points of contact (X and Y) between the arc/circle to be
constructed and the two given circles.
5) Use H as center, XH/YH as radius and construct the required arc/circle.

59 Lubega_Ashraf
 Ra
Rb

Ra Ra
Rb Rb
A B

+R
A B A B X

RB
R
R
+
a
Y

R
Y

+R
X

R
X

RB
H

RB

R
R

+
+

a
R
a
R
1 2 3
H
H

To find the center of an arc/circle of radius R, which blends with two circles (external arc)
1) Construct the given circles.
2) Use A as center, the radius to be used will be got by subtracting the radius (ra) from the radius (R) of the arc/circle to
be constructed, i.e., (R - ra) and swing an arc.
3) Use B as center, the radius to be used will be got by subtracting the radius (rb) from the radius (R) of the arc/circle to
be constructed, i.e., (R - rb) and swing an arc to intersect the previous one at H.
4) Draw lines from D through center A and B to intersect their circumferences at X and Y respectively. X and Y are the
points of contact between the arc/circle to be constructed and the two given circles. Use D as center, XD/YD as radius
and construct the required arc/circle.

60 Lubega_Ashraf
 Ra Ra
Rb Rb

R
A B A B

a
a

-R
-R
b

R
b

R
-R -R
R
R

1 2

Ra
Rb
R

A B
a
-R

b
R

-R
R

To construct an arc or a circle of given radius to touch two given circles the arc/circle is to include one and exclude the
other.
1) Construct the given two circles, whose centers are A and B.
2) Use A as center, the radius to be used will be got by adding the radius (R) of the arc/circle to be constructed and the
radius (ra) of the circle whose center is A i.e., (R + ra) and swing an arc.
3) Use B as center, the radius to be used will be got by subtracting the radius (rb) from the radius (R) of the arc/circle to
be constructed, i.e., (R - rb) and swing an arc to intersect the previous one at D. X and Y are the points of tangency.
4) Use D as center, YD or XD as radius and draw the required arc or circle.

61 Lubega_Ashraf
Ra Ra
A Y
Rb
Rb

A B A B

A X

a
a

R
R - Rb

R
R - Rb

+
+

R
R
1
2
Ra
Y

Rb

A B

a
R
R - Rb

+
R

62 Lubega_Ashraf
Use the knowledge you have acquired to draw the figures below.(use a suitable scale.)

Ø16 Ø38
50

R15
R60 R8
R19

60 102

R14
R3
20
R22 35
30
R19
R22
20 R50 5.0

R10.0
R31.0
R
45 R36.0
Ø16

19°
R10.0
R5.0
R2.5 R9.0
R15
15°

54.0

63 Lubega_Ashraf
Solution to first figure.
1. Draw centreline AG and locate centres J, H, D, E & A and draw circles/ arcs as shown.
2. Next step is to draw tangential lines/arcs onto the circles and arcs we drew in step (1). Use knowledge of tangency to
do this. Follow through.

50

G G
G

J H J H J
r(60-15) r(60-15) H
r(60-15)
60

F project F
G points
20
30
50 50 of 50

C B
J C B
D E
D E
45

A
A F
tangency to the other side.
r(50+15)
C
Outline the opener.
B C
D r(50+15) D

64 Lubega_Ashraf
 G G

J H J H
r(60-15)

F F

50

C B C B
D E D E

A A

65 Lubega_Ashraf
Development of shapes.
Have you cared to dismantle a box before? The plane surfaces you get once the box is dismantled is what we call the “box’s
development.”
It is these surfaces that you have to put together to form the box.

All shapes can be dismantled just like the box we just talked about.
This is what development is about. We want to cut a figure at a suitable seam and dismantle it and see how the surface it is
made of looks like.
Seam simply refers to a cutting edge or a point of dismantling.

Basically, there are three types of development;

1) Parallel line development
2) Radial line development and,
3) Development by triangulation
For our course, we shall concentrate on the first two.

Enough with the introduction, lets dive in.

66 Lubega_Ashraf
Parallel line development.
This method is used in cylinders, cubes & prisms, where we only use parallel lines to develop our piece.
Development of a cylinder.
1. Draw the elevation and plan.
2. Project a horizontal line from the elevation and mark off the circumference.
3. Follow procedure to understand the solution.

h h

0 x 1 2 3 4 5 6 7 8 9 10 11 12
ELEVATION x
CIRCUMFERENCE OF CIRCLE
4 3 2
r DEVELOPMENT OF CYLINDER
5 r1

6 0

11
7
x
8 10
9

67 Lubega_Ashraf
Development of a truncated cylinder.

0 x 1 2 3 4 5 6 7 8 9 10 11 12
ELEVATION x
CIRCUMFERENCE OF CIRCLE
4 3 2
r DEVELOPMENT OF TRUNCATED CYLINDER
5 r1
seam

6 0

11
7
x
8 10
9

PLAN

68 Lubega_Ashraf
Elbows.

h1
h2 Follow through
carefully to
h3 understand
h4 procedure.
B
h5
h6 A

h7

0 1 2 3 4 5 6 7 8 9 10
ELEVATION x x 11 12

4 3 2 CIRCUMFERENCE OF CIRCLE
5 r
1 DEVELOPMENT OF A
r

6 0

11
7
h1

h2

x
h3

8 10
9
h4

h5

PLAN h6

0 x 1 2 3 4 h7
5 6 7 8 9 10 11 12
x
CIRCUMFERENCE OF CIRCLE
DEVELOPMENT OF B

69 Lubega_Ashraf
Development of a square bend, three-piece elbow piece.
This development is basically made out of three different cylinders that have been truncated. Follow the steps taken to
construct the three-piece elbow piece and its development.

C seam
r

R DEVELOPMENT OF A, B AND C
g
1
f
R

2,12

e 3,11 A

g
f
A

e
d
c
d b
a

4,10
g

B
f

c
e

5,9
d

g
f
6,8 B

e
c

d
a 7
c
b

b
a

C
a

ELEVATION

70 Lubega_Ashraf
 g
f
e
d
c
b
a
seam

0 1 2 3 4 5 6 7 8 9 10
ELEVATION x x 11 12
3 2 CIRCUMFERENCE OF CIRCLE
4
5
1
seam DEVELOPMENT OF PART B

6 0

11
7
x
8 10
9
g

e
f

PLAN
d
c

0 1 2 3 4 5 6 7 8 9 10
x x 11 12

CIRCUMFERENCE OF CIRCLE
DEVELOPMENT OF PART A

71 Lubega_Ashraf
To draw

m
development of part Y

k
A, draw line XY

i g
that is

n
l
perpendicular to the

e c

j
h
line along part A,

f
a
seam

d
use this line as a

b
reference line for
A
taking dimensions
as you develop the
piece.

X
B

0 1 2 3 4 5 6 7 8 9 10
ELEVATION x x 11 12
3 2 CIRCUMFERENCE OF CIRCLE
4
5
1
seam DEVELOPMENT OF PART B

6 0

11
7

k
i
x

g
8 10

e
9

a
X Y
PLAN

b
d
f
h
j
n

0 1 2 3 4 5 6 7 8 9 10 11 12
x
CIRCUMFERENCE OF CIRCLE
DEVELOPMENT OF PART A

72 Lubega_Ashraf
Radial line development.
This method is used in cones & pyramids, where the piece is developed using lines radiating from a common center.
Development of a cone.
 Follow through to understand procedure.

O
true length

z 0
x
1 12
x
11
2
10
6 3
ELEVATION 4 9
3 5 8
4 2 6 7
5 r
r1

6 0

11
7
x
8 10
9 DEVELOPMENT OF CONE

PLAN

73 Lubega_Ashraf
Development of a truncated cone.
1. Construct the front elevation and plan (complete the plan showing the cut surface) and true shape if required.

h3 2. Divide the plan into 12 equal parts & project them to the
h2
h1 base of the front elevation.
3. Draw lines from the vertex V to the points on the base of
O the front elevation. Where the radiating lines cut the section,
h5 h4
project them horizontally to touch the edge as shown in the
figure.
TRUE SHAPE
4. Use V as center & each of the points on the edge as radius
i.e., (V-a), (V-b), (V-c) …, and draw arcs.
5. On the largest arc whose radius is (V-1), step off 12 arcs of
radius x, and connect them to V. the intersections between the
radiating lines and the arcs give the points of intersections for
the top shape.
6
ELEVATION To draw the true shape, project perpendicular lines from the cut
4 3 2 surface as shown. (Follow procedure).
5
1

h5 h2 h3
h4
6 0 h1

11
7

8 10
9

PLAN

74 Lubega_Ashraf
 O

z a
cb
d
e
g f

a
z b
c 0
d
e x
f
g 1 12

x
11
2
seam 10
6 3
ELEVATION 4 9
3 5 8
4 2 6 7
5
1
DEVELOPMENT OF TRUNCATED CONE

6 0

11
7
x
8 10
9

PLAN

75 Lubega_Ashraf
 x1
x2
h5 a
h4
h3 a'
h2 1
TRUE SHAPE. o h
f'

h2
h3

h4

h5
h1
d' e' f
d' a'
c',e' b' c'
b',f' a

a' e
seam b
x
a b,f c,e d x d

x
ELEVATION c
DEVELOPMENT
f e

f' e'
x1

x2

a o
a' d' d
c'
b'

b c
PLAN

76 Lubega_Ashraf
Development of a funnel.
A funnel can take multiple shapes depending on the purpose for which it is made. So, we can’t commit to a single shape of a
funnel but rather, we can simply understand the concept employed in developing any funnel shape.
SHAPE 1.

O
y
z
y

0
z

x
1 12

x
11
2
10
6 3
ELEVATION 4 9
3 5 8
4 2 6 7
r
5 DEVELOPMENT OF FUNNEL
1
r

6 0

7 11

x
8 10
9

PLAN

77 Lubega_Ashraf
SHAPE 2.

A
We can settle with
these two shapes for

y
now,
We shall look at
other complex
z

circumference of cylinder
part A funnel shapes after
DEVELOPMENT OF PART A handling
interpenetration.
B
y
z
6
ELEVATION
4 3 2 0
r
5
x

1
r

1 12
x

6 0 11
2
10
11 3
7 4 9
5 8
x 6 7
8 10
9 DEVELOPMENT OF CONE
PLAN part B

78 Lubega_Ashraf
Interpenetration of solid figures.
When solids are connected together such that they intersect each other, there is formed a line or curve of intersection/
interpenetration between them.
This line or curve is the point of intersection between these intersecting solids.
As water engineers, we always see intersections of pipe joints, this is basically what interpenetration is about. Solids
intersecting to form a single unit just like the pipe elbow.

In this topic, we shall be tasked with drawing this line/curve of interpenetration.

 But, to draw this line/curve, we need to first determine common points between the solids connecting and join these
points together.
In order to draw the line/curve of intersection, we always need to first draw the elevation, plan and sometimes the end
elevation of the solids. Sometimes even the auxiliary view.
We shall later develop these pieces.
Enough with the introduction, lets dive in and do some interpenetration. Ready? let’s go.

79 Lubega_Ashraf
The interpenetration between two cylinders of the same diameter.
 Follow arrows to obtain the curve of interpenetration.

h1 1
2,12
h2
3,11
h3

4,10
h4
h

h
A B
5,9

6,8
7

ELEVATION CIRCUMFERENCE OF A
10 DEVELOPMENT OF A
9,11

8,12

1,7

h4
h3
2,6
h2
h1

3,5
PLAN 4 1 2 3 4 5 6 7 8 9 10 11 12 1
CIRCUMFERENCE OF B

DEVELOPMENT OF B

80 Lubega_Ashraf
 The interpenetration between two cylinders of the same diameter connected at angle 45°.
 Follow arrows to obtain the curve of interpenetration.

12

11
2,

3,

10
4,
1

9
5,
8
6,
B

7
h

h
A

ELEVATION CIRCUMFERENCE OF A
10
9,11
DEVELOPMENT OF A
8,12

1,7

2,6

3,5
PLAN 4

81 Lubega_Ashraf
 12

11
2,

3,

10
4,
h1

9
5,
h2

8
6,
h3

h4

h5
h4

h6

h7
h5

h6

h3
h7

h2
h1 1 2 3 4 5 6 7 8 9 10 11 12 1

CIRCUMFERENCE OF B

DEVELOPMENT OF B

82 Lubega_Ashraf
Interpenetration between two cylinders of different diameters.
1. draw the front elevation & plan.
2. Divide the smaller circle into 12 equal
parts both in plan and front elevation. h1
1
2,12
3. Project the divisions in the plan to h2
meet the surface of the bigger cylinder 3,11
h3
and project them vertically upwards to 4,10
meet the corresponding points in the h h4
front elevation. This gives the curve of 5,9
intersection. A B
6,8
7

Use knowledge of development to

develop parts A & B.
ELEVATION
seam

y
10
z 9,11
8,12
x
1,7
2,6
3,5
4

PLAN

83 Lubega_Ashraf
 h1
1
2,12
h2
3,11
h3
h h4
4,10 h
5,9
A B
6,8
7

ELEVATION x
y
z DEVELOPMENT OF A

y
10
z 9,11
8,12
x
1,7 h4 h3
h2 h1
2,6
3,5 1 2 3 4 5 6 7 8 9 10 11 12 1
4
DEVELOPMENT OF B
PLAN

84 Lubega_Ashraf
 Interpenetration between two cylinders of different diameters and one is positioned at 45°.
 Follow the same steps discussed in previous example.
 Development for part B is on next page.

12
11
2,
3,

10
1

4,
9
6, 5,
8
7
B

h h

A 45°

ELEVATION x
y
z DEVELOPMENT OF A

y
10
z 9,11
8,12
x
1,7
2,6
3,5
4

PLAN

85 Lubega_Ashraf
 12
11
2,
3,

10
1

4,
h1
h2

9
6, 5,
8
h3
h4

h7
h6
B

h5
h5

h4
h6

h3
A

h2
h7

h1
1 2 3 4 5 6 7 8 9 10 11 12 1

DEVELOPMENT OF B

86 Lubega_Ashraf
Interpenetration between two cylinders of different diameters but is tangential to the larger cylinder,
and one is positioned at 45° and is tangential to the larger cylinder. (Develop both pieces A & B)

2
11
1
2,
3,

10
1

4,
45°

9
6, 5,
1
2,12

8
3,11
4,10

7
B
5,9
A B A
6,8
7

ELEVATION ELEVATION

10 10
9,11 9,11
8,12 8,12
1,7 1,7
2,6 2,6
3,5 3,5
4 4
PLAN PLAN

87 Lubega_Ashraf
EXERCISE.
i. Draw the given views.
ii. Complete the elevation with the lines/curves of interpenetration.
iii. Produce developments for both A & B pieces.

Ø60
Ø60

Ø
60
30

B
B

120
Ø60
85

45°
A A

2
10
42.5

25
ELEVATION ELEVATION

1 2

88 Lubega_Ashraf
 Ø60 Ø60

30

30

Ø
30
Ø30

85
85

B
A A

25

45°
ELEVATION ELEVATION
10

PLAN PLAN
3 4

89 Lubega_Ashraf
Principles of isometric projection.
Isometric projection is a method for visually representing three-dimensional objects in two dimensions in engineering
drawings.

120°
12

0°
12
0°

Isometric drawings are built on a framework of three lines

representing the three edges of the cuboid(box).
90
°

30°
30°

These three lines form three equal angles of 120° and are called
“isometric axes.” One is drawn vertically while the other two are
drawn at 30° as shown in the figure.

ISOMETRIC VIEW OF BOX

90 Lubega_Ashraf
Isometric rectangle.
1. Draw the rectangle ABCD.
2. For the isometric of the rectangle in vertical position (drawing B);
a) Draw a horizontal line and mark off point a, and draw a vertical line and another line inclined at 30° as shown.
b) Mark off the length of the rectangle ab & width ac,
c) Complete the figure as shown.
4) For the isometric of the rectangle in the horizontal position (drawing A)
a) Follow the previous steps.

TH c
LENGTH NG
LE

WIDTH
d c

d
B
WIDTH

90
a b

30°
a °

c
WI
TH DT
NG H
LE

b
A
d
30°
30°

91 Lubega_Ashraf
Isometric of a cuboid.

c
Draw the isometric rectangle as discussed previously, then drop the heights at
the edges as shown.
b
Some edges will be hidden and these a re represented with dotted lines as
d shown.

30°
30°

HEIGHT
Isometric circle.

 Follow the
same steps used
for the rectangle
and follow up to

b
understand
procedure.

a
b
a

a
b

a
b

30°
a
b
b
a

30°

30°
92 Lubega_Ashraf
Isometric of a cylinder.

1. Draw the isometric axes and the box which encloses the cylinder.
2. Draw the circles in isometric projection for the ends and complete the
cylinders by drawing in the sides.

30°

30°
HEIGHT

HT
IG
HE

30°

30°
93 Lubega_Ashraf
Non isometric lines.
These are lines that are not parallel to any of the isometric axes. Such lines will not show in their true length and cannot be
measured. They are drawn by locating their two ends.

In the figure shown, line ab, is an example of a non-isometric line, as it

is not parallel to any of the isometric axes.
a

 To draw line ab, locate points a & b and simply join them.

c
In way of practicing the concepts discussed in isometric, attempt all the
e numbers on the next page.
Go ahead and draw their isometric views and then transform these
views into orthographic projection.

94 Lubega_Ashraf
 20

20

20
°
35

10
50
20

10

95 Lubega_Ashraf
96 Lubega_Ashraf
Orthographic projection.
This projection shows the object as it looks from the front, right, left, top, bottom, or back, and are typically positioned relative
to each other according to the rules of either first angle or third angle projection. In orthographic projection, the views are seen
in directions that make right angles (i.e., 900) with each other. The number of views needed should be sufficient to represent
the object completely and conveniently, but it should be kept to the minimum. For most purposes, three views are usually
sufficient.
The Front View (ABCD) – abbreviated as FV, is the view of
utmost importance in representing the object (normally the
most complicated of all the views) as seen when the object is
placed directly in front of the viewer.

The First Angle Projection

Here the plan is drawn below the front elevation. The essence is that the other views are viewed as ‘shadows’ of the front
elevation.
Symbol:

The Third Angle Projection

Here the plan is drawn above the front elevation. The essence is that other views are viewed as ‘reflections’ of the front
elevations.
Symbol:

97 Lubega_Ashraf
Let’s explore an example.

Figure shows a box with views from different directions.

LENGTH

A view of the plan as seen from the

WIDTH
direction of arrow Z.
the height is not used.

LENGTH
Z
A view of the end as seen
from the direction of arrow Y.
the length is not used.

HEIGHT
WIDTH PLAN

HEIGHT
FRONT
END
HEIGHT

A view of the front as seen

Y from the direction of arrow X.
WI H the width is not used.
DT N GT
H LE X

98 Lubega_Ashraf
Exploring another example.
Use hatching to understand where the views have been picked from.

PLAN PLAN
as seen in direction Z

Y FRONT
END
X

FRONT
END
as seen in direction X
as seen in direction Y

99 Lubega_Ashraf
QUESTION.

Represent the figure shown below in both first angle and third angle projections.

44 Represent the figure in both first angle and third angle orthographic
projection.

43
23

22
48
14 18
F.E

NOTE:
For most cases, three views are sufficient to completely define the component (i.e., Front View, Left Hand side View, and Top
View (Plan)).
When drawing these views, one may start with any view. However, the common practice is to start with the FV. Once the first
view has been correctly drawn, with all the dimensions correctly measured, the second view is drawn much faster by projecting
(i.e., transforming) most of the dimensions from the first view. The third view is drawn without taking any new measurements.
The use of specially designed drawing facilities (e.g., a drawing table, a T-Square, set Squares etc, or drafting machine) simplifies
this task.

It should be noted that corresponding views are identical in both methods of projection except for their relative positions on the
drawing paper.

100 Lubega_Ashraf
 FIRST ANGLE PROJECTION. THIRD ANGLE PROJECTION.

12 44

18
14 44
22
23

65
45°
FRONT ELEVATION END ELEVATION PLAN
45°
44 12

18
14 44
22
23

65
PLAN END ELEVATION FRONT ELEVATION

Procedure for producing the views.

 First, interpret the figure and locate the front elevation, the side elevation and the plan.
 Visualize the views in your head.
 Then draw the front elevation, project lines from the front elevation to the spaces for the other views.
 Position the plan, and finally the side elevation.
 Label and dimension the drawings.

101 Lubega_Ashraf
Assignment.
Transform the following figures into their 1st and 3rd angle projections.

20

20

20
°
35

10
50
20

10

102 Lubega_Ashraf
103 Lubega_Ashraf
 Orthographic views to Isometric projection.
Draw the given block in isometric with X as the lowest point.

Stage 1, Draw a box using the overall dimensions of the block.

Stage 2, Mark off the thickness of the horizontal part.
Stage 3, locate the vertical part, and finally outline the edges.
45

20
45
35 20
90

45
45

20
90 1 35
2
20

3 4

104 Lubega_Ashraf
For the views given in first angle projection below, draw the;
a) The orthographic views and add the third missing view,
b) An isometric drawing with X in the fore ground.

30

30 30
85

20
60

55
30

30

25
70 x 90 x

15
15

45
55

15
x
15

x 2
1

105 Lubega_Ashraf
 10

10
10 40A/C
4
Ø2
R3

60
0

24

50
10

10
10 24 12 10
x 10
50 x
20
10

50
40
22
0
R2
10

10
80
x x
3 4

106 Lubega_Ashraf
Oblique projections.
Oblique projection is an alternative pictorial view of isometric projection.

RECEEDING AXIS

HORIZONTAL AXIS In oblique projection, the front face of the object appears in its true size and shape, as

45°
it is placed parallel to the picture plane. The receding lines representing the other two
faces are usually drawn at 30°,45° or 60° to the horizontal, 45° being the most
common practice.

VERTICAL AXIS

TRUE SHAPE
RECEEDING LINES

There are two types of oblique projections i.e., Cavalier and Cabinet.

°
This is shown below.

90

45°
°
90

90
°
OBLIQUE PROJECTION OF
CUBOID

107 Lubega_Ashraf
 To reduce the amount of distortion and to have a more realistic appearance, the
length of the receding lines is reduced. This is known as cabinet oblique.

75 37.5
full length half length

45°
54 54

CAVALIER OBLIQUE CABINET OBLIQUE

Oblique projection has the following advantages over Isometric drawing:

1. Circular or irregular features on the front face appear in their true shape.
2. Distortion may be reduced by fore-shortening the measurement along the receding axis, and,
3. A greater choice is permitted in the selection of the position of the axes.

108 Lubega_Ashraf
Oblique circle.

Ordinate spacing ab reduced by half, (cabinet oblique)

/2
ab
/2
DC
Diameter DC is too reduced by half.

de
de de & ef remain full.
ef

cd

ab

45°
DC

Note: use the knowledge of isometric to draw any

figure in oblique.

45°
109 Lubega_Ashraf
3D sketches and views
The ability to produce drawings of objects without instruments is often taken for granted by students. It comes in handy to
enable ideas and solutions to be quickly put down on paper.
While freehand sketching is a fairly quick technique, it must be understood that sketches should be neat and in good
proportion.
Freehand sketching is one of the effective methods to communicate ideas irrespective of the branch of study. The basic
principles of drawing used in freehand sketching are similar to those used in drawings made with instruments.
A 3D projection is a design technique used to display a three-dimensional object on a two-dimensional surface.
When sketching, the following should be kept in mind.
1. Use a free hand, no instrument should be used.
2. Be proportionate, this affects dimensions. Let a smaller dimension appear smaller than a big one.
3. Outline your final sketch.
4. Be neat.
Procedure.
 Sketch your subject lightly.
 Refine your shape.
 Keep refining
 Define the shape.

110 Lubega_Ashraf
Sketch of a building. Sketches of building tools.

111 Lubega_Ashraf