Engineering Graphics Course

A Course Material on
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Dr.V.KANDAVEL, B.E.,M.E.,Ph.D.,
Department of Mechanical Engineering
SSM INSTITUTE OF ENGINEERING AND TECHNOLOGY
(Approved by AICTE, Affiliated to Anna University, Accredited by NAAC)
Dindigul – Palani Highway, Dindigul – 624 002.

GE8152 ENGINEERING GRAPHICS L T P C
2 0 4 4
OBJECTIVES:
To develop in students, graphic skills for communication of concepts, ideas and design of
Engineering products.
T o expose them to existing national standards related to technical drawings.
CONCEPTS AND CONVENTIONS (Not for Examination) 1
Importance of graphics in engineering applications – Use of drafting instruments – BIS
conventions and specifications – Size, layout and folding of drawing sheets – Lettering and
dimensioning.
UNIT I PLANE CURVES AND FREEHAND SKETCHING 7+12
Basic Geometrical constructions, Curves used in engineering practices: Conics – Construction of
ellipse, parabola and hyperbola by eccentricity method – Construction of cycloid – construction of
involutes of square and circle – Drawing of tangents and normal to the above curves.
Visualization concepts and Free Hand sketching: Visualization principles –Representation of Three
Dimensional objects – Layout of views- Freehand sketching of multiple views from pictorial views of
objects
UNIT II PROJECTION OF POINTS, LINES AND PLANE SURFACE 6+12
Orthographic projection - principles-Principal planes-First angle projection-projection of points.
Projection of straight lines (only First angle projections) inclined to both the principal planes -
Determination of true lengths and true inclinations by rotating line method and traces
Projection of planes (polygonal and circular surfaces) inclined to both the principal planes by
rotating object method.
UNIT III PROJECTION OF SOLIDS 5+12
Projection of simple solids like prisms, pyramids, cylinder, cone and truncated solids when the axis is
inclined to one of the principal planes by rotating object method.
UNIT IV PROJECTION OF SECTIONED SOLIDS AND DEVELOPMENT OF
SURFACES 5+12
Sectioning of above solids in simple vertical position when the cutting plane is inclined to the one of
the principal planes and perpendicular to the other – obtaining true shape of section.
Development of lateral surfaces of simple and sectioned solids – Prisms, pyramids cylinders and cones.
UNIT V ISOMETRIC AND PERSPECTIVE PROJECTIONS 6+12
Principles of isometric projection – isometric scale –Isometric projections of simple solids and
truncated solids - Prisms, pyramids, cylinders, cones- combination of two solid objects in simple
vertical positions
Perspective projection of simple solids-Prisms, pyramids and cylinders by visual ray method.
TOTAL: 90 PERIODS

OUTCOMES:
On successful completion of this course, the student will be able to
 familiarize with the fundamentals and standards of Engineering graphics
 perform freehand sketching of basic geometrical constructions and multiple views of
objects.
 project orthographic projections of lines and plane surfaces.
 draw projections and solids and development of surfaces.
 visualize and to project isometric and perspective sections of simple solids.
TEXT BOOK:
1. Natrajan K.V., ―A text book of Engineering Graphics‖, Dhanalakshmi Publishers, Chennai, 2009.
2. Venugopal K. and Prabhu Raja V., ―Engineering Graphics‖, New Age International (P) Limited,
2008.
REFERENCES:
1. Bhatt N.D. and Panchal V.M., ―Engineering Drawing‖, Charotar Publishing House, 50
th
Edition, 2010.
2. Basant Agarwal and Agarwal C.M., ―Engineering Drawing‖, Tata McGraw Hill Publishing Company
Limited, New Delhi, 2008.
3. Gopalakrishna K.R., ―Engineering Drawing‖ (Vol. I&II combined), Subhas Stores, Bangalore, 2007.
4. Luzzader, Warren.J. and Duff,John M., ―Fundamentals of Engineering Drawing with an
introduction to Interactive Computer Graphics for Design and Production, Eastern Economy
Edition, Prentice Hall of India Pvt. Ltd, New Delhi, 2005.
5. N S Parthasarathy and Vela Murali, ―Engineering Graphics‖, Oxford University, Press, New Delhi,
2015.
6. Shah M.B., and Rana B.C., ―Engineering Drawing‖, Pearson, 2nd
Edition, 2009.
Publication of Bureau of Indian Standards:
1. IS 10711 – 2001: Technical products Documentation – Size and lay out of drawing
sheets.
2. IS 9609 (Parts 0 & 1) – 2001: Technical products Documentation – Lettering.
3. IS 10714 (Part 20) – 2001 & SP 46 – 2003: Lines for technical drawings.
4. IS 11669 – 1986 & SP 46 – 2003: Dimensioning of Technical Drawings.
5. IS 15021 (Parts 1 to 4) – 2001: Technical drawings – Projection Methods.
Special points applicable to University Examinations on Engineering Graphics:
1. There will be five questions, each of either or type covering all units of the syllabus.
2. All questions will carry equal marks of 20 each making a total of 100.
3. The answer paper shall consist of drawing sheets of A3 size only. The
students will be permitted to use appropriate scale to fit solution within A3 size.
4. The examination will be conducted in appropriate sessions on the same day

CONCEPTS AND CONVENTIONS (Not for Examination)
GRAPHIC LANGUAGE
A technical person can use the graphic language as powerful means of communication with
others for conveying ideas on technical matters. However, for effective exchange of ideas with
others, the engineer must have proficiency in (i) language, both written and oral, (ii) symbols
associated with basic sciences and (iii) the graphic language. Engineering drawing is a suitable
graphic language from which any trained person can visualize the required object. As an
engineering drawing displays the exact picture of an object, it obviously conveys the same ideas to
every trained eye.
Irrespective of language barriers, the drawings can be effectively used in other countries, in
addition to the country where they are prepared. Thus, the engineering drawing is the universal
language of all engineers.
Engineering drawing has its origin sometime in 500 BC in the regime of King Pharos of
Egypt when symbols were used to convey the ideas among people.
IMPORTANCE OF GRAPHIC LANGUAGE
The graphic language had its existence when it became necessary to build new structures
and create new machines or the like, in addition to representing the existing ones. In the absence of
graphic language, the ideas on technical matters have to be conveyed by speech or writing, both
are unreliable and difficult to understand by the shop floor people for manufacturing. This method
involves not only lot of time and labour, but also manufacturing errors. Without engineering
drawing, it would have been impossible to produce objects such as aircrafts, automobiles,
locomotives, etc., each requiring thousands of different components.
NEED FOR CORRECT DRAWINGS
The drawings prepared by any technical person must be clear, unmistakable in meaning
and there should not be any scope for more than one interpretation, or else litigation may arise. In a
number of dealings with contracts, the drawing is an official document and the success or failure of
a structure depends on the clarity of details provided on the drawing. Thus, the drawings should
not give any scope for mis-interpretation even by accident.
It would not have been possible to produce the machines/automobiles on a mass scale
where a number of assemblies and sub-assemblies are involved, without clear, correct and accurate
drawings. To achieve this, the technical person must gain a thorough knowledge of both the
principles and conventional practice of draughting. If these are not achieved and or practiced, the

drawings prepared by one may convey different meaning to others, causing unnecessary delays
and expenses in production shops.
DRAFTING INSTRUMENTS
Engineering drawings are prepared with the help of a set of drawing instruments. Accuracy
and speed in the execution of drawings depend upon the quality of instruments. It is desirable for
students to procure instruments of good quality. The following instruments are commonly used:
1. Drawing board
2. Minidrafter
3. Precision instrument box
4. 45° set square and 30°-60° set square
5. Engineers’ Scales or Scales of Engineering Drawing
6. Protractor
7. Irregular or French curves
8. Drawing pins or clips
9. Drawing paper
10. Pencils
11. Eraser
12. Duster
Size of Drawing Sheet
Engineering drawings are prepared on standard size drawing sheets. The correct shape and
size of the object can be visualized from the understanding of not only its views but also from the
various types of lines used, dimensions, notes, scale etc. To provide the correct information about

the drawings to all the people concerned, the drawings must be prepared, following certain
standard practices, as recommended by Bureau of Indian Standards (BIS).
The standard drawing sheet sizes are arrived at on the basic Principle of X : Y = 1 : √2and
XY = 1 where X and Y are the sides of the sheet. For example AO, having a surface area of 1
Sq.m; X = 841 mm and Y = 1189 mm. The successive sizes are obtained by either by halving
along the length or doubling the width, the area being in the ratio 1 : 2. Designation of sizes is
given in Fig.l and their sizes are given in Table.1. For class work use of A3 size drawing sheet is
preferred.
Table.1 Preferred drawing sheet sizes (First choice) ISO-A Series
Fig. 1 Drawing sheet formats
Title Block
The title block should lie within the drawing space at the bottom right hand comer of the
sheet. The title block can have a maximum length of 170 mm providing the following information.
1. Title of the drawing.
2. Drawing number.
3. Scale.
4. Symbol denoting the method of projection.

5. Name of the firm, and
6. Initials of staff who have designed, checked and approved.
Fig. 2 Title block used in shop floor
Fig.3 Title block suggested for students
Fig. 4 Symbol for projection method
Drawing Sheet Layout (Is 10711 : 2001)
The layout of a drawing sheet used on the shop floor is shown in Fig.5

Fig. 5 Drawing sheet layout
Folding of Drawing Sheets
IS : 11664 - 1999 specifies the method of folding drawing sheets. Two methods of folding
of drawing sheets, one suitable for filing or binding and the other method for keeping in filing
cabinets are specified by BIS. In both the methods of folding, the Title Block is always visible.
Fig. 6 (a) Folding of drawing sheet for filing or binding

Fig. 6 (b) Folding of drawing sheet for storing in filing cabinet
LETTERING
The essential features of lettering on technical drawings are, legibility, uniformity and
suitability for microfilming and other photographic reproductions. In order to meet these
requirements, the characters are to be clearly distinguishable from each other in order to avoid any
confusion between them, even in the case of slight mutilations. The reproductions require the
distance between two adjacent lines or the space between letters to be at least equal to twice the
line thickness (Fig. 7). The line thickness for lower case and capital letters shall be the same in
order to facilitate lettering.
Fig. 7 Dimensions of lettering
Dimensions
The following specifications are given for the dimensions of letters and numerals:
(i) The height of capital letters is taken as the base of dimensioning (Tables.2 and 3).
(ii) The two standard ratios for d/h, 1/14 and 1/10 are the most economical, as they result in a
minimum number of line thicknesses.
(iii) The lettering may be inclined at 15° to the right, or may be vertical.

Table.2 Lettering A (d = h/14)
Table.3 Lettering B (d = h/10)
Fig. 8 Inclined lettering
Figures 8 and 9 show the specimen letters of type A, inclined and vertical and are given only as a
guide to illustrate the principles mentioned above.

DIMENSIONING
Drawing of a component, in addition to providing complete shape description, must also
furnish information regarding the size description. These are provided through the distances
between the surfaces, location of holes, nature of surface finish, type of material, etc. The
expression of these features on a drawing, using lines, symbols, figures and notes is called
dimensioning.
Fig. 10 Elements of dimensioning
Principles of Dimensioning
Some of the basic principles of dimensioning are given below.
1. All dimensional information necessary to describe a component clearly and completely shall be
written directly on a drawing.
2. Each feature shall be dimensioned once only on a drawing, i.e., dimension marked in one view
need not be repeated in another view.
3. Dimension should be placed on the view where the shape is best seen (Fig.11)
4. As far as possible, dimensions should be expressed in one unit only preferably in millimeters,
without showing the unit symbol (mm).
5. As far as possible dimensions should be placed outside the view (Fig.12).
6. Dimensions should be taken from visible outlines rather than from hidden lines (Fig.13).

Fig. 11 Placing the Dimensions where the Shape is Best Shown
Fig. 12 Placing Dimensions Outside the View
Fig. 13 Marking the dimensions from the visible outlines
7. No gap should be left between the feature and the start of the extension line (Fig.14).
8. Crossing of centre lines should be done by a long dash and not a short dash (Fig.15).
Fig. 14 Marking of Extension Lines
Fig. 15 Crossing of Centre Lines

Fig. 16 Marking of Arrow Head
Execution of Dimensions
1. Projection and dimension lines should be drawn as thin continuous lines.
2. Projection lines should extend slightly beyond the respective dimension lines.
3. Projection lines should be drawn perpendicular to the feature being dimensioned. Where
necessary, they may be drawn obliquely, but parallel to each other (Fig.17). However, they must
be in contact with the feature.
4. Projection lines and dimension lines should not cross each other, unless it is unavoidable
(Fig. 18).
5. A dimension line should be shown unbroken, even where the feature to which it refers, is shown
broken (Fig. 19).
6. A centre line or the outline of a part should not be used as a dimension line, but may be used in
place of projection line (Fig. 15).
Fig. 17 Fig. 18 Fig. 19
Termination and Origin Indication
Dimension lines should show distinct termination, in the form of arrow heads or oblique
strokes or where applicable, an origin indication. Two dimension line terminations and an origin
indication are shown in Fig. 20. In this,
1. The arrow head is drawn as short lines, having an included angle of 15°, which is closed and
filled-in.
2. The oblique stroke is drawn as a short line, inclined at 45°.
3. The origin indication is drawn as a small open circle of approximately 3 mm in diameter.

Fig.20 Fig.21 Fig.22
The size of the terminations should be proportionate to the size of the drawing on which
they are used. Where space is limited, arrow head termination may be shown outside the intended
limits of the dimension line that is extended for that purpose. In certain other cases, an oblique
stroke or a dot may be substituted (Fig. 21).
Where a radius is dimensioned, only one arrow head termination, with its point on the arc
end of the dimension line, should be used (Fig. 22). However, the arrow head termination may be
either on the inside or outside of the feature outline, depending upon the size of feature.
Methods of Indicating Dimensions
Dimensions should be shown on drawings in characters of sufficient size, to ensure
complete legibility. They should be placed in such a way that they are not crossed or separated by
any other line on the drawing. Dimensions should be indicated on a drawing, according to one of
the following two methods. However, only one method should be used on any one drawing.
METHOD–1 (Aligned System)
Dimensions should be placed parallel to their dimension lines and preferably near the
middle, above and clear-off the dimension line (Fig. 23). An exception may be made where
superimposed running dimensions are used (Fig. 30 b).
Dimensions may be written so that they can be read from the bottom or from the right side
of the drawing. Dimensions on oblique dimension lines should be oriented as shown in Fig. 24.
Angular dimensions may be oriented as shown in Fig. 25.
Fig. 23 Fig. 24 Oblique dimensioning Fig. 25 Angular dimensioning

METHOD–2 (Uni-directional System)
Dimensions should be indicated so that they can be read from the bottom of the drawing
only. Non-horizontal dimension lines are interrupted, preferably near the middle, for insertion of
the dimension (Fig. 26).
Angular dimensions may be oriented as in Fig. 27.
Fig. 26 Fig.27 Angular dimensioning
The following indications (symbols) are used with dimensions to reveal the shape
identification and to improve drawing interpretation. The symbol should precede the dimensions
(Fig. 28).
ϕ : Diameter Sϕ : Spherical diameter R : Radius SR : Spherical radius □ : Square
Fig. 28 Shape identification symbols
ARRANGEMENT OF DIMENSIONS
The arrangement of dimensions on a drawing must indicate clearly the design purpose. The
following are the ways of arranging the dimensions.
Chains dimensions: Chains of single dimensions should be used only where the possible
accumulation of tolerances does not endanger the functional requirement of the part (Fig. 29).

Fig. 29
Parallel dimension: In parallel dimensioning, a number of dimension lines, parallel to one another
and spaced-out are used. This method is used where a number of dimensions have a common
datum feature (Fig. 30a).
Fig.30a Parallel Dimension Fig. 30b Super imposed running Dimensions
Super imposed running Dimensions:
These are simplified parallel dimensions and may be used where there are space limitations
(Fig. 30b).
Combined Dimensions:
These are the result of simultaneous use of chain and parallel dimensions (Fig. 31).
Fig. 31 Combined Dimensions Fig. 32 Co-ordinate Dimensions
Co-ordinate Dimensions:
The sizes of the holes and their co-ordinates may be indicated directly on the drawing; or
they may be conveniently presented in a tabular form, as shown in Fig. 32.
Termination of Leader Line
A leader is a line referring to a feature (dimension, object, outline, etc.).
Leader lines should terminate (Fig. 33),
(a) With a dot, if they end within the outlines of an object,

(b) With an arrow head, if they end on the outline of an object,
(c) Without dot or arrow head, if they end on a dimension line.
Fig. 33 Termination of leader lines
LINES
Lines of different types and thicknesses are used for graphical representation of objects.
The types of lines and their applications are shown in Table.4.
Table.4 Types of lines and their applications

UNIT – I
PLANE CURVES AND FREE HAND SKETCHING
1.1 BASIC GEOMETRICAL CONSTRUCTIONS
Many geometrical methods are used to construct the geometrical shapes such as polygons
and circles. The commonly used methods to construct the polygons are discussed in the following
examples.
Method-I Method-II
Method-I Method-II

ENGINEERING CURVES
Frequently required engineering curves are conics, cycloids, involutes, and spirals.
1.3 CONIC SECTIONS
The sections obtained by the intersection of a right circular cone by a cutting plane in
different positions are called conic sections or conics. Refer Figure. 1.1 and Figure 1.2
Figure. 1.1 Right CircularCone

Figure. 1.2 Conic Sections
Circle: When the cutting plane is parallel to the base or perpendicular to the axis, then the true
shape of the section is circle.
Ellipse: An ellipse is obtained when a section plane A–A, inclined to the axis cuts all the
generators of the cone.
Parabola: A parabola is obtained when a section plane B–B, parallel to one of the generators cuts
the cone. Obviously, the section plane will cut the base of the cone.
Hyperbola: A hyperbola is obtained when a section plane C–C, inclined at a very small angle to
the axis cuts the cone on one side of the axis. A rectangular hyperbola is obtained when a section
plane D–D, parallel to the axis cuts the cone.
Isosceles Triangle: When the cutting plane is passing through the apex and the base of the cone,
the curve of intersection obtained is an isosceles triangle.
Conic Sections as Loci of a Moving Point
A conic section may be defined as the locus of a point moving in a plane such that the ratio
of its distance from a fixed point (Focus) and fixed straight line (Directrix) is always a constant.
The ratio is called eccentricity. The line passing through the focus and perpendicular to the
directrix is the axis of the curve. The point at which the conic section intersects the axis is called
the vertex or apex of the curve.
Eccentricity, e =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑣𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑓𝑟𝑜𝑚 𝑓𝑜𝑐𝑢𝑠
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑣𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑓𝑟𝑜𝑚 𝑑𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥
The eccentricity value is less than 1 for ellipse, equal to 1 for parabola and greater than 1
for hyperbola (Fig. 1.3).

Figure 1.3 Locus of a point
Ellipse
An ellipse is also defined as a plain curve generated by a point which moves in such a way
that, at any position the sum of its distance from two fixed points is always a constant. The fixed
points are called foci and the constant is equal to the major axis of the ellipse.
Problem 1: Draw the locus of a point P moving so that the ratio of the distance from a fixed point
F to its distance from a fixed straight line AB is
3
4
. Point F is at a distance of 70 mm from AB. Also
draw a tangent and normal to the curve.
Solution: To construct the ellipse using focus and directrix
1. Draw the directrix and axis as shown.
2. Mark F on axis such that CF= 70 mm.
3. Divide CF into 3 + 4 = 7 equal parts and mark V at the fourth division from C.
Now, e = VF/ CV = 3/4.
4. At V, erect a perpendicular VB = VF. Join CB.
5. Through F, draw a line at 45° to meet CB produced at D. Through D, drop a perpendicular DV’
on CC’. Mark O at the midpoint of V– V’.
6. Mark a few points, 1, 2, 3,… on V– V’ and erect perpendiculars though them meeting CD at 1’,
2’, 3’…. Also erect a perpendicular through O.

7. With F as a centre and radius = 1–1’, cut two arcs on the perpendicular through 1 to locate P1
and P1’. Similarly, with F as a centre and radii = 2–2’, 3–3’, etc., cut arcs on the corresponding
perpendiculars to locate P2 and P2’, P3 and P3’, etc. Also, cut similar arcs on the perpendicular
through O to locate V1 and V1’.
To draw tangent and normal to the ellipse
1. Mark the given point P on the curve and join PF.
2. At F draw a line perpendicular to PF to cut AB at T.
3. Join TP and extend it. TT is the tangent at P.
4. Through P, draw a line NN perpendicular to TT. NN is the normal at P.
Parabola
A parabola is also defined as a plain curve generated by a point which moves in such a way
that at any position, its distance from a fixed point (focus) is always equal to its distance from a
fixed straight line (directrix).
Problem 2: Construct a parabola when the distance between the focus and directrix is 60 mm.
Draw the tangent and normal at any point on the curve.
Solution: To construct the parabola using focus and directrix.
1. Draw directrix AB and axis CC’ as shown.
2. Mark F on CC’ such that CF = 60 mm.

3. Mark V at the midpoint of CF. Therefore, e = VF/CV = 1.
5. Mark a few points, say, 1, 2, 3, … on VC’ and erect perpendiculars through them meeting CB
produced at 1’, 2’, 3’, …
perpendiculars to locate P2 and P2’, P3 and P3’, etc.
7. Draw a smooth curve passing through V, P1, P2, P3 … P3’, P2’, P1’ to complete the parabola.
To draw tangent and normal to the parabola
Hyperbola
A hyperbola is also defined as a plain curve generated by a point which moves in such a
way that at any position, the difference of its distance from two fixed points is always a constant.
The fixed points are called as foci and the constant is equal to the transverse axis of the hyperbola.

Problem 3: Draw a hyperbola of e = 3/2 if the distance of the focus from the directrix = 50 mm.
Solution: To construct the hyperbola using focus and directrix.
1. Draw directrix AB and axis CC’ as shown.
2. Mark F on CC’ such that CF = 50 mm.
3. Divide CF into 3 + 2 = 5 equal parts and mark V at the second division from C.
Now, e = VF/ CV = 3/2.
5. Mark a few points, say, 1, 2, 3, … on VC’ and erect perpendiculars through them meeting CB
produced at 1’, 2’, 3’, …
perpendiculars to locate P2 and P2’, P3 and P3’, etc.
7. Draw a smooth curve passing through V, P1, P2, P3 … P3’, P2’, P1’ to complete the hyperbola
To draw tangent and normal to the hyperbola

1.4 CYCLOIDAL CURVES
A cycloid is a curve generated by a point on the circumference of a circle rolling along a
straight line without slipping. The rolling circle is called a generating circle and the straight line is
called a directing line or base line. The point on the generating circle which traces the curve is
called the generating point.
The cycloid is called the epicycloid when the generating circle rolls along another circle
outside it.
Hypocycloid, opposite to the epicycloid, is obtained when the generating circle rolls along
another circle inside it.
The other circle along which the generating circle rolls is called the directing circle or the
base circle.
Problem 1: A wheel of diameter 60 cm rolls on a straight horizontal road. Draw the locus of a
point P on the periphery of the wheel, for one revolution of the wheel, if P is initially on the road.
Solution: To construct the cycloid
1. Draw the base line P’–P” equal to the circumference of generating circle, i.e., π D = π x 60 cm =
188 cm. Take Scale: 1 : 10.

2. Draw the generating circle with C as a centre and radius = 30 cm, tangent to P’–P” at P’. Point P
is initially at P’.
3. Draw C–C’ parallel and equal to P’–P” to represent the locus of the centre of the generating
circle.
4. Obtain 12 equal divisions on the circle. Number the divisions as 1, 2, 3, etc., starting from P’ as
shown. Through 1, 2, 3, etc., draw lines parallel to P’ –P”.
5. Obtain 12 equal divisions on C–C’ and name them as C1, C2, C3, etc.
6. With C1, C2, C3, etc. as the centres and radius = CP’ = 30 mm, cut the arcs on the lines through
1, 2, 3, etc., to locate respectively P1, P2, P3, etc.
7. Join P’, P1, P2, P3, etc. by a smooth curve to complete the cycloid.
To draw tangent and normal to cycloid
1. With P as a centre and radius = CP’ (i.e., radius of generating circle), cut an arc on C–C’ at M.
2. From M, draw a normal MN to P’–P”.
3. Join NP for the required normal. Draw tangent T–T perpendicular to NP at P.
Problem 2: Draw an epicycloid, if a circle of 40 mm diameter rolls outside another circle of 120
mm diameter for one revolution. Draw the tangent and normal at any point on the curve.
Solution: We know that for one revolution of the generating circle along the directing circle, it
covers an arc of length equal to the circumference of the generating circle. It is impossible to
measure the length of an arc, so the subtended angle of it is calculated as follows.
Subtended angle, θ =
𝑟
𝑅
x 360o
r – Radius of generating circle
R – Radius of directing circle
∴ Included angle of the arc, θ =
20
60
x 360o
= 120°

1. With O as a centre and radius = 60 mm, draw the directing arc P’–P” of the included angle 120°.
2. Produce OP’ and locate C on it such that CP’ = radius of generating circle = 20 mm. With C as
centre and radius = CP’, draw a circle.
3. With O as a centre and radius = OC, draw an arc C–C’ such that ∠COC’ = 120°. Arc C–C’
represents the locus of centre of generating circle.
4. Divide the circle into 12 equal parts. With O as a centre and radii = O–1, O–2, O–3, etc., draw
the arcs through 1, 2, 3, etc., parallel to arc P’–P”.
5. Divide arc P’–P” into12 equal parts and produce them to cut the locus of centres at C1, C2,….
6. Taking C1 as centre, and radius equal to 20mm, draw an arc cutting the arc through 1 at P1.
Repeat this procedure with centres C2, C3,…, C12 to obtain points P2, P3,…., P12.
7. Join P1, P2, P3, etc by drawing a smooth curve to complete the epicycloid.
To draw tangent and normal to epicycloid
2. Join MO and then locate N at the intersection of P’–P”.
Problem 3: A circle of diameter 40 mm rolls inside another circle of radius 60 mm. Draw the
hypocycloid traced by a point on the rolling circle initially in contact with the directing circle for
one revolution. Draw the tangent and normal at any point on the curve.
Solution: We know that for one revolution of the generating circle along the directing circle, it
covers an arc of length equal to the circumference of the generating circle. It is impossible to
measure the length of an arc, so the subtended angle of it is calculated as follows.
Subtended angle, θ =
𝑟
𝑅
x 360o
r – Radius of generating circle

R – Radius of directing circle
∴ Included angle of the arc, θ =
20
60
x 360o
= 120°
1. With O as a centre and radius = 60 mm, draw the directing arc P’–P” of included angle 120°.
2. On OP’, locate C such that CP’ = 20 mm. With C as a centre and radius = CP’, draw a circle.
3. With O as a centre and radius = OC, draw an arc C–C’ such that ∠COC’ = 120°. Arc C–C’
represents the locus of centre of generating circle.
4. Divide the circle into 12 equal parts. With O as a centre and radii = O–1, O–2, O–3, etc., draw
the arcs through 1, 2, 3, etc., parallel to arc P’–P”.
5. Divide arc P’–P” into12 equal parts and produce them to cut the locus of centres at C1, C2,….
6. Taking C1 as centre, and radius equal to 20mm, draw an arc cutting the arc through 1 at P1.
Repeat this procedure with centres C2, C3,…, C12 to obtain points P2, P3,…., P12.
7. Join P1, P2, P3, etc by drawing a smooth curve to complete the hypocycloid.
To draw tangent and normal to hypocycloid
2. Join MO and then extend it to the directing circle to locate N at the intersection of P’–P”.
1.5 INVOLUTE
An involute is a spiral curve generated by a point on a cord or thread as it unwinds from a
polygon or a circle, the thread being kept tight and tangential to the circle or the sides of the
polygon.
Depending on whether the involute is traced over a circle or a polygon, the involute is
called an involute of circle or involute of polygon.
Problem 1: Draw the involute of a square of side 25 mm.
Steps for Construction of Involute of square:

1. Draw a Square ABCD of side 25mm. Assume a thread is unwound from end D in clockwise
direction.
2. A as center and radius 25 mm (= length of 1 side), draw an arc to cut BA produced at P1.
3. B as center and BP1 as radius (= length of 2 side), draw an arc from P1 to get P2.
4. C as center and CP2 as radius (= length of 3 side), draw an arc from P2 to get P3.
5. Similarly, D as center and DP3 as radius (= length of 4 side), draw an arc from P3 to get P4.
Normal and Tangent:
1. The given point M lies in the arc P3 P4 .The center of the arc P3 P4 is point B. Join B and M.
2. Extend BM which is required Normal. At M draw a perpendicular to the normal to obtain the
Tangent TT.
Problem 2: Draw the curve traced out by an end of a thin wire unwound from a regular hexagon
of side 20 mm, the wire being kept taut. Draw a tangent and normal at any point on the curve.

Steps for Construction of Involute of hexagon:
1. Draw a hexagon ABCDEF of side 20mm. Assume a thread is unwound from end F in clockwise
direction.
2. With centres A, B, etc., and radius 20 mm, 40 mm, etc., draw arcs to get P1, P2, etc.,
3. Join P1, P2, etc., to complete the hexagon.
Normal and Tangent:
1. Mark the point M on the arc P4 P5 .The center of the arc P4 P5 is point F. Join F and M.
2. Extend FM which is required Normal. At M draw a perpendicular to the normal to obtain the
Tangent TT.
Problem 3: Construct one convolution of an involute of a circle of diameter 40 mm. Draw tangent
and normal at a point on the involute 100 mm distant from the centre of the circle.

Steps for Construction of Involute of circle:
1. With 0 as centre and radius R of 20 mm, draw the given circle.
2. Taking P as the starting point, draw the tangent PQ equal in length to the circumference of the
circle.
3. Divide the line PQ and the circle into the same number of equal parts and number the points.
4. Draw tangents to the circle at the points 1, 2, 3 etc., and locate the points P1, P2, P3 etc., such
that 1-P1 = P-1’, 2-P2 = P-2’ etc.
5. A smooth curve through the points P, P1, P2 etc., is the required involute.
Normal and Tangent:
1.Mark the point M at a distance of 100 mm from the centre of the circle. Join it with centre O of
the circle.
2. Draw a semi-circle with diameter OM, i.e., with centre C which is the midpoint of OM to get the
intersection point N on the circle.
3. Draw a line from N passing through M which is the normal NMN to the curve.
4. Draw a line through M, perpendicular to NMN to get the tangent TT to the curve.
1.6 FREEHAND SKETCHING
An engineer or designer conceives an idea of a non-existing object three dimensionally,
which can be conveyed to another person only through a drawing. Initially the object is sketched
on paper as an isometric or perspective drawing, then that will be given dimensions.

Good practice in freehand sketching helps the engineer to think about the new design rather
than to think about the method of preparing the drawing. When the designer’s mind thinks an idea
that is sketched by him in freehand. An engineer who has a thorough idea in isometric and
orthographic projections can prepare the sketches easily.
Freehand Sketching Practice
1. For freehand sketching practice, soft grade pencil (HB), eraser and paper are required.
2. Papers with thin cross-section guidelines (ruled paper) may be used for sketching which helps
the inexperienced person to draw straight lines satisfactorily.
Fig. 1.4 Sketch on a ruled paper
Sketching a straight line
1. Mark the end points of the line to be sketched.
2. Observe the two points carefully and make some trial movements between the points without
sketching any line.
3. Sketch light and thin trial lines connecting the two points.
4. Complete the straight line by drawing a straight and dark line over the thin line.
Fig. 1.5 Steps to sketch a straight line
Sketching a small circle
1. Sketch the center lines for the circle horizontally and vertically and mark four points on them
approximately equal to the radius of the circle.
2. Sketch the light circle passing through these points.
3. Complete the circle with dark lines.

Fig. 1.6Sketching a small circle
Sketching a big circle
1. Sketch the center lines for the circle horizontally and vertically and mark four points on them
approximately equal to the radius of the circle.
2. Sketch more diagonal lines in addition to the center lines for big circle.
3. Sketch the light circle passing through these points.
4. Complete the circle with dark lines.
Sketching Orthographic views from pictorial view
The methods and rules followed to prepare orthographic projections is used to sketch the
multiviews of an object. Consider the following steps to draw the orthographic projections.
1. Observe the shape of the object carefully.
2. Decide the number of views to be selected.
3. Sketch thin and light rectangles for the views to be drawn.
4. Sketch and complete the views, then darken the lines.
5. Sketch the dimensions, notes, drawing name, title, etc.
6. Check the correctness of the lines and details given in various views.

Fig. 1.7 sketching a large circle
General Procedure to draw free hand sketches
1. Mark their directions in the Pictorial View.
2. Measure the overall Length L, Width W and Height H of the object.
3. Using L, W and H, mark spaces for the front and Right Side Views in the form of rectangles [2H
Pencil].
4. Mark the axes at the appropriate places [2H Pencil].
5. Sketch the details simultaneously in both Front and Right Side Views by drawing the
corresponding projects [2H Pencil].
6. Darken all the visible lines [HB Pencil].
7. Add the dimensions.
Problem 1: Sketch the front, top and left side views of an object shown in the figure.

Solution:
Problem 2: Sketch the front, top and right side views of an object shown in the figure.

Solution:
Problem 3: Sketch the front, top and left side views of the machine component shown in the
figure.

Solution:
Problem 4: Make free hand sketch of the front, top and right side views of the object shown in the
figure.

Solution:
Problem 5: Make free hand sketch of the front, top and right side views of the object shown in the
figure.

Solution:
Problem 6: Draw the plan, elevation and left side views for the given isometric view.

Solution:
1.7 ASSIGNMENT PROBLEMS
PLAIN SCALE
1. Construct a plain scale to show meters when 1 cm represents 4 meters and long enough to
measure up to 50 meters. Find the R.F and mark the distance of 36 meter and 28 meter.
2. A room of 1000 m3
volume is represented by block of 125 cm3
volume. Find R.F and construct
a plain scale to measure up to 30 m. measure the distance of 18 m on the scale.

DIAGONAL SCALE
1. A distance between Coimbatore and Madurai is 200 km and its equivalent distance on the map
measures 10 cm. Draw the diagonal scale to indicate 223 km and 135 km.
2. A rectangular plot of land measuring 1.28 hectors represented on a map by a similar rectangle
of 8 sq.cm. Calculate R.F of scale. Draw diagonal scale to read single meter. Show a distance
of 438 m on it.
VERNIER SCALE
1. Draw a vernier scale of R.F = 1/25 to read centimeters up to 4 meters and on it,show lengths
2.39 m and 0.91 m.
2. A map of size 500 cm X 50 cm wide represents an area of 6250 sq.kms. Construct a vernier
scale to measure kilometers, hectometers and decameters and long enough to measure upto
7km. Indicate on it (a) 5.33 km (b) 59 decameters.
ELLIPSE
1. Draw the locus of a point P moving so that the ratio of the distance from a fixed point F to its
distance from a fixed straight line DD’ is ¾. Point F is at a distance of 35 mm from DD’. Also
draw a tangent and normal to the curve.
PARABOLA
2. Construct a parabola when the distance between the focus and directrix is 30 mm. Draw the
tangent and normal at any point on the curve.
HYPERBOLA
3. The vertex of the hyperbola is 65 mm from its focus. Draw the curve if the eccentricity is 3/2.
Draw also a tangent and normal at any point on the curve.
CYCLOID
4. A circle of 50 mm diameter rolls on a straight line without slipping. Trace the locus of a point
‘P’ on the circumference of the circle rolling for one revolution. Name the curve. Draw normal and
tangent to the curve at any point on the curve.
EPICYCLOID
5. Construct an epicycloids generated by a rolling circle of diameter 50 mm and a directing circle
of diameter 150 mm. Draw the tangent and normal to the curve at any point on the epicycloid.
HYPOCYCLOID
6. Draw a hypocycloid of a circle of 40 mm diameter which rolls inside another circle of 200 mm
diameter for one revolution. Draw a tangent and normal at any point on it.

7. Draw a hypocycloid when the radius of the directing circle is twice the radius of the generating
circle. The radius of the generating circle is equal to 35 mm.
SQUARE INVOLUTE
8. Draw the involute of a square of side 40 mm.
CIRCLE INVOLUTE
9. Construct one convolution of an involute of a circle of diameter 30 mm. Draw tangent and
normal at appoint on the involute 65 mm distant from the centre of the circle.
FREEHAND SKETCHING
10. Draw the front, top and any one of the side views for all the objects shown below.
1 2
3 4

5 6
1.8 UNIVERSITY QUESTIONS
SCALES
1. A water tank of size 27 m3
was represented in the drawing by 216 cm3
size. Construct a vernier
scale for the same to measure up to 5 m. Show on it, the following lengths (i) 3.95 m
(ii) 0.27 m (iii) 0.042 m. (Nov 2014)
2. The distance between Chennai and Madurai is 400 Km. It is represented by a distance of 8 cm
on a railway map. Find the R.F and construct a diagonal scale to read kilometers. Show on it
the distances of 543 km, 212 km and 408 km. (Jan 2014)
ELLIPSE
3. Draw the locus of a point “P” which moves in a plane in such a way that the ratio of its
distances from a fixed point F and a fixed straight line AB is always 2/3. The distance between
the fixed point F and fixed straight line is 50 mm. Also draw a tangent and normal on the point
on the locus at a horizontal distance of 55 mm from the fixed straight line. (Nov 2011),(Jan
2012)
4. Construct an ellipse when the distance b/w the focus and the directrix is 40mm and the
eccentricity is 3/4. Draw the tangent and normal at any point P on the curve using directrix.
5. Construct an Ellipse when the distance between the fixed line and moving point is 20 mm and
the eccentricity is 4/5, and also draw the tangent and normal curve at any point.
PARABOLA
6. Construct a parabola, with the distance of the focus from the directrix as 50 mm. Also, draw a
normal and tangent to the curve at a point 40 mm from the directrix. (Nov 2014)

7. The head lamp reflector of a motor car has a maximum rim diameter of 130 mm and maximum
depth of 100 mm. Draw the profile of the reflector and name it. (Jan
2013)
8. The focus of a conic is 50 mm from the directrix. Draw the locus of a point “P” moving in such
a way that its distance from the directrix is equal to its distance from the focus. Name the
curve. Draw the tangent to curve at a point 60 mm from the directrix.
(June 2011)
HYPERBOLA
9. Construct a hyperbola when the distance between the focus and the directrix is 40mm and the
eccentricity is 4/3.Draw a tangent & normal at any point on the hyperbola. (Jan 2013)
10. Draw a hyperbola when the distance of the focus from the directrix is 70 mm and the
eccentricity e is 1.5. Draw the tangent and normal to the curve at a point P distance 50 mm
from the directrix. (June 2012)
CYCLOID
11. A circle of 40mm diameter rolls over a horizontal table without slipping. A point on the
circumference of the circle is in contact with the table surface in the beginning and after one
complete revolution. Draw the path traced by the point. Draw a tangent and normal at any
point on the curve.
12. A roller of 35mm diameter rolls on a straight line without slip. In the initial position the
diameter PQ of the circle is parallel to the line on which it rolls. Draw the locus of the point.
EPICYCLOID
13. Construct the path traced by a point on a circular disc radius of 30 mm rolls on a circular path
of radius 100mm. Draw the tangent and normal curve at any point on the curve.
14. A circular disc of radius 30 mm rolls outside another circle of radius 90 mm for one complete
revolution. Draw the locus of the point also draw the tangent and normal curve at any point on
the curve.
HYPOCYCLOID
15. A circus man rides a motor bike inside a globe of 6 m diameter. The motor bike has the wheel
of 1 m diameter. Draw the locus of the point on the circumference of motorbike wheel for one
complete revolution. Adopt suitable scale. (June 2014)
16. A circular disc of radius 40 mm rolls inside another circle of radius 90 mm for one revolution.
Draw the locus of a point also draw the tangent and normal curve at any point on the curve.
CIRCLE INVOLUTE
17. Draw the involute of a circle of 40 mm diameter. Also, draw a tangent and a normal to the
curve at a point 95mm from the center of the circle. (Jan 2013) (Apr 2015)

18. Draw the involute of a circle of diameter 50 mm when string is unwound in the clockwise
direction. Draw tangent and normal at a point located on the involute. (Jan 2012)
19. A coir is unwound from a drum of 30mm diameter, draw the locus of free end of the coir for
unwinding through an angle of 360°
20. A string of length 135 mm is wound around a circle diameter of 36 mm. It is unwound from
the circle, trace path of the end of string. Draw a tangent and normal curve at any point on the
curve.
SQUARE INVOLUTE
21. Draw the involute of a square of side 30 mm. Draw a tangent and normal at any point on the
involute. (Jan 2014)
22. Draw an involute of a square of side 35mm, Draw the tangent and normal at any point.
FREEHAND SKETCHING
Draw the front, top and any one of the side views for all the objects shown below.
APR 2015, JUNE 2005 JAN 2014, NOV 2010
NOV 2014 NOV 2014

JUNE 2009, JAN 2010 JAN 2012
NOV 2010 MAY 2012

MAY 2010 JUNE 2009
JAN 2006 JAN 2013

UNIT – II
PROJECTION OF POINTS, LINES AND PLANE SURFACES
2.1 PRINCIPLES OF PROJECTION
Projecting the image of an object to the plane of projection is known as projection. The
object may be a point, line, plane, solid, machine component or a building. The figure or view
JAN 2010 Additional 1
Additional 2 Additional 3

formed by joining, in correct sequence, the points at which these lines meet the plane is called the
projection of the object. (It is obvious that the outlines of the shadow are the projections of an
object).
Figure 2.1 Projection of an object
Projectors
The lines or rays drawn from the object to the plane are called projectors.
Plane of Projections
The transparent plane on which the projections are drawn is known as plane of projectors.
2.2 TYPES OF PROJECTIONS
1. Pictorial Projections
a) Perspective Projection
b) Isometric Projections
c) Oblique Projections
2. Orthographic Projections
1) Pictorial Projections
The projection in which the description of the object is completely understood in one view
is known as Pictorial Projection. The Pictorial projections have the advantage of conveying an
immediate impression of the general shape and details of the object, but no its true dimensions or
sizes.
Note: Isometric projection gives true shape of the object, while Perspective and Oblique
Projections do not.
a) Perspective Projection
Imagine that the observer looks at the object form an infinite distance. The rays will now be
parallel to each other and perpendicular in both the front surface of the object and the plane, when
the observer is at a finite distance from the object, the rays converge to the eye as in the case of
Perspective Projection.
The observer looks from the front. The front surface F of the block is seen in its true shape
and size.

Note: Orthographic Projection is the standard drawing form of the industrial world. The form is
unreal in that we do not see an object as it is drawn orthographically. Pictorial drawing however
has photographic realism.
Figure 2.2 Perspective Projection
Perspective View
If any imaginary transparent plane is introduced such that the object is in between the
observer and the plane. The image obtained on the plane/screen is as shown. This is called
perspective view of the object.
b) Isometric Projection
“Iso” means “equal” and “metric projection” means “a projection to a reduced measure”.
An Isometric Projection is one type of pictorial projection in which the three dimensions of a solid
are not only shown in one view, but also their dimension can be scaled from this drawing.
Figure 2.3 Isometric Projection
c) Oblique Projection
The word “oblique” means “slanting” There are three axes-vertical, horizontal and oblique.
The oblique axis, called receding axis is drawn either at 30o
or 45o
. Thus an oblique drawing can
be drawn directly without resorting to projection techniques.

Figure 2.4 Oblique Projection
2) Orthographic Projection
'ORTHO' means right angle and orthographic means right angled drawing. When the
projectors are perpendicular to the plane on which the projection is obtained, it is known as
orthographic projection.
2.3 PRINCIPAL PLANES
In engineering drawing practice, two principal planes are used to get the projection of an
object as shown in the figure 2.5. They are,
(i) VERTICAL PLANE (VP) which is assumed to be placed vertically. The front view of the
object is projected onto this plane.
(ii) HORIZONTAL PLANE (HP) which is assumed to be placed horizontally. The top view
of the object is projected onto this plane.
These principal planes are also known as reference planes or co-ordinate planes. The planes
considered are imaginary, transparent and dimensionless. The reference planes VP and HP are
placed in such a way that they intersect each other at right angles. As a result of intersection, an
intersection line is obtained which is known as the reference line or XY line.
Four Quadrants
When the planes of projections are extended beyond their line of intersection, they form
Four Quadrants. These quadrants are numbered as I, II, III and IV in clockwise direction when
rotated about reference line XY as shown in Figure 2.5.

Figure 2.5 Principal Planes and Four Quadrants
First Angle Projection
When the object is situated in First Quadrant, that is, in front of V.P and above H.P, the
projections obtained on these planes is called First angle projection.
(i) The object lies in between the observer and the plane of projection.
(ii) The front view is drawn above the XY line and the top view below XY. (Above XY line is V.P
and below XY line is H.P).
(iii) In the front view, H.P coincides with XY line and in top view V.P coincides with XY line.
(iv) Front view shows the length (L) and height (H) of the object and Top view shows the length
(L) and breadth (B) or width (W) or thickness (T) of it.
Figure 2.5 First Angle and Third Angle Projection
Third Angle Projection
In this, the object is situated in Third Quadrant. The Planes of projection lie between the object and
the observer. The front view comes below the XY line and the top view about it.

Notation followed:
1. Actual points in space are denoted by capital letters A, B, C.
2. Their front views are denoted by their corresponding lower case letters with dashes a’, b’, c’
etc., and their top views by the lower case letters a, b, c etc.
3. The projections in top and front views are drawn in thick lines. Projectors are always drawn as
continuous thin lines.
Note:
1. Students are advised to make their own paper/card board/perplex model of H.P and V.P. The
model will facilitate developing a good concept of the relative position of the points lying in any of
the four quadrants.
2. Since the projections of points, lines and planes are the basic chapters for the subsequent topics
on solids viz, projection of solids, development, pictorial drawings and conversion of pictorial to
orthographic and vice versa, the students should follow these basic chapters carefully to draw the
projections.
PROJECTION OF POINTS IN SPACE
2.4 VARIOUS POSITIONS OF POINTS IN SPACE
In FIRST Quadrant (Above H.P., In front of V.P.)
In SECOND Quadrant (Above H.P., Behind V.P.)

In THIRD Quadrant (Below H.P., Behind V.P.)
In FOURTH Quadrant (Below H.P., In front of V.P.)

In PLANE (On V.P., Above H.P.)
In PLANE (On H.P., Behind V.P.)

In PLANE (On V.P., Below H.P.)
In PLANE (On H.P., In front of V.P.)

In PLANE (On both H.P. & V.P.)
Problem 1: Draw the orthographic projections of the following points.
(a) Point P is 30 mm above H.P and 40 mm in front of VP
(b) Point Q is 25 mm above H.P and 35 mm behind VP
(c) Point R is 32 mm below H.P and 45 mm behind VP
(d) Point Sis 35 mm below H.P and 42 mm in front of VP
(e) Point T is in H.P and 30 mm is behind VP
(f) Point U is in VP and 40 mm below HP

(g) Point V is in VP and 35 mm above H.P
(h) Point W is in H.P and 48 mm in front of VP
Solution: The location of the given points is the appropriate quadrants are shown in above figure
(a) and their orthographic projections are shown figure (b).
Problem 2: Draw the projections of the following points on a common reference line. Take 30 mm
distance between the projectors.
1) A, 35 mm above HP and 25 mm in front of VP.
2) B, 40 mm below HP and 15 mm behind VP.
3) C, 50 mm above HP and 25 mm behind VP.
4) D, 45 mm below HP and 20 mm in front of VP.
5) E, 30 mm behind VP and on HP.
6) F, 35 mm below HP and on VP.
7) G, on both HP and VP.
8) H, 25 mm below HP and 25 mm in front of VP.

PROJECTION OF STRAIGHT LINES
2.5 BASICS OF STRAIGHT LINE
A line is a geometric primitive that has length and direction, but no thickness. Straight line
is the Locus of a point, which moves linearly. A straight line is the shortest distance between two
points. Projections of the ends of any line can be drawn using the principles developed for
projections of points. Top views of the two end points of a line, when joined, give the top view of
the line. Front views of the two end points of the line, when joined, give the front view of the line.
Both these Projections are straight lines.
Figure 2.6 A straight Line
The projections of a straight line are obtained by joining the top and front views of the
respective end points of the line. The actual length of the straight line is known as true length (TL).
2.6 PROJECTIONS OF STRAIGHT LINE
A straight line is placed with reference to the planes of projections in the following
positions.
1. Line perpendicular to HP and parallel to VP
2. Line perpendicular to VP and parallel to HP
3. Line parallel to both HP and VP
4. Line inclined to HP and parallel to VP
5. Line inclined to VP and parallel to HP
6. Line inclined to both HP and VP
In first angle projection method, the line is assumed to be placed in the first quadrant. The
projections of the line in the above mentioned positions are discussed below.
Projections of a Line kept Perpendicular to HP and Parallel to VP
Consider a straight line AB kept perpendicular to HP and parallel to VP (Fig 2.7a). Its front
view is projected onto VP which is a line having true length. The top view is projected onto HP
which is a point, one end b of which is visible while the other end ‘a’is invisible and is enclosed

within ( ).
Figure 2.7 Projections of a Line kept Perpendicular to HP and Parallel to VP
Now the HP is rotated in the clockwise direction through 90° and is obtained in the vertical
position. The projections obtained are seen as given in Fig. 2.7b. It is drawn with reference to the
XY line as follows.
1. Draw the XY line.
2. Draw the front view a′b′, which is a line perpendicular to XY and having true length (TL).
3. Project the top view ab. The end b is visible and the invisible end ‘a’ is marked inside ( ).
Projections of a Line kept Perpendicular to VP and Parallel to HP
Consider a straight line AB kept perpendicular to VP and parallel to HP (Fig. 2.8a). Itstop
view is projected onto HP which is a line having true length (TL). The front view isprojected onto
VP which is a point, the end b′ of which is visible and the other end a′ isinvisible which is shown
enclosed in ( ).
Now the HP is rotated in the clockwise direction through 90° and is obtained in thevertical
position. The projections obtained are seen as given in Fig. 2.8b. It is drawnwith reference to the
XY line as follows.
2. Draw the top view ab, a line perpendicular to XY and having true length (TL).
3. Project the front view a′b′. The end b′ is visible and the invisible end a′ is markedinside ( ).

Figure 2.8 Projections of a Line kept Perpendicular to VP and Parallel to HP
Projections of a Line kept parallel to Both HP and VP
Consider a straight line AB kept parallel to both HP and VP (Fig. 2.9a). Its front viewis
projected onto VP which is a line having true length (TL). The top view is projected onto HP
which is also a line having true length.
Figure 2.9 Projections of a Line kept parallel to Both HP and VP
Now the HP is rotated in the clockwise direction through an angle of 90° andis obtained in
the vertical position. The projections obtained are seen as given inFig. 2.9b. It is drawn with
references to the XY line as follows.
2. Draw the front view a′b′, a line parallel to XY and having true length (TL).
3. Project the top view ab which is also a line parallel to XY having true length (TL).
Projections of a Line kept Inclined to HP and Parallel to VP

Consider a straight line AB kept inclined to HP and parallel to VP (Fig. 2.10a).
Figure 2.10 Projections of a Line kept Inclined to HP and Parallel to VP
Its front view is projected onto VP which is an inclined line at an angle θ to XY and having
true length (TL). The top view is projected onto HP which is also a line but smaller than the true
length and parallel to XY. The inclination of the line with HP is always represented by the symbol
θ.
Now the HP is rotated in the clockwise direction through 90° and is obtained in the vertical
position. The projections obtained are seen as given in Fig. 2.10b. It is drawn with reference to the
XY line as follows.
2. Draw the front view a′b′, a line inclined at an angle θ to XY and having true length (TL).
3. Project the top view ab which is also a line parallel to XY and smaller than true length.
Projections of a Line kept Inclined to VP and Parallel to HP
Consider a straight line AB kept inclined to VP and parallel to HP (Fig. 2.11a). Its topview
is projected onto HP which is a line inclined at an angle ɸ to XY and having truelength (TL). The
front view is projected onto VP which is also a line but smaller thantrue length and is parallel to
XY. The inclination of the line with VP is always representedby the symbolɸ.

Figure 2.11 Projections of a Line kept Inclined to VP and Parallel to HP
Now the HP is rotated in the clockwise direction through an angle of 90° andis obtained in
the vertical position. The projections obtained are seen as given inFig. 2.11b. It is drawn with
reference to the XY line as follows.
2. Draw the top view ab, a line inclined at an angle ɸ to XY and having true length(TL).
3. Project the front view a′b′, which is also a line parallel to XY but smaller thantrue length.
2.7 TRACE OF A LINE
The point of intersection or meeting of a line with the reference plane, extended
ifnecessary, is known as the trace of a line. The point of intersection of a line with the HPis known
as the horizontal trace, represented by HT and that with the VP is known asthe vertical trace,
represented by VT. No trace is obtained when a line is kept parallel toa reference plane. Note that
HT always lies on plan or extended plan, VT always lies onelevation or extended elevation.
Problem 1:A line AB 60 mm long has its end A 20 mm above HP and 30 mm infront of VP. The
line is kept perpendicular to HP and parallel to VP. Draw its projections.Also mark the traces.
Solution: Assume that end A of the line is nearer to HP. The front view a′b′ is a linehaving true
length. The top view is a point, the end b of which is visible and a is invisiblewhich is enclosed in (
).The line is extended to meet HP to obtain the horizontal trace (HT). No vertical trace(VT) is
obtained because the line is kept parallel to VP.
The projections obtained are drawn with reference to XY line.

1. Mark the projections of the end A by considering it as a point. Its front view a′is 20 mm above
XY and the top view a is 30 mm below XY.
2. The front view of the line a′b′ is obtained by drawing a line perpendicular to XYfrom a′ and
having a length of 60 mm.
3. Top view of the line is obtained by projecting the other end b which coincideswith a. The
invisible end a is enclosed in ( ).
4. The horizontal trace (HT) is marked coinciding with the top view of the line. Novertical trace
(VT) is obtained.
Problem 2: A line AB 60 mm long has its end A 20 mm above HP and 30 mm infront of VP. The
line is kept perpendicular to VP and parallel to HP. Draw its projections.Also mark the traces.
Solution:Assume that end A of the line is nearer to VP. The top view ab is a line havingtrue
length. The front view is a point, the end b′ of which is visible and a′ is invisiblewhich is enclosed
in ( ).The line is extended to meet VP to obtain the vertical trace (VT). No horizontal trace(HT) is
obtained because the line is kept parallel to HP.
The projections obtained are drawn with reference to the XY line
1. Mark the projections of the end A by considering it as a point. Its front view a′ is20 mm above
XY and top view a is 30 mm below XY.
2. Top view of the line ab is obtained by drawing a line of length 60 mm perpendicularto XY from
a.
3. Front view of the line is obtained by projecting the other end b′ which coincideswith a′. The
invisible end a′ is enclosed in ( ).
4. The vertical trace (VT) is marked coinciding with the front view of the line. Nohorizontal trace
(HT) is obtained.

line is kept parallel to both HP and VP. Draw its projections.
Solution: The front view a′b′ and top view ab are lines having true lengths. No horizontaland
vertical traces are obtained because the line is kept parallel to both HP and VP.
The projections obtained are drawn with reference to the XY line.
1. Mark the projections of end A by considering it as a point. Its front view a′ is20 mm above XY
and top view a is 30 mm below XY.
2. The front view of the line a′b′ is obtained by drawing a line parallel to XY from a′having a
length of 60 mm.
3. The top view of the line ab is obtained by drawing another line parallel to XY froma, also of
length 60 mm.
4. No traces are marked because the line is kept parallel to both HP and VP.
line is kept inclined at 40° to HP and parallel to VP. Draw its projections,also mark the traces.

Solution:The front view a′b′ is a line inclined at an angle of 40° to XY having true length.Top
view ab is parallel to XY and smaller than true length.The line is extended to meet HP to obtain
the horizontal trace (HT). No vertical trace(VT) is obtained because the line is kept parallel to VP.
The projections obtained are drawn with reference to the XY.
2. The front view of the line a′b′ is obtained by drawing a line inclined at 40° to XYfrom a′ and
having a length of 60 mm.
3. The top view of the line ab is obtained by drawing a line parallel to XY from a anddrawing a
vertical line (projector) from b′. It is parallel to XY and smaller than thetrue length.
4. To mark the horizontal trace (HT), the front view of the line a′b′ is extended tointersect with the
XY line at h′. Then by drawing a vertical line from h′ and ahorizontal line from the top view ab,
the HT is located. No vertical trace (VT) isobtained.
line is inclined at 40° to VP and parallel to HP. Draw its projections.Also mark the traces.
Solution: The top view ab is a line inclined at an angle of 40° to XY and having truelength. Its
front view a′b′ is parallel to XY and smaller than true length.The line is extended to meet VP to
obtain the vertical trace (VT). No horizontal trace(HT) is obtained because the line is kept parallel
to HP. The projections obtained are drawn with reference to XY line.
2. The top view of the line ab is obtained by drawing a line inclined at an angle 40°to XY from a
and having a length of 60 mm.
3. The front view of the line a′b′ is obtained by drawing a line parallel to XY from a′and drawing a
vertical line (projector) from b. It is parallel to XY and smaller thanthe true length.

4. To mark the vertical trace (VT) the top view of the line ab is extended to intersectwith XY line
at v. Then by drawing a vertical line from v and a horizontal linefrom a′b′, the VT is located. No
horizontal trace (HT) is obtained.
 Tips to solve problems
When a line is inclined to one plane and parallel to the other plane.
(a) When a line is inclined to HP and parallel to VP: The projections are usuallyobtained as
follows:
There are three variables namely TL, θ and TV marked in the drawing. In a problemusually
any two variable values will be given and the third variable value can be obtainedgraphically by
completing the drawing as mentioned below. Draw the projections a′ anda of the given end A.
(i) When TL and θ are given. Draw the front view a′b′ from a′ using TL and θ. Topview (TV) ab is
projected and obtained by drawing a line parallel to the XY lineand a vertical line (projector) from
b′.
(ii) When TL and TV are given. Draw the top view ab using TV parallel to XY line.Draw the
vertical line (projector) from b. Using TL as radius and a′ as centre, marka point in the vertical line
to get b′. Join a′ and b′ to complete the front view a′b′of the line. The inclination of a′b′ with XY is
measured to get θ.
(iii) When TV and θ are given. Draw the top view using the length of TV parallel to theXY line.
Draw the vertical line (projector) from b. Using the angle θ, draw a linewhich intersects the
projector at b′. Join a′ and b′ to complete the front view a′b′of the line. The length of a′b′ is
measured to get TL.

(b) When a line is inclined to VP and parallel to HP: The projections are usuallyobtained as
follows.
There are three variables namely TL, θ and FV marked in the drawing. In a problemusually
any two variable values will be given and the third variable value can be obtainedgraphically by
completing the drawing as mentioned below. Draw the projections a′ anda of the given end A.
(i) When TL and ɸ are given. Draw the top view ab using TL and ɸ. The front view(FV) a′b′ is
projected and obtained by drawing a line parallel to XY and a verticalline (projector) from b.
(ii) When TL and FV are given. Draw the front view a′b′ using FV parallel to the XYline. Draw
the vertical line (projector) from b′. Using TL as radius and a as centre,mark a point in the vertical
line to get b. Join a and b to complete the top view abof the line. The inclination of ab with XY is
measured to get ɸ.
(iii) When FV and ɸ are given. Draw the front view a′b′ using FV parallel to the XYline. Draw the
vertical line (projector) from b′. Using the angle ɸ, draw a line whichintersects with the projector
at b. Join a and b to complete the top view ab of theline. The length of ab is measured to get TL.
Problem 6: A line PQ 80 mm long has its end P 30 mm above HP and 15 mm in frontof VP. Its
top view (plan) has a length of 50 mm. Draw its projections when the line iskept parallel to VP and
inclined to HP. Also find the inclination of the line with HP.
Solution:
The projections of the line are drawn with reference to the XY line as follows.
1. Mark the projections of end P by considering it as a point. Its front view p’ is30 mm above XY

and top view p is 15 mm below XY.
2. The top view of the line pq is drawn parallel to XY to the given length of 50 mm.
3. Draw a vertical line (projector) from q.
4. Using true length 80 mm as radius and p’ as center, mark a point in the verticalline to get q’.
Join p’ and q’ to complete the front view p’q’ of the line.
5. The inclination of p’q’ with XY is measured to get θ.
Problem 7: A line GH 60 mm long has its end G 15 mm above HP and 20 mm infront of VP. Its
front view has a length of 40 mm. The line is kept parallel to HP andinclined to VP. Draw its
projections and find the true inclination of the line with VP.
Solution: The projections of the line are drawn with reference to the XY line as follows.
1. Mark the projections of end G by considering it as a point. Its front view g’ is15 mm above XY
2. The front view g’h’ of the line is drawn parallel to XY for a length of 40 mm.
3. Draw a vertical line (projector) from h’.
4. Using true length 60 mm as radius and g as centre, mark a point in the verticalline to get h. Join
g and h to complete the top view gh of the line.
5. The inclination of gh with XY is measured to get ɸ.

Problem 8: A line RS 70 mm long has its end R 20 mm above HP and 25 mm in frontof VP. The
line is inclined to HP and parallel to VP. Draw its projections when thedistance between the
projectors is 45 mm. Also mark the traces of the line.
Solution: The projections of the line are drawn with reference to the XY line as follows.
1. Mark the projections of end R by considering it as a point. Its front view r’ is15 mm above XY
and top view r is 25 mm below XY.
2. The top view of the line rs is drawn parallel to XY to the given length of 45 mm.
Note that the distance between end projectors is equal to the length of top view.
3. Draw a vertical line (projector) from s.
4. Using true length 70 mm as radius and r’ as centre, mark a point in the verticalline to get s’. Join
r’ and s’ to complete the front view r’s’ of the line.
5. The inclination of r’s’ with XY line is measured to get q.
Problem 9: A line PQ has its end P 25 mm above HP and 15 mm in front of VP. Itsplan has a
length of 45 mm. The line is inclined at 45° to HP and parallel to VP. Drawits projections and find
the true length of the line.
Solution:The projections of the line are drawn with reference to the XY line as follows.

1. Mark the projections of end P by considering it as a point. Its front view p’ is25 mm above XY
and top view p is 15 mm below XY.
2. The top view of the line pq is drawn parallel to XY to the given length of 45 mm.
3. Draw a vertical line (projector) from q.
4. Using the angle 45°, draw a line from p’ to get the front view p’q’ of the line.
5. The length of p’q’ is measured to get the true length of the line.
Problem 10: A line EF 50 mm long is in VP and inclined to HP. The top view measures30 mm.
The end E is 10 mm above HP. Draw the projections of the line.
Solution: The projections obtained are drawn with reference to the XY line as follows.
1. Mark the projections of end E by considering it as a point. Its front view e’ is10 mm above XY
and top view e is on the XY line.
2. The top view of the line ef is obtained by drawing a line on XY from e to the givenlength of 30
mm.
3. Draw a vertical line (projector) from f.
4. Using true length 50 mm as radius end e’ as centre, mark a point in the verticalline to get f ’.
Join e’ and f ’ to complete the front view e’f ’ of the line.
5. The inclination of e’f ’ with XY line is measured to get θ.
Projections of a Line kept Inclined to Both VP and HP

When a line is placed inclined to both HP and VP, its projections obtained in top and front
views are smaller than the true length of the line and inclined to the XY line. So it is impossible to
project and draw the top or front view of the line directly. Any one of the following methods may
be used to draw the projections.
(i) Rotating line method
(ii) Rotating trapezoidal plane method
(iii) Auxiliary plane method
ROTATING LINE METHOD
Consider a line AB is placed inclined at θ to HP and ɸ to VP. Draw its projections
assuming that the line is placed in the first quadrant. The following steps are to find the top view
(plan) and front view (elevation) lengths and then, they are rotated to the required position to
represent the projections of the line in the given position.
Mark the projections of the end A by considering it as a point. Its front view a’ will be
obtained above XY and top view a will be obtained below the XY line.
Step 1: Assume that the line is kept inclined to HP and parallel to VP. Draw the front view a’b1’,
it is a line inclined at q to XY and having true length (TL). Project and get the top view ab1 length
which is parallel to XY line. Then this will be rotated to the required position (Fig. 2.12 (i)).
Step 2: Assume the line is kept inclined to VP and parallel to HP. Draw the top view ab2, it is a
line inclined atɸto XY and having true length (TL). Project and get the front view ab2’ length
which is parallel to XY line. Then this will be rotated to the required position (Fig. 2.12 (ii)).
Figure 2.12 (i) Figure 2.12 (ii)
Step 3: Draw the locus of the other end B of the line in top and front views. Draw the locus of the
front view b’ as a line passing through b1’ and parallel to XY line. Draw the locus of the top view
b as a line passing through b2 and parallel to XY line (Fig. 2.12 (iii)). Note that step 1 and step 2
are shown together.

Figure 2.12 (ii)
Step 4: Rotate the top view ab1 and front view a’b2’ to the required position. Take a ascentre, top
view length ab1 as radius, draw an arc to intersect with the locus of b at b.Join a and b to get the
top view ab of the line in required position. Taking a’ as centre,front view a’ b2’ as radius, draw an
arc to intersect the locus of b’ at b’. Join a’ and b’ toget the front view a’ b’ of the line in required
position (Fig. 2.12 (iv)).
Check the drawings obtained, by drawing the projector for the end B by joining b’and b
which is a line always perpendicular to XY line.

Figure 2.12 (iv)
Note:
1. The front view a’ b’ is inclined to XY and is known as the apparent inclinationwith HP,
represented by symbol α. It is always greater than the true inclination ofthe line with HP, denoted
by θ.
2. The top view ab is inclined to XY and is known as the apparent inclination withVP, represented
by symbol β. It is always greater than the true inclination of theline with VP, denoted by ɸ.
3. Check the result obtained by drawing the projector joining b’ and b which shouldbe a vertical
line (perpendicular to XY).
Tips to Solve Problems when a Line is Inclined to Both HP and VP
1. You require thorough knowledge in solving problems when a line is placed inclinedto one plane
and parallel to the other plane.
2. Understand the steps suggested in rotating line method, place any object havinglength (for
example, your drawing pencil) with reference to the reference planes(for example, your note book
in 90° open position to represent VP and HP) as shownin Fig. 2.13.
Position the line (pencil) as given in the steps and understand the projections inorder to solve
problems when a line is placed inclined to both HP and VP.
3. Steps (1) to (4) can be done in any order to get required projections of the line.

Figure 2.13
SPECIAL CASES
1.Projections of a Line when θ+ ɸ= 90o
When the sum of inclinations with HP and VP is equal to 90° (θ+ ɸ= 90°), then the
linecontained by a plane will be perpendicular to both HP and VP. The top view aband frontview
a’b’of the line will be obtained perpendicular to XY line.
2.Projections of a Line when One End of the Line is in HP and the other in VP
This is considered a special case, when only one condition is given for both ends, oneend in
HP and another end in VP.
Consider a line AB having its end A in HP, (its position from VP not given) and
anotherend B in VP (its position from HP is not given).
The drawing procedure for this line is the same as in the previous problems, butsteps
involved in drawing the projections are drawn separately to get the final projections.
The vertical trace (VT) will coincide with the end touching the VP and the horizontaltrace

(HT) will coincide with the end touching the HP.
Problem 11:A line AB 80 mm long has its end A 20 mm above HP and 25 mm infront of VP. The
line is inclined at 45° to HP and 35° to VP. Draw its projections.
Solution: Mark the projections of end A by considering it as a point. Its front view a’ is20 mm
above XY and top view a is 25 mm below the XY line.
1. Assume that the line is kept inclined to HP and parallel to VP. Draw the frontview a’b1’, a line
inclined at 45° to XY line and having a length of 80 mm. Projectand get the top view ab1 length
which is parallel to XY line.
2. Assume that the line is kept inclined to VP and parallel to HP. Draw the top viewab2, a line
inclined at 35° to XY line and having a length of 80 mm. Project andget the front view ab2’ length
which is also parallel to XY line.
3. Draw the locus of the other end B of the line in top and front views. Draw thelocus of b’ which
is a line passing through b1’ and parallel to XY line. Also drawthe locus of b which is a line
passing through b2 and parallel to XY line.
4. Rotate the top view ab1 and front view a’b2’ to the required position. Take a ascentre, top view
length ab1 as radius, draw an arc to intersect the locus ofb at b.Join a and b to get the top view ab
of the line. Take a’ as centre, front view lengtha’b2’ as radius, draw an arc to intersect the locus of
a’ at b’. Join a’ and b’ to getthe front view a’b’ of the line.
line.

Problem 12:A line LM 70 mm long has its end L 10 mm above HP and 15 mm infrontof VP. The
top view and front view measures 60 mm and 40 mm respectively. Draw theprojections of the line
and determine its inclination with HP and VP.
Solution: Mark the projections of the end L by considering it as a point. Its front view l’ is 10 mm
above XY and top view l is 15 mm below XY line.
1. Assume that the line is kept inclined to HP and parallel to VP. In this case,considering the given
data the top view lm1 can be drawn parallel to XY and havinga length of 60 mm. Draw a vertical
line (projector) through m1. Using true length70 mm as radius and l’ as centre, draw an arc to
intersect the vertical line throughm1 to get m1’. Join l’ and m1’ which represents the true length of
the line. Theinclination of l’m1’ to XY is the inclination of the line with HP (ϴ).
2. Assume that the line is kept inclined to VP and parallel to HP. In this case,considering the given
data, the front view l’m2’ can be drawn parallel to XY andhaving a length of 40 mm. Draw a
vertical line (projector) through m2’. Using truelength 70 mm as radius and l as centre, draw an arc
to intersect the vertical linethrough m2’ to get m2. Join l and m2 which represents the true length of
the line.The inclination of lm2 to XY is the inclination of the line with VP (ɸ).

3. Draw the locus of the other end M of the line in top and front views. Draw thelocus of m’ which
is a line passing through m1’ and parallel to XY line. Also drawthe locus of m which is a line
passing through m2 and parallel to XY line.
4. Rotate the top view lm1 and front view l’m2’ to the required position. Take l ascentre, top view
lm1 as radius, draw an arc to intersect with the locus of m at m.Join l and m to get the top view lm
of the line. Take l’ as centre, front view l’m2’ asradius, draw an arc to intersect the locus of m’ at
m’. Join l’ and m’ to get thefront view l’m’ of the line.
5. Check the result obtained by drawing the projector joining m’ and m which shouldbe a vertical
line.
Problem 13: A line PQ 65 mm long has its end P in the horizontal plane and 15 mminfront of the
vertical plane. The line is inclined at 30∞ to the horizontal plane and is at60o
to the vertical plane.
Draw its projections.
Solution: Mark the projections of end P. Its front view p’ is on XY line and top view p is15 mm
below XY line.
1. Assume that the line is kept inclined to HP and parallel to VP. Draw the frontview p’q1’ which
is inclined at 30° to XY and has a length of 65 mm. The top viewlength pq1 is projected and
obtained parallel to XY line.

2. Assume that the line is kept inclined to VP and parallel to HP. Draw the top viewpq2 which is
inclined at 60° to XY line and has a length of 65 mm. The front viewlength p’q2’ is projected and
obtained parallel to XY line.
3. Draw the locus of q’ passing through q1’ and parallel to XY line. Also draw thelocus of q
passing through q2 and parallel to XY line.
4. Rotate the top view pq1 by taking p as centre, pq1 as radius to get q, which touchesthe locus of q.
Join p and q to complete the top view pq of the line. Rotate thefront view p’q2’ by taking p’ as
centre, p’q2’ as radius to get q’ which touches thelocus of q’. Join p’ and q’ to get the front view
p’q’ of the line.
Note: The projections obtained are perpendicular to the XY line.
Problem 14: A line AB, 75 mm long, is in the first quadrant with end A in HP and endB in VP.
The line is inclined at 35° to HP and 45° to VP. Draw the projections of thestraight line AB and
indicate the projections of the mid-point M of the line. Also markthe traces.
Solution:
1. Mark the front view a’ of the end A on XY line arbitrarily. Assume that the line iskept inclined
to HP and parallel to VP. Draw the front view a’b1’ of the line whichis inclined at 35° to XY and
has a length of 75 mm. The top view ab1 length isprojected and obtained on the XY line.
2. Mark the top view b of the end B on XY line arbitrarily. Assume that the line iskept inclined to
VP and parallel to HP. Draw the top view ba2 of the line which isinclined at 45° to XY and has a

length of 75 mm. The front view ba2’ length isprojected and obtained on XY line.
3. Draw the locus of b’ passing through b1’ and parallel to XY line. Also draw thelocus of a
passing through a2 and parallel to XY line.
4. Mark the front view a’ on XY line arbitrarily in another position to get theprojections.
Considering a’ as centre, front view length ba2’ as radius, draw anarc to get b’ in the locus of b’.
Join a’ and b’ to get the front view a’b’ of the line.Draw the projector passing through b’ to mark
the top view b on XY line.Considering b as centre, top view length a’b1 as radius, draw an arc to
get a inthe locus of a. Join a and b to get the top view ab of the line.
5. Check the result obtained by drawing the projector joining a’ and a which shouldbe a vertical
line.
6. To draw the projections of the mid-point of the line, mark m1’ on the mid-point ofa’b1’ and draw
its locus parallel to XY line to get m’ on a’b’. Similarly mark m1 onthe mid-point ba2 and draw its
locus parallel to XY line to get m on ab. Draw theprojector joining m’ and m which should be a
vertical line.
The end A is in HP, so the horizontal trace (HT) is marked coinciding with the topview a of the
line. The end B is in VP, so the vertical trace (VT) is marked coinciding withthe front view b’ of
the line.
Problem 15: A line AB 70 mm long has its end A 15 mm above the HP and 20 mm infront of VP.
The end B is 40 mm above HP and 50 mm in front of VP. Draw theprojections and find its
inclination with HP and VP.
Solution: Mark the projections of end A. Its front view a’ is 15 mm above XY and topview a is 20
mm below the XY line.
1. Draw the locus of the other end B in front and top views. Locus of b’ is drawn ata distance of 40
mm above the XY line and parallel to it. Locus ofb is drawn at adistance of 50 mm below XY line

and parallel to it.
2. Assume that the line is kept inclined to HP and parallel to VP. Draw the frontview a’b1’ by
considering a’ as centre and true length 70 mm as radius, cut anarc in the locus of b’ to mark b1’.
The top view length ab1 is projected and obtainedparallel to XY line. The inclination of front view
a’b1’ with XY is the inclination ofthe line with HP (ϴ).
3. Assume that the line is kept inclined to VP and parallel to HP. Draw the top viewab2 by
considering a as centre and true length 70 mm as radius, cut an arc inthe locus of b to mark b2. The
front view length a’b2’ is projected and obtainedparallel to XY line. The inclination of top view ab2
with XY is the inclination of the line with VP (ɸ).
4. Rotate the top view ab1 to the required position by taking a as centre, ab1 as radiusto get the
intersection point b with the locus of b. Join a and b to complete thetop view ab of the line. Rotate
the front view a’b2’ by taking a’ as centre, a’b2’ asradius to get the intersection point b’ with the
locus ofb’. Join a’ and b’ to get thefront view a’b’ of the line.
line.
Problem 16: One end P of a line PQ is 15 mm above HP and 20 mm infront of VP whilethe end Q
is 50 mm above HP and 45 mm infront of VP. If the end projectors are at adistance of 60 mm,
draw the projections of the line. Find the true length and itsinclinations with HP and VP.
Solution: Mark the projections of the end P. Its front view p’ is 15 mm above XY lineand top view
p is 20 mm below the XY line.
1. Draw the projector for the other end Q at a distance of 60 mm from p-p’. Markthe projection of
end Q, its front view q’ is 50 mm above XY line and top view q is45 mm below XY line. Join p’
and q’ to get front view p’q’ of the line. Join p andq to get the top view pq of the line.
2. Draw the locus of the end Q in front and top views. Locus of q’ is drawn passingthrough q’ and

parallel to XY line. Locus of q is drawn passing through q andparallel to XY line.
3. Rotate top view pq in the reverse order, take p as centre, top view length pq asradius, draw an
arc to get q1, on a line drawn parallel to XY line. Project q1 tolocus of q’ to get q1’. Join p’q1’
which has true length (TL) of the line. The inclinationof pq1’ with XY line is the true inclination of
the line with HP (ϴ).
4. Rotate the front view p’q’ in the reverse order, take p’ as centre, front view lengthp’q’ as radius,
draw an arc to get q2’, on a line drawn parallel to XY line. Projectq2’ to the locus of q to get q2.
Join pq2 which has true length (TL) of the line. Theinclination of pq2 with XY line is the
inclination of the line with VP (ɸ).
Problem 17: The end A of a line AB is 10 mm in front of VP and 20 mm above HP. Theline is
inclined at 30° to HP and front view is 45° with XY. Top view is 60 mm long.Complete the two
views. Find the true length and inclination with VP. Locate the traces.
Solution: Mark the projections of end A. Its front view a’ is 20 mm above XY and topview a is 10
mm below XY line.
1. Assume that the line is kept inclined to HP and parallel to VP. Draw its top viewab1 from a
which is parallel to the XY line for a length of 60 mm. Draw a verticalline (projector) from b1 and
draw a line from a’ inclined at 30° to XY line,intersecting at b1’. The front view length a’b1’ is the
true length (TL) of the line.
2. Draw the locus of b’ passing through b1’ and parallel to XY line. Draw the frontview a’b’ of the
line inclined at 45° to the XY line from a’ and intersecting thelocus of b’ at b’.
3. Draw the vertical line (projector) passing through b’. Rotate the top view ab1 bytaking a as
centre and ab1 as radius to intersect with the projector at b. Drawthe locus of b passing through b
and parallel to XY line.

4. Rotate the front view a’b’ in the reverse order. Take a’ as centre, front view a’b’ asradius and
draw an arc to get b2’, parallel to XY line. Project b2’ to the locus of bto get b2. Join ab2 which has
the true length (TL) of the line. The inclination ofab2 with the XY line is the true inclination of the
line with VP (ɸ).
To mark the traces
1. Extend the front view a’b’ to get the intersection point h’ with XY line.
2. Produce the top view ab to get the intersection point v with XY line.
3. Draw a vertical line from h’ to intersect with the top view to get horizontal trace(HT).
4. Draw another vertical line from v to intersect with the front view to get the verticaltrace (VT).
PROJECTION OF PLANE SURFACES
2.8 PLANE SURFACE
A plane figure has two dimensions viz. the length and breadth. It may be of any shape such
as triangular, square, pentagonal, hexagonal, circular etc. The possible orientations of the planes
with respect to the principal planes H.P and V.P of projection are:
1. Plane parallel to one of the principal planes and perpendicular to the other,

2. Plane perpendicular to both the principal planes,
3. Plane inclined to one of the principal planes and perpendicular to the other,
4. Plane inclined to both the principal planes.
1. Plane parallel to one of the principal planes and perpendicular to the other
When a plane is parallel to V.P the front view shows the true shape of the plane. The top
view appears as a line parallel to XY. Figure 2.14.a shows the projections of a square plane
ABCD, when it is parallel to V.P and perpendicular to H.P. The distances of one of the edges
above H.P and from the V.P are denoted by d1 and d2 respectively.
Figure 2.14.b shows the projections of the plane. Figure 2.14.c shows the projections of the plane,
when its edges are equally inclined to H.P.
Figure 2.14
Figure 2.15 shows the projections of a circular plane, parallel to H.P and perpendicular to V.P
Figure 2.15

2. Plane perpendicular to both H.P and V.P
When a plane is perpendicular to both H.P. and V.P, the projections of the plane appear as
straight lines. Figure 2.16 shows the projections of a rectangular plane ABCD, when one of its
longer edges is parallel to H.P. Here, the lengths of the front and top views are equal to the true
lengths of the edges.
Figure 2.16
3. Plane inclined to one of the principal planes and perpendicular to the other
When a plane is inclined to one plane and perpendicular to the other, the projections are
obtained in two stages.
Problem: Projections of a pentagonal plane ABCDE, inclined at ϴ to H.P and perpendicular to
V.P and resting on one of its edges on H.P.
Construction: (Fig.2.17)
Figure 2.17

4. Plane inclined to both H.P and V.P
If a plane is inclined to both H.P and V.P, it is said to be an oblique plane. Projections of
oblique planes are obtained in three stages.
Problem: A rectangular plane ABCD inclined to H.P by an angle ϴ, its shorter edge being parallel
to H.P and inclined to V.P by an angle ɸ. Draw its projections.
Construction: (Fig.2.18)
Figure 2.18
Problem 1: A square plane of side 40 mm has its surface parallel to VP and perpendicular to HP.
Draw its projections when one of the sides is inclined at 300
to HP.
Solution: When a plane is placed with its surface parallel to VP and perpendicular to HP, draw its
front view which will have the true shape and size. Project the top view which will be a line parallel
to XY.To draw the projections

1. Draw a line inclined at 30o
to XY. Arbitrarily mark the side 40mm of the square on this line
and construct the square. Name the corners as a’b’c’d’.
2. Project the top view of the plane by projecting all the corners from the front view which is a
line a(b)cd drawn parallel to the XY line.
Problem 2: A circular plate of diameter 50 mm has its surface parallel to HP and perpendicular to
VP. Its center is 20 mm above HP and 30 mm in front of VP. Draw its projections.
Solution: When a plane is placed with its surface parallel to HP and perpendicular to VP, draw its
top view which will have the true shape and size. Project the front view which will be a line
parallel to XY.To draw the projections
1. Mark the projections of the centre of the circle – its front view o’ is 20mm above XY and top
view o is 30mm below XY line.
2. Draw the top view of the plane with o as centre and 25mm radius. Project the front view of the
plane by projecting the top view which is a line passing through o’ and parallel to XY line.

Problem 3: A hexagonal plate of size 30 mm is placed with a side on VP and surface inclined at
450
to VP and perpendicular to HP. Draw the projections.
Solution:When a plane is placed with its surface inclined to VP and perpendicular to HP, its
projections are obtained in two steps.
Step 1:Assume that the plate has its surface parallel to VP and perpendicular to HP. Draw its e a
line parallel to XY.
Step 2:Reproduce the top view tilted to the given angle to VP and project the front view of the
plate which will be smaller than the true shape and size.
To draw the projections
1. Draw the top view of the rectangle considering that one of the shorter sides is perpendicular
to XY. Then, only while titling the surface to the required inclination with VP, this side of the
plate will rest on VP.
2. Project the top view of the plate by projecting the front view a’b’c’d’e’f’ to get the top view
a(b)f(c)e(d) as a line on XY line.

3. Tilt and reproduce the top view a(b)f(c)e(d) to get the required angle 45o
with XY in such a
way that the end a(b) is on the XY line.
4. Draw horizontal lines from a’b’c’d’e’and f’ and vertical lines from top view a1,b1,c1,d1,e1,and
f1 to get the required front view 𝑎1
′
, 𝑏1
′
, 𝑐1
′
, 𝑑1
′
, 𝑒1
′
𝑎𝑛𝑑 𝑓1
′
.
5. Join 𝑎1
′
, 𝑏1
′
, 𝑐1
′
, 𝑑1
′
, 𝑒1
′
𝑎𝑛𝑑 𝑓1
′
to get the front view of the hexagonal plate smaller than the true
shape and size.
Problem 4: A circular plate of diameter 50 mm is resting on HP on a point on the circumference
with its surface inclined at 450
to HP and perpendicular to VP. Draw its projections.
Solution: To draw the projections
1. Assume that the plate has its surface parallel to HP and perpendicular to VP. Draw its top view.
It is a circle of radius 25mm.
2. Project and get the front view which is a line on XY.
3. As the circle does not have any corners, divide the circle into equal parts, say 8, (students are
asked to divide the circle into a minimum of 12 parts) in such a way that 8 points are marked
on its circumference and project them to the front view.
4. Tilt and reproduce front view to the given angle of 45o
with XY line, in such a way that the end
a’ is on XY line.
5. Draw horizontal lines from a, b, c, etc. and vertical lines from𝑎1
′
, 𝑏1
′
, 𝑐1
′
, etc. to get the required
top view a1, b1, c1, etc.
6. Join a1, b1, c1, etc. by drawing a smooth curve to get the top view of the circle as an ellipse.
Problem 5: A rectangular plate of side 50 x 25 mm is resting on its shorter side on HP and
inclined at 30̊ to VP. Its surface is inclined at 60̊ to HP. Draw the projections.
Solution:In this position, the surface of the plane is inclined to both HP and VP, its projections are
obtained in three steps.

Step1: Assume that the plate has its surface parallel to HP and perpendicular to VP. Draw its top
view which will have the true shape and size. Project the front view which will be a line parallel to
the XY line.
Step 2: Reproduce the front view tilted to the given angle 𝜃 to HP and project the top view of the
plate which will be smaller than the shape and size.
Step 3: Reproduce the top view by considering the side of the plate the makes, the given angle
with VP. Project the front view of the plate which is also smaller than the true shape and size.
1. Draw the top view of the rectangle considering that one of the shorter sides is perpendicular to
XY. Then only while tilting the surface to the required angle with HP, this side of the plate will
rest on HP.
2. The front view of the plate is projected and obtained on XY as a line a’(d’)b’(c’).
3. Tile and reproduce the front view a’(d’)b’(c’) to the given angle 60o
with XY in such a way
that the end a’(d’) is o XY line.
4. Draw horizontal lines from to view a, b, c and d vertical lines from front view 𝑎1
′
, 𝑏1
′
, 𝑐1
′
𝑎𝑛𝑑 𝑑1
′
to get the top view a1, b1, c1, d1 smaller than the true shape and size.
5. Reproduce the top view a1, b1, c1, d1 in such a way that the side a1, d1 is inclined to the given
angle 30o
to VP.
6. Draw horizontal lines from 𝑎1
′
, 𝑏1
′
, 𝑐1
′
𝑎𝑛𝑑 𝑑1
′
and vertical lines from top view a2, b2, c2 and d2
to get the required front view 𝑎2
′
, 𝑏2
′
, 𝑐2
′
𝑎𝑛𝑑 𝑑2
′
of the plate smaller than the true shape and size.
Problem 6: A hexagonal plate of side 30mm is resting on one of its sides on VP and inclined at
40o
to HP. Its surface is inclined at 35o
to VP. Draw its projections.

Solution:In this position, the surface of the plate is inclined to both VP and HP. Its projections are
obtained in three steps.
Step 1: Assume that the plate has its surface parallel to VP and perpendicular to HP. Draw its front
view which will have the true shape and size. Project the top view which will be a line parallel to
XY line.
Step 2: Reproduce the top view tilted to the given angle to VP and project the front view of the
plate which will be smaller than the true shape and size.
Step 3:Reproduce the front view by considering the side of the plate that makes the given angle
with HP. Project the top view of the plate which is also smaller than the true shape and size.
1. Draw the front view of the hexagon considering one of the sides perpendicular to XY. Then
only while tilting the surface to the required angle with VP, this side of the plate will rest on
VP.
2. The top view of the plate is projected and obtained on XY as a line.
3. Tilt and reproduce the top view line to the given angle 35o
with XY in such a way that the end
a(b) is on the XY line.
4. Draw horizontal lines from front view a’, b’, c’, etc. and vertical lines from top view a1, b1, c1,
etc, to get the front view of the plate which is smaller than the true shape and size.
5. Reproduce the front view in such a way that the side a1b1 is inclined to the given angle 40o
to
HP. (Note that side d1e1 is also at the same angle).
6. Draw horizontal lines from a1, b1, c1, etc., and a`2, b`2, c`2, etc., to get the required top view of
the hexagonal plate which is smaller than the true shape and size.

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