Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Engineering Graphics Full Notes
Description: This Notes Provides A Full Explaination To Engineering Graphics Course. It Is Written By Me and It Is Verified By My Senior. Sailent Features:- Course Is Explained In A Very Easy Simple Language. Drawings Are Made Colorful In Order To Make The Book Very Interesting To Students. Ample Number Of Solved Examples With Detailed Explaination Of The Problem Is Also Included.
Description: This Notes Provides A Full Explaination To Engineering Graphics Course. It Is Written By Me and It Is Verified By My Senior. Sailent Features:- Course Is Explained In A Very Easy Simple Language. Drawings Are Made Colorful In Order To Make The Book Very Interesting To Students. Ample Number Of Solved Examples With Detailed Explaination Of The Problem Is Also Included.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Contents
1
...
Engineeri
3
...
Engineering Curves - II
4
...
Orthographic Projections - Basics
6
...
Projections of Points and Lines
8
...
Projection of Solids
10
...
Intersection of Surfaces
12
...
Exercises
14
...
Basic Information
2
...
Plain Scales (3 Problems)
4
...
Diagonal Scales (3 Problems)
6
...
Vernier Scales - information
8
...
Scales of Cords - construction
10
...
Classification
2
...
Common Definition
4
...
Parabola – ( Three methods of construction)
6
...
Methods of drawing Tangents & Normals ( four cases)
Engineering Curves – II
1
...
Definitions
3
...
Cycloid
5
...
Epic cycloid and Hypo - cycloid
7
...
Helix – on cylinder & on cone
9
...
Definitions - Classifications
2
...
Oscillating links (two problems)
4
...
Drawing – The fact about
2
...
Orthographic (Definitions and Important terms)
4
...
Pattern of planes & views
6
...
1st angle and 3rd angle method – two illustrations
Conversion of pictorial views in to orthographic views
...
Explanation of various terms
2
...
3rd angle method – illustration
4
...
Seven illustrations (no
...
Total nineteen illustrations ( no
...
2
...
4
...
6
...
Projections – Information
Notations
Quadrant Structure
...
Projections of a Point – in 1st quadrant
...
Simple Objective & Types
...
9
...
11
...
13
...
15
...
17
...
19
...
21
...
Lines inclined to one plane
...
Imp
...
Group A problems 1 to 5
Traces of Line ( HT & VT )
To locate Traces
...
6 to 8
HT-VT additional information
...
9 to 11
Group B1 problems: No
...
12 & 13
Applications of Lines:: Information
Group D: Application Problems: 14 to 23
Projections of Planes:
1
...
Illustration of surface & side inclination
...
Procedure to solve problem & tips:
4
...
Problems:6 to 11: Indirect inclinations:
6
...
Problems: 12 & 13
8
...
Problems: 14 to 17
Projections of Solids:
1
...
Important parameters:
3
...
Pattern of Standard Solution
...
Problem no 1,2,3,4: General cases:
6
...
Problem no 7 : Freely suspended:
8
...
Problem no 9 : True length case:
10
...
Problem no 12 : Frustum & auxiliary plane:
Section & Development
1
...
Sectioning a solid: Information:
3
...
Typical shapes of sections & planes:
5
...
Development of diff
...
Development of Frustums:
8
...
1 & 2
9
...
3 & 4
10
...
5
11
...
6 to 9
Intersection of Surfaces:
1
...
Display of Engineering Applications:
3
...
Case 1: Cylinder to Cylinder:
5
...
Case 3: Cone to Cylinder
7
...
8
...
Case 6: Prism to Prism: Axis Skew
10
...
Case 8: Cylinder to Cone:
Isometric Projections
1
...
Important Terms
3
...
4
...
5
...
Isometric of a part of circle
7
...
Isometric of solids & frustums (no
...
Isometric of sphere & hemi-sphere (no
...
Isometric of Section of solid
...
19)
11
...
20 to 38)
OBJECTIVE OF THIS CD
Sky is the limit for vision
...
Anything in the jurisdiction of vision can be memorized for a long period
...
So vision helps visualization and both help in memorizing an event or situation
...
Every effort has been done in this CD, to bring various planes, objects and situations
in-front of observer, so that he/she can further visualize in proper direction
and reach to the correct solution, himself
...
Then success is yours !!
Go ahead confidently! CREATIVE TECHNIQUES wishes you best luck !
SCALES
DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE
ON STANDARD SIZE DRAWING SHEET
...
SUCH A SCALE IS CALLED REDUCING SCALE
AND
THAT RATIO IS CALLED REPRESENTATIVE FACTOR
...
HENCE THIS SCALE IS CALLED ENLARGING SCALE
...
USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC
...
F
...
V ACTUAL VOLUME
B
LENGTH OF SCALE = R
...
X
MAX
...
FOR FULL SIZE SCALE
R
...
=1 OR ( 1:1 )
MEANS DRAWING
& OBJECT ARE OF
SAME SIZE
...
1 KILOMETRE
1 HECTOMETRE
1 DECAMETRE
1 METRE
1 DECIMETRE
1 CENTIMETRE
= 10 HECTOMETRES
= 10 DECAMETRES
= 10 METRES
= 10 DECIMETRES
= 10 CENTIMETRES
= 10 MILIMETRES
TYPES OF SCALES:
1
...
3
...
5
...
PROBLEM NO
...
Show on it a distance of 4 m and 6 dm
...
F
...
F
...
F
...
distance
= 1/100 X 600 cm
= 6 cms
b) Draw a line 6 cm long and divide it in 6 equal parts
...
c) Sub divide the first part which will represent second unit or fraction of first unit
...
Number the units on right side of Zero and subdivisions
on left-hand side of Zero
...
e) After construction of scale mention it’s RF and name of scale as shown
...
4 M 6 DM
10
DECIMETERS
0
1
2
3
R
...
= 1/100
PLANE SCALE SHOWING METERS AND DECIMETERS
...
2:- In a map a 36 km distance is shown by a line 45 cms long
...
F
...
12 km
...
3 km on it
...
F
...
F
...
1000
...
F
...
distance
= 1/ 80000
12 km
= 15 cm
b) Draw a line 15 cm long and divide it in 12 equal parts
...
c) Sub divide the first part which will represent second unit or fraction of first unit
...
Number the units on right side of Zero and subdivisions
on left-hand side of Zero
...
e) After construction of scale mention it’s RF and name of scale as shown
...
3 km on it as shown
...
F
...
3:- The distance between two stations is 210 km
...
Construct a plain scale to measure time up to a single minute
...
CONSTRUCTION:a) 210 km in 7 hours
...
F
...
distance per hour
= 1/ 2,00,000
30km
= 15 cm
b) 15 cm length will represent 30 km and 1 hour i
...
60 minutes
...
Each part will represent 5 km and 10 minutes
...
Each smaller part will represent distance traveled in one minute
...
Number the units on right side of Zero and subdivisions
on left-hand side of Zero
...
e) Show km on upper side and time in minutes on lower side of the scale as shown
...
f) Show the distance traveled in 29 minutes, which is 14
...
DISTANCE TRAVELED IN 29 MINUTES
...
5 KM
KM 5
MIN 10
2
...
F
...
KM
We have seen that the plain scales give only two dimensions,
such as a unit and it’s subunit or it’s fraction
...
The principle of construction of a diagonal scale is as follows
...
From Y draw a perpendicular YZ to a suitable height
...
Divide YZ in to 10 equal parts
...
From geometry we know that similar triangles have
their like sides proportional
...
x X Y = 0
...
Similarly
1’ – 1 = 0
...
2 XY
Thus, it is very clear that, the sides of small triangles,
which are parallel to divided lines, become progressively
shorter in length by 0
...
The solved examples ON NEXT PAGES will
make the principles of diagonal scales clear
...
4 : The distance between Delhi and Agra is 200 km
...
Find it’s R
...
Draw a diagonal scale to show single km
...
Indicate on it following distances
...
It will represent 600 km
...
( each will represent 100 km
...
Each will represent 10 km
...
Name those parts 0 to 10 as shown
...
Then draw parallel lines to this line from remaining sub divisions and
complete diagonal scale
...
F
...
400
500 KM
PROBLEM NO
...
28 hectors is represented on a map by a similar rectangle
of 8 sq
...
Calculate RF of the scale
...
Show a distance of 438 m on it
...
meters
1
...
28 X 10, 000 sq
...
28 X 104 X 104 sq
...
cm area on map represents
= 1
...
cm on land
1 cm sq
...
28 X 10 4 X 104 / 8 sq cm on land
1 cm on map represent
= 1
...
DIAGONAL
SCALE
Draw a line 15 cm long
...
Divide it in six equal parts
...
)
Divide first division in ten equal parts
...
Draw a line upward from left end and
mark 10 parts on it of any distance
...
Join 9th sub-division
of horizontal scale with 10th division of the vertical divisions
...
M
438 meters
10
9
8
7
6
5
4
3
2
1
0
M 100
50
0
100
200
300
R
...
= 1 / 4000
DIAGONAL SCALE SHOWING METERS
...
6:
...
F
...
5, showing centimeters
and millimeters and long enough to measure up to 20 centimeters
...
F
...
5
Length of scale = 1 / 2
...
= 8 cm
...
Draw a line 8 cm long and divide it in to 4 equal parts
...
)
2
...
(Each will show 1 cm
...
At the left hand end of the line, draw a vertical line and
on it step-off 10 equal divisions of any length
...
Complete the scale as explained in previous problems
...
4 cm on it
...
4 CM
CM
10
9
8
7
6
5
4
3
2
1
0
5 4 3 2 1 0
5
10
R
...
= 1 / 2
...
15
CENTIMETRES
COMPARATIVE SCALES:
These are the Scales having same R
...
but graduated to read different units
...
SOLUTION STEPS:
Scale of Miles:
40 miles are represented = 8 cm
: 80 miles = 16 cm
R
...
= 8 / 40 X 1609 X 1000 X 100
= 1 / 8, 04, 500
CONSTRUCTION:
Take a line 16 cm long and divide it into 8 parts
...
Subdivide the first part and each sub-division will measure single mile
...
90 cm
10
0
10
20
EXAMPLE NO
...
Construct a plain scale to read 80 miles
...
609 km )
CONSTRUCTION:
On the top line of the scale of miles cut off a distance of 14
...
Each part will represent 10 km
...
Each subdivision will show single km
...
F
...
length of scale = RF X 60 km
= 1 / 4,00,000 X 60 X 105
= 15 cm
...
( each part will represent 10 km
...
( each will represent 1 km
...
8 :
A motor car is running at a speed of 60 kph
...
Time Scale:
Same 15 cm line will represent 60 minutes
...
It will show minimum 1 minute & max
...
47 MINUTES
10
MIN
...
F
...
9 :
A car is traveling at a speed of 60 km per hour
...
Construct a suitable comparative scale up to 10 hours
...
Show the time required to cover 476 km and also distance in 4 hours and 24 minutes
...
This length of scale will also represent 600 kms
...
Divide it in TEN equal parts
...
( Each will represent 10 km)
At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length
...
Time scale:
Draw a line 20 cm long
...
( Each will show 1 hour) Sub-divide 1st part in SIX subdivisions
...
And complete the diagonal scale to read minimum ONE minute
...
10
5
MIN
...
( 264 kms )
476 kms ( 7 hrs 56 min
...
It consists of two parts – a primary scale and a vernier
...
As it would be difficult to sub-divide the minor divisions in ordinary way, it is done with the help of the vernier
...
Figure to the right shows a part of a plain scale in
which length A-O represents 10 cm
...
Now it would
not be easy to divide each of these parts into ten equal
divisions to get measurements in millimeters
...
1 cm
...
1 – 1
...
1 cm or 1 mm
...
Minimum this distance can be measured by this scale
...
The combination of
plain scale and the vernier is vernier scale
...
9
7
...
5
3
...
1 0
A 9 8 7 6 5 4 3 2 1
0
Example 10:
Draw a vernier scale of RF = 1 / 25 to read centimeters upto
4 meters and on it, show lengths 2
...
91 m
SOLUTION:
Length of scale = RF X max
...
Divide it in 4 equal parts
...
( each will represent decimeter )
Name those properly
...
Each will show 0
...
1 dm or 11 cm and construct a rectangle
Covering these parts of vernier
...
39 m : Subtract 0
...
39 i
...
2
...
99 = 1
...
99 ( left of Zero) and 1
...
39 m
(2) For 0
...
11 from 0
...
e
...
91 – 0
...
80 m
The distance between 0
...
80 (both left side of Zero) is 0
...
39 m
0
...
1
...
77
...
33
...
0
...
8
...
6
...
4
...
2
...
4
2
3 METERS
Example 11: A map of size 500cm X 50cm wide represents an area of 6250 sq
...
Construct a vernier scaleto measure kilometers, hectometers and decameters
and long enough to measure upto 7 km
...
33 km b) 59 decameters
...
Divide it in 7 equal parts
...
( each will represent hectometer )
Name those properly
...
V 6250 km sq
...
Distance
= 2 / 105 X 7 kms
= 14 cm
TO MEASURE GIVEN LENGTHS:
a) For 5
...
33 from 5
...
e
...
33 - 0
...
00
The distance between 33 dm
( left of Zero) and
5
...
33 k m
(b) For 59 dm :
Subtract 0
...
59
i
...
0
...
99 = - 0
...
4 km is 59 dm
(both left side of Zero)
CONSTRUCTION: ( vernier)
Take 11 parts of hectometer part length
and divide it in 10 equal parts
...
1 hm m or 11 dm and
Covering in a rectangle complete scale
...
33 km
Decameters
99
90
77
70
55
50
10
HECTOMETERS
33
11
30
10
0
1
2
3
4
5
6
KILOMETERS
700
800
900
SCALE OF CORDS
600
500
400
300
200
100
00
O
A
0
10
20
30
40
50
60
70 80 90
CONSTRUCTION:
1
...
( ‘OA’ ANY CONVINIENT DISTANCE )
2
...
3
...
4
...
AS CORD LENGTHS ARE USED TO MEASURE & CONSTRUCT
DIFERENT ANGLES IT IS CALLED SCALE OF CORDS
...
CONSTRUCTION:
First prepare Scale of Cords for the problem
...
( You are supposed to measure angles x, y and z)
To measure angle at x:
Take O-A distance in compass from cords scale and mark it on lower side of triangle
as shown from corner x
...
Then O as center, O-A radius
draw an arc upto upper adjacent side
...
Take A-B cord in compass and place on scale of cords from Zero
...
Draw arc with radius O1A1
...
To measure angle at z:
Subtract the SUM of these two angles from 1800 to get angle at z
...
SCALE OF CORDS
CONSTRUCTION:
First prepare Scale of Cords for the problem
...
Mark point O on it
...
Take O-A distance in compass from cords scale and mark it on on the line drawn, from O
500
Name O & A as shown
...
400
Take cord length of 250 angle from scale of cords in compass and
300
from A cut the arc at point B
...
The angle AOB is thus 250
To construct 1150 angle at O
...
Hence Subtract 1150 from 1800
...
O
A
Extend previous arc of OA radius and taking cord length of 750 in compass cut this arc
00
0 10 20 30 40 50 60 70 80 90
at B1 with A as center
...
Now angle AOB1 is 750 and angle COB1 is 1150
...
1150
O
To construct 1150 angle at O
...
Concentric Circle Method
1
...
Rectangular Hyperbola
(coordinates given)
2
...
Oblong Method
4
...
Basic Locus Method
(Directrix – focus)
2 Rectangular Hyperbola
(P-V diagram - Equation given)
3
...
Rhombus Metho
6
...
CONIC SECTIONS
ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS
BECAUSE
THESE CURVES APPEAR ON THE SURFACE OF A CONE
WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES
...
Ellipse
Section Plane
Through Generators
Section Plane
Parallel to Axis
...
Hyperbola
COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA:
These are the loci of points moving in a plane such that the ratio of it’s distances
from a fixed point And a fixed line always remains constant
...
(E)
A) For Ellipse
E<1
B) For Parabola E=1
C) For Hyperbola E>1
Refer Problem nos
...
9 & 12
SECOND DEFINATION OF AN ELLIPSE:It is a locus of a point moving in a plane
such that the SUM of it’s distances from TWO fixed points
always remains constant
...
}
These TWO fixed points are FOCUS 1 & FOCUS 2
Refer Problem no
...
ELLIPSE
BY CONCENTRIC CIRCLE METHOD
Problem 1 :Draw ellipse by concentric circle method
...
3
2
Steps:
1
...
2
...
3
...
4
...
5
...
6
...
7
...
It is
required ellipse
...
2
...
3
...
Divide
vertical small side and horizontal
long side into same number of
equal parts
...
Name those as shown
...
Now join all vertical points
1,2,3,4, to the upper end of minor
axis
...
e
...
6
...
Similarly
extend C-2, C-3, C-4 lines up to
D-2, D-3, & D-4 lines
...
Mark all these points properly
and join all along with ends A
and D in smooth possible curve
...
along with lower half of
the rectangle
...
It is required ellipse
...
Take major axis 100 mm and minor axis 70 mm long
...
Draw a parallelogram of 100 mm and 70 mm long
sides with included angle of 750
...
BY OBLONG METHOD
STEPS ARE SIMILAR TO
THE PREVIOUS CASE
(RECTANGLE METHOD)
ONLY IN PLACE OF RECTANGLE,
HERE IS A PARALLELOGRAM
...
MAJOR AXIS AB & MINOR AXIS CD ARE
100 AMD 70MM LONG RESPECTIVELY
...
STEPS:
1
...
Name the
ends & intersecting point
2
...
e
...
( focus 1 and 2
...
On line F1- O taking any distance,
mark points 1,2,3, & 4
4
...
Name the point p1
5
...
Name the point p2
6
...
With same steps positions of P can be
located below AB
...
Join all points by smooth curve to get
an ellipse/
BY ARCS OF CIRCLE METHOD
As per the definition Ellipse is locus of point P moving in
a plane such that the SUM of it’s distances from two fixed
points (F1 & F2) remains constant and equals to the length
of major axis AB
...
1+ B
...
2 + B
...
DRAW RHOMBUS OF 100 MM & 70 MM LONG
DIAGONALS AND INSCRIBE AN ELLIPSE IN IT
...
Draw rhombus of given
dimensions
...
Mark mid points of all sides &
name Those A,B,C,& D
3
...
4
...
5
...
6
...
7
...
ELLIPSE
BY RHOMBUS METHOD
2
A
B
4
3
C
D
1
PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB
...
{ ECCENTRICITY = 2/3 }
STEPS:
1
...
2
...
3
...
It is 20mm
and 30mm from F and AB line resp
...
e 20/30
4 Form more points giving same ratio such
as 30/45, 40/60, 50/75 etc
...
Taking 45,60 and 75mm distances from
line AB, draw three vertical lines to the
right side of it
...
Now with 30, 40 and 50mm distances in
compass cut these lines above and below,
with F as center
...
Join these points through V in smooth
curve
...
It is an ELLIPSE
...
Draw the path of the ball (projectile)-
STEPS:
1
...
Consider left part for construction
...
Join vertical 1,2,3,4,5 & 6 to the
top center of rectangle
4
...
5
...
Join all in sequence
...
...
8: Draw an isosceles triangle of 100 mm long base and
110 mm long altitude
...
C
Solution Steps:
1
...
2
...
of
equal parts
...
Name the parts in ascending and
descending manner, as shown
...
Join 1-1, 2-2,3-3 and so on
...
Draw the curve as shown i
...
tangent to
all these lines
...
A
B
PARABOLA
PROBLEM 9: Point F is 50 mm from a vertical straight line AB
...
DIRECTRIX-FOCUS METHOD
PARABOLA
SOLUTION STEPS:
1
...
This will be initial
point P and also the vertex
...
Mark 5 mm distance to its right side,
name those points 1,2,3,4 and from
those
draw lines parallel to AB
...
Mark 5 mm distance to its left of P and
name it 1
...
Take O-1 distance as radius and F as
center draw an arc
cutting first parallel line to AB
...
(FP1=O1)
5
...
6
...
It will be the locus of P equidistance
from line AB and fixed point F
...
10: Point P is 40 mm and 30 mm from horizontal
and vertical axes respectively
...
THROUGH A POINT
OF KNOWN CO-ORDINATES
Solution Steps:
1) Extend horizontal
line from P to right side
...
3) On horizontal line
from P, mark some points
taking any distance and
name them after P-1,
2,3,4 etc
...
Let them cut
part [P-B] also at 1,2,3,4
points
...
40 mm
7) Line from 1
horizontal and line from
1 vertical will meet at
P1
...
8) Repeat the procedure
by marking four points
on upward vertical line
from P and joining all
those to pole O
...
and join them by smooth
curve
...
11: A sample of gas is expanded in a cylinder
from 10 unit pressure to 1 unit pressure
...
If initial volume being 1 unit, draw the
curve of expansion
...
10
Form a table giving few more values of P & V
10
5
4
2
...
5 =
4 =
5 =
10 =
10
10
10
10
10
10
Now draw a Graph of
Pressure against Volume
...
Take pressure on vertical axis and
Volume on horizontal axis
...
A POINT P IS MOVING IN A PLANE
SUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINS CONSTANT
AND EQUALS TO 2/3 DRAW LOCUS OF POINT P
...
Draw a vertical line AB and point F
50 mm from it
...
Divide 50 mm distance in 5 parts
...
Name 2nd part from F as V
...
It is first point giving ratio of it’s
distances from F and AB 2/3 i
...
5
...
6
...
7
...
This is required locus of P
...
A
30mm
(vertex)
B
V
F ( focus)
HYPERBOLA
DIRECTRIX
FOCUS METHOD
ELLIPSE
TANGENT & NORMAL
Problem 13:
TO DRAW TANGENT & NORMAL
TO THE CURVE FROM A GIVEN POINT ( Q )
1
...
BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL
A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE
...
p4
p3
C
p2
p1
A
F1
1
2
3
4
O
Q
D
B
F2
ELLIPSE
TANGENT & NORMAL
Problem 14:
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )
ELLIPSE
A
1
...
2
...
EXTEND THE LINE TO MEET DIRECTRIX
AT T
4
...
THIS IS
TANGENT TO ELLIPSE FROM Q
5
...
IT IS NORMAL TO CURVE
...
JOIN POINT Q TO F
...
CONSTRUCT 900 ANGLE WITH
THIS LINE AT POINT F
3
...
JOIN THIS POINT TO Q AND EXTEND
...
TO THIS TANGENT DRAW PERPENDICULAR
LINE FROM Q
...
VERTEX V
900
F
( focus)
N
Q
B
N
T
HYPERBOLA
TANGENT & NORMAL
Problem 16
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )
1
...
2
...
EXTEND THE LINE TO MEET DIRECTRIX AT T
4
...
THIS IS
TANGENT TO CURVE FROM Q
5
...
IT IS NORMAL TO CURVE
...
Involute of a circle
a)String Length = D
b)String Length > D
c)String Length < D
CYCLOID
1
...
Trochoid
( superior)
3
...
Epi-Cycloid
SPIRAL
HELIX
1
...
1
...
On a Cone
2
...
2
...
5
...
Rod Rolling over
a Semicircular Pole
...
DEFINITIONS
CYCLOID:
IT IS A LOCUS OF A POINT ON THE
PERIPHERY OF A CIRCLE WHICH
ROLLS ON A STRAIGHT LINE PATH
...
:
IF IT IS INSIDE THE CIRCLE
IT IS A LOCUS OF A FREE END OF A STRING
WHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID
IF THE CIRCLE IS ROLLING ON
ANOTHER CIRCLE FROM OUTSIDE
SPIRAL:
IT IS A CURVE GENERATED BY A POINT
WHICH REVOLVES AROUND A FIXED POINT
AND AT THE SAME MOVES TOWARDS IT
...
IF THE CIRCLE IS ROLLING FROM
INSIDE THE OTHER CIRCLE,
HELIX:
IT IS A CURVE GENERATED BY A POINT WHICH
MOVES AROUND THE SURFACE OF A RIGHT CIRCULAR
CYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION
AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION
...
String length is equal to the circumference of circle
...
Means if this string is wound round
the circle, it will completely cover
given circle
...
2) Divide D (AP) distance into 8
number of equal parts
...
4) Name after A, 1, 2, 3, 4, etc
...
5) To radius C-1, C-2, C-3 up to C-8
draw tangents (from 1,2,3,4,etc to
circle)
...
7) Name this point P1
8) Take 2-B distance in compass
and mark it on the tangent from
point 2
...
9) Similarly take 3 to P, 4 to P, 5 to
P up to 7 to P distance in compass
and mark on respective tangents
and locate P3, P4, P5 up to P8 (i
...
A) points and join them in smooth
curve it is an INVOLUTE of a given
circle
...
String length is MORE than the circumference of circle
...
But remember!
Whatever may be the length of
string, mark D distance
horizontal i
...
along the string
and divide it in 8 number of
equal parts, and not any other
distance
...
Draw
the curve completely
...
INVOLUTE OF A CIRCLE
String length is LESS than the circumference of circle
...
But remember!
Whatever may be the length of
string, mark D distance
horizontal i
...
along the string
and divide it in 8 number of
equal parts, and not any other
distance
...
Draw
the curve completely
...
ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER
DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY
...
)
P1
P
P2
1 to P
SOLUTION STEPS:
Draw pole shape as per
dimensions
...
Calculate perimeter
length
...
On this line mark 30mm
from A
Mark and name it 1
Mark D/2 distance on it
from 1
And dividing it in 4 parts
name 2,3,4,5
...
P3
3 to P
4
3
2
5
1
A
6
1
P4
P5
P6
2
3
D/2
4
5
6
P
PROBLEM 21 : Rod AB 85 mm long rolls
over a semicircular pole without slipping
from it’s initially vertical position till it
becomes up-side-down vertical
...
B
A4
4
Solution Steps?
If you have studied previous problems
properly, you can surely solve this also
...
Means when one end is approaching,
other end will move away from poll
...
Take Circle diameter as 50 mm
p4
4
3
p3
5
C
2
p2
C1
6
p5
C2
C3
C4
p1
1
CYCLOID
C5
C6
C7
p6 C
8
p7
7
p8
P
D
Solution Steps:
1)
2)
3)
4)
5)
6)
7)
From center C draw a horizontal line equal to D distance
...
Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8
...
(parallel to locus of C)
With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1
...
Repeat this procedure from C2, C3, C4 upto C8 as centers
...
Join all these points by curve
...
SUPERIOR TROCHOID
PROBLEM 23: DRAW LOCUS OF A POINT , 5 MM AWAY FROM THE PERIPHERY OF A
CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH
...
Draw circle by CP radius, as in this case CP is larger than radius of circle
...
PROBLEM 24: DRAW LOCUS OF A POINT , 5 MM INSIDE THE PERIPHERY OF A
CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH
...
Draw circle by CP radius, as in this case CP is SHORTER than radius of circle
...
This curve is called Inferior Trochoid
...
Take diameter of rolling Circle 50 mm
EPI CYCLOID :
And radius of directing circle i
...
curved path, 75 mm
...
2) Calculate by formula = (r/R) x
3600
...
4) Divide this sector into 8 number of
equal angular parts
...
5) Divide smaller circle (Generating
circle) also in 8 number of equal parts
...
6) With O as center, O-1 as radius
draw an arc in the sector
...
Take fixed distance C-P in
compass, C1 center, cut arc of 1 at P1
...
This is EPI –
CYCLOID
...
Take diameter of
HYPO CYCLOID
rolling circle 50 mm and radius of directing circle (curved path) 75 mm
...
It has to rotate
anticlockwise to move
ahead
...
Only change is
in numbering direction of
8 number of equal parts
on the smaller circle
...
4) Further all steps are
that of epi – cycloid
...
P
7
P1
6
P2
1
P3
5
2
3
4
P4
P5
=
r
3600
R
O
OC = R ( Radius of Directing Circle)
CP = r (Radius of Generating Circle)
P8
P6
P7
Problem 27: Draw a spiral of one convolution
...
SPIRAL
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS
...
With PO radius draw a circle
and divide it in EIGHT parts
...
up to 8
2
...
3
...
Name the point P1
4
...
It is a SPIRAL of one convolution
...
It starts moving towards O and reaches it in two
revolutions around
...
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS
...
Just divide both in same parts i
...
Circle in EIGHT parts
...
Rest steps are similar to the previous
problem
...
Given 80 mm pitch and 50 mm diameter of a cylinder
...
Divide circle and axis in to same no
...
( 8 )
Name those as shown
...
Join all points by smooth possible curve
...
P5
5
4
P4
3
P3
2
P2
1
P1
P
6
5
7
P
4
1
3
2
PROBLEM: Draw a helix of one convolution, upon a cone,
diameter of base 70 mm, axis 90 mm and 90 mm pitch
...
of equal parts
...
Mark initial position of point ‘P’
Mark various positions of P as shown in animation
...
Make upper half dotted, as it is going behind the solid
and hence will not be seen from front side
...
MARK POINT Q ON IT AS DIRECTED
...
CONSIDERING CQ DIAMETER, DRAW
A SEMICIRCLE AS SHOWN
...
Q
THIS WILL BE NORMAL TO INVOLUTE
...
IT WILL BE TANGENT TO INVOLUTE
...
MARK POINT Q ON IT AS DIRECTED
...
CUT THE
POINT ON LOCUS OF C AND JOIN IT TO Q
...
THIS WILL BE NORMAL TO
CYCLOID
...
IT WILL BE TANGENT TO CYCLOID
...
Method of Drawing
Tangent & Normal
SPIRAL (ONE CONVOLUSION
...
P3
=
P4
4
O
P5
7 6 5 4 3 2 1
P7
P6
7
5
6
P
OP – OP2
/2
=
OP – OP2
1
...
185 m
...
STEPS:
*DRAW SPIRAL AS USUAL
...
* LOCATE POINT Q AS DISCRIBED IN PROBLEM AND
THROUGH IT DRAW A TANGENTTO THIS SMALLER
CIRCLE
...
*DRAW A LINE AT RIGHT ANGLE
*TO THIS LINE FROM Q
...
LOCUS
It is a path traced out by a point moving in a plane,
in a particular manner, for one cycle of operation
...
A} Basic Locus Cases
...
Then one point will be allowed to move in a plane maintaining specific relation
with above objects
...
Oscillating & Rotating Link:
Here a link oscillating from one end or rotating around it’s center will be described
...
And now studying
the situation carefully you will be asked to draw it’s locus
...
: Point F is 50 mm from a vertical straight line AB
...
P7
SOLUTION STEPS:
1
...
This will be initial
point P
...
Mark 5 mm distance to its right side,
name those points 1,2,3,4 and from those
draw lines parallel to AB
...
Mark 5 mm distance to its left of P and
name it 1
...
Take F-1 distance as radius and F as
center draw an arc
cutting first parallel line to AB
...
5
...
6
...
A
P3
P1
p
1 2 3 4
4 3 2 1
P2
P4
B
It will be the locus of P equidistance
from line AB and fixed point F
...
Draw locus of point P, moving in a plane such that
it always remains equidistant from given circle and line AB
...
Locate center of line, perpendicular to
AB from the periphery of circle
...
2
...
3
...
4
...
Name upper point P1 and
lower point P2
...
Similarly repeat this process by taking
again 5mm to right and left and locate
P3P4
...
Join all these points in smooth curve
...
P7
P6
P8
75 mm
C
Basic Locus
PROBLEM 3 :
Center of a circle of 30 mm diameter is 90 mm away from center of another circle of 60 mm diameter
...
SOLUTION STEPS:
1
...
Name it P
...
2
...
3
...
4
...
5
...
Cases:
60 D
P7
P5
30 D
P3
P1
p
C1
4 3 2 1
1 2 3 4
P2
P4
P6
P8
It will be the locus of P
equidistance from given two circles
...
Draw one circle touching these three objects
...
(As per solution of
case D above)
...
(as per solution of case B
above)
...
Name it x
...
4) Take x as centre and its
perpendicular distance on
AB as radius, draw a circle
which will touch given two
circles and line AB
...
There is a point P, moving in a plane such that the
difference of it’s distances from A and B always
remains constant and equals to 40 mm
...
p7
p5
p3
p1
Solution Steps:
1
...
2
...
On both sides of P mark points 5
mm apart
...
4
...
5
...
e
...
It will be locus of P
P
A
4 3 2 1
1 2 3 4
p2
p4
p6
p8
70 mm
30 mm
B
FORK & SLIDER
Problem 6:-Two points A and B are 100 mm apart
...
Draw locus of point P
...
2) Divide line (M initial
and M lower most ) into
eight to ten parts and mark
them M1, M2, M3 up to the
last position of M
...
4) Mark point P1 on M1N1
with same distance of MP
from M1
...
It will be locus of P
...
7:
A Link OA, 80 mm long oscillates around O,
600 to right side and returns to it’s initial vertical
Position with uniform velocity
...
Draw locus of point P
Solution Steps:
OSCILLATING LINK
O
p
1
Point P- Reaches End A (Downwards)
1) Divide OA in EIGHT equal parts and from O to A after O
name 1, 2, 3, 4 up to 8
...
e
...
2) Divide 600 angle into four parts (150 each) and mark each
point by A1, A2, A3, A4 and for return A5, A6, A7 andA8
...
3) Take center O, distance in compass O-1 draw an arc upto
OA1
...
1) Similarly O center O-2 distance mark P2 on line O-A2
...
( It will be thw desired locus of P )
p1
p2
p3
p4
2
3
A4
p5
4
5
p6
A3
6
7
8
A
p8
A8
p7
A1
A7
A2
A6
A5
OSCILLATING LINK
Problem No 8:
A Link OA, 80 mm long oscillates around O,
600 to right side, 1200 to left and returns to it’s initial
vertical Position with uniform velocity
...
Draw locus of point P
Op
16
15
1
14
Solution Steps:
( P reaches A i
...
moving downwards
...
e
...
Here distance traveled by point P is PA
...
Hence divide it into eight equal parts
...
2
...
Hence
total angular displacement is 2400
...
(150 each
...
(A, A1 A2 etc)
3
...
and complete the problem
...
Meanwhile point P, initially on A starts moving towards B and reaches B
...
1) AB Rod revolves around
center O for one revolution and
point P slides along AB rod and
reaches end B in one
revolution
...
3) Distance traveled by point P
is AB mm
...
4) Initially P is on end A
...
Mark it from A1 and
name the point P1
...
Mark it as above
from A2
...
7) Similarly locate P4, P5, P6,
P7 and P8 which will be eight
parts away from A8
...
8) Join all P points by smooth
curve
...
Meanwhile point P, initially on A starts moving towards B, reaches B
And returns to A in one revolution of rod
...
ROTATING LINK
A2
Solution Steps
1) AB Rod revolves around center O
for one revolution and point P slides
along rod AB reaches end B and
returns to A
...
3) Distance traveled by point P is AB
plus AB mm
...
4) Initially P is on end A
...
Mark it from A1 and name the point
P1
...
Mark it as above from A2
...
7) Similarly locate P4, P5, P6, P7
and P8 which will be eight parts away
from A8
...
8) Join all P points by smooth curve
...
A1
A3
p5
p1
p4
A
P
p8
p2
1+7
2+6 p
6
+ 5
3
4
p7 p3
A7
A5
A6
+B
A4
DRAWINGS:
( A Graphical Representation)
The Fact about:
If compared with Verbal or Written Description,
Drawings offer far better idea about the Shape, Size & Appearance of
any object or situation or location, that too in quite a less time
...
Drawings
(Some Types)
Botanical Drawings
( plants, flowers etc
...
)
Drawings
( maps etc
...
Orthographic Projections
(Fv,Tv & Sv
...
Engg terms)
(Plan, Elevation- Civil Engg
...
)
Zoological Drawings
(creatures, animals etc
...
)
Machine component Drawings
Isometric ( Mech
...
Term
...
Term)
(Actual Object Drawing 3-D)
ORTHOGRAPHIC PROJECTIONS:
IT IS A TECHNICAL DRAWING IN WHICH DIFFERENT VIEWS OF AN OBJECT
ARE PROJECTED ON DIFFERENT REFERENCE PLANES
OBSERVING PERPENDICULAR TO RESPECTIVE REFERENCE PLANE
Different Reference planes are
Horizontal Plane (HP),
Vertical Frontal Plane ( VP )
Side Or Profile Plane ( PP)
And
Different Views are Front View (FV), Top View (TV) and Side View (SV)
FV is a view projected on VP
...
SV is a view projected on PP
...
2 Pattern of planes & Pattern of views
3 Methods of drawing Orthographic Projections
PLANES
1
PRINCIPAL PLANES
HP AND VP
AUXILIARY PLANES
Auxiliary Vertical Plane
(A
...
P
...
I
...
)
A
...
P
...
P
...
ARROW DIRECTION IS A NORMAL WAY OF OBSERVING THE OBJECT
...
THE OTHER PLANES AND VIEWS ON THOSE CAN NOT BE SEEN
...
THIS WAY BOTH PLANES ARE BROUGHT IN THE SAME PLANE CONTAINING VP
...
PP
VP
Y
FV
LSV
Y
X
X
TV
HP
900
HP IS ROTATED DOWNWARD
AND
BROUGHT IN THE PLANE OF VP
...
ACTUAL PATTERN OF PLANES & VIEWS
OF ORTHOGRAPHIC PROJECTIONS
DRAWN IN
FIRST ANGLE METHOD OF PROJECTIONS
3
Methods of Drawing Orthographic Projections
First Angle Projections Method
Here views are drawn
by placing object
Third Angle Projections Method
Here views are drawn
by placing object
in 1st Quadrant
in 3rd Quadrant
...
FV
X
Y
Tv above X-y, Fv below X-y )
TV
NOTE:-
X
1st
TV
HP term is used in
Angle method
&
For the same
Ground term is used
in 3rd Angle method of projections
Y
FV
G
L
FIRST ANGLE
PROJECTION
FOR T
...
IN THIS METHOD,
THE OBJECT IS ASSUMED TO BE
SITUATED IN FIRST QUADRANT
MEANS
ABOVE HP & INFRONT OF VP
...
PP
VP
FV
LSV
Y
X
TV
HP
ACTUAL PATTERN OF
PLANES & VIEWS
IN
FIRST ANGLE METHOD
OF PROJECTIONS
THIRD ANGLE
PROJECTION
IN THIS METHOD,
THE OBJECT IS ASSUMED TO BE
SITUATED IN THIRD QUADRANT
( BELOW HP & BEHIND OF VP
...
TV
X
Y
LSV
FV
ACTUAL PATTERN OF
PLANES & VIEWS
OF
THIRD ANGLE PROJECTIONS
FOR T
...
ORTHOGRAPHIC PROJECTIONS
{ MACHINE ELEMENTS }
OBJECT IS OBSERVED IN THREE DIRECTIONS
...
AND NOW PROJECT THREE DIFFERENT VIEWS ON THOSE PLANES
...
FRONT VIEW IS A VIEW PROJECTED ON VERTICAL PLANE ( VP )
TOP VIEW IS A VIEW PROJECTED ON HORIZONTAL PLANE ( HP )
SIDE VIEW IS A VIEW PROJECTED ON PROFILE PLANE ( PP )
FIRST STUDY THE CONCEPT OF 1ST AND 3RD ANGLE
PROJECTION METHODS
AND THEN STUDY NEXT 26 ILLUSTRATED CASES CAREFULLY
...
V
...
OBJECT IS INBETWEEN
OBSERVER & PLANE
...
)
PLANES BEING TRANSPERENT
AND INBETWEEN
OBSERVER & OBJECT
...
V
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
H
...
V
...
V
...
V
...
V
...
H
...
V
...
V
...
V
...
V
...
V
...
SLOT
10
50
35
10
X
20 D
TV
60 D
30 D
TOP VIEW
O
Y
20
21
ORTHOGRAPHIC PROJECTIONS
10
25
15
25
40
10
25
25
O
O
80
F
...
PICTORIAL PRESENTATION IS GIVEN
DRAW FV AND SV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
25
S
...
FOR T
...
ORTHOGRAPHIC PROJECTIONS
450
30
FV
40
X
Y
30 D
TV
40
O
PICTORIAL PRESENTATION IS GIVEN
DRAW FV AND TV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
40
15
22
23
ORTHOGRAPHIC PROJECTIONS
HEX PART
30
20
20
40
O
20
50
20
15
O
30
60
100
PICTORIAL PRESENTATION IS GIVEN
DRAW FV ABD SV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
FRONT VIEW
L
...
SIDE VIEW
FOR T
...
ORTHOGRAPHIC PROJECTIONS
24
40
20
FRONT VIEW
F
...
30
10
X
Y
O
10
30
10
30
O
PICTORIAL PRESENTATION IS GIVEN
DRAW FV AND TV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
80
T
...
TOP VIEW
25
ORTHOGRAPHIC PROJECTIONS
10
10
15
25
25
Y
X
O
50
FV
10
LSV
PICTORIAL PRESENTATION IS GIVEN
DRAW FV AND LSV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
PICTORIAL PRESENTATION IS GIVEN
DRAW FV AND SV OF THIS OBJECT
BY FIRST ANGLE PROJECTION METHOD
26
ORTHOGRAPHIC PROJECTIONS
30
10
20
20
15
10
15
O
30
15
15
Y
X
50
F
...
LEFT S
...
ORTHOGRAPHIC PROJECTIONS
OF POINTS, LINES, PLANES, AND SOLIDS
...
}
B) OBSERVER
{ ALWAYS OBSERVING PERPENDICULAR TO RESP
...
PLANE}
...
P
...
P
...
P
...
P
FORM 4 QUADRANTS
...
IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV )
OF THE OBJECT WITH RESP
...
STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS
...
BECAUSE IT’S ALL VIEWS ARE JUST POINTS
...
OBJECT
POINT A
LINE AB
IT’S TOP VIEW
a
ab
IT’S FRONT VIEW
a’
a’ b’
IT’S SIDE VIEW
a”
a” b”
SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED
INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED
...
Quad
...
4th Quad
...
POINT A IN
Point A is
ND QUADRANT
Placed In
2
different
A
quadrants
and it’s Fv & Tv
are brought in
same plane for
Observer to see
HP
clearly
...
But as Tv is
is a view on Hp,
it is rotated
downward 900,
In clockwise
direction
...
Observe and
note the
process
...
POINT A ABOVE HP
& INFRONT OF VP
For Tv
For Tv
PICTORIAL
PRESENTATION
a’
a’
For Tv
PICTORIAL
PRESENTATION
A
A
Y
Y
X
POINT A IN HP
& INFRONT OF VP
POINT A ABOVE HP
& IN VP
a’
a
a
X
X
a
Y
A
ORTHOGRAPHIC PRESENTATIONS
OF ALL ABOVE CASES
...
Fv above xy,
Tv on xy
...
VP
a’
X
VP
a’
Y
X
a
Y
a’
X
a
HP
a
HP
HP
Y
PROJECTIONS OF STRAIGHT LINES
...
AIM:- TO DRAW IT’S PROJECTIONS - MEANS FV & TV
...
A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)
2
...
3
...
4
...
5
...
STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE
SHOWING CLEARLY THE NATURE OF FV & TV
OF LINES LISTED ABOVE AND NOTE RESULTS
...
a’
A
1
...
P
...
P
...
b’
A Line
// to Hp
&
// to Vp
B
a’
A
V
...
Note:
Fv & Tv both are
// to xy
&
both show T
...
a’
Fv
b’
X
Y
b
Y
a
X
a
H
...
Tv
b
b’
3
...
P
...
b’
B
a’
a’
Y
X
(Pictorial presentation)
A
Y
a
b
T
...
b
X
a
H
...
Orthographic Projections
4
...
P
...
a’
b’
Fv
b’
a’
A
Ø
B
(Pictorial presentation)
X
Y
a
a
Ø
Ø
Tv
b
b
H
...
For Tv
For Tv
5
...
e
...
Tv as a image on Hp,
a’
A
X
T
...
Y
b
a
T
...
b
V
...
b’
FV
a’
X
Y
Orthographic Projections
Fv is seen on Vp clearly
...
(No view is parallel to xy)
Both Fv & Tv are reduced lengths
...
H
...
b
Note the procedure
When Fv & Tv known,
How to find True Length
...
Orthographic Projections
Means Fv & Tv of Line AB
are shown below,
with their apparent Inclinations
&
V
...
V
...
V
...
b’
b’
FV
a’
a’
Y
TL
a’
X
Y
a
TV
H
...
b’ b1’
b 1’
FV
X
a
Note the procedure
When True Length is known,
How to locate Fv & Tv
...
H
...
b
In this sketch, TV is rotated
and made // to XY line
...
H
...
b
b1
Here a -1 is component
of TL ab1 gives length of Fv
...
a’ b’ will be Fv
...
The most important diagram showing graphical relations
among all important parameters of this topic
...
1) True Length ( TL) – a’ b1’ & a b
2) Angle of TL with Hp 3) Angle of TL with Vp –
4) Angle of FV with xy –
5) Angle of TV with xy –
V
...
Distance between
End Projectors
...
b & b1 on same locus
...
P
...
It’s horizontal component
is drawn & it is further rotated to locate view
...
Line AB is 75 mm long and it is 300 &
400 Inclined to Hp & Vp respectively
...
Draw projections
...
b’
FV
SOLUTION STEPS:
1) Draw xy line and one projector
...
3) Take 300 angle from a’ & 400 from
a and mark TL I
...
75mm on both
lines
...
4) Join both points with a’ and a resp
...
6) Draw horizontal component of TL
a b1 from point b1 and name it 1
...
)
7) Extend it up to locus of a’ and
rotating a’ as center locate b’ as
shown
...
8) From b’ drop a projector down
ward & get point b
...
e
...
b’1
TL
a’
X
Y
a
LFV
Ø
1
TV
TL
b
b1
PROBLEM 2:
Line AB 75mm long makes 450 inclination with Vp while it’s Fv makes 550
...
If line is in 1st quadrant
draw it’s projections and find it’s inclination with Hp
...
Draw x-y line
...
Draw one projector for a’ & a
3
...
4
...
Take 550 angle from a’ for Fv
above xy line
...
Draw a vertical line from b1
up to locus of a and name it 1
...
7
...
This a’ b’
line is Fv
...
Drop a projector from b’ on
locus from point b1 and
name intersecting point b
...
9
...
Join a’ b1’ as TL and measure
it’s angle at a’
...
550
a’
y
X
LFV
a
1
LOCUS OF b
b
b1
PROBLEM 3:
Fv of line AB is 500 inclined to xy and measures 55
mm long while it’s Tv is 600 inclined to xy line
...
SOLUTION STEPS:
1
...
2
...
3
...
4
...
5
...
6
...
X
a’
b’
b’1
500
y
a
600
b
b1
PROBLEM 4 :Line AB is 75 mm long
...
End A is 10 mm above Hp and 15 mm in front of Vp
...
Find angle with Hp and Vp
...
Draw xy line and one projector
...
Locate a’ 10 mm above xy and
a 15 mm below xy line
...
Draw locus from these points
...
Cut 60mm distance on locus of a’
& mark 1’ on it as it is LTV
...
Similarly Similarly cut 50mm on
locus of a and mark point 1 as it is LFV
...
From 1’ draw a vertical line upward
and from a’ taking TL ( 75mm ) in
compass, mark b’1 point on it
...
7
...
With same steps below get b1 point
and draw also locus from it
...
Now rotating one of the components
I
...
a-1 locate b’ and join a’ with it
to get Fv
...
Locate tv similarly and measure
Angles &
a’
1’
LTV
X
Y
a
LFV
b
1
b1
PROBLEM 5 :-
T
...
of a 75 mm long Line CD, measures 50 mm
...
End D is 15 mm in front of Vp and it is above Hp
...
SOLUTION STEPS:
1
...
2
...
3
...
4
...
Cut 50mm & 75 mm distances on
locus of d from c and mark points
d & d1 as these are Tv and line CD
lengths resp
...
6
...
e
...
7 Then draw one projector from d to
meet this arc in d’ point & join c’ d’
8
...
Measure Angles
&
X
c’
d’
d’1
Y
d
c
LOCUS OF d’ & d’1
d1
LOCUS OF d & d1
GROUP (B)
PROBLEMS INVOLVING TRACES OF THE LINE
...
A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES H
...
,
THAT POINT IS CALLED TRACE OF THE LINE ON H
...
( IT IS CALLED H
...
)
SIMILARLY, A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES V
...
,
THAT POINT IS CALLED TRACE OF THE LINE ON V
...
( IT IS CALLED V
...
)
V
...
:-
It is a point on Vp
...
Hence it’s Tv comes on XY line
...
T
...
Hence it is called Tv of a point in Hp
...
( Here onward named as ’h’ )
b’
STEPS TO LOCATE HT
...
)
1
...
3
...
Begin with FV
...
Name this point h’
( as it is a Fv of a point in Hp)
Draw one projector from h’
...
This point is HT
STEPS TO LOCATE VT
...
)
1
...
Extend TV up to XY line
...
Name this point v
( as it is a Tv of a point in Vp)
Draw one projector from v
...
This point is VT
3
...
a’
v
h’
VT’
HT
x
y
a
Observe & note :-
1
...
b
2
...
3
...
4
...
5
...
These points are used to
solve next three problems
...
Line’s Tv makes 300 with XY line
...
Draw projections of line AB,determine inclinations with Hp & Vp and locate HT, VT
...
10
0 angle from a’ and
Take 45
VT’
marking 60 mm on it locate point b’
...
as fv-h’-vt’ lie on one st
...
Draw projector from vt, locate v on xy
...
Draw projector from b’ and locate b I
...
Tv point
...
Name extension of Fv, touching xy as h’
and below it, on extension of Tv, locate HT
...
It’s Fv is 450 inclined to xy while it’s HT & VT are 45mm and 30 mm below xy respectively
...
b’
b’
1
a’
10
X
v
LOCUS OF b’ & b’1
450
h’
Y
30
45
VT’
HT
SOLUTION STEPS:Draw xy line, one projector and
locate a’ 10 mm above xy
...
Take 450 angle from a’ and extend that line backward
to locate h’ and VT, & Locate v on xy above VT
...
Then join v – HT – and extend to get top view end b
...
Now as usual rotating views find TL and it’s inclinations
...
End A is 10 mm above Hp, VT is 35 mm below Hp
while it’s HT is 45 mm in front of Vp
...
VT
b’
SOLUTION STEPS:1
...
2
...
3
...
b’1
55
Locus of a’
10
X
a’
50
h’
b
35
a
HT
80
y
v
b1
Instead of considering a & a’ as projections of first point,
if v & VT’ are considered as first point , then true inclinations of line with
Hp & Vp i
...
angles & can be constructed with points VT’ & V respectively
...
&
From point VT’ & h’
angles & can be drawn
...
a
b
b1
PROBLEM 9 :Line AB 100 mm long is 300 and 450 inclined to Hp & Vp respectively
...
Draw projections of the line and it’s HT
...
20
Draw locus of a’ 10 mm above xy
...
VT’
Where it intersects with locus of a’
name it a1’ as it is TL of that part
...
e
...
Draw it’s component on locus of VT’
& further rotate to get other end of Fv i
...
b’
Join it with VT’ and mark intersection point
(with locus of a1’ ) and name it a’
Now as usual locate points a and b and h’ and HT
...
It’s Fv & Tv make 450 and 600 inclinations with X-Y line resp
End A is 15 mm above Hp and VT is 20 mm below Xy line
...
Draw projections, find inclinations with Hp & Vp
...
b’
b1’
FV
a’
Locus of a & a1’
15
X
h’
v
600
20
Y
450
VT’
SOLUTION STEPS:Similar to the previous only change
is instead of line’s inclinations,
views inclinations are given
...
a1’
HT
a
a1
TV
b
b1
PROBLEM 11 :- The projectors drawn from VT & end A of line AB are 40mm apart
...
VT of line is 20 mm below Hp
...
40mm
b
b1
GROUP (C)
CASES OF THE LINES IN A
...
P
...
I
...
& PROFILE PLANE
...
It’s FV (a’b’) is shown projected on Vp
...
V
...
A
B
Line AB is in AVP as shown in above figure no 2
...
(Looking in arrow direction)
Here one can clearly see that the
Inclination of AVP with VP = Inclination of TV with XY line
a
b
LINE IN A PROFILE PLANE ( MEANS IN A PLANE PERPENDICULAR TO BOTH HP & VP)
For T
...
ORTHOGRAPHIC PATTERN OF LINE IN PROFILE PLANE
a’
VP
PP
VT
a’
A
FV
b’
a”
LSV
b’
X
b”
HT
Y
a
B
TV
a
b
b
HP
Results:1
...
2
...
Sum of it’s inclinations with HP & VP equals to 900 ( + = 900 )
4
...
OBSERVE CAREFULLY ABOVE GIVEN ILLUSTRATION AND 2nd SOLVED PROBLEM
...
Vertical Plane 450 inclined to Vp
...
Draw projections, fine angle with Vp and Ht
...
Other end B is 15 mm above Hp
and 50 mm in front of Vp
...
Find true angles with ref
...
Front view
After drawing xy line and one projector
Locate top view of A I
...
e
...
and Tv of B i
...
b, 50 mm
below xy asit is 50 mm in front of Vp
Draw side view structure of Vp and Hp
top view
and locate S
...
of point B i
...
b’’
From this point cut 75 mm distance on Vp and
b
Mark a’’ as A is in Vp
...
)
From this point draw locus to left & get a’
HT
Extend SV up to Hp
...
As it is a Tv
Rotate it and bring it on projector of b
...
(VT)
a”
Side View
( True Length )
VP
SOLUTION STEPS:-
VT
a’
HP
b”
(HT)
Y
APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES
IN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS
...
It’s relation with Ground ( HP )
And
a Wall or some vertical object ( VP ) will be given
...
Here various problems along with
actual pictures of those situations are given
for you to understand those clearly
...
CHECK YOUR ANSWERS
WITH THE SOLUTIONS
GIVEN IN THE END
...
Flower A is 1M & 5
...
Orange B is 4M & 1
...
Drawing projection, find distance between them
If flower is 1
...
5 M above the ground
...
TV
B
A
FV
Wall Q
PROBLEM 15 :- Two mangos on a tree A & B are 1
...
00 m above ground
and those are 1
...
5 m from a 0
...
If the distance measured between them along the ground and parallel to wall is 2
...
TV
B
A
0
...
All equally inclined and the shortest
is vertical
...
is TV of three rods OA, OB and OC
whose ends A,B & C are on ground and end O is 100mm
above ground
...
TV
O
C
A
FV
45 mm
B
PROBLEM 17:- A pipe line from point A has a downward gradient 1:5 and it runs due East-South
...
Pipe line from B runs
200 Due East of South and meets pipe line from A at point C
...
5
1
A
12 M
B
E
C
PROBLEM 18: A person observes two objects, A & B, on the ground, from a tower, 15 M high,
At the angles of depression 300 & 450
...
Draw projections of situation and find distance of objects from
observer and from tower also
...
5m and 7
...
The poles are 10 M apart
...
TV
C
15 M
A
300
4
...
5M
PROBLEM 20:- A tank of 4 M height is to be strengthened by four stay rods from each corner
by fixing their other ends to the flooring, at a point 1
...
7 M from two adjacent walls respectively,
as shown
...
TV
4M
PROBLEM 21:- A horizontal wooden platform 2 M long and 1
...
Draw projections of the objects and determine length of each chain along with it’s inclination with ground
...
A room is of size 6
...
5m high
...
A switch is placed in one of the corners of the room, 1
...
Draw the projections an determine real distance between the bulb and switch
...
IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS
...
5 M ABOVE WALL RAILING
...
24
T
...
of a 75 mm long Line CD, measures 50 mm
...
End D is 15 mm in front of Vp and it is above Hp
...
SOME CASES OF THE LINE
IN DIFFERENT QUADRANTS
...
d’
d’1
LOCUS OF d’ & d’1
X
Y
c’
c
d
d1
LOCUS OF d & d1
PROBLEM NO
...
End B in Vp
...
Distance between projectors is 70mm
...
b’
a
X
a’
LOCUS OF b’ & b’1
b’1
b
b1
Y
LOCUS OF b & b1
70
PROBLEM NO
...
Line is 300 inclined to Hp
...
Draw projections, find inclination with Vp and traces
...
27
End A of a line AB is 25mm above Hp and end B is 55mm behind Vp
...
If both it’s HT & VT coincide on xy in a point,
35mm from projector of A and within two projectors,
b
Draw projections, find TL and angles and HT, VT
...
What is usually asked in the problem?
To draw their projections means F
...
V
...
V
...
Description of the plane figure
...
It’s position with HP and VP
...
Inclination of it’s SURFACE with one of the reference planes will be given
...
Inclination of one of it’s EDGES with other reference plane will be given
(Hence this will be a case of an object inclined to both reference Planes
...
CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS
...
Assume suitable conditions & draw Fv & Tv of initial position
...
Now consider surface inclination & draw 2nd Fv & Tv
...
After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv
...
If in problem surface is inclined to HP – assume it // HP
Or If surface is inclined to VP – assume it // to VP
2
...
And If surface is assumed // to VP – It’s FV will show True Shape
...
Hence begin with drawing TV or FV as True Shape
...
While drawing this True Shape –
keep one side/edge ( which is making inclination) perpendicular to xy line
( similar to pair no
...
Now Complete STEP 2
...
(Ref
...
By making side inclined to the resp plane & project it’s other view
...
3nd pair
C on previous page illustration )
APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS
Read problem and answer following questions
1
...
Assumption for initial position? ------// to HP
3
...
Which side will be vertical? ---One small side
...
Problem 1:
Rectangle 30mm and 50mm
sides is resting on HP on one
small side which is 300 inclined
to VP,while the surface of the
plane makes 450 inclination with
HP
...
Surface // to Hp
a’ b’
Surface inclined to Hp
b
d’1
c’ d’
a’ b’ 450
X
a
c’1
d’ c’
b’1
d
a1
d1
c
b1
c1
300
a’1
Y
Side
Inclined
to Vp
Problem 2:
A 300 – 600 set square of longest side
100 mm long, is in VP and 300 inclined
to HP while it’s surface is 450 inclined
to VP
...
Surface inclined to which plane? ------VP
2
...
So which view will show True shape? --- FV
4
...
(Surface & Side inclinations directly given)
a’
Hence begin with FV, draw triangle above X-Y
keeping longest side vertical
...
Surface inclined to which plane? ------VP
2
...
So which view will show True shape? --- FV
4
...
Problem 3:
A 300 – 600 set square of longest side
100 mm long is in VP and it’s surface
450 inclined to VP
...
Draw it’s projections
Hence begin with FV, draw triangle above X-Y
keeping longest side vertical
...
Side inclination indirectly given)
a’
First TWO steps are similar to previous problem
...
End A 35 mm above Hp & End B is 10 mm above Hp
...
a’1
c’
c’1
c’1
a’1
35
X
10
a
a
b
b’1
b’1
b’
c
b
a1
450
c
b1
c1
Y
Problem 4:
Read problem and answer following questions
A regular pentagon of 30 mm sides is
resting on HP on one of it’s sides with it’s
surface 450 inclined to HP
...
Surface inclined to which plane? ------HP
2
...
So which view will show True shape? --- TV
4
...
Hence begin with TV,draw pentagon below
X-Y line, taking one side vertical
...
d’1
d’
c’e’
X
b’ a’
c’e’
d’
c’1
e’1
b’ a’
a’1
450
e1
e
d1
d1
b1
c
300
b1
d
b
a1
e1
a1
a
b’1 Y
c1
c1
Read problem and answer following questions
1
...
Assumption for initial position? ------ // to HP
3
...
Which side will be vertical? --------any side
...
Problem 5:
A regular pentagon of 30 mm sides is resting
on HP on one of it’s sides while it’s opposite
vertex (corner) is 30 mm above HP
...
SURFACE INCLINATION INDIRECTLY GIVEN
SIDE INCLINATION DIRECTLY GIVEN:
ONLY CHANGE is
the manner in which surface inclination is described:
One side on Hp & it’s opposite corner 30 mm above Hp
...
Keep a’b’ on xy & d’ 30 mm above xy
...
Draw it’s projections
...
Surface inclined to which plane? ------HP
2
...
So which view will show True shape? --- TV
4
...
Draw it’s projections
...
In problem no
...
While in no
...
e
...
Hence here angle of TL is taken,locus of c1
Is drawn and then LTV I
...
a1 c1 is marked and
final TV was completed
...
a’
b’ d’
c’1
b’1
c’
d
Note the difference in
construction of 3rd step
in both solutions
...
Diameter AB, 50 mm long is
300 & 600 inclined to HP & VP respectively
...
Read problem and answer following questions
1
...
Assumption for initial position? ------ // to HP
3
...
Which diameter horizontal? ---------AB
Hence begin with TV,draw CIRCLE below
X-Y line, taking DIA
...
9
...
Like 9th problem True Length inclination of dia
...
Means Line AB lies in a Profile Plane
...
So do the construction accordingly AND note
X
the case carefully
...
Draw it’s projections
...
Read problem and answer following questions
1
...
Assumption for initial position? ------ // to HP
3
...
Which diameter horizontal? ---------AC
Hence begin with TV,draw rhombus below
X-Y line, taking longer diagonal // to X-Y
ONLY CHANGE is the manner in which surface inclination
is described:
One side on Hp & it’s opposite side 25 mm above Hp
...
Keep a’b’ on xy & d’e’ 25 mm above xy
...
Because it is in VP
as said in problem
...
IMPORTANT POINTS
1
...
2
...
3
...
4
...
(Here keep line joining point of contact & centroid of fig
...
Always begin with FV as a True Shape but in a suspended position
...
Problem 12:
An isosceles triangle of 40 mm long
base side, 60 mm long altitude Is
freely suspended from one corner of
Base side
...
Draw it’s projections
...
Y
B
b
a,g
c
450
Similarly solve next problem
of Semi-circle
IMPORTANT POINTS
Problem 13
:A semicircle of 100 mm diameter
is suspended from a point on its
straight edge 30 mm from the midpoint
of that edge so that the surface makes
an angle of 450 with VP
...
1
...
2
...
3
...
4
...
(Here keep line joining point of contact & centroid of fig
...
Always begin with FV as a True Shape but in a suspended position
...
A
a’
20 mm
p’
P
b’
G
g’
CG
c’
d’
X
First draw a given semicircle
With given diameter,
Locate it’s centroid position
And
join it with point of suspension
...
BY USING AUXILIARY PLANE METHOD
WHAT WILL BE THE PROBLEM?
Description of final Fv & Tv will be given
...
Follow the below given steps:
1
...
2
...
L
...
Draw x1-y1 perpendicular to this line showing T
...
4
...
Draw x2-y2 // to this line view & project new view on it
...
e
...
The facts you must know:If you carefully study and observe the solutions of all previous problems,
You will find
IF ONE VIEW IS A LINE VIEW & THAT TOO PARALLEL TO XY LINE,
THEN AND THEN IT’S OTHER VIEW WILL SHOW TRUE SHAPE:
NOW FINAL VIEWS ARE ALWAYS SOME SHAPE, NOT LINE VIEWS:
SO APPLYING ABOVE METHOD:
WE FIRST CONVERT ONE VIEW IN INCLINED LINE VIEW
...
plane)
THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE
...
Ab is 50 mm long, angle cab is 300 and angle cba is 650
...
a’ is 25 mm, b’ is 40 mm and c’ is 10 mm above Hp respectively
...
As per the procedure1
...
2
...
So draw x1y1 perpendicular to it
...
Project view on x1y1
...
b) from xy take distances of a,b & c( Tv) mark on these projectors from x1y1
...
c) This line view is an Aux
...
Draw x2y2 // to this line view and project Aux
...
for that from x1y1 take distances of a’b’ & c’ and mark from x2y= on new projectors
...
Name points a’1 b’1 & c’1 and join them
...
Y1
a1b1
Y
2
b’
15
b’1
a’
15
C1
C’
10
X
c
a
Y
650
300
50 mm
b
X1
a’1
X2
c’1
ALWAYS FOR NEW FV TAKE
DISTANCES OF PREVIOUS FV
AND FOR NEW TV, DISTANCES
OF PREVIOUS TV
REMEMBER!!
Problem 15: Fv & Tv of a triangular plate are shown
...
50
USE SAME PROCEDURE STEPS
OF PREVIOUS PROBLEM:
25
BUT THERE IS ONE DIFFICULTY:
c’
15
a’
NO LINE IS // TO XY IN ANY VIEW
...
1’
20
IN SUCH CASES DRAW ONE LINE
// TO XY IN ANY VIEW & IT’S OTHER
VIEW CAN BE CONSIDERED AS TL
FOR THE PURPOSE
...
HENCE it’s Tv a-1 becomes TL
...
(STUDY THE ILLUSTRATION)
y1
c’1
y2
c1
b’1
x2
ALWAYS FOR NEW FV TAKE
DISTANCES OF PREVIOUS FV
AND FOR NEW TV, DISTANCES
OF PREVIOUS TV
REMEMBER!!
b1
d1
PROBLEM 16: Fv & Tv both are circles of 50 mm diameter
...
ADOPT SAME PROCEDURE
...
Then a’c’ becomes TL for the purpose
...
50D
y1
b’
b1
Study the illustration
...
c
b
ac1
1
c’
d
X1 1
Y
y2
c’1
b’1
X2
a’1
d’1
TRUE SHAPE
Problem 17 : Draw a regular pentagon of
30 mm sides with one side 300 inclined to xy
...
Determine it’s true shape
...
X1
BUT ACTUALLY WE DONOT REQUIRE
TL TO FIND IT’S TRUE SHAPE, AS ONE
VIEW (FV) IS ALREADY A LINE VIEW
...
a’
e1
d1
b’
e’
c’
d’
450
X
300
e
ALWAYS FOR NEW FV
TAKE DISTANCES OF
PREVIOUS FV AND FOR
NEW TV, DISTANCES OF
PREVIOUS TV
REMEMBER!!
d
a
b
c
Y1
Y
SOLIDS
To understand and remember various solids in this subject properly,
those are classified & arranged in to two major groups
...
Cylinder
Cone
Prisms
Triangular
Square
Cube
( A solid having
six square faces)
Pentagonal Hexagonal
Pyramids
Triangular
Square
Tetrahedron
( A solid having
Four triangular faces)
Pentagonal Hexagonal
SOLIDS
Dimensional parameters of different solids
...
Frustum of cone & pyramids
...
P
On it’s base
...
)
F
...
X
X
RESTING ON H
...
(Axis inclined to Hp
And // to Vp)
LYING ON H
...
(Axis inclined to Hp
And // to Vp)
F
...
F
...
While observing Fv, x-y line represents Horizontal Plane
...
(Vp)
T
...
T
...
STANDING ON V
...
Axis perpendicular to Vp
And // to Hp
T
...
RESTING ON V
...
Axis inclined to Vp
And // to Hp
LYING ON V
...
Axis inclined to Vp
And // to Hp
Y
Y
STEPS TO SOLVE PROBLEMS IN SOLIDS
Problem is solved in three steps:
STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION
...
BEGIN WITH THIS VIEW:
IT’S OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS):
IT’S OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS):
DRAW FV & TV OF THAT SOLID IN STANDING POSITION:
STEP 2: CONSIDERING SOLID’S INCLINATION ( AXIS POSITION ) DRAW IT’S FV & TV
...
GENERAL PATTERN ( THREE STEPS ) OF SOLUTION:
GROUP B SOLID
...
CYLINDER
GROUP B SOLID
...
CYLINDER
AXIS
er
TO VP
AXIS
INCLINED
VP
Three steps
If solid is inclined to Vp
AXIS
INCLINED HP
AXIS
er
TO VP
AXIS
INCLINED
VP
Three steps
If solid is inclined to Vp
Study Next Twelve Problems and Practice them separately !!
CATEGORIES OF ILLUSTRATED PROBLEMS!
PROBLEM NO
...
5 & 6
CASES OF CUBE & TETRAHEDRON
PROBLEM NO
...
PROBLEM NO
...
9
CASE OF TRUE LENGTH INCLINATION WITH HP & VP
...
10 & 11
CASES OF COMPOSITE SOLIDS
...
12
CASE OF A FRUSTUM (AUXILIARY PLANE)
Problem 1
...
Draw its projections
...
Assume it standing on Hp
...
It’s Tv will show True Shape of base( square)
3
...
( a triangle)
4
...
5
...
e
...
And project it’s Tv
...
Make visible lines dark and hidden dotted, as per the procedure
...
Then construct remaining inclination with Vp
( Vp containing axis ic the center line of 2nd Tv
...
o’
a’1
X a’b’
c’
d’
d
a
b
c’1
d’1
d1
a1
b’1
o’1
Y
a1
o1
o
c
c1
b1
For dark and dotted lines
1
...
2
...
3
...
4
...
(APEX
NEARER
TO V
...
(APEX
AWAY
FROM V
...
)
Solution Steps:
Resting on Hp on one generator, means lying on Hp:
1
...
2
...
Draw 40mm dia
...
( a triangle)
4
...
5
...
e
...
And
project it’s Tv below xy
...
Make visible lines dark and hidden dotted,
as per the procedure
...
Then construct remaining inclination with Vp
( generator o1e1 300 to xy as shown) & project final Fv
...
For dark and dotted lines
1
...
2
...
3
...
4
...
o’
h’1
a’1
b’1
g’1
X
a’ h’b’
c’ g
’
g
h
o’
f’ d’ e’
g1
f
f1
f’1
e’1
g1
h1
c’
d’1 1
a
a1
e e1
a1
o1
b1
e1
b
d
c
b1
d1
c1
d1
o1
30
o1
h1
f1
Y
c1
Problem 3:
A cylinder 40 mm diameter and 50 mm
axis is resting on one point of a base
circle on Vp while it’s axis makes 450
with Vp and Fv of the axis 350 with Hp
...
d’
4’d’
3’
c’ a’
1’ a’
X
bd
b’
c
4’
c’
2’ b’
a
Solution Steps:
Resting on Vp on one point of base, means inclined to Vp:
1
...
It’s Fv will show True Shape of base & top( circle )
3
...
Circle as Fv & taking 50 mm axis project Tv
...
Name all points as shown in illustration
...
Draw 2nd Tv making axis 450 to xy And project it’s Fv above xy
...
Make visible lines dark and hidden dotted, as per the procedure
...
Then construct remaining inclination with Hp
( Fv of axis I
...
center line of view to xy as shown) & project final Tv
...
Assume it standing on Hp but as said on apex
...
2
...
Draw a corner case square of 30 mm sides as Tv(as shown)
Showing all slant edges dotted, as those will not be visible from top
...
taking 50 mm axis project Fv
...
Name all points as shown in illustration
...
Draw 2nd Fv keeping o’a’ slant edge vertical & project it’s Tv
7
...
8
...
e
...
Then as usual project final Fv
...
Draw it’s projections
...
1
...
Project Fv and name all points of FV & TV
...
Draw a body-diagonal joining c’ with 3’( This can become // to xy)
3
...
Draw 2nd Fv in which 1’-p’ line is vertical means c’-3’ diagonal
must be horizontal
...
6
...
Then as usual project final FV
...
Begin with Tv , an equilateral triangle as side case as shown:
First project base points of Fv on xy, name those & axis line
...
In 2nd Fv make face o’b’c’ vertical as said in problem
...
Problem 6:A tetrahedron of 50 mm
long edges is resting on one edge on
Hp while one triangular face containing
this edge is vertical and 450 inclined to
Vp
...
IMPORTANT:
Tetrahedron is a
special type
of triangular
pyramid in which
base sides &
slant edges are
equal in length
...
Like cube it is also
described by One X
dimension only
...
o’1
o’
TL
900
b’
c’
a’
c’1
450
c
a
o
c1
a1
b
a’1
o1
b1
b’1
Y
FREELY SUSPENDED SOLIDS:
Positions of CG, on axis, from base, for different solids are shown below
...
Draw it’s three views
...
1
...
2
...
) and name g’ and
Join it with corner d’
3
...
4
...
LINE
o’
d’g’ VERTICAL
d’
d”
c’e’
g’
FOR SIDE VIEW
H
a’b’
g’
IMPORTANT:
When a solid is freely
suspended from a
corner, then line
joining point of
contact & C
...
remains vertical
...
)
So in all such cases,
assume solid standing
on Hp initially
...
Assuming it standing on Hp begin with Tv, a square of corner case
...
Project corresponding Fv
...
3
...
Project it’s Tv drawing dark and dotted lines as per the procedure
...
With standard method construct Left-hand side view
...
( Draw a 450 inclined Line in Tv region ( below xy)
...
After this, draw
horizontal lines, from all points of Fv, to meet these
lines
...
For dark & dotted lines
locate observer on left side of Fv as shown
...
Draw it’s projections
...
7 & 9 from projections of planes topic
...
Like
Tv of axis is inclined to Vp etc
...
Means here TL inclination is expected
...
See carefully the final Tv and inclination taken there
...
o’1
o’
a’1
h’1
b’1
g’1
X
a’ h’b’
c’ g’
f’ d’ e’
h
a
h1
e
d
c
400
g1
f
b
f’1
450
g
a1
c’1
d’1
e’1
y
Axis True Length
f1
1
b1
e1
Axis Tv Length
o1
1
d1
c1
Axis Tv Length
Locus of
Center 1
Problem 10: A triangular prism,
40 mm base side 60 mm axis
is lying on Hp on one rectangular face
with axis perpendicular to Vp
...
It’s base side is
30 mm & axis is 60 mm long resting on Hp
on one edge of base
...
Project another FV
on an AVP 450 inclined to VP
...
Project Tv of both solids
...
Fv on it
...
Fv from x1y1 line
...
F
...
X
Y
450
T
...
Aux
...
V
...
A tetrahedron is placed centrally
on the top of it
...
Draw projections of both solids
...
STEPS:
Draw a regular hexagon as Tv of
standing prism With one side // to xy
and name the top points
...
Now join it’s alternate corners
a-c-e and the triangle formed is base
of a tetrahedron as said
...
and complete Fv
of tetrahedron
...
Tv on it by using similar
Steps like previous problem
...
Tv
Y
e
f
Tva
o
o1
e1
450
f1
d
d1
c1
a1
b1
b
c
X1
Problem 12: A frustum of regular hexagonal pyramid is standing on it’s larger base
On Hp with one base side perpendicular to Vp
...
Project it’s Aux
...
Base side is 50 mm long , top side is 30 mm long and 50 mm is height of frustum
...
e
...
Tv
d1
e
d
Tv
e1
c1
5
4
a
2
X1
1
3
2
b
c
a1
b1
ENGINEERING APPLICATIONS
OF
THE PRINCIPLES
OF
PROJECTIONS OF SOLIDES
...
SECTIONS OF SOLIDS
...
DEVELOPMENT
...
INTERSECTIONS
...
An object ( here a solid ) is cut by
some imaginary cutting plane
to understand internal details of that object
...
Two cutting actions means section planes are recommended
...
( This is a definition of an Aux
...
e
...
I
...
)
NOTE:- This section plane appears
as a straight line in FV
...
( This is a definition of an Aux
...
e
...
V
...
)
NOTE:- This section plane appears
as a straight line in TV
...
After launching a section plane
either in FV or TV, the part towards observer
is assumed to be removed
...
As far as possible the smaller part is
assumed to be removed
...
For TV
SECTION
PLANE
TRUE SHAPE
Of SECTION
x
y
Apparent Shape
of section
SECTION LINES
(450 to XY)
SECTIONAL T
...
Typical Section Planes
&
Typical Shapes
Of
Sections
...
Ellipse
Cylinder through
generators
...
Trapezium
Sq
...
MEANING:ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET
...
THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLED
DEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID
...
ENGINEERING APLICATION:
THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY
CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES
...
THERE IS A VAST RANGE OF SUCH OBJECTS
...
WHAT IS
OUR OBJECTIVE
IN THIS TOPIC ?
But before going ahead,
note following
Important points
...
1
...
2
...
3
...
4
...
Study illustrations given on next page carefully
...
(Lateral surface is the surface excluding top & base)
Cylinder:
A Rectangle
Cone: (Sector of circle)
Pyramids: (No
...
L=Slant height
...
of Rectangles
=
L= Slant edge
...
Tetrahedron: Four Equilateral Triangles
All sides
equal in length
FRUSTUMS
DEVELOPMENT OF
FRUSTUM OF CONE
DEVELOPMENT OF
FRUSTUM OF SQUARE PYRAMID
Base side
Top side
=
R 3600
L
R= Base circle radius of cone
L= Slant height of cone
L1 = Slant height of cut part
...
STUDY NEXT NINE PROBLEMS OF
SECTIONS & DEVELOPMENT
Problem 1: A pentagonal prism , 30 mm base side & 50 mm axis
is standing on Hp on it’s base whose one side is perpendicular to Vp
...
Draw Fv, sec
...
Side view
...
C
B
D
A
Y1
A
E
d”
X1
a”
e
B
C
D
E
A
c”
b”
e”
X
Solution Steps:for sectional views:
Draw three views of standing prism
...
plane in Fv as described
...
Join those points in sequence and show
Section lines in it
...
Y
DEVELOPMENT
d
For True Shape:
a
Draw x1y1 // to sec
...
Mark distances of points
of Sectioned part from Tv,
on above projectors from
x1y1 and join in sequence
...
It is required true shape
...
Name from
cut-open edge I
...
A
...
Mark the cut points on respective edges
...
lines
...
dark
...
It cut by a section plane 450 inclined
to Hp through base end of end generator
...
Y1
Solution Steps:for sectional views:
Draw three views of standing cone
...
plane in Fv as described
...
Join those points in
sequence and show Section lines in it
...
A
SECTIONAL S
...
plane
Draw projectors on it from
cut points
...
Draw section lines in it
...
h
E
g” h”f”
a”e”
b”d” c”
Y
F
G
f
a
e
b
d
c
SECTIONAL T
...
Name from cut-open edge i
...
A
...
Mark the cut
points on respective edges
...
Make existing parts dev
...
H
A
Problem 3: A cone 40mm diameter and 50 mm axis is resting on one generator on Hp( lying on Hp)
which is // to Vp
...
It is cut by a horizontal section plane through it’s base
center
...
Follow similar solution steps for Sec
...
V
(SHOWING TRUE SHAPE OF SECTION)
a’ b’
c’ f’
d’ e’
3
SECTIONAL F
...
4
Use similar steps for sec
...
2
6
Y
8
f
7
f1
a1 e1
e
a
A
...
P300 inclined to Vp
Through mid-point of axis
...
plane appears as a line
...
Hence
it is opened from c and named C,D,E,F,A,B,C
...
30 mm base side &
55 mm axis is lying on Hp on it’s rect
...
It is cut by a section plane normal to Hp and
300 inclined to Vp bisecting axis
...
Views, true shape & development
...
Fv:
Those are transferred to
1st TV, then to 1st Fv and
Then on 2nd Fv
...
V
...
c1
c X1
C
D
E
F
Y1
DEVELOPMENT
A
B
C
2
Problem 5:A solid composed of a half-cone and half- hexagonal pyramid is
shown in figure
...
Draw F
...
, sectional T
...
,true shape of section and
development of remaining part of the solid
...
)
3
4
Y1
5
1
Note:
O’
A
6
Fv & TV 8f two solids
sandwiched
Section lines style in both:
Development of
half cone & half pyramid:
B
7
X1
2’ 6’
3’
5’
4’
C
F
...
4
2
3
X
d’e’
c’f’
f
a’
Y
O
E
7
g
5
SECTIONAL 7
TOP VIEW
...
If the semicircle is development of a cone and inscribed circle is some
curve on it, then draw the projections of cone showing that curve
...
L=Slant height
...
d
e
Solution Steps:
Draw semicircle of given diameter, divide it in 8 Parts and inscribe in it
a largest circle as shown
...
Semicircle being dev
...
( L )
Then using above formula find R of base of cone
...
Take o -1 distance from dev
...
e
...
Then project all on Tv
on respective generators and join by smooth curve
...
Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largest
rhombus
...
Solution Steps:
Similar to previous
Problem:
o’
E
D
F
C
2’ 6’
X
a’ h’ b’
c’ g’
B
3’ 5’
f’ d’ e’
1’ 7’
4’
g
7
H
Y
A
O
L
6
h
f
5
a
4
3
b
2
1
G
c
d
e
R=Base circle radius
...
R 3600
L
=
A
Problem 8: A half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat face
parallel and nearer to VP
...
If the string is of shortest length, find it and show it on the projections of the cone
...
o’
A
B
C
1
1’
2’
3’
4’
D
2
3
X
a’
c’
o
b’
a
b
1
d’
e’
e
4
Y
E
4
O
3
2
d
c
A
Concept: A string wound
from a point up to the same
Point, of shortest length
Must appear st
...
Solution steps:
Hence draw development,
Name it as usual and join
A to A This is shortest
Length of that string
...
On dev
...
Bring
Those on Fv & Tv and join
by smooth curves
...
Problem 9:
A particle which is initially on base circle of a cone, standing
on Hp, moves upwards and reaches apex in one complete turn around the cone
...
Take base circle diameter 50 mm and axis 70 mm long
...
as usual
On all form generators & name
...
Draw horizontal lines from those
points on both end generators
...
b’o’ intersect
...
Line & gen
...
In this way locate all points on Fv
...
Join in curvature
...
generator
from apex, Mark on development
& join
...
THIS CURVE IS CALLED CURVE OF INTERSECTION
AND
IT IS A RESULT OF INTERPENETRATION OF SOLIDS
...
Curves of Intersections being common to both Intersecting solids,
show exact & maximum surface contact of both solids
...
Minimum Surface Contact
...
Square Pipes
...
Square Pipes
...
Circular Pipes
...
BY WHITE ARROWS
...
A Feeding Hopper
In industry
...
Intersection of two cylinders
...
Two Cylindrical
surfaces
...
Pump lid having shape of a
hexagonal Prism and
Hemi-sphere intersecting
each other
...
REFFER ILLUSTRATIONS
AND
NOTE THE COMMON
CONSTRUCTION
FOR ALL
1
...
2
...
PRISM TO CYLINDER
3
...
TRIANGULAR PRISM TO CYLNDER
5
...
PRISM TO SQ
...
SQ
...
PRISM
( SKEW POSITION)
7
...
CYLINDER TO CONE
COMMON SOLUTION STEPS
One solid will be standing on HP
Other will penetrate horizontally
...
Name views as per the illustrations
...
On it’s S
...
mark number of points
And name those(either letters or nos
...
V
...
And other
points from SV should be brought to
Tv first and then projecting upward
To Fv
...
Accordingly those should be joined
by curvature or straight lines
...
Similarly in case of penetration from top it is not required
...
and 70mm axis is completely penetrated
by another of 40 mm dia
...
Draw projections showing curves of intersections
...
CYLINDER STANDING
&
CYLINDER PENETRATING
f”
c”
d”
e”
Y
4
3
1
2
Problem: A cylinder 50mm dia
...
and 70 mm axis, horizontally
...
All faces of prism are equally inclined to Hp
...
2’ 4’
1’
4”
3’
a’
a’
b’
d’
c’
X
2”
a”
d”
b’
d’
c’
1”3”
CASE 2
...
PRISM PENETRATING
b”
c”
Y
4
3
1
2
CASE 3
...
Both axes intersect & bisect
each other
...
7’
6’ 8’
1’ 5’
2’ 4’
3’
X
Y
1
28
37
46
5
Problem: A sq
...
and 70mm axis is completely penetrated
CASE 4
...
and 70 mm axis, horizontally
...
PRISM STANDING
Intersects & bisect each other
...
&
Draw projections showing curves of intersections
...
PRISM PENETRATING
2’4’
1’
3’
a’
4”
a’
b’
d’
1”3”
a”
d”
b’
d’
c’
c’
X
2”
b”
c”
Y
4
3
1
2
Problem: A cylinder 50mm dia
...
and 70 mm axis, horizontally
...
Draw projections showing curves of intersections
...
CYLINDER STANDING & TRIANGULAR PRISM PENETRATING
2’4’
1’
3’
a
a
4”
1”
3”
a
b
b
b
2”
c
c
d
e
X
d
e
f
e
f
f
Y
4
3
1
2
CASE 6
...
PRISM STANDING
&
SQ
...
prism 30 mm base sides
...
and 70 mm axis, horizontally
...
Two faces of penetrating prism are 300 inclined to Hp
...
2’4’
1’
3’
a’
1”3”
4”
2”
f”
f’
e’
b’
c’
c”
d’
300
X
4
3
1
2
Y
CASE 7
...
PRISM PENETRATING
(BOTH AXES VERTICAL)
2’
1’
3’
5’
4’
6’
X
a’ b’h’
c’g’
d’f’
e’
Y
g
8
h
f
7
9
10
6
1
a
e
2
5
3
4
b
d
c
5 mm OFF-SET
Problem: A cone70 mm base diameter and 90 mm axis
is completely penetrated by a square prism from top
with it’s axis // to cone’s axis and 5 mm away from it
...
Take all faces of sq
...
Base Side of prism is 0 mm and axis is 100 mm long
...
Problem: A vertical cone, base diameter 75 mm and axis 100 mm long,
CASE 8
...
The axis of the
CONE STANDING
cylinder is parallel to Hp and Vp and intersects axis of the cone at a point
&
28 mm above the base
...
CYLINDER PENETRATING
o’
o”
1
1
2
3
3
7,
a’ b’h’
8
c’g’
d’f’
e’
g”
e
b
d
c
4
5
5
f
a
2
3
6
g
h
1
7
64
4
5
X
8,2
g”h”
a”e”
b”d”
c”
Y
ISOMETRIC DRAWING
IT IS A TYPE OF PICTORIAL PROJECTION
IN WHICH ALL THREE DIMENSIONS OF
AN OBJECT ARE SHOWN IN ONE VIEW AND
IF REQUIRED, THEIR ACTUAL SIZES CAN BE
MEASURED DIRECTLY FROM IT
...
ALL THESE DRAWINGS MAY BE CALLED
3-DIMENSIONAL DRAWINGS,
OR PHOTOGRAPHIC
OR PICTORIAL DRAWINGS
...
TYPICAL CONDITION
...
( 1200)
NOW OBSERVE BELOW GIVEN DRAWINGS
...
ISO MEANS SAME, SIMILAR OR EQUAL
...
EACH IS 1200 INCLINED WITH OTHER TWO
...
SOME IMPORTANT TERMS:
ISOMETRIC AXES, LINES AND PLANES:
The three lines AL, AD and AH, meeting at point A and making
1200 angles with each other are termed Isometric Axes
...
A
The planes representing the faces of of the cube as well as
other planes parallel to these planes are called Isometric Planes
...
This reduction is 0
...
) It forms a reducing scale which
Is used to draw isometric drawings and is called Isometric scale
...
This is conveniently done by constructing an isometric scale as described
on next page
...
SCALE
...
Mark divisions of true length and from
each division-point draw vertical lines
upto AC line
...
1
IF THE FIGURE IS
FRONT VIEW, H & L
AXES ARE REQUIRED
...
3
2
Shapes containing
Inclined lines should
be enclosed in a
rectangle as shown
...
of that rectangle and
then inscribe that
shape as it is
...
Isometric view if the Shape is
F
...
or
T
...
SHAPE
ISOMETRIC
OF
PLANE FIGURES
A
A
1
2
2
4
H
PENTAGON
E
1
4
D
A
E
1
D
4
D
E
A
1
C
2
B
C
3
2
B
3
3
C
A
B
2
STUDY
Z
ILLUSTRATIONS
2
DRAW ISOMETRIC VIEW OF A
CIRCLE IF IT IS A TV OR FV
...
IT’S ISOMETRIC IS A RHOMBUS WITH
D & L AXES FOR TOP VIEW
...
FOR CONSTRUCTION USE RHOMBUS
METHOD SHOWN HERE
...
2
B
A
4
3
C
D
1
3
STUDY
Z
ILLUSTRATIONS
25 R
DRAW ISOMETRIC VIEW OF THE FIGURE
SHOWN WITH DIMENTIONS (ON RIGHT SIDE)
CONSIDERING IT FIRST AS F
...
AND THEN T
...
50 MM
IF FRONT VIEW
100 MM
IF TOP VIEW
ISOMETRIC
OF
PLANE FIGURES
AS THESE ALL ARE
2-D FIGURES
WE REQUIRE ONLY
TWO ISOMETRIC
AXES
...
V
...
V
...
IF THE FIGURE IS
TOP VIEW, D & L
For Isometric of Circle/Semicircle use Rhombus method
...
of sides equal to Diameter of circle always
...
topic ENGG
...
)
SEMI CIRCLE
For Isometric of
Circle/Semicircle
use Rhombus method
...
( Ref
...
)
5
STUDY
Z
ILLUSTRATIONS
ISOMETRIC VIEW OF
PENTAGONAL PYRAMID
STANDING ON H
...
(Height is added from center of pentagon)
ISOMETRIC VIEW OF BASE OF
PENTAGONAL PYRAMID
STANDING ON H
...
4
4
D
D
E
E
1
3
C
A
B
1
3
C
A
2
B
2
6
STUDY
Z
ILLUSTRATIONS
ISOMETRIC VIEW OF
PENTAGONALL PRISM
LYING ON H
...
4
H
E
1
D
A
C
ISOMETRIC VIEW OF
HEXAGONAL PRISM
STANDING ON H
...
2
B
3
7
STUDY
Z
ILLUSTRATIONS
CYLINDER STANDING ON H
...
CYLINDER LYING ON H
...
8
STUDY
Z
ILLUSTRATIONS
HALF CYLINDER
STANDING ON H
...
( ON IT’S SEMICIRCULAR BASE)
HALF CYLINDER
LYING ON H
...
( with flat face // to H
...
)
STUDY
Z
ILLUSTRATIONS
60
FV
X
40
Y
20
TV
ISOMETRIC VIEW OF
A FRUSTOM OF SQUARE PYRAMID
STANDING ON H
...
ON IT’S LARGER BASE
...
DRAW IT’S ISOMETRIC VIEW
...
y
x
1
4
D
E
THEN DRAWSAME SHAPE
AS TOP, 60 MM ABOVE THE
BASE PENTAGON CENTER
...
A
TV 40 20
B
2
C
3
ISOMETRIC VIEW OF
11
A FRUSTOM OF CONE
STANDING ON H
...
ON IT’S LARGER BASE
...
DRAW ISOMETRIC VIEW OF THE PAIR
...
DRAW ISOMETRIC VIEW OF THE PAIR
...
TO DRAW ISOMETRIC OF A CUBE IS SIMPLE
...
BUT FOR PYRAMID AS IT’S BASE IS AN EQUILATERAL TRIANGLE,
IT CAN NOT BE DRAWN DIRECTLY
...
SO DRAW TRIANGLE AS A TV, SEPARATELY AND NAME VARIOUS POINTS AS SHOWN
...
THEN ADD HEIGHT FROM IT’S CENTER AND COMPLETE IT’S ISOMETRIC AS SHOWN
...
IT’S FV & TV ARE SHOWN
...
14
15
STUDY
Z
ILLUSTRATIONS
FV
PROBLEM:
A CIRCULAR PLATE IS PIERCED THROUGH CENTRALLY
BY A SQUARE PYRAMID WHICH COMES OUT EQUALLY FROM BOTH FACES
OF PLATE
...
DRAW ISOMETRIC VIEW
...
V
...
V
...
Draw it’s isometric view
...
Scale
P
C
r
R
r
C
R
R
r
P
TO DRAW ISOMETRIC PROJECTION
OF A HEMISPHERE
P
C = Center of Sphere
...
TO DRAW ISOMETRIC PROJECTION OF A SPHERE
1
...
2
...
NAME IT P
...
FROM PDRAW VERTICAL LINE UPWARD, LENGTH ‘ r mm’
AND LOCATE CENTER OF SPHERE “C”
4
...
THIS IS ISOMETRIC PROJECTION OF A SPHERE
...
Draw lower semicircle only
...
For this use iso-scale
...
18
PROBLEM:
A HEMI-SPHERE IS CENTRALLY PLACED
ON THE TOP OF A FRUSTOM OF CONE
...
STUDY
ILLUSTRATIONS
Z
r
50 D
r
R
30 D
r
50
P
50 D
FIRST CONSTRUCT ISOMETRIC SCALE
...
450
300
19
STUDY
Z
ILLUSTRATIONS
A SQUARE PYRAMID OF 40 MM BASE SIDES AND 60 MM AXIS
IS CUT BY AN INCLINED SECTION PLANE THROUGH THE MID POINT
OF AXIS AS SHOWN
...
3’ 4’
4
1’2’
3
X
Y
a
1
d
1
2
4
o
b
2
3
c
20
STUDY
Z
ILLUSTRATIONS
F
...
& T
...
of an object are given
...
50
X
O
Y
20
25
O
25
20
21
STUDY
Z
ILLUSTRATIONS
F
...
& T
...
of an object are given
...
35
FV
35
x
10
O
20
10
40
70
O
TV
30
y
STUDY
Z
ILLUSTRATIONS
22
F
...
& T
...
and S
...
of an object are given
...
FV
30
10
30
SV
30
10
30
x
y
TV
ALL VIEWS IDENTICAL
STUDY
Z
ILLUSTRATIONS
24
F
...
& T
...
and S
...
of an object are given
...
ALL VIEWS IDENTICAL
FV
SV
y
x
10
40
60
40
TV
60
STUDY
Z
ILLUSTRATIONS
F
...
& T
...
and S
...
of an object are given
...
25
ALL VIEWS IDENTICAL
FV
SV
y
x
10
40
40
60
TV
60
26
F
...
& T
...
and S
...
of an object are given
...
ORTHOGRAPHIC PROJECTIONS
FRONT VIEW
L
...
SIDE VIEW
20
20
x
20
O
y
50
20
30
20
20
TOP VIEW
20
O
STUDY
Z
ILLUSTRATIONS
27
STUDY
Z
ILLUSTRATIONS
F
...
and S
...
of an object are given
...
30 SQUARE
40
20
50
20
10
O
30
F
...
O
60
S
...
28
STUDY
Z
ILLUSTRATIONS
F
...
& T
...
of an object are given
...
FV
40
10
O
45
30 D
10
50
O
80
TV
STUDY
Z
ILLUSTRATIONS
F
...
& T
...
of an object are given
...
40
FV
O
X
10
Y
100
10
25
TV
10
30
10
25
O
30 R
20 D
29
STUDY
Z
ILLUSTRATIONS
30
F
...
& T
...
of an object are given
...
30
FV
RECT
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
V
...
V
...
Draw it’s isometric view
...
DRAW ISOMETRIC ACCORDINGLY
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
V
...
V
...
Draw it’s isometric view
...
V
...
V
...
A line AB is in first quadrant
...
The distance between the end projectors is 75mm
...
Draw the projections of AB and determine its true length
and HT and inclination with HP
...
A line AB measures 100mm
...
The point A is 35mm above HP and 25mm in front VP
...
Draw the
projections of line and determine its HT and Inclinations with HP and VP
...
Draw the three views of line AB, 80mm long, when it is lying in profile plane and inclined
at 350 to HP
...
Determine also its traces
...
A line AB 75 mm long, has its one end A in VP and other end B 15mm above HP and
50mm in front of VP
...
Determine the true angles of inclination and show traces
...
A line AB is 75mm long and lies in an auxiliary inclined plane (AIP) which makes an
angle of 450 with the HP
...
The end A is in VP and
20mm above HP
...
6
...
End A is 10
mm above Hp
...
Distance between projectors is 50 mm
...
Locate traces also
...
An electric point hang in the center of ceiling and 1m
below it
...
Draw the projections of the and its length and slope angle with the floor
...
5m high
...
consider appropriate scale
9) Two pegs A and B are fixed in each of the two adjacent side walls of the rectangular room 3m x
4m sides
...
5m above the floor, 1
...
3m
from the wall
...
2m from the
wall
...
Also find the height of the roof if the
shortest distance between peg A and and center of the ceiling is 5m
...
Determine graphically the distance between the motors
...
11) Two mangos on a two tree are 2m and 3m above the ground level and 1
...
5m from a
0
...
Distances being measured from the center line of the
wall
...
Determine the real distance between the ranges
...
12)Three vertical poles AB, CD and EF are lying along the corners of equilateral triangle lying on the
ground of 100mm sides
...
Draw their projections and find
real distance between their top ends
...
Another straight road from B due 300 east of north to a point C is also 4 kms long but going
downward and has slope of 150
...
14) An electric transmission line laid along an uphill from the hydroelectric power station due west to a
substation is 2km long and has a slop of 300
...
Determine the length and slope of the proposed
telephone line joining the the power station and village
...
The other end of one wire rope
is attached to the top of the vertical pole 5m high and the rope makes an angle of depression of 45 0
...
The pole in the top
view are 2m apart
...
16) Two hill tops A and B are 90m and 60m above the ground level respectively
...
From C angles and elevations for A and B are 45 0 and 300
respectively
...
Determine the two distances between A, B and C
...
A thin regular pentagon of 30mm sides has one side // to Hp and 30 0 inclined to Vp while its surface is 45 0
inclines to Hp
...
2
...
Draw its
projections if
a) the TV of same diameter is 450 inclined to Vp, OR b) Diameter AB is in profile plane
...
A thin triangle PQR has sides PQ = 60mm
...
and RP = 50mm
...
Side PQ rest on
ground and makes 300 with Vp
...
Draw its
projections
...
An isosceles triangle having base 60mm long and altitude 80mm long appears as an equilateral triangle of
60mm sides with one side 300 inclined to XY in top view
...
5
...
Draw
projections of it and find its inclination with Hp
...
A rhombus of 60mm and 40mm long diagonals is so placed on Hp that in TV it appears as a square of 40mm
long diagonals
...
7
...
8
...
Determine it’s true shape
...
Draw a rectangular abcd of side 50mm and 30mm with longer 35 0 with XY, representing TV of a quadrilateral plane
ABCD
...
Draw a suitable Fv and determine its true shape
...
Draw a pentagon abcde having side 500 to XY, with the side ab =30mm, bc = 60mm, cd =50mm, de = 25mm and
angles abc 1200, cde 1250
...
Complete the projections and determine the true shape of the plane figure
...
Draw the projections of a square prism of 25mm sides base and 50mm long axis
...
2
...
above ground
...
Draw an another front view on an AVP inclined at 30 0 to edge on which it is
resting so that the base is visible
...
A square pyramid of side 30mm and axis 60 mm long has one of its slant edges inclined at 45 0 to
HP and a plane containing that slant edge and axis is inclined at 30 0 to VP
...
4
...
Its axis makes an angle of 600 with the HP
...
Draw another top view on an auxiliary plane inclined at 500 to the HP
...
Draw the three views of a cone having base 50 mm diameter and axis 60mm long It is resting on a
ground on a point of its base circle
...
6
...
The axis makes an angle
of 300 with VP and 450 with HP
...
7
...
Draw the projections (three
views) when the edge of base apposite to the point of suspension makes an angle of 30 0 to VP
...
s
8
...
Draw the front view and the top view when the plane containing its axis is perpendicular to
HP and makes an angle of 450 with VP
...
9
...
A right circular cone base 25mm diameter and height 50mm is placed centrally on the top
of the cube so that their axis are in a straight line
...
Project another top view on an AIP making 450 with the HP
10
...
Draw the front view, top view and side view of the solid when the axis of the bar is
inclined at 300 to HP and parallel to VP, the sides of a bar being 450 to VP
...
A cube of 50mm long edges is resting on the ground with its vertical faces equally inclined to
VP
...
Draw
the front view and the top view of the solids, project another top view on an AIP making an angle
of 450 with the HP
...
A circular block, 75mm diameter and 25mm thick is pierced centrally through its flat faces by
a square prism of 35mm base sides and 125mm long axis, which comes out equally on both sides
of the block
...
Draw side view also
...
Edges of base is equally
inclined to VP
...
The plane cuts the axis at 10mm
above the base
...
2) A hexagonal pyramid, edge of base 30mm and axis 75mm, is resting on its edge on HP which is perpendicular toVP
...
the solid is cut by a section plane perpendicular to both HP and VP, and passing
through the mid point of the axis
...
3) A cone of base diameter 60mm and axis 80mm, long has one of its generators in VP and parallel to HP
...
Draw the sectional FV, true shape of section and develop the lateral
surface of the cone containing the apex
...
A horizontal section plane passing through midpoint of vertical solid diagonal
cuts the cube
...
5) A vertical cylinder cut by a section plane perpendicular to VP and inclined to HP in such a way that the true shape of a
section is an ellipse with 50mm and 80mm as its minor and major axes
...
Draw the projections and show the true shape of the section
...
Draw the development of the lower half of the cylinder
...
It is cut by a section plane perpendicular to
VP such that the true shape of section is regular hexagon
...
Draw the
sectional top view and true shape of section
...
The length
of axis is 80mm
...
A plane parallel to VP
cuts the solid at a distance of 10mm from the top view of the axis
...
Also develop the lateral surface of the cut solid
...
It is cut by a section plane, perpendicular to VP and passing through a point on the
axis 10mm from the base
...
Also draw the lateral surface development of the pyramid
...
It is cut by a section
plane perpendicular to VP in such a way that the true shape of a section is a parabola having base 40mm
...
10) A hexagonal pyramid, base 50mm side and axis 100mm long is lying on ground on one of its triangular
faces with axis parallel to VP
...
Draw the top view, sectional front view
and the development of surface of the cut pyramid containing apex
...
It is cut by a
section plane parallel to HP and passing through a point on the axis 25mm from the apex
...
A particle P, initially at the mid point of edge of base, starts moving over the surface and
reaches the mid point of apposite edge of the top face
...
Also show the path in front and top views
12) A cube of 65 mm long edges has its vertical face equally inclined to the VP
...
PROBLEM 14:-Two objects, a flower (A) and an orange (B) are within a rectangular compound wall,
whose P & Q are walls meeting at 900
...
5M & 1 M from walls P & Q respectively
...
5M & 5
...
Drawing projection, find distance between
them If flower is 1
...
5 M above the ground
...
b’
b’1
3,5M
a’
1
...
5M
a
3
...
5M
Wall Q
F
...
Wall P
PROBLEM 15 :- Two mangos on a tree A & B are 1
...
00 m above ground
and those are 1
...
5 m from a 0
...
If the distance measured between them along the ground and parallel to wall is 2
...
b’
TV
b1’
B
3
...
5m
(GL)
A
WALL
X
Y
b
0
...
5m
Wall thickness
1
...
6m
REAL DISTANCE BETWEEN
MANGOS A & B IS = a’ b1’
0
...
All equally inclined and the shortest
is vertical
...
is TV of three rods OA, OB and OC
whose ends A,B & C are on ground and end O is 100mm
above ground
...
Tv
O
o’
C
A
TL2
x
b1’ b’
TL1
a’
a 1’
c’
c1’
Fv
y
B
a
o
Answers:
TL1 TL2 & TL3
b
c
PROBLEM 17:- A pipe line from point A has a downward gradient 1:5 and it runs due South - East
...
Pipe line from B runs
150 Due East of South and meets pipe line from A at point C
...
12m
b’
5
a’
1
5
1
FV
c’
x
W
c’1
N
a
450
b
c’2
A
12 M
B
E
y
EAST
C
150
TV
c
= Inclination of pipe line BC
SOUTH
PROBLEM 18: A person observes two objects, A & B, on the ground, from a tower, 15 M high,
At the angles of depression 300 & 450
...
Draw projections of situation and find distance of objects from
observer and from tower also
...
5m and 7
...
The poles are 10 M apart
...
c1’
c’
c’2
TV
C
a’
300
b’
15M
450
15 M
7
...
5M
A
4
...
5M
c
Answers:
Length of Rope BC= b’c’2
Length of Rope AC= a’c’1
Distances of poles from building = ca & cb
PROBLEM 20:- A tank of 4 M height is to be strengthened by four stay rods from each corner
by fixing their other ends to the flooring, at a point 1
...
7 M from two adjacent walls respectively,
as shown
...
FV
a’
TV
True Length
Answers:
Length of each rod
= a’b’1
Angle with Hp
...
5 M wide is supported by four chains
from it’s corners and chains are attached to a hook 5 M above the center of the platform
...
h’
TV
TL
x
d’1
H
5M
a’d’
Hook
b’c’ y (GL)
d
c
D
1
...
=
B
PROBLEM 22
...
5m L ,5m D,3
...
An electric bulb hangs 1m below the center of ceiling
...
5m above the flooring
...
6
...
5m
b’
Ceiling
TV
b’1
Bulb
a’
1
...
IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS
...
5 M ABOVE WALL RAILING
...
5M
350
A
B
1M
c’d’
X
a1
D
(wall railing)
Y
ad
C
(frame)
h
(chains)
b1
bc
Answers:
Length of each chain= hb1
True angle between chains =
Wall railing
Title: Engineering Graphics Full Notes
Description: This Notes Provides A Full Explaination To Engineering Graphics Course. It Is Written By Me and It Is Verified By My Senior. Sailent Features:- Course Is Explained In A Very Easy Simple Language. Drawings Are Made Colorful In Order To Make The Book Very Interesting To Students. Ample Number Of Solved Examples With Detailed Explaination Of The Problem Is Also Included.
Description: This Notes Provides A Full Explaination To Engineering Graphics Course. It Is Written By Me and It Is Verified By My Senior. Sailent Features:- Course Is Explained In A Very Easy Simple Language. Drawings Are Made Colorful In Order To Make The Book Very Interesting To Students. Ample Number Of Solved Examples With Detailed Explaination Of The Problem Is Also Included.