The document describes different types of planes based on their orientation relative to reference planes. It provides examples of planes that are perpendicular to both reference planes, perpendicular to one plane and parallel/inclined to the other, and perpendicular to one plane and inclined to the other. Diagrams show how to draw projections of planes in different positions, including determining front and top views. Examples include triangles, rectangles, pentagons, hexagons and circles oriented in various ways relative to horizontal and vertical planes.
Download the original presentation for animation and clear understanding. This Presentation describes the concepts of Engineering Drawing of VTU Syllabus. However same can also be used for learning drawing concepts. Please write to me for suggestions and criticisms here: hareeshang@gmail.com or visit this website for more details: www.hareeshang.wikifoundry.com.
Download the original presentation for animation and clear understanding. This Presentation describes the concepts of Engineering Drawing of VTU Syllabus. However same can also be used for learning drawing concepts. Please write to me for suggestions and criticisms here: hareeshang@gmail.com or visit this website for more details: www.hareeshang.wikifoundry.com.
Section of solids - ENGINEERING DRAWING/GRAPHICSAbhishek Kandare
Section of solids
THIS SLIDE CONTAINS WHOLE SYLLABUS OF ENGINEERING DRAWING/GRAPHICS. IT IS THE MOST SIMPLE AND INTERACTIVE WAY TO LEARN ENGINEERING DRAWING.SYLLABUS IS RELATED TO rajiv gandhi proudyogiki vishwavidyalaya / rajiv gandhi TECHNICAL UNIVERSITY ,BHOPAL.
Projection of solids - ENGINEERING DRAWING/GRAPHICSAbhishek Kandare
Projection of solids
HIS SLIDE CONTAINS WHOLE SYLLABUS OF ENGINEERING DRAWING/GRAPHICS. IT IS THE MOST SIMPLE AND INTERACTIVE WAY TO LEARN ENGINEERING DRAWING.SYLLABUS IS RELATED TO rajiv gandhi proudyogiki vishwavidyalaya / rajiv gandhi TECHNICAL UNIVERSITY ,BHOPAL.
introduction of engineering graphics ,projection of points,lines,planes,solids,section of solids,development of surfaces,isometric projection,perspective projection
Section of solids - ENGINEERING DRAWING/GRAPHICSAbhishek Kandare
Section of solids
THIS SLIDE CONTAINS WHOLE SYLLABUS OF ENGINEERING DRAWING/GRAPHICS. IT IS THE MOST SIMPLE AND INTERACTIVE WAY TO LEARN ENGINEERING DRAWING.SYLLABUS IS RELATED TO rajiv gandhi proudyogiki vishwavidyalaya / rajiv gandhi TECHNICAL UNIVERSITY ,BHOPAL.
Projection of solids - ENGINEERING DRAWING/GRAPHICSAbhishek Kandare
Projection of solids
HIS SLIDE CONTAINS WHOLE SYLLABUS OF ENGINEERING DRAWING/GRAPHICS. IT IS THE MOST SIMPLE AND INTERACTIVE WAY TO LEARN ENGINEERING DRAWING.SYLLABUS IS RELATED TO rajiv gandhi proudyogiki vishwavidyalaya / rajiv gandhi TECHNICAL UNIVERSITY ,BHOPAL.
introduction of engineering graphics ,projection of points,lines,planes,solids,section of solids,development of surfaces,isometric projection,perspective projection
A plane is a two dimensional object having length and breadth only. Its thickness is always neglected. Various shapes of plane figures are considered such as square, rectangle, circle, pentagon, hexagon, etc.
There are two types of planes
Perpendicular planes which have their surface perpendicular to any one of the reference planes and parallel or inclined to the other reference plane.
2. Oblique planes which have their surface inclined to both the reference planes.
HEAP SORT ILLUSTRATED WITH HEAPIFY, BUILD HEAP FOR DYNAMIC ARRAYS.
Heap sort is a comparison-based sorting technique based on Binary Heap data structure. It is similar to the selection sort where we first find the minimum element and place the minimum element at the beginning. Repeat the same process for the remaining elements.
NUMERICAL SIMULATIONS OF HEAT AND MASS TRANSFER IN CONDENSING HEAT EXCHANGERS...ssuser7dcef0
Power plants release a large amount of water vapor into the
atmosphere through the stack. The flue gas can be a potential
source for obtaining much needed cooling water for a power
plant. If a power plant could recover and reuse a portion of this
moisture, it could reduce its total cooling water intake
requirement. One of the most practical way to recover water
from flue gas is to use a condensing heat exchanger. The power
plant could also recover latent heat due to condensation as well
as sensible heat due to lowering the flue gas exit temperature.
Additionally, harmful acids released from the stack can be
reduced in a condensing heat exchanger by acid condensation. reduced in a condensing heat exchanger by acid condensation.
Condensation of vapors in flue gas is a complicated
phenomenon since heat and mass transfer of water vapor and
various acids simultaneously occur in the presence of noncondensable
gases such as nitrogen and oxygen. Design of a
condenser depends on the knowledge and understanding of the
heat and mass transfer processes. A computer program for
numerical simulations of water (H2O) and sulfuric acid (H2SO4)
condensation in a flue gas condensing heat exchanger was
developed using MATLAB. Governing equations based on
mass and energy balances for the system were derived to
predict variables such as flue gas exit temperature, cooling
water outlet temperature, mole fraction and condensation rates
of water and sulfuric acid vapors. The equations were solved
using an iterative solution technique with calculations of heat
and mass transfer coefficients and physical properties.
We have compiled the most important slides from each speaker's presentation. This year’s compilation, available for free, captures the key insights and contributions shared during the DfMAy 2024 conference.
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Understanding Inductive Bias in Machine LearningSUTEJAS
This presentation explores the concept of inductive bias in machine learning. It explains how algorithms come with built-in assumptions and preferences that guide the learning process. You'll learn about the different types of inductive bias and how they can impact the performance and generalizability of machine learning models.
The presentation also covers the positive and negative aspects of inductive bias, along with strategies for mitigating potential drawbacks. We'll explore examples of how bias manifests in algorithms like neural networks and decision trees.
By understanding inductive bias, you can gain valuable insights into how machine learning models work and make informed decisions when building and deploying them.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
CW RADAR, FMCW RADAR, FMCW ALTIMETER, AND THEIR PARAMETERSveerababupersonal22
It consists of cw radar and fmcw radar ,range measurement,if amplifier and fmcw altimeterThe CW radar operates using continuous wave transmission, while the FMCW radar employs frequency-modulated continuous wave technology. Range measurement is a crucial aspect of radar systems, providing information about the distance to a target. The IF amplifier plays a key role in signal processing, amplifying intermediate frequency signals for further analysis. The FMCW altimeter utilizes frequency-modulated continuous wave technology to accurately measure altitude above a reference point.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
DESIGN AND ANALYSIS OF A CAR SHOWROOM USING E TABS
Projections of planes.pptx
1. Types of Planes
• On the basis of position with respect to reference
planes
• Perpendicular to both the reference planes
• Perpendicular to one plane and parallel to the other
• Perpendicular to one plane and inclined to the other
Perpendicular planes
• Inclined to both the reference planes
Oblique planes
1
SBT
08-Jul-09
2. Projections of different positions
of planes
• FV and TV (both edge views) are obtained in straight line
perpendicular to x-y line
• FV Coincide with Vertical trace (VT)
• TV Coincide with Horizontal trace (HT)
Perpendicular to both ref. planes
VP
HP
Square
plate x y
FV
TV
2
SBT
08-Jul-09
3. Plane perpendicular to one plane and parallel to other
• Plane perpendicular to HP and parallel to VP
• Plane perpendicular to VP and parallel to HP
• FV shows true shape of plane
• TV shows edge view (line parallel to x-y line) of plane
A. Plane perpendicular to HP and parallel to VP
• TV shows true shape of plane
• FV shows edge view (line parallel to x-y line) of plane
B. Plane perpendicular to VP and parallel to HP
x x
y y
FV
TV
Edge view
Edge view
FV
TV
True shape
True shape
3
SBT
08-Jul-09
4. 08-Jul-09 SBT 4
y
HP
VP
x
a
b
c
b’ c’
a’
5
10
45˚
Plane parallel to one plane and perpendicular to other plane: Illustrative Examples
Plane: equilateral triangle
Surface: parallel to HP &
perpendicular to VP
Position: corner A is 5 mm
away from VP & edge AB
makes angle of 45˚ with VP
Plane: regular pentagon
Surface: in VP
Position: edge AB
perpendicular to HP
a’
b’
c’
d’
e’
b
a e
c
d
Plane: square
Surface: parallel to VP and
10 mm from VP
Position: corner B is in HP
& edge BA & BC equally
inclined at 45˚ with HP
b’
a’ c’
d’
10
a b
d
c
45˚
5. 08-Jul-09 SBT 5
y
HP
VP
x
Plane parallel to one plane and perpendicular to other plane: Illustrative Examples
Plane: regular hexagon
Surface: parallel to & 5
mm above HP
Position: corner F is 10 mm
from VP and edge AB & DE
are perpendicular to VP
10
f
a
b
c
d
e
5
a’
b’
f’
c’
e’
d’
Plane: regular pentagon
Surface: parallel to & 10
mm above HP
Position: corner A in VP &
edge AE makes an angle of
30˚ with VP
30˚
a
e
b
c
d
10
b’ c’ d’ e’
a’
Plane: circle of dia. 40 mm
Surface: in VP
Position: centre is 25 mm
above HP
25
1’
2’
3’
4’
1
4
3
2
6. Plane perpendicular to one plane and inclined to other
• A. Plane perpendicular to VP and inclined to HP
• B. Plane perpendicular to HP and inclined to VP
6
SBT
08-Jul-09
7. Initially plane is
resting on HP
OR llel to HP
TV1
FV1
Initially plane is
llel to VP
TV
FV FV
TV
TV
FV FV
TV
FV1
FV1
FV2
FV2
x y
x y
Reorient
previous FV
Reorient
previous TV
FV 2
TV 2
start
start end
end
7
SBT
08-Jul-09
8. y
HP
VP
x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
1. A rectangle of side 20 mm and 40 mm is having shorter edge parallel to VP and
surface is perpendicular to HP inclined at angle 30˚ with VP
Initially plane is assumed to be parallel
to VP. The FV of plane is drawn first
showing true shape & size of the plane
with shorter edge is perpendicular to xy
Stage I
a’
b’ c c’
d’
a
b
d
c
In next stage plane making 30˚ with
VP. Corresponding projection is
obtained by reorienting the earlier
TV at 30˚ with xy
a
b
d
c
a’ d’
b’ c’
30˚
Then , obtain corresponding
TV i.e. edge view (line)
parallel to xy
Stage II
Then , obtain
corresponding TV which
shows an apparent shape
& size of plane
20
50
9. y
VP
x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
Initially plane is assumed to be parallel
to VP. The FV of plane is drawn first
showing true shape & size of the plane
with shorter edge is perpendicular to xy
Stage I
a’
b’ c c’
d’
a
b
d
c
Then , obtain corresponding
TV i.e. edge view (line)
parallel to xy
20
50
2. A rectangle of 20 and 40 mm side is so placed that its surface is perpendicular to HP
and inclined to VP at such an angle that FV of plane appears to be a square of 20 mm
sides. Draw projection and fined out the angle of surface with VP.
Stage II
In next stage plane is
inclined to VP at such an
angle that FV appears to be
a square of 20 mm side
a’ d’
b’ c’
a
b
d
c
20
δ = 60˚
Then , obtain corresponding
TV (edge view) having same
length of earlier TV as shown
in coming step and measure
angle δ
HP
10. y
VP
x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
3. A regular hexagon of 20 mm side is having as edge in VP and surface is perpendicular
to HP and makes 30˚ with VP.
HP
f’
a’
b’
c’
e’
a
b
f
c
e
d b
a
d
30˚
e
c
f
f’
a’ e’
b’
d’
d’
c’
Initially it is assumed plane is in VP. The
FV of plane is drawn first showing true
shape & size of the plane with one edge
is perpendicular to xy
Then , obtain corresponding
TV i.e. edge view (line) on
xy
In next stage plane making 30˚ with
VP. Corresponding projection is
obtained by reorienting the earlier
TV at 30˚ with xy
Then , obtain
corresponding TV which
shows an apparent shape
& size of plane
Stage I Stage II
11. y
VP
x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
4. A regular hexagon of 20 mm edges is having a corner in VP and surface is
perpendicular to HP and makes 30˚ with VP.
HP
a’
b’
e’
f’
a
f
b
e
c d a 30˚
d
c
e
b
f
f’ e’
a’
d’ d’
b’
c’
c’
Initially it is assumed plane is in VP. The
FV of plane is drawn first showing true
shape & size of the plane with one edge
is parallel to xy
Then , obtain corresponding
TV i.e. edge view (line) on
xy
In next stage plane making 30˚ with
VP with a corner is on VP
Corresponding projection is
obtained by reorienting the earlier
TV at 30˚ with xy with pt ‘a’ on xy
Stage I Stage II
Then , obtain
corresponding FV which
shows an apparent shape
& size of plane
12. y
HP
VP
x
5. A rectangle of side 20 mm and 40 mm is having shorter edge parallel to VP and
surface is perpendicular to HP inclined at angle 30˚ with VP
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
a
b
c
d
e
108˚
a’
e’
b’
d’ c’
Stage I
Then , obtain corresponding
TV i.e. edge view (line) on
xy
a’
e’
d’
d
b’
40˚
c’
e
c
a
b
Then , obtain
corresponding TV which
shows an apparent shape
& size of plane
Initially it is assumed plane is in HP. The
TV of plane is drawn first showing true
shape & size of the plane with one edge
is perpendicular to xy
Stage II
In next stage plane making 40˚ with
HP with an edge is on HP.
Corresponding projection is
obtained by reorienting the earlier
FV at 40˚ with xy with pt a’ & e’ on
xy
13. y
HP
VP
x
6. A regular pentagon having a corner in HP and surface is perpendicular to VP and
inclined at 40˚ with HP.
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
b
c
d
a’
b’
e’ c’
e
d’
a
Initially it is assumed plane is in HP. The
TV of plane is drawn first showing true
shape & size of the plane with one edge is
perpendicular to xy (set farthest corner
of this edge on RHS )
Stage I
a’
e’
b’
40˚
c’
Then , obtain corresponding
TV i.e. edge view (line) on
xy
Stage II
In next stage plane making 40˚ with
HP with an edge is on HP.
Corresponding projection is
obtained by reorienting the earlier
FV at 40˚ with xy with pt a’ & e’ on
xy
e
d
d’
a
b
c
Then , obtain
corresponding TV which
shows an apparent shape
& size of plane
14. y
HP
VP
x
7. A circle of 40 mm diameter is having its surface perpendicular to VP and inclined at
45˚ with HP.
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
2
4
6
8
10
1’
2’ 3’ 4’ 5’ 6’
7’
8’
9’
10’
11’
12’
1’
7
7’
10’
4’
11’
12’
9’
5’
6’
8’
3’
2’
10
11 9
12 8
1 7
2 6
3 5
4
45˚
Stage I Stage II
1
3
5
12
11 9
15. y
HP
VP
x
8. A circle of 40 mm diameter is so placed that its surface is perpendicular to VP and is
inclined to HP such a manner that is TV appears to be an ellipse of 40 mm major axis and
30 mm minor axis. Draw the projection and find out the angle of surface of plane with HP.
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
2
4
6
8
10
1’
2’ 3’ 4’ 5’ 6’
7’
8’
9’
10’
11’
12’
1’
7
7’
10’
4’
11’
12’
9’
5’
6’
8’
3’
2’
10
11 9
12 8
1 7
2 6
3 5
4
45˚
Stage I Stage II
1
3
5
12
11 9
16. y
VP
x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
3. (Q.14 Pg.21) A thin 45⁰ set square has its longest side 250 mm long in VP and
inclined at 30⁰ to HP. Its surface makes an angle of 45⁰ with VP. Draw projections.
(W-2003, 7 marks)
HP
a’
b’
45˚
a
b c
a
b
c
a’
c’
c’
b’
a’
c’
b’
a’
c’
b’
a’
c’
b’
30˚
a b
c
Initially it is assumed plane is in VP.
The FV of plane is drawn first showing
true shape & size of the plane with
longest edge is perpendicular to xy
Stage I
Then , obtain corresponding
TV i.e. edge view (line) on
xy In next stage plane making 45˚ with
VP with a longest edge is in VP.
Corresponding projection is obtained
by reorienting the earlier TV at 45˚
with xy with pt. ‘a’ & ’b’ on xy
Stage II
In last stage longest edge of plane
which is in VP making 30˚ with HP.
Corresponding projection is obtained
by reorienting the earlier FV in such a
manner that FV of the above
mentioned edge makes 30˚ with xy .
Stage III
Then , obtain
corresponding final TV .
45˚
Then , obtain
corresponding FV which
shows an apparent shape
& size of plane
17. x
Plane perpendicular to one plane and inclined to other plane: Illustrative Examples
3. A 30⁰ - 60⁰ triangle PQR has its longest side PR is 50 mm long and is contained in VP.
P and R are 20 mm and 45 mm respectively from HP, while point Q is 10 mm away from
VP. Draw projection. (W – 92, 7 marks)
20
VP
HP
30˚
60˚
p’
r’
p
r
q
Then , obtain corresponding
TV i.e. edge view (line) on
xy
In next stage pt. Q is 10 mm away
from VP with longest edge (PR) is in
VP. Corresponding projection is
obtained by reorienting the earlier TV
in such a manner pt. ‘q’ 10 mm below
xy
10
p
r
Initially it is assumed plane is in VP. The
FV of plane is drawn first showing true
shape & size of the plane with longest
edge PQ is perpendicular to xy and at
given position with respect to xy
Stage I
p’
q’
q’
r’
In last stage longest edge of plane
which is in VP making 30˚ with HP.
Corresponding projection is obtained
by reorienting the earlier FV in such a
manner that FV of the above
mentioned edge makes 30˚ with xy .
Stage III
Stage II
p’
q’
r’
20
r’
45
p’
q’
20
p’
Then , obtain
corresponding final TV .
r P
q
q’
y
18. y
x
3. A 30⁰ - 60⁰ - 90⁰ set square is having its surface making an angle of 45⁰ with HP and
is having its longer edge PQ which is 125 mm long is in HP and making an angle of 30⁰
with VP.
HP
30˚
60˚
p
r’
VP p’
q’
p’
q’
45˚
r’
p
r r
q q
p
r
q
30˚
p
r
q
p’ q’
r’
19. y
x
3. A thin triangular plane PQR has sides PQ 60 mm, QR = 80 mm and RP = 50
respectively. Side PQ rests on ground and makes an angle of 30⁰ with VP. Point P is 20
mm away from VP and point R is 40 mm above ground. Draw projection of plane.
p
q
r
60
p’
q’
r’
40
p’
q’
r’
r
p
q
r
p
q
20
r
p
q
30˚
r’
q’ p’
20. Plane perpendicular to one plane and inclined to other
• A. Plane perpendicular to VP and inclined to HP
• B. Plane perpendicular to HP and inclined to VP
• Projection is obtained in TWO stages
• In first stage, initially plane is consider to be llel to HP with given
edge condition and obtain TV (true shape) and then
corresponding FV (edge view)
• In next stage, previous FV is reoriented at given angle and
corresponding TV is obtained
A. Plane perpendicular to VP and inclined to HP
• Projection is obtained in TWO stages
• In first stage, initially plane is consider to be llel to VP with
given edge condition and obtain FV (true shape) and then
corresponding TV (edge view)
• In next stage previous TV is reoriented at given angle and
corresponding FV is obtained
B. Plane perpendicular to HP and inclined to VP
20
SBT
08-Jul-09
21. Plane inclined to both the plane (Oblique plane)
• A. Plane is inclined to HP with edge or dia. or diagonal is llel to HP
and inclined to VP
• B. Plane is inclined to VP with edge or dia. or diagonal is llel to VP
and inclined to HP
21
SBT
08-Jul-09
22. A. Plane is inclined to HP with edge or diameter
or diagonal is llel to HP and inclined to VP
TV
FV FV
TV
start
TV
FV
end
22
SBT
08-Jul-09
Initially plane is
resting on HP
OR llel to HP
FV1
FV2
x
y
Reorient
previous FV
Reorient
previous edge
FV1
TV 3
TV 2
FV 3
θ˚
φ˚
23. B. Plane is inclined to VP with edge or diameter
or diagonal is llel to VP and inclined to HP
FV1
x y
Initially plane is
llel to VP
Reorient
previous TV
Reorient
previous FV
TV
FV FV
TV
start
TV
FV
end
TV1
23
SBT
08-Jul-09
φ˚
θ˚
FV 2 FV3
FV3
TV2
26. y
x
3. A plate having shape of isosceles triangle has base 50 mm long and altitude 75 mm.
It is so placed that in Front View it is seen as an equilateral triangle of 50 mm sides and
one side inclined at 45⁰ to xy. Draw its Top View.
HP
VP
a
b c
a’ a’
b’
c’ c’
a
b
c
δ
a’
b’
c’
a’
b’
c’
a’
b’
c’
45˚
a b
c
b’
27. y
x
3. A thin rectangular plate of size 70 mm X 40 mm appears as a square of 40 mm sides
in TV with one of its sides inclined at 30⁰ to VP and parallel to HP. Draw the projections
of plate and determine its inclined with HP.
HP
VP
70
40
a
b c
d
a’
b’
d’
c’
a
b c
d
a’
b’
d’
c’
γ
30˚
b
c
d
a
40
40
b’ a’
c’ d’
28. y
x
3. A 30⁰ - 60⁰ set square has its shortest side 40 mm long in HP. TV of set square is an
isosceles triangle. Draw projection of plane and find its inclination with HP.
HP
VP
60˚
a
b
c
a’
b’
c’
a
b
c
a’
b’
c’
γ
29. y
x
3. A The TV of a 45⁰ set square with side BC in HP and side AB in VP, is a triangle abc.
The side bc = 20 mm being perpendicular to xy and angle bca = 25⁰. Draw TV and FV
and measure the inclination of set square with HP. Draw also Side View.
HP
VP
b
c
a
45˚
b
c
25˚
a
b’c’ a’ b’c’
a’
γ
x1
y1
c’’
b’’
a’’
30. 08-Jul-09 SBT 30
x y
1
2
3
4
5
6
8
10
11
12
HP
VP 1’
2’ 3’ 4’ 5’ 6’
7’
8’
9’
10’
11’
12’
1’
7
9
7’
10’
4’
11’
12’
9’
5’
6’
8’
3’
2’
10
11 9
12 8
1 7
2 6
3 5
4
45˚
1
7
4
10
2
6
12
8
11
9
3
5
1’
2’ 12’
3’
11’
4’
10’
5’
9’
6’
8’
7’
60˚
30˚
Reorient FV1 i.e. edge view
at an angle of 45°
Reorient TV2 so that TV2 of
diagonal AB (a’b’) at an angle
of 45°
Stage I Stage II Stage III
3. A circle of 40 mm diameter is having its surface making an angle of 45⁰ with HP and end A of
diameter AB is in HP and TV of diameter AB makes an angle of 30⁰ with VP. Draw projections.
31. 08-Jul-09 SBT 31
x y
1
2
3
4
5
6
8
10
11
12
HP
VP 1’
2’ 3’ 4’ 5’ 6’
7’
8’
9’
10’
11’
12’
1’
7
9
7’
10’
4’
11’
12’
9’
5’
6’
8’
3’
2’
10
11 9
12 8
1 7
2 6
3 5
4
45˚
30˚
a
b
b1
β˚
1
7
4
10
3
5
11
9
2
6
12
8
1’
2’ 12’
3’
4’
5’
6’
7’
8’
9’
10’
11’
Reorient FV1 i.e. edge view
at an angle of 45°
Reorient TV2 so that TV2 of
diagonal AB (a’b’) at an angle
of 45°
Stage I Stage II Stage III
3. A circle of 40 mm diameter is having its surface making an angle of 45⁰ with HP and end A of
diameter AB is in HP and diameter AB makes an angle of 30⁰ with VP. Draw projections.
33. x
y
3. A rhombus of diagonal 125 mm and 50 mm is so placed that it appears as a square of 50 mm
diagonal in TV. The smaller diagonal of rhombus is parallel to both HP and VP.
HP
VP
b
d
a’
b’
d’ c’
d
b
a
a c c
a’
b’
d
b
a c
c
a
d b
a’
d’ b’
c’
d’
c’
34. 08-Jul-09
x
y
3. A equilateral triangle ABC having side length as 50 mm is suspended from a point O on side AB
15 mm from A in such a way that plane of triangle makes an angle of 60⁰ with VP. Point O is 50 mm
above HP and 40 mm in front of VP. Draw projection of triangle.
a’
HP b’
60˚
40
b a c
50
g’
o’
VP
c’
a’
b’
g’
o’
c’
50
a’
b’ 50
g’
o’
c’
a’
b’
g’
o’
50
c’
a
b c
o
g
b
c
o
g
a
60˚
b’
c’
a’
35. 08-Jul-09 SBT 35
x
y
VP 2’
4’
5’
6’
7’
3’
1’
HP
3. A semicircle having diameter 100 mm is suspended from a point ‘O’ on straight edge 30 mm
from centre of that edge so that the surface makes an angle of 45⁰ with VP. Point ‘O’ is 90 mm
above HP and 50 mm in front of VP. Draw projections. (W – 05, 7 marks)
m’ g’
o’
1
7
6
2
5
3
4
m g
o
50
2’
5’
6’
7’
3’
1’
m’ g’
o’
90
2’
5’
6’
7’
3’
1’
m’
g’
o’
90
2’
5’
6’
7’
3’
1’
m’
g’
o’
30
4’
1 2 3 4 5
6
7
g
o
m
1
4
5
6
2
3
7
g
o
m
45˚
1’
2’
3’
4’
5’
6’
7’
37. x y
HP
VP
a’
b’
c’
d’
e’
f’
b
a
c
f
d
e
40˚
b
a
c
f
d
e
b’ d’
c’
a’ e’
f’
b’ d’
c’
a’ e’
f’
b’
c’
a’
e’
f’
30˚
b’
c’
a’
e’
f’
a b
c
f
e d
3. A regular hexagon of 30 mm edge is having its surface inclined at 40⁰ with VP and is
having an edge in VP and inclined at an angle 30⁰ with HP. Draw projections.
38. x y
HP
VP
a’
b’
c’
d’
e’
f’
b
a
c
f
d
e
δ
b
a
c
f
d
e
b’ d’
c’
a’ e’
f’
b’ d’
c’
a’ e’
f’
b’
c’
a’
e’
f’
30˚
b’
c’
a’
e’
f’
a b
c
f
e d
3. One side of a regular hexagon of 30 mm side is in VP, while opposite side is 35 mm
in front of VP and inclined at 30⁰ to HP. Draw projection of plane and find its surface
inclination with VP.
35
40. x y
HP
VP
c
e
f
a
f
c
e
d
b
40˚
d
f
f’ e’
b’ c’
f’ e’
a’ d’
b’ c’
f’
a’
d’
b’
c’
f’
b’
30˚
a’
d’
c’
a
f b
e c
d
b
e
c
a d a’ d’
3. A regular hexagon of 25 mm edge is having its surface making an angle of 40⁰ with
VP and is having a corner in VP. The FV of diagonal containing this corner makes an
angle of 30⁰ with HP. Draw the projections.
b a
41. x HP
VP
c
e
f
a b
f
c
e
d
b
40˚
a
d
f
f’ e’
b’ c’
f’ e’
a’ d’
b’ c’
30˚
b
e
c
d’ a’ d’
3. A regular hexagon of 25 mm edge is having its surface making an angle of 40⁰ with
VP and is having a corner in VP. The diagonal containing this corner makes an angle of
30⁰ with HP. Draw the projections.
d1’
d’
α
f’
e’
b’
c’
a’
a
f
b
e c
d
42. x y
c
e
f
a’ b’ c’
e’
d’
a d
b
a’
d’
b’
35˚
f e
a d
b c
f
e
a
d
b
c
f
a
d
b
c
e’
f’ b’
a’
x1
y1
a’’
f’’
e’’
d’’
3. A hexagonal plane ABCDEF of side30 mm has corner A in HP and opposite corner D
in VP. Draw three views o plane when diagonal AD inclined at 35⁰ to HP and parallel to
profile plane. Determine its surface inclination with VP.
e
c’ c’’
c’
e’
f’
f’
b’’
d’
43. x y
HP
VP
3. A thin pentagonal plate of negligible thickness and sides 25 mm long having one of
its corner in VP and surface makes an angle of 30⁰ with VP and side opposite to that
corner makes an angle of 60⁰ with HP. Draw its projection. (S – 2005)
d’
c’
108˚
a’
e’
b’
a b
e
c
d
30˚
a
c
d
b
e
e’
d’
a’
b’
c’
e’
d’
a’
b’
c’
d’
a’
b’
c’
60˚
a
b
e
c d
44. y
3. A composite plane consist of square ABCD of 75 mm sides and a semi circle along
CD as diameter. Draw the TV and FV of plane if its surface is perpendicular to HP and
makes an angle of 45⁰ with VP. Then draw auxiliary view on an AIP which makes an
angle of 30⁰ with edge AB which is perpendicular to HP.
6’
3’
2’
1’
5’
7’
a’
b’
4’
d’
c’
a
b
d
c
1
7
2
6
3
5
4
a
b
d
c 1
7
6
3
5
4
a’ d’
3’
4’
b’ c’ 7’
45˚
x1
b1’
7’
d1’
1’
2’
3’
4’
4’
c1’
6’
5’
6’
y1
2’
1’
a1’
45. x y
HP
VP
a
108˚
c
e
b’
a’
c
e
d b’
a’
5. A A thin regular pentagonal plate of 60 mm long edge has one of its edge in the HP
and perpendicular to VP, while its farthest corner is 60 mm above HP. Draw projection
of plate. Project another FV on an AVP making angle of 45⁰ with VP. Also project an
auxiliary view (ATV) of the same plane on and AIP which makes an angle of 60⁰ with HP.
60
d’
c’
e’
d
d
b b
c
e
a
x1
y1
60˚
a1
b1
e1
c1
d1
30˚
x2
y2
a1’
b1’
c1’ d1’
e1’
46. 3. Determine a true shape of a plane where surface is perpendicular to VP and is
inclined at an angle of 45⁰ with HP and whose TV is a regular pentagon of 30 mm sides
with one side making an angle of 30⁰ with xy line in TV.
x y
HP
VP
108˚
a
c
b
d
30˚
e
a’
b’
e’
d’
c’
45˚
x2
y2
b1’
c1’
a’
d1’
e1’
47. 3. abc is an equilateral triangle of altitude 50 mm with ab in xy and c bellow xy. abc’ is
an isosceles triangle of altitude 75 mm and c’ above xy. Determine the true shape of
triangle of ABC of which abc is TV and abc’ is FV.
x HP
VP
30˚
30˚
50
a b
c
c’
75
x1
y1
a’’
b’’
c’’
δ
γ
x2
y2
b2
a2
c2
y
48. 3. The FV and TV of a plane is squares of 40 mm sides, with two sides equally inclined to xy.
Draw the true shape of the plane. Also determine the inclined angle between the sides.
x HP
VP
y
b’
d’
b
45˚ 45˚
c
d
x1
y1
b’’
a
c’
a’
a’’
c’’
d’’
γ
45˚ 45˚
δ
x2
y2
a2
b2
c2
d2
49. 3. Lines AB and AC appears to make an angle of 120⁰ between them in their FV and TV line
AB is parallel to both the plane. Assume suitable view length. Determine the real angle
between lines AB and AC.
x y
HP
VP
a’
b’
c’
120˚
b
a
c
120˚
x1
y1
a’’
b’’
c’’
x2
y2
a2
b2
c2
ψ = 112⁰
50. x y
HP
3. The A picture frame 2 m wide and 1 mm high is to be fixed on a wall railing by two
straight wires attached to the top corner. The frame is to make an angle of 40⁰ with the
wall and the wires are to be fixed to a hook on a wall on center line of frame and 1.5 m
above railing. Find the length of wires and angles between them.
x1
y1
40˚
a’
b’ c’
d’
c’’
a’’
d’’
h’ h’’
b’’
1.5
m
a d
VP b c
2 m
h
x2
h2
a2
ψ = 91⁰
d2
y2
51. 3. An isosceles triangle ABC is having its corners A, B and C respectively 30 mm, 60 mm and
90 mm in front of VP. Draw projection of plane and determine its surface inclination with
HP and VP. Take base of triangle AB equal to 60 mm and altitude equal to 90 mm .
x y
HP
VP
30
30
b
a
b
ci
a
b’ a’ ci’
30
c
c
x1
y1
a1’
b1’
ci1’
c1
γ = 55⁰
c’
52. 3. An isosceles triangle ABC is having, base 60 mm and altitude 40 mm has its base AC in
HP and inclined at 30° to VP. Corner A and B are in VP. Draw its projection.
x HP
VP
Initially it is assumed ∆ ABC is in HP with pt
A is in VP and edge AC making angle of 30°
with VP. Corresponding projection is
obtained by drawing TV (showing True
Shape ) and corresponding FV (showing
Edge View)
c
30˚
bi
y
x1
y1
c’ bi’
b a
a1’
b1’
bi1’
b1’
γ = 64⁰
b’
53. 3. An isosceles triangle PQR having base PQ 50 mm long and altitude 75 mm has its corner
P, Q and R, 25 mm, 50 mm and 75 mm respectively above ground. Draw its projection.
x HP
VP
y
ri’
r’
30
30
q’
r’
30
p’
x1
y1
q p ri
q’
p’ p1
q1
ri1
r1
γ = 55⁰
r
54. 3. Draw TV and FV of a hexagonal lamina of 30 mm side, having two of the edges parallel to
both HP and VP and the nearest edge is 12 mm from each plane. The surface of lamina is
inclined at 60° to HP.
x y
HP
VP
c
d
e
a
12
b’
a’
c’
d’
d’
e’
b’
a’
d’
e’
d’
d
a e
d
b b d
c
60⁰
12
d
a e
b d
c
d
a
e
b
d
c
d
a
e
c
b
d
d’
d’
c’
a’ b’
x1
y1
a’’
b’’
c’’
d’’
d’’
e’’
e’
c’
55. Try to understand
P
Q
R
S
A
B
C
D
R
P
30
C
A
60
R C
P A
30 60
Corners P and A on the ground and
plane parallel to VP
Corners P and A on the ground and
plane perpendicular to VP with
corners R and C touches Each other
such that diagonal PR and AC
perpendicular to each other
x = 67
x2 = 302 + 602
3. An PQRS and ABCD are two square planes of diagonals 30 mm and 60 mm respectively
having their corners P and A on ground and corners R and C touches each other. The
surface of both the planes are perpendicular to each other and are perpendicular to VP.
Draw projection of planes and also find out the actual length of the side of square. (In
other words obtained the true shape of each plane.)
56. 3. An PQRS and ABCD are two square planes of diagonals 30 mm and 60 mm respectively
having their corners P and A on ground and corners R and C touches each other. The
surface of both the planes are perpendicular to each other and are perpendicular to VP.
Draw projection of planes and also find out the actual length of the side of square. (In
other words obtained the true shape of each plane.)
HP y
p’ 67 a’
r’ c’
q’
s’ b’
d’
p r
q
s
c
a
b
x1
y1
x VP
c1
b1
a1
d1
x2
y2
r1
s1
q1
p1
58. 2. A thin 45⁰ set square has its longest side 250 mm long in VP and inclined
at 30⁰ to HP. Its surface makes an angle of 45⁰ with VP. Draw projections.
(W-2003, 7 marks)
3. A 30⁰ - 60⁰ triangle PQR has its longest side PR is 50 mm long and is
contained in VP. P and R are 20 mm and 45 mm respectively from HP, while
point Q is 10 mm away from VP. Draw projection. (W – 92, 7 marks)