Projection of Planes : Projection of Planes Hareesha N G
Dept of Aeronautical Engg
Dayananda Sagar College of Engg
Bangalore-78 3/27/2012 1 Hareesha N G, DSCE, Blore
Slide 2: 3/27/2012 2 Hareesha N G, DSCE, Blore
Slide 3: FV-3 T V-3 FV-1 T V-1 FV-2 T V-2 SURFACE PARALLEL TO HP
PICTORIAL PRESENTATION SURFACE INCLINED TO HP
PICTORIAL PRESENTATION ONE SMALL SIDE INCLINED TO VP
PICTORIAL PRESENTATION ORTHOGRAPHIC
TV-True Shape
FV- Line // to xy ORTHOGRAPHIC
FV- Inclined to XY
TV- Reduced Shape ORTHOGRAPHIC
FV- Apparent Shape
TV-Previous Shape 3/27/2012 3 Hareesha N G, DSCE, Blore
Slide 4: PROCEDURE OF SOLVING THE PROBLEM:
IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration )
STEP 1. Assume suitable conditions & draw Fv & Tv of initial position.
STEP 2. Now consider surface inclination & draw 2nd Fv & Tv.
STEP 3. After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv.
ASSUMPTIONS FOR INITIAL POSITION:
(Initial Position means assuming surface // to HP or VP)
1.If in problem surface is inclined to HP – assume it // HP
Or If surface is inclined to VP – assume it // to VP
2. Now if surface is assumed // to HP- It’s TV will show True Shape.
And If surface is assumed // to VP – It’s FV will show True Shape.
3. Hence begin with drawing TV or FV as True Shape.
4. While drawing this True Shape –
keep one side/edge ( which is making inclination) perpendicular to xy line
( similar to pair no. on previous page illustration ). APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS 3/27/2012 4 Hareesha N G, DSCE, Blore
Slide 5: X Y a b c d a’ b’ c’ d’ a1 b1 c1 d1 a1 b1 c1 d1 a’ b’ d’ c’ c’1 d’1 b’1 a’1 450 300 Surface // to Hp Surface inclined to Hp Side
Inclined
to Vp 3/27/2012 5 Hareesha N G, DSCE, Blore
Slide 6: c1 X Y 300 450 a’1 b’1 c’1 a c a’ a b1 b’ b a1 b c a’1 b’1 c’1 c’ Hence begin with FV, draw triangle above X-Y
keeping longest side vertical. Surface // to Vp Surface inclined to Vp side inclined to Hp 3/27/2012 6 Hareesha N G, DSCE, Blore
Slide 7: c c1 X Y 450 a’1 b’1 c’1 a c a’ a b1 b’ b a1 b a’1 b’1 c’1 c’ 35 10 Problem 3:
A 300 – 600 set square of longest side
100 mm long is in VP and it’s surface
450 inclined to VP. One end of longest
side is 10 mm and other end is 35 mm
above HP. Draw it’s projections
(Surface inclination directly given.
Side inclination indirectly given) Read problem and answer following questions
1 .Surface inclined to which plane? ------- VP
2. Assumption for initial position? ------// to VP
3. So which view will show True shape? --- FV
4. Which side will be vertical? ------longest side.
Hence begin with FV, draw triangle above X-Y
keeping longest side vertical. First TWO steps are similar to previous problem.
Note the manner in which side inclination is given.
End A 35 mm above Hp & End B is 10 mm above Hp.
So redraw 2nd Fv as final Fv placing these ends as said. 3/27/2012 7 Hareesha N G, DSCE, Blore
Slide 8: Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which side will be vertical? -------- any side.
Hence begin with TV,draw pentagon below
X-Y line, taking one side vertical. Problem 4:
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.
Draw it’s projections when the side in HP makes 300 angle with VP a’ b’ d’ b1 d c1 a c’e’ b c d1 b’1 a1 e’1 c’1 d’1 a1 b1 c1 d1 d’ a’ b’ c’e’ e1 e1 a’1 X Y 450 300 e SURFACE AND SIDE INCLINATIONS
ARE DIRECTLY GIVEN. 3/27/2012 8 Hareesha N G, DSCE, Blore
Slide 9: 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.
Draw projections when side in HP is 300 inclined to VP. Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which side will be vertical? --------any side.
Hence begin with TV,draw pentagon below
X-Y line, taking one side vertical. X Y a’ b’ d’ c’e’ SURFACE INCLINATION INDIRECTLY GIVEN
SIDE INCLINATION DIRECTLY GIVEN: 3/27/2012 9 Hareesha N G, DSCE, Blore
Slide 10: 450 300 a’ b’1 c’1 d’ a’ b’ d’ c’ a c b d a1 b1 c1 d1 b’ c’ a’1 d’1 d1 b1 c1 a1 X Y 300 b’1 c’1 c1 a’1 d’1 d1 b1 c1 a1 TL Problem 6: A rhombus of diagonals 40 mm and 70 mm long respectively has one end of it’s longer diagonal in HP while that diagonal is 350 inclined to HP. If the top-view of the same diagonal makes 400 inclination with VP, draw it’s projections. Problem 7: A rhombus of diagonals 40 mm and 70 mm long respectively having one end of it’s longer diagonal in HP while that diagonal is 350 inclined to HP and makes 400 inclination with VP. Draw it’s projections. Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which diagonal horizontal? ---------- Longer
Hence begin with TV,draw rhombus below
X-Y line, taking longer diagonal // to X-Y c2 3/27/2012 10 Hareesha N G, DSCE, Blore
Slide 11: T L 450 300 300 Problem 8: A circle of 50 mm diameter is
resting on Hp on end A of it’s diameter AC
which is 300 inclined to Hp while it’s Tv
is 450 inclined to Vp.Draw it’s projections. Problem 9: A circle of 50 mm diameter is
resting on Hp on end A of it’s diameter AC
which is 300 inclined to Hp while it makes
450 inclined to Vp. Draw it’s projections. Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which diameter horizontal? ---------- AC
Hence begin with TV,draw rhombus below
X-Y line, taking longer diagonal // to X-Y 3/27/2012 11 Hareesha N G, DSCE, Blore
Slide 12: Problem 10: End A of diameter AB of a circle is in HP
A nd end B is in VP.Diameter AB, 50 mm long is
300 & 600 inclined to HP & VP respectively.
Draw projections of circle. TL X Y 300 600 Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which diameter horizontal? ---------- AB
Hence begin with TV,draw CIRCLE below
X-Y line, taking DIA. AB // to X-Y 3/27/2012 12 Hareesha N G, DSCE, Blore
Slide 13: X Y Problem 11:
A hexagonal lamina has its one side in HP and
Its apposite parallel side is 25mm above Hp and
In Vp. Draw it’s projections.
Take side of hexagon 30 mm long. Read problem and answer following questions
1. Surface inclined to which plane? ------- HP
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV
4. Which diameter horizontal? ---------- AC
Hence begin with TV,draw rhombus below
X-Y line, taking longer diagonal // to X-Y 3/27/2012 13 Hareesha N G, DSCE, Blore
Slide 14: A B C H H/3 G X Y a’ b’ c’ g’ b a,g c b a,g c 450 a’1 c’1 b’1 g’1 FREELY SUSPENDED CASES. Problem 12:
An isosceles triangle of 40 mm long
base side, 60 mm long altitude Is
freely suspended from one corner of
Base side.It’s plane is 450 inclined to
Vp. Draw it’s projections. 3/27/2012 14 Hareesha N G, DSCE, Blore