Projection of PlanesProjection of Planes Plane figures or surfaces have only two Plane figures or surfaces have only two

dimensions, viz. length & breadth. They do dimensions, viz. length & breadth. They do not have thickness.not have thickness.

A plane figure, extended if necessary, will A plane figure, extended if necessary, will meet the reference planes in lines, unless it meet the reference planes in lines, unless it is parallel to any one of them. These lines is parallel to any one of them. These lines are called the traces of the plane.are called the traces of the plane.

The line in which the plane meets the HP is The line in which the plane meets the HP is called the HT (Horizontal Trace) of the plane called the HT (Horizontal Trace) of the plane & the line in which it meets the VP, is called & the line in which it meets the VP, is called its VT( Vertical Trace). its VT( Vertical Trace).

A plane is usually represented by its traces.A plane is usually represented by its traces.

Projection of PlanesProjection of Planes

Planes are divided into two main types:Planes are divided into two main types:1.1. Perpendicular PlanesPerpendicular Planes

2.2. Oblique PlanesOblique Planes

Perpendicular Planes can be sub-divide into the Perpendicular Planes can be sub-divide into the following 3 sub-types:following 3 sub-types:

1.1. Perpendicular to both the reference planesPerpendicular to both the reference planes

2.2. Perpendicular to one reference plane and parallel to the other.Perpendicular to one reference plane and parallel to the other.

3.3. Perpendicular to one reference plane and inclined to the other.Perpendicular to one reference plane and inclined to the other.

Planes which are inclined to both the reference planes, Planes which are inclined to both the reference planes, are called oblique planesare called oblique planes

Projection of Planes Perpendicular Projection of Planes Perpendicular to to both the reference planesboth the reference planes

In the figure, a square In the figure, a square ABCDABCD is perpendicular is perpendicular to both the reference to both the reference planes.planes.

Its HT & VT are in a Its HT & VT are in a straight line straight line perpendicular to the perpendicular to the xy.xy.

The front view The front view b’c’b’c’ and and the top view the top view abab of the of the square are both straight square are both straight lines coinciding with the lines coinciding with the VT and the HT VT and the HT respectively.respectively.

Projection of Planes Perpendicular Projection of Planes Perpendicular to to HP parallel to VPHP parallel to VP

In the figure, a triangle In the figure, a triangle PQRPQR is perpendicular to the HP is perpendicular to the HP and parallel to the VP.and parallel to the VP.

Its HT is parallel to Its HT is parallel to xyxy. It does not have any VT. It does not have any VT.. The front view The front view p’q’r’p’q’r’ shows the exact shape & size of shows the exact shape & size of

the triangle. The top view the triangle. The top view pqrpqr is a line, parallel to is a line, parallel to xyxy. It . It coincides with the HT. coincides with the HT.

Projection of Planes Perpendicular Projection of Planes Perpendicular to to VP parallel to HPVP parallel to HP

In the figure, a square In the figure, a square ABCDABCD is perpendicular to the is perpendicular to the VP and parallel to the HP.VP and parallel to the HP.

Its VT is parallel to Its VT is parallel to xyxy. It does not have any HT. It does not have any HT.. The top view The top view abcdabcd shows the true shape & size of the shows the true shape & size of the

square. The front view square. The front view a’b’a’b’ is a line, parallel to is a line, parallel to xyxy. It . It coincides with the VT. coincides with the VT.

Projection of Planes Perpendicular Projection of Planes Perpendicular to to the HP inclined to the VPthe HP inclined to the VP

A square A square ABCDABCD is perpendicular to the HP and inclined at an is perpendicular to the HP and inclined at an angle angle ΦΦ to the VP. to the VP.

Its VT is perpendicular to Its VT is perpendicular to xyxy. Its HT is inclined at an angle . Its HT is inclined at an angle ΦΦ to to the the xyxy..

The top view The top view abab is a line inclined at an angle is a line inclined at an angle ΦΦ to the to the xyxy . . The front view The front view a’b’c’d’a’b’c’d’ is smaller than the true size of is smaller than the true size of ABCDABCD..

Projection of Planes Perpendicular Projection of Planes Perpendicular to to the VP inclined to the HPthe VP inclined to the HP

A square A square ABCDABCD is perpendicular to the VP and inclined at an is perpendicular to the VP and inclined at an angle angle θθ to the HP. to the HP.

Its HT is perpendicular to Its HT is perpendicular to xyxy. Its VT is inclined at an angle . Its VT is inclined at an angle θθ to the to the xyxy..

The top view The top view abcd abcd is a rectangle which is smaller than the square is a rectangle which is smaller than the square ABCDABCD..

The front view The front view a’b’a’b’ is a line inclined at an angle is a line inclined at an angle θθ to the to the xyxy . .

Projection of Planes – General Projection of Planes – General ConclusionsConclusions

Traces:Traces:1.1. When a plane is perpendicular to both the reference planes, its When a plane is perpendicular to both the reference planes, its

traces lie on a straight line perpendicular to xy.traces lie on a straight line perpendicular to xy.2.2. When a plane is perpendicular to one of the reference planes, its When a plane is perpendicular to one of the reference planes, its

trace upon the other plane is perpendicular to xy 9except when it trace upon the other plane is perpendicular to xy 9except when it is parallel to the other plane.is parallel to the other plane.

3.3. When a plane is parallel to a reference plane, it has no trace on When a plane is parallel to a reference plane, it has no trace on that plane. Its trace on the other reference plane , to which it is that plane. Its trace on the other reference plane , to which it is perpendicular, is parallel to xy.perpendicular, is parallel to xy.

4.4. When a plane is inclined to the HP and perpendicular to the VP, When a plane is inclined to the HP and perpendicular to the VP, its inclination is shown by the angle which its VT makes with xy. its inclination is shown by the angle which its VT makes with xy. When it is inclined to the VP and perpendicular to the HP, its When it is inclined to the VP and perpendicular to the HP, its inclination by the angle which its HT makes with xy.inclination by the angle which its HT makes with xy.

5.5. When a plane has two traces, they, produced if necessary, When a plane has two traces, they, produced if necessary, intersect in xy (except when both are parallel to xy as in case of intersect in xy (except when both are parallel to xy as in case of some oblique lines.)some oblique lines.)

Projection of Planes – General Projection of Planes – General ConclusionsConclusions

Projections:Projections:1.1. When a plane is perpendicular to a reference plane, its When a plane is perpendicular to a reference plane, its

projection on that plane is a straight line.projection on that plane is a straight line.2.2. When a plane is parallel to a reference plane, its When a plane is parallel to a reference plane, its

projection on that plane shows its true shape & size.projection on that plane shows its true shape & size.3.3. When a plane is perpendicular to one of the reference When a plane is perpendicular to one of the reference

planes and inclined to the other, its inclination is shown planes and inclined to the other, its inclination is shown by the angle which its projection on the plane to which by the angle which its projection on the plane to which it is perpendicular, makes with the xy. Its projection on it is perpendicular, makes with the xy. Its projection on the plane to which it is inclined, is smaller than the the plane to which it is inclined, is smaller than the plane itself.plane itself.

Projection of Planes Parallel to the HP.Projection of Planes Parallel to the HP.Prob1#Prob1# An equilateral triangle of 50 mm side has its VT An equilateral triangle of 50 mm side has its VT

parallel to and 25 mm above parallel to and 25 mm above xyxy. It has no HT. Draw its . It has no HT. Draw its projections when one of its sides is inclined at 45projections when one of its sides is inclined at 45°° to to the VP, its end-point nearest to the VP being 5 mm in the VP, its end-point nearest to the VP being 5 mm in front of the VP . front of the VP .

a’b’

a

b

c

c’

45°

25

5• As the VT is parallel to xy, and there is no HT, the triangle is parallel to the HP.• Therefore, begin with the TV.• Draw an equilateral triangle abc of 50 mm side, keeping one side ac inclined at 45° to xy.• Project the FV parallel to and 25 mm above xy, as shown.

VT

Projection of Planes Parallel to the VPProjection of Planes Parallel to the VPProb2#Prob2# A square ABCD of 40 mm side has a corner on the HP and 20 A square ABCD of 40 mm side has a corner on the HP and 20

mm in front of the VP. All the sides of the square are equally mm in front of the VP. All the sides of the square are equally inclined to the HP and parallel to the VP. Draw its projections and inclined to the HP and parallel to the VP. Draw its projections and show its traces.show its traces.

a’

b’

c’

d’

a (d, b) c

45°

20

40• As all the sides of the square ABCD are parallel to the VP, the surface of the square would also be parallel to the VP.•The FV would show the true shape & position of the square.• First draw a’b’c’d’ in the FV with one of the corners in xy and all its sides inclined at 45° to xy. • Project the TV keeping the line ac parallel to xy and 30 mm below it. The TV is the HT and there will be no VT.HT

Projection of Planes Projection of Planes inclined to the HPinclined to the HP & perpendicular & perpendicular to the VPto the VP

In the initial stage, the plane is assumed to be parallel In the initial stage, the plane is assumed to be parallel to the HP and to the HP and perpendicular to the VP. perpendicular to the VP.

The top view will show the true shape. The top view will show the true shape. THE front view will be a straight line parallel to xy.THE front view will be a straight line parallel to xy. Now the plane is tilted so that it becomes inclined to the HP.Now the plane is tilted so that it becomes inclined to the HP. The new front view will be inclined to the xy at the true inclination.The new front view will be inclined to the xy at the true inclination. In the top view, the corners will move along their respective paths In the top view, the corners will move along their respective paths

parallel to xy.parallel to xy.

Projection of Planes Projection of Planes inclined to the HPinclined to the HP & perpendicular & perpendicular to the VPto the VP

Problem# 3. A regular pentagon of side 25 mm has one side on the Problem# 3. A regular pentagon of side 25 mm has one side on the ground. Its plane is inclined at 45° to the HP and perpendicular to the ground. Its plane is inclined at 45° to the HP and perpendicular to the VP. Draw its projections and show its traces.VP. Draw its projections and show its traces.

Assume the plane of the pentagon parallel to the HP and draw the Assume the plane of the pentagon parallel to the HP and draw the pentagon in the top view. pentagon in the top view.

Project the front view from the top view. It will be the line a’c’ Project the front view from the top view. It will be the line a’c’ contained in the line xy.contained in the line xy.

Tilt the front view about point a’, so that it makes 45° with the line xy.Tilt the front view about point a’, so that it makes 45° with the line xy.

• Project the new top view ab1c1d1e upwards from this front view.• It will be more convenient if the front view is reproduced in a new position separately and the top view is projected from it.• The VT coincides with the FV and the HT is perpendicular to xy, through the point of intersection between xy and the FV produced.

Projection of Planes Projection of Planes inclined to the VPinclined to the VP & perpendicular & perpendicular to the HPto the HP

In the initial stage, the plane is assumed to be parallel to the VP In the initial stage, the plane is assumed to be parallel to the VP and perpendicular to the VP. and perpendicular to the VP.

The front view will show the true shape. The front view will show the true shape. The top view will be a straight line parallel to xy.The top view will be a straight line parallel to xy. Now the plane is tilted so that it becomes inclined to the VP.Now the plane is tilted so that it becomes inclined to the VP. The new top view will be inclined to the xy at the true inclination.The new top view will be inclined to the xy at the true inclination. In the front view, the corners will move along their respective In the front view, the corners will move along their respective

paths parallel to xy.paths parallel to xy.

Projection of Planes inclinedProjection of Planes inclined toto thethe HP & HP & perpendicular toperpendicular to thethe VPVP

Problem# 4. Draw the projections of a circle of 5 cm diameter, having its plane Problem# 4. Draw the projections of a circle of 5 cm diameter, having its plane vertical and inclined at 30° to the VP. Its centre is 3 cm above the HP and 2 cm in vertical and inclined at 30° to the VP. Its centre is 3 cm above the HP and 2 cm in front of the VP. Show also the traces.front of the VP. Show also the traces.

• A circle has no corners to project one view from the other. However, a number of A circle has no corners to project one view from the other. However, a number of points, say 12, at equal distances apart, may be marked on its circumference. points, say 12, at equal distances apart, may be marked on its circumference.

• Assume the circle to be parallel to the VP, draw its projections. Assume the circle to be parallel to the VP, draw its projections. • The FV will be a circle having its centre 20 mm above xy. The TV will be a line The FV will be a circle having its centre 20 mm above xy. The TV will be a line

parallel to and 30 mm below xy.parallel to and 30 mm below xy.

• The TV will be a line, parallel to and 20 mm below xy.• Divide the circumference into 12 equal parts and mark the points as shown.• Project these points in the TV. The centre O will coincide the point 4.• When the circle is tilted so as to make an angle of 30° with the VP, 30° with the VP, its TV will become inclined at 30° to its TV will become inclined at 30° to the xy line.the xy line.• In the FV all these points would In the FV all these points would move along their respective paths move along their respective paths parallel to xy.parallel to xy.

Projection of Planes inclinedProjection of Planes inclined toto thethe HP & HP & perpendicular toperpendicular to thethe VPVP

Reproduce the TV keeping the centre a at the same Reproduce the TV keeping the centre a at the same distance, i.e., 20 mm from xy and inclined at 30distance, i.e., 20 mm from xy and inclined at 30° to xy. ° to xy.

For the final FV, project all these points upwards from this For the final FV, project all these points upwards from this final TV and horizontally from the initial FV.final TV and horizontally from the initial FV.

Draw a free-hand curve through the twelve points 1’, 2’ etc. Draw a free-hand curve through the twelve points 1’, 2’ etc. This curve will be an ellipse.This curve will be an ellipse.

Projection of Oblique PlanesProjection of Oblique PlanesWhen a plane has its surface inclined to one plane and an edge When a plane has its surface inclined to one plane and an edge or a diameter or a diagonal parallel to that plane and inclined to or a diameter or a diagonal parallel to that plane and inclined to the other plane, its projection is drawn in three stages:the other plane, its projection is drawn in three stages:

A.A. If the surface of a plane is inclined to the HP and an edge (or a If the surface of a plane is inclined to the HP and an edge (or a diameter or a diagonal) is parallel to the HP and inclined to the diameter or a diagonal) is parallel to the HP and inclined to the VP:VP:

I.I. in the initial position, the plane is assumed to be parallel to the in the initial position, the plane is assumed to be parallel to the HP and an edge perpendicular to the VP.HP and an edge perpendicular to the VP.

II.II. then it is tilted so as to make the required angle with the HP. Its then it is tilted so as to make the required angle with the HP. Its FV, in this position, will be a straight line while its TV will be FV, in this position, will be a straight line while its TV will be smaller in size.smaller in size.

III.III. In the final position, the plane is turned to the required inclination In the final position, the plane is turned to the required inclination with the VP and only the position of the TV will change while its with the VP and only the position of the TV will change while its shape and size would not change. In the FV, the corresponding shape and size would not change. In the FV, the corresponding distances of all the corners from xy will remain the same as in the distances of all the corners from xy will remain the same as in the second FV.second FV.

If an edge is in the HP or on the ground, in the initial position, the If an edge is in the HP or on the ground, in the initial position, the plane is assumed to be lying in the HP or on the ground, with the plane is assumed to be lying in the HP or on the ground, with the edge perpendicular to the VP.edge perpendicular to the VP.

If a corner is in the HP or on the ground, the line joining that If a corner is in the HP or on the ground, the line joining that corner with the centre of the plane is kept parallel to the VP, in the corner with the centre of the plane is kept parallel to the VP, in the initial position.initial position.

Projection of Oblique PlanesProjection of Oblique PlanesB.B. If the surface of a plane is inclined to the VP and an edge If the surface of a plane is inclined to the VP and an edge

(or a diameter or a diagonal) is parallel to the VP and inclined to (or a diameter or a diagonal) is parallel to the VP and inclined to the HP:the HP:

I.I. in the initial position, the plane is assumed to be parallel to the in the initial position, the plane is assumed to be parallel to the VP and an edge perpendicular to the HP.VP and an edge perpendicular to the HP.

II.II. then it is tilted so as to make the required angle with the VP. Its then it is tilted so as to make the required angle with the VP. Its TV, in this position, will be a straight line while its FV will be TV, in this position, will be a straight line while its FV will be smaller in size.smaller in size.

III.III. In the final position, the plane is turned to the required inclination In the final position, the plane is turned to the required inclination with the HP and only the position of the FV will change while its with the HP and only the position of the FV will change while its shape and size would not change. In the TV, the corresponding shape and size would not change. In the TV, the corresponding distances of all the corners from xy will remain the same as in the distances of all the corners from xy will remain the same as in the second TV.second TV.

If an edge is in the VP, in the initial position, the plane is assumed If an edge is in the VP, in the initial position, the plane is assumed to be lying in the VP, with the edge perpendicular to the HP.to be lying in the VP, with the edge perpendicular to the HP.

If a corner is in the VP, the line joining that corner with the centre If a corner is in the VP, the line joining that corner with the centre of the plane is kept parallel to the HP, in the initial position.of the plane is kept parallel to the HP, in the initial position.

#5. A square ABCD of side 50 mm has its corner A in the HP, its #5. A square ABCD of side 50 mm has its corner A in the HP, its diagonal AC inclined at 30diagonal AC inclined at 30°° with the HP and the diagonal BD inclined at with the HP and the diagonal BD inclined at 4545°° to the VP and parallel to the HP. Draw its projections. to the VP and parallel to the HP. Draw its projections.

In the initial position, assume the square In the initial position, assume the square ABCD to be lying in the HP with AC parallel ABCD to be lying in the HP with AC parallel to VP.to VP.

Draw the TV and the FV.Draw the TV and the FV. When the square is tilted about the corner When the square is tilted about the corner

A, so that AC makes an angle of 30A, so that AC makes an angle of 30°° with with the HP. BD remains perpendicular to VP the HP. BD remains perpendicular to VP and parallel to the HPand parallel to the HP

Draw the second FV a’c’ inclined at 30Draw the second FV a’c’ inclined at 30°° to to xy, keeping a’ in xy.xy, keeping a’ in xy.

Project the second TV. The square is now Project the second TV. The square is now turned so that BD makes an angle of 45turned so that BD makes an angle of 45° ° with the VP and remains parallel to HP. with the VP and remains parallel to HP. Only the position of the TV will change and Only the position of the TV will change and its shape and size would remain the same.its shape and size would remain the same.

Reproduce the TV so that b1d1 is inclined Reproduce the TV so that b1d1 is inclined at 45° to xy.at 45° to xy.

Project the final FV upwards from this final Project the final FV upwards from this final TV and horizontally from the second TVTV and horizontally from the second TV

#6. Draw the projections of a regular hexagon of 25 mm side, having #6. Draw the projections of a regular hexagon of 25 mm side, having one of its sides in the HP and inclined at 60one of its sides in the HP and inclined at 60°° to the VP, and its surface to the VP, and its surface making an angle of 45making an angle of 45° to the HP.° to the HP.

In the initial position, In the initial position, assume the hexagon assume the hexagon ABCDEFABCDEF to be lying in the to be lying in the HP with one side HP with one side AFAF perpendicular to VP.perpendicular to VP.

Draw the hexagon in the Draw the hexagon in the TV with one side TV with one side afaf perpendicular to perpendicular to xyxy..

Project the FV Project the FV a’c’a’c’ in in xyxy.. Draw Draw a’c’a’c’ inclined at 45 inclined at 45° to ° to

xyxy keeping keeping a’a’ in in xyxy and and project the second TV.project the second TV.

Reproduce this TV making Reproduce this TV making aa11ff11 inclined at 60 inclined at 60° to ° to xyxy and project the final FV. and project the final FV.

#7. Draw the projections of a circle of 50 mm diameter, resting in the #7. Draw the projections of a circle of 50 mm diameter, resting in the HP with on a point A on the circumference, its plane inclined at 45HP with on a point A on the circumference, its plane inclined at 45° to ° to the HPthe HP and (a) the TV of the diameter AB making 30 and (a) the TV of the diameter AB making 30° with the VP; ° with the VP; (b) the diameter making (b) the diameter making 3030° angle with the VP.° angle with the VP.

Draw the projections of the circle with A in the HP and its plane inclined at 45Draw the projections of the circle with A in the HP and its plane inclined at 45° to ° to the HP and perpendicular to the VP.the HP and perpendicular to the VP.

In the second TV, the line In the second TV, the line aa11bb11 is the TV of the diameter is the TV of the diameter ABAB.. Reproduce this TV so that Reproduce this TV so that aa11bb1 1 makesmakes an angle 30 an angle 30° with xy. ° with xy. Project the required FV.Project the required FV. If the diameter, which makes an angle of 45If the diameter, which makes an angle of 45° with the HP, is inclined at 30° to the ° with the HP, is inclined at 30° to the

VP also, its TV VP also, its TV aa11bb1 1 would make an angle greater than 30would make an angle greater than 30° with ° with xyxy. This apparent . This apparent angle of inclination is determined as follows. angle of inclination is determined as follows.

Draw any line Draw any line aa11bb2 2 equal to AB and inclined at equal to AB and inclined at 30° to xy.30° to xy.

#7. Draw the projections of a circle of 50 mm diameter, resting in the #7. Draw the projections of a circle of 50 mm diameter, resting in the HP with on a point A on the circumference, its plane inclined at 45HP with on a point A on the circumference, its plane inclined at 45° to ° to the HPthe HP and (a) the TV of the diameter AB making 30 and (a) the TV of the diameter AB making 30° with the VP; ° with the VP; (b) the diameter making (b) the diameter making 3030° angle with the VP.° angle with the VP.

Draw any line Draw any line aa11bb2 2 equal to equal to ABAB and inclined at and inclined at 30° to 30° to xyxy.. With With aa11 as centre and radius equal to the TV of as centre and radius equal to the TV of ABAB, viz. , viz. aa11bb11, ,

draw an arc cutting draw an arc cutting rsrs ( the path of ( the path of BB in the TV) at in the TV) at bb33.. Draw the line joining Draw the line joining aa11 with with bb33, and around it reproduce the , and around it reproduce the

second TV.second TV. Project the final FV. It is evident that Project the final FV. It is evident that aa11bb33 is inclined at an is inclined at an

angle which is greater than 30°angle which is greater than 30°

#8. A thin 30#8. A thin 30°-°-6060° set-square has its longest edge in the VP and ° set-square has its longest edge in the VP and inclined at inclined at 3030° to the HP. Its surface makes an angle of 45° with the VP. ° to the HP. Its surface makes an angle of 45° with the VP. Draw its projections. Draw its projections.

Draw the FV Draw the FV a’b’c’a’b’c’ and project the TV and project the TV acac in line in line xyxy.. Tilt the arc about the end-point Tilt the arc about the end-point aa so that it makes an angle of 45° with so that it makes an angle of 45° with xyxy and and

project the FV project the FV aa11’b’b11’c’c11’’.. Reproduce the second FV into the third FV Reproduce the second FV into the third FV aa22’b’b22’c’c22’ ’ such that such that aa22’ c’ c22’’ makes an makes an

angle of 30° with xy. angle of 30° with xy. Project the final TV Project the final TV aa22bb22cc22’.’.

a’

c’

b’

(a, c) bx y

(a, c)(a1, c1)

b1

a1’

c1’

b1’

a2’

c2’

b2’

c2a2

b2

45°