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Projection of Planes Plane figures or surfaces have only two dimensions, viz. length & breadth. They do not have thickness. A plane figure, extended if.

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Presentation on theme: "Projection of Planes Plane figures or surfaces have only two dimensions, viz. length & breadth. They do not have thickness. A plane figure, extended if."— Presentation transcript:

1 Projection of Planes Plane figures or surfaces have only two dimensions, viz. length & breadth. They do not have thickness. A plane figure, extended if necessary, will meet the reference planes in lines, unless it is parallel to any one of them. These lines are called the traces of the plane. The line in which the plane meets the HP is called the HT (Horizontal Trace) of the plane & the line in which it meets the VP, is called its VT( Vertical Trace). A plane is usually represented by its traces.

2 Projection of Planes Planes are divided into two main types:
Perpendicular Planes Oblique Planes Perpendicular Planes can be sub-divide into the following 3 sub-types: Perpendicular to both the reference planes Perpendicular to one reference plane and parallel to the other. Perpendicular to one reference plane and inclined to the other. Planes which are inclined to both the reference planes, are called oblique planes

3 Projection of Planes Perpendicular to both the reference planes
In the figure, a square ABCD is perpendicular to both the reference planes. Its HT & VT are in a straight line perpendicular to the xy. The front view b’c’ and the top view ab of the square are both straight lines coinciding with the VT and the HT respectively.

4 Projection of Planes Perpendicular to HP parallel to VP
In the figure, a triangle PQR is perpendicular to the HP and parallel to the VP. Its HT is parallel to xy. It does not have any VT. The front view p’q’r’ shows the exact shape & size of the triangle. The top view pqr is a line, parallel to xy. It coincides with the HT.

5 Projection of Planes Perpendicular to VP parallel to HP
In the figure, a square ABCD is perpendicular to the VP and parallel to the HP. Its VT is parallel to xy. It does not have any HT. The top view abcd shows the true shape & size of the square. The front view a’b’ is a line, parallel to xy. It coincides with the VT.

6 Projection of Planes Perpendicular to the HP inclined to the VP
A square ABCD is perpendicular to the HP and inclined at an angle Φ to the VP. Its VT is perpendicular to xy. Its HT is inclined at an angle Φ to the xy. The top view ab is a line inclined at an angle Φ to the xy . The front view a’b’c’d’ is smaller than the true size of ABCD.

7 Projection of Planes Perpendicular to the VP inclined to the HP
A square ABCD is perpendicular to the VP and inclined at an angle θ to the HP. Its HT is perpendicular to xy. Its VT is inclined at an angle θ to the xy. The top view abcd is a rectangle which is smaller than the square ABCD. The front view a’b’ is a line inclined at an angle θ to the xy .

8 Projection of Planes – General Conclusions
Traces: When a plane is perpendicular to both the reference planes, its traces lie on a straight line perpendicular to xy. When a plane is perpendicular to one of the reference planes, its trace upon the other plane is perpendicular to xy 9except when it is parallel to the other plane. 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 perpendicular, is parallel to xy. 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. When it is inclined to the VP and perpendicular to the HP, its inclination by the angle which its HT makes with xy. When a plane has two traces, they, produced if necessary, intersect in xy (except when both are parallel to xy as in case of some oblique lines.)

9 Projection of Planes – General Conclusions
Projections: When a plane is perpendicular to a reference plane, its projection on that plane is a straight line. When a plane is parallel to a reference plane, its projection on that plane shows its true shape & size. When a plane is perpendicular to one of the reference planes and inclined to the other, its inclination is shown by the angle which its projection on the plane to which it is perpendicular, makes with the xy. Its projection on the plane to which it is inclined, is smaller than the plane itself.

10 Projection of Planes Parallel to the HP.
Prob1# An equilateral triangle of 50 mm side has its VT parallel to and 25 mm above xy. It has no HT. Draw its projections when one of its sides is inclined at 45° to the VP, its end-point nearest to the VP being 5 mm in front of the VP . a’ VT c’ b’ 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. 25 c 5 45° a b

11 Projection of Planes Parallel to the VP
Prob2# 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 inclined to the HP and parallel to the VP. Draw its projections and show its traces. b’ 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. 40 a’ c’ 45° d’ 20 HT (d, b) a c

12 Projection of Planes inclined to the HP & perpendicular to the VP
In the initial stage, the plane is assumed to be parallel to the HP and perpendicular to the VP. The top view will show the true shape. THE front view will be a straight line parallel to xy. 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. In the top view, the corners will move along their respective paths parallel to xy.

13 Projection of Planes inclined to the HP & perpendicular to the VP
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 VP. Draw its projections and show its traces. Assume the plane of the pentagon parallel to the HP and draw the pentagon in the top view. Project the front view from the top view. It will be the line a’c’ contained in 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.

14 Projection of Planes inclined to the VP & perpendicular to the HP
In the initial stage, the plane is assumed to be parallel to the VP and perpendicular to the VP. The front view will show the true shape. The top view will be a straight line parallel to xy. 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. In the front view, the corners will move along their respective paths parallel to xy.

15 Projection of Planes inclined to the HP & perpendicular to the VP
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 front of the VP. Show also the traces. 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. 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 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, its TV will become inclined at 30° to the xy line. In the FV all these points would move along their respective paths parallel to xy.

16 Projection of Planes inclined to the HP & perpendicular to the VP
Reproduce the TV keeping the centre a at the same distance, i.e., 20 mm from xy and inclined at 30° to xy. For the final FV, project all these points upwards from this final TV and horizontally from the initial FV. Draw a free-hand curve through the twelve points 1’, 2’ etc. This curve will be an ellipse.

17 Projection of Oblique Planes
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 the other plane, its projection is drawn in three stages: 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 VP: in the initial position, the plane is assumed to be parallel to the HP and an edge perpendicular to the VP. 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 smaller in size. 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 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 second FV. 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 edge perpendicular to the VP. 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 initial position.

18 Projection of Oblique Planes
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 the HP: in the initial position, the plane is assumed to be parallel to the VP and an edge perpendicular to the HP. 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 smaller in size. 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 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 second TV. 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. 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.

19 #5. A square ABCD of side 50 mm has its corner A in the HP, its diagonal AC inclined at 30° with the HP and the diagonal BD inclined at 45° to the VP and parallel to the HP. Draw its projections. In the initial position, assume the square ABCD to be lying in the HP with AC parallel to VP. Draw the TV and the FV. When the square is tilted about the corner A, so that AC makes an angle of 30° with the HP. BD remains perpendicular to VP and parallel to the HP Draw the second FV a’c’ inclined at 30° to xy, keeping a’ in xy. Project the second TV. The square is now turned so that BD makes an angle of 45° with the VP and remains parallel to HP. Only the position of the TV will change and its shape and size would remain the same. Reproduce the TV so that b1d1 is inclined at 45° to xy. Project the final FV upwards from this final TV and horizontally from the second TV

20 Draw the hexagon in the TV with one side af perpendicular to xy.
#6. Draw the projections of a regular hexagon of 25 mm side, having one of its sides in the HP and inclined at 60° to the VP, and its surface making an angle of 45° to the HP. In the initial position, assume the hexagon ABCDEF to be lying in the HP with one side AF perpendicular to VP. Draw the hexagon in the TV with one side af perpendicular to xy. Project the FV a’c’ in xy. Draw a’c’ inclined at 45° to xy keeping a’ in xy and project the second TV. Reproduce this TV making a1f1 inclined at 60° to xy and project the final FV.

21 #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 45° to the HP and (a) the TV of the diameter AB making 30° with the VP; (b) the diameter making 30° angle with the VP. Draw the projections of the circle with A in the HP and its plane inclined at 45° to the HP and perpendicular to the VP. In the second TV, the line a1b1 is the TV of the diameter AB. Reproduce this TV so that a1b1 makes an angle 30° with xy. Project the required FV. If the diameter, which makes an angle of 45° with the HP, is inclined at 30° to the VP also, its TV a1b1 would make an angle greater than 30° with xy. This apparent angle of inclination is determined as follows. Draw any line a1b2 equal to AB and inclined at 30° to xy.

22 Draw any line a1b2 equal to AB and inclined at 30° to xy.
#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 45° to the HP and (a) the TV of the diameter AB making 30° with the VP; (b) the diameter making 30° angle with the VP. Draw any line a1b2 equal to AB and inclined at 30° to xy. With a1 as centre and radius equal to the TV of AB, viz. a1b1, draw an arc cutting rs ( the path of B in the TV) at b3. Draw the line joining a1 with b3, and around it reproduce the second TV. Project the final FV. It is evident that a1b3 is inclined at an angle which is greater than 30°

23 #8. A thin 30°-60° set-square has its longest edge in the VP and inclined at 30° to the HP. Its surface makes an angle of 45° with the VP. Draw its projections. Draw the FV a’b’c’ and project the TV ac in line xy. Tilt the arc about the end-point a so that it makes an angle of 45° with xy and project the FV a1’b1’c1’. Reproduce the second FV into the third FV a2’b2’c2’ such that a2’ c2’ makes an angle of 30° with xy. Project the final TV a2b2c2’. a2’ a1’ a’ b’ b1’ b2’ c2’ c’ c1’ a2 c2 y x (a, c) (a, c) b 45° (a1, c1) b1 b2


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