Problem 6.11 Projection of Solids – ABCD is a tetrahedron of 60 mm long edges. The edge AB is in the H.P. The edge CD is inclined at an angle of 30^{0} to the H.P. and 45^{0} to the V.P. Draw the projections of the tetrahedron.

Tetrahedron: It is a triangular pyramid shape with all six edges of equal length, here the length is given as 60 mm and height is not given, so we need to find it. Height can be found by one of the slant edges which will show true length in elevation and a parallel line to XY (reference line) in plan. Here it is CD slant edge which is shown as c'd' in elevation and as cd in plan.

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*Procedure:*

* Step-1* Draw a horizontal x-y reference line of some suitable length.

**Step-2**** **To draw edge CD inclined to H.P. at 30^{0}, we need to assume that tetrahedron is resting on H.P. through one its base edges AB. Draw plan and elevation of tetrahedron resting on H.P. through one of the edge AB which will be the first stage as shown in figure above.

**Step-3**** **To find point d in plan we need to joint corners with their opposite edge's mid point. At a point where all joint lines will be intersected should be denoted as "d". And all rest points of plan and elevation can be founded by general projection procedure that we have already discussed in earlier projection of solids problems.

* Step-4* Now its time to draw second stage of projection in which slant edge CD is making an angle of 30

^{0 }with the H.P. Draw edge c'd' from 1st stage elevation inclined to H.P. at 30

^{0 }so that perpendicular bisector of c'd' makes 60

^{0 }with H.P. In such a way we'll get an elevation for second stage in which tetrahedron's edge cd will be inclined to H.P. by 30

^{0}.

* Step-5* To find out plan of second stage we need to project all points towards plan area or towards H.P. for reference. And reference of 1st stage plan will be used to track all point of 2nd stage plan. Join a' with a, b' with b and similarly find all point of 2nd stage plan.

* Step-6* Except ab edge all other edges will be full thick and ab will be hidden in plan as it cant be seen from top when cd is inclined to H.P. by 30

^{0}. This will complete second stage as both plan and elevation drawn.

* Step-7* Now to draw last stage, we need to find out locus of c as actual slant edge is making an angle of 45

^{0 }with V.P. not a plan of it.

If plan of that slant edge (cd) makes an angle of 45Note:^{0 }with V.P. then we can draw second stage's plan length of cd directly inclined to V.P. by 45^{0 }but here in current data it is given as slant edge is making an angle of 45^{0 }with V.P. so we need to follow locus procedure.

* Step-8* First draw a line with true length of slant edge (T.L. = 60 mm given) inclined to V.P. by 45

^{0 }, give notations as per shown in figure above. Then extend locus of c2 parallel to XY reference line upto suitable distance. Now with d as center and radius equal to plan length of edge cd (cd line in second stage plan) make an arc on locus that gives intersection point c on locus of c2.

* Step-7* Now the angle of line cd with V.P. will be denoted as β as it will shows virtual angle of actual slant edge when it is inclined to V.P. by 45

^{0}.

* Step-8* To find out last stage elevation we need to project all point of third stage plan towards elevation or towards V.P. area. Then find all projection intersection points as per general procedure. give notations as a', b', c'… as shown in figure.

* Step-9* Except ac edge all other edges will be full thick and ac will be hidden in elevation as it cant be seen from front when CD is inclined to V.P. by 45

^{0}. This will complete third stage as both plan and elevation drawn.

* Step-10* Give the dimensions by any one method of dimensions and give the notations as shown into the figure.