line cutting an angle of the solid to the bisection of one of its shortest sides.

The mere inspection of the diagram should demonstrate its construction.

A right pyramid, 1.8 inches high, has one of the sides of its base, which forms a regular pentagon of 1.1 inch side, making an angle of 26° with the vertical plane. Construct its projection, also a transverse section passing through its axis perpendicular to the vertical plane, and inclined to the horizon at an angle of 31°. (Fig. 27.)

Construct the plan and elevation of the solid as in the preceding examples, and on the elevation determine the section lines according to conditions, and cutting the edges of the pyramid in a' b' c' d' and e'; from these points drop perpendiculars so as to intersect on the plan the corresponding edges of the solid in a b c d and e: join these points, and the sectional plan of the pyramid will be produced.

To show the real section; on the plan, draw a ƒ parallel to xy, and from 1, 2, 3 and 4 draw ordinates perpendicular to it, meeting the angles of the section; at b'c' d' and e' draw likewise perpendiculars 1' 2' 3′ and 4' to a' e', and transfer on these the length of the ordinates on the plan; connect a 1 3 4 2 and a', which will give the real section reduced to the vertical plane.

A pentagonal pyramid 1 inch side and 1.75 inches high has one of the edges of its base forming an angle of 24° with the vertical plane: construct its projections, and also a section cutting 4 of its sides perpendicularly to the horizon, but not passing through the apex. (Fig. 28.)

After having obtained the plan and elevation in the usual manner, repeat the operations already explained (Fig. 25) to obtain the section, viz.

G

Draw the sectional line s intersecting 4 sides and 3

edges.

Draw a a' bb' c c' perpendicular to a y, and draw u a'' b b'' c c'' perpendicular to s l, and of the same altitude as their corresponding heights on the elevation; join s a'' b'' c'' and I for the section required.

The inspection of the two following diagrams should be sufficient for their apprehension.

The profile section of a right prism is an equilateral triangle of 1 inch side, the prism is 1.75 inches long and rests on one of its faces: construct its plan and elevation, showing one face and one end of the solid; also a section on any plane cutting both faces and both ends. (Fig. 29.)

A prism 1.75 inches long, having for its ends regular hexagons of .5 inch side is laid on one of its faces: show its plan, elevation, and a section taken on a line cutting all the sides and one end of the solid. (Fig. 30.)

OBSERVATION.-Before going on with the following problems, the general student, the military candidate whose time for preparation is not too closely limited, is referred to the article on "Descriptive Geometry" (page 112), the proper acquisition of whose elements is recommended as a safe, although slower foundation for his future progress the following problems being mostly practical applications based on questions given at the examinations.

A circle of 1.5 inches diameter is placed on a plane forming an angle of 46° with the horizon. (Fig. 31.)

(N.B. The intersections of a plane with the two intersecting planes are termed 'the traces of the plane,' and a plane is determined when the traces of that plane are given.)

Let a b represent the horizontal trace acting as an axis of revolution to the plane b c, so as to determine its given inclination with the horizon upon the vertical plane.

Let d e f g represent the two diameters and k the centre of the given circle; draw the perpendiculars d d', g k' and e e' meeting xy, and from b as a centre describe the arcs d' d'', k' k'' and e' e''. Lines drawn perpendicularly to zy from d'' to n, k'' to m and e'' to o, will determine the projections of the two diameters of the circle, which, in their new position, will form the major axis Im and the minor axis no of an ellipse, which is to be completed by drawing the curve In mo by the hand.

It is evident from the foregoing problem that the oblique projection of a circle is an ellipse. The solution of the next exercise will therefore be easy, and require no further explanation.

The side of a cylinder 1.75 inches long and 1 inch diameter, forms an angle of 23° with the vertical plane. Construct its projections. (Fig. 32.)

A plane forming an angle of 40° with the horizon, contains on its surface a line 1.25 inches long, inclined to the horizon at an angle of 30°. Contrast its projections. (Fig. 33.)

A plane can contain a line having an inclination less or equal to its own, but never a line having an inclination greater than that of the plane, for in order to be contained by a plane the two extremities of the line must coincide with it. And should the inclination of the line be greater than that of the plane, the line could only touch the plane at one point.

Draw the traces a b and b c, forming the given angle (40°) with the horizon.

Draw anywhere the line de 1.25 inches long, forming with x y the given angle (30°.) f will be the height above the received by the given plane.

Draw e f' parallel to x y, horizon of the line d e when From centre b, with radius.

bf describe the arc f' f', and draw f'' k parallel to a b. The distance bf being equal to the distance bf'', the inclined plane a bf' will evidently be reduced to the horizon in f''k: take in the compass the distance de, and transfer it, so that the two extremities may coincide any where on the two horizontals a band f'' k in g and f. Draw fl parallel to x y and f' l parallel to a b, and intersecting it,

g

I will be the horizontal projection of the given line.

A square 1.25 inches side is placed on a plane forming an angle of 30° with the horizon, and one of its sides is inclined at an angle of 15°. Show its plan. (Fig. 34.)

The first part of this exercise (excepting the difference in the value of the angles) is a mere repetition of the last problem, and, therefore, requires no fresh demonstration.

On gf complete the square m n, and draw mo and a p perpendicular to a y, from centre b describe the arcs o m' and pn', and draw m' m'' and n' n'', meeting the lines m m'' n n'', drawn parallel to x y.

The side of a pentagonal prism 1 inch long and .35 inch high is inclined at an angle of 20° with the horizon, and stands on a plane, forming an angle of 50° with the horizon. (Fig. 35.)

No matter what may be the shape of the figure, the line first obtained will ever serve as a base of construction to it. On the line gf, therefore, construct the pentagon and project it as before. On b c make b d equal to the given height (.35) and lines drawn parallel to it in h 1 m, and also to b c in d and e will lead to the completion of the horizontal projection of the solid.

To obtain the horizontal projection of a regular solid it is not always necessary to complete its vertical projection. The position and heights of its different parts being given will be sufficient, as

A heptagonal prism .25 inch high and 75 inch side has

one of its sides forming an angle of 20°, and is placed on a plane inclined at an angle of 60° with the horizon. The prism is surmounted by a heptagonal pyramid 1 inch high, whose base coincides exactly with the upper surface of the prism. Show its horizontal projection.

The heights a b .25 and c d 1.25 showing the position and combined height of the prism, and of the axis of the pyramid will suffice, as the inspection of the diagram will show ; to determine the horizontal projection required. (Fig. 36.)

Two sides of a triangle, respectively .75 and 1 inch long, meet on the horizontal plane, the first has its extremity at .25, and the second at .5 inch above the horizon. They are contained by a plane inclined to the horizon at an angle of 50°. Complete the triangle, and determine the real angle contained by the two lines as well as their horizontal projection. (Fig. 37.)

Construct the angle a b c, and draw any where the perpendicular de equal to .25 inch, and d f equal to .5 inch. Make f h equal to .75 inch and e g equal to 1 inch. The plane a b c being reduced to the horizon will contain the two lines f'h' and f' g, equal to fh and eg, the real angle between which has to be measured, and the horizontal projection of the triangle to be completed as usual.

A plane forming an angle of 42°, contains an isosceles triangle of 1.5 inches side, whose base of 1 inch is inclined to the horizon at an angle of 29°, show its projection. (Fig. 38.)

This exercise requires no explanation, being a mere repetition.

When the inclination of two sides of an object are given, to determine its horizontal projection and the inclination of the plane containing it, as,

Two adjacent sides of a square 1.25 inches side are respectively inclined to the horizon at angles of 34° and 20°.