dc'; draw kl perpendicular to da, and from k as a centre with radius kl, describe the arc 1m: join mg, m h, and the triangle g mh shows the dihedral angle required, reduced to the horizon.

The student will see the analogy between the latter part of this problem and the preceding one.

33. The traces pqr of a plane being given, to find the angles which the plane makes with the two planes of projection, and also the angle between the traces. (Fig. 40.)

1st. Find on the vertical plane the angle the plane makes with the horizontal plane.

Draw b c' perpendicular to x y and b a perpendicular to pq; from centre b, with radius ba, describe the arc a v, join v c' and the angle c' vb is the angle required.

2nd. To find the same angle upon the horizontal plane. Perpendicular to a b draw b c equal to b c', and join a c : cab will be the angle required.

To find on the vertical plane the angle which the plane makes with the vertical plane, draw be perpendicular to xy and b d perpendicular to qr, make bt equal to be, and t db will be the required angle. To find the same upon the horizontal plane: on xy make b z equal to bd, zeb will be the angle required.

To find the angle between the traces.

As the hypothenuse a c, to the right angle triangle cb a, determines the length of the plane ac' q, if we suppose that plane to revolve on a q till it is reduced to the horizon and developed in a a" q, the angle aq a" will give the angle of the traces.

34. To find the angle contained between a plane and a straight line. (Fig. 41.)

Let a bb' be the traces of the plane, and cd, c' d' the projections of the line. If from any point on the line a perpendicular is drawn to the plane, the angle contained

by the given line and that perpendicular will be the complement of the required angle; therefore, from any point a a' on the line draw cc' perpendicular to x y, and draw ce c'e' perpendicular to a b and a' b', construct the angle between the two lines, which will be def; draw f k perpendicular to fe, and the required angle will be dfk.

35. To construct the angle contained by two straight lines, intersecting in space, and whose given projections are a b c and a' b' c'. (Fig. 42.)

Produce a' b' and a' c' to d' and e' in x y, and draw d d e e' perpendicular to x y. Produce likewise a b a c to d and e, and join d e. Draw a a' perpendicular through xy and h g, perpendicular to d e. Make f k equal to a h', and join a' k: make hg equal to k a', and join d g, ge. The angle dg e is the angle required, reduced to the horizon.

36. To reduce a given angle to the horizon, having given the angles which its sides make with it. (Fig. 43.)

Let a represent the horizontal projection of the vertex, ab the horizontal projection of one of the sides, the projection of the side a e is required.

Draw indefinitely the perpendicular a a', and make the angle a b d equal to that of the first side (say 40°.)

Draw dc, so that the angle a c d may equal that made by the second side with the horizon, (say 55°.) From a, with radius a c, describe indefinitely the arc cf, and make cdg equal to the given vertical angle (say 42°.) Draw bg perpendicular to x y. From d, with radius d c, describe the arc c g.

From bas centre, with radius b g, determine that distance in e, intersecting the arc cf. Join a e: a e will determine the horizontal projection of the side required, and the angle a e b will be equal to the observed angle reduced. to the horizon.

HORIZONTAL PROJECTIONS.

Topographic, horizontal, or, as it is sometimes termed, single projection, is a sister branch of Descriptive Geometry, which enables us to determine and ascertain by means of numbers or indices on one single ground plan, the position, shape, and altitude of the objects it contains. That method is very convenient for civil and military engineering purposes, and most especially applicable to military topography, whether for illustrating the accidents and irregularities of large tracts of country, or for determining the position, shape, and relief of military works, as well as the importance they bear to each other: for that method enables us, with the help of Descriptive Geometry, to construct elevations, sections, &c., of the planes containing these works, either for the purpose of defilading them, or for other operations.

Indices are numbers refering to the altitudes of points above the horizontal plane. That plane is usually assumed to be on a level with points (0 or zero) below the lowest parts of the plane. Therefore on a plan, a point whose index is 30, signifies that it is 30 units above the Zero plane, or lowest plane of level.

In some instances military men find it more convenient to assume a plane parallel to the horizon, and passing at some distance above the highest point of ground shown on the plan (usually 10 yards). In this case the indices show a decrease of height below that plane instead of an increase above the Zero plane. In the following problems we will, however, follow the first method.

As the position of points are determined by their projections, the letters, numerals, or indices which accompany them denote also their height. Their notation follow the rules already given.

In a similar manner, the position of a straight line will be determined by the projection and indices of its two extremities. (Fig. 1.)

Let, for instance, a 3 and 6 6 represent the two extremities of the line a b, it is clear that the point b will be 3 units higher than the point a, that being the difference of height between a and b above the horizontal plane. Erect, therefore, at b the perpendicular 6 c, equal in height to the difference of units between the indices 3 and 6, and join a c, which will determine the slope or inclination of the line, b c will be the scale of the line, obtained according to the conventional scale of the drawing, and a b will be the scale of slope or inclination. The divisions on the scale of slope or inclination, and those on the slope, or inclination, are always determined by those on the scale of the line. To complete the scale, through 4 and 5, draw the horizontals 4-4' and 5-5' parallel to a b, then draw the perpendiculars 4' 4" and 5' 5".

Now, suppose the triangle a b c to revolve on a b, as on an axis, until c is perpendicular to the plane of the paper. The real position of the line a b will be determined.

N.B. When scales of slope or inclination refer to lines, they are represented by single straight lines. When they refer to planes they are shown by double lines. (Fig. 2.)

As one index determines the position of a point, two indices that of a line, we will find that three indices can determine the position of a plane.

A plane is represented by its trace on the horizontal plane of comparison, and by its inclination to that plane.

As the scale of the plane is that which determines its angle of inclination by its altitude between two given points, a plane is likewise determined by a vertical trace showing its angle of slope or declivity.

K

The comparative altitudes of the several parts of a plane are often shown by horizontal contours.

Contours are equidistant horizontal planes supposed to pass through the object, parallel to the plane of comparison, and showing the respective height of the different parts of the object above the horizon.

Let a 3, b 15 and c 12, not in the same straight line, determine the position and altitude of the several points of a plane above the plane of comparison. (Fig. 2.)

Join a b, the two extreme indices, and divide that distance into as many units as are contained in the difference between a and b (12) from c 12, draw a straight line, cutting a b at the point similarly indicated on it; c 12 and c' 12 becoming therefore points on the same level. These will necessarily indicate the position of one horizontal or contour of the plane containing them: to complete the plane, through 5, 6, 7, 8, 9, 10, 11, 13, 14 and 15, draw lines parallel to c 12, c' 12, and the direction of the levels of the plane will be defined. To find the scale of inclination of that plane draw any where the line de perpendicular to c 12, c' 12, and make e ƒ perpendicular to d e, and equal in altitude to the difference of units (by conventional scale) between the indices a and b. Join df, and the angle e df will be the profile angle or angle of greatest inclination of the plane a bc: ef will become the scale of the plane, and de its scale of inclination. If we conceive the triangle def to revolve on d e till it stands perpendicularly to the plane of the plan, and another plane intersecting them perpendicularly through the diagonal d e, and the horizontals of the plan transferred to them, we will obtain a clear conception of the position of the inclined plane.

As a plane can be determined by three points, the indices of a triangle will evidently determine likewise its scale and inclination.