intersections will be an approximate representation of the surface itself. It is evident that the greater the number of touching planes, the more nearly will they coincide with the curved surface, and the more correctly will the projection of their intersections with the equidistant horizontal planes represent the surface on the plan. The projections therefore of the intersections of the surface itself with the equidistant horizontal planes will best represent that surface, dispensing with the touching planes, except where it may be necessary to determine the scale of slope on a limited portion of the surface. As the scale of slope on a plane is a perpendicular to the projection of a horizontal straight line in the plane, the scale of slope at any point on a surface is a perpendicular at that point to the projection of the horizontal line passing through the point. Figure 9 thus represents a curved surface to which the six planes represented in figure 8 may be considered an approximation; and figure 10 represents a more complex surface having flexures in various directions, with some of the lines of greatest slope in directions approximately perpendicular to the projections of the curved intersections of the surface with the equidistant horizontal planes. Such a plan may represent a portion of country, without other details of the surface being shown on it, and immediately presents an idea of the general nature of that surface as regards its slopes, the greater or less steepness of which is shown by the approach of the horizontal curves or their receding from each other. 16. The horizontal curves which thus represent a country are called the horizontal Contours of the country, and a plan or map in which these contours are represented is called a Contoured Plan or Map. The advantages of such plans and maps, whether for the purposes of military or civil engineering, are incalculable. They at once speak to the eye, conveying a knowledge of the most important feature of the country as regards the operations in both those branches of service, and, besides, they furnish data for the details of these operations. It is not here a question how materials are to be obtained for the construction of such plans, but we may remark that they must be furnished by the operations of leveling, whether by actually tracing the levels at the required intervals of height, or by any other mode which may be more convenient. 17. The principles of this method of representation as applied to points, lines, whether straight or curved, and surfaces, whether plane or curved, having been explained, we proceed to some problems which most commonly occur in the application of the method. PROBLEM I. A straight line AB being given by its projection ab, and the indices a, ẞ of two points in it whose projections are a and b :— 1. To find the index of another point C in it whose projection is c; 2. To find the projection of a point in it whose index y is given. 1. It is evident that if, in the vertical projecting plane of AB, a straight line be drawn through B, supposed to be the lower point, parallel to the horizontal projection ab, this line, with the vertical projecting lines of A and C, will form two similar right-angled triangles of which the bases are equal to ba and be, and the perpendiculars to the excess of A's index above B's, and the excess of C's above B's. Geometrically, the excess of C's index above B's, is immediately obtained by constructing these right-angled triangles according to the scale of the plan. Arithmetically, C's index is thus obtained. a, ß being the indices of A and B, let y be that of C; then by the similar triangles Y-B bc bc ñá 2. When the index y of the point C is given, to find its projection, we have Geometrically, the distance bc is the base of a right-angled triangle of which the perpendicular is y-ß, the excess of the given index of C above that of B, and which is similar to the right-angled triangle of which the base is ab, and perpendicular is a-ß, the excess of A's index above B's. Arithmetically, we have by similar triangles— In the solution of the problem, the point C has been assumed between A and B, but when the point is beyond A or B, the solution is evidently obtained on the same principle. When the straight line AB is horizontal, the index of every point in it is the same, and consequently when the projection of the point C is given, the point is given, since its index is so; but the projection of a point in the line cannot be determined from its index. When AB is vertical, then the projection of every point in it being the same point, any point in the line is only known by its index. PROBLEM II. A straight line AB being given by its projection and the indices of the points A, B, to find its true length and its inclination to the horizon. The true length of the line AB is evidently the hypothenuse of a right-angled triangle of which the base is the projection of AB, and its perpendicular is the difference of the indices of A and B; whence it is immediately found, either Geometrically or Arithmetically, whichever may be most convenient. The inclination of the straight line to the horizontal plane is the angle which it makes with its projection, and this angle is the angle at the base of the above right-angled triangle. If the slope of the straight line, as that of a plane expressed in its scale of slope, be required, this slope is expressed by a fraction whose numerator is the difference of the indices of A and B, and the denominator is the projection of AB. If the converse of the former part of this problem were proposed, namely, a true length in a straight line, given by its projection and the indices of two points in it, being given, to find its projection; or, which amounts to the same, from a given indefinite straight line to cut off a given length; we should then have to find the base of a right-angled triangle similar to that which determines the scale of slope of the line, and the hypothenuse of which is the given length to be cut off. Arithmetically, we should have to find a fourth proportional to, the true distance between two points whose projections are given, the distance between the projections of those points, and the given length to be cut off. PROBLEM III. Through a given point in a given vertical plane, to draw a straight line that shall have a given inclination to the horizon. The projection of the required straight line will coincide with the straight line which is the projection of the given vertical plane ; and the projection of the given point being in the same straight line, if another point in it be assumed as the projection of a second point through which the line passes, its index will readily be found in the following manner. Let mn (fig. 11) be the projection of the given vertical plane, a the projection of the given point A, and BAC the given angle. In mn take another point b; make AC equal to ab, and draw CB perpendicular to AC. Let the index of a equal a, that of b equal ß, and suppose BC equal y; then if ẞ=a+y, the straight line passing through the points A and B whose projections are a and b, and indices 2 and 6 will make the given angle BAC with the horizon. PROBLEM IV. To determine the intersection of two given straight lines AB and CD in space, in the case of their meeting: that is, aa b and c,ds being two given straight lines, it is required to find the projection and the index of their point of intersection, in the case of their meeting. The projection of the point of intersection of the two straight lines being the point of intersection e of their projections ab, cd (fig. 12), it is only necessary to determine (Prob. I.) the index of the point e in either of the lines abs or cds. Thus being the index of the point of intersection, we have If the lines intersect, it is evident that the index of the point e must be the same by whichever of the two given straight lines it is determined. When the two given straight lines are in the same vertical plane, their projections coincide, and they necessarily intersect, except in the case of their being parallel. In order to determine the projection and index of the point of intersection, it is only necessary to draw the straight lines as they exist in the vertical plane. Let AB, CD (fig. 13) be the two given straight lines, of which the projections are ab and cd, and the indices of their extremities are Aa=a, Bb ß, Cc=y, Dd=8; then the projection and index of their intersection is thus found: Geometrically. By measuring ae and eE on the scale of the plan, the values of these lines will be obtained to the degree of accuracy generally required. Arithmetically. Drawing AF and DG parallel to ab, we have Putting ab=b, ac=c, ad=d, AH=ae=x, and HE=y, these proportions become x= XC b d-x d C = From these equations we obtain ( y −8) d + (d—a) (d—c) .b; y=· (y-8)b+ (Ba) (d-c) (v−8)d+ (8—a)(d—c) (B-a). Substituting in these the numerical values of the letters, we have those of aex, and Ee=a+y. When the straight lines are both horizontal, they can only intersect when they have the same index. In this case, the projection of their intersection will be the intersection of their projections, and its index will be the common index of the straight lines. PROBLEM V. Through a given point A whose projection is a, and index a, to draw a straight line parallel to a given straight line BC whose projection is bc (fig. 14), and the indices of two points B, C in it are ß, y. The projection of the line required will be parallel to that of the given line (Props. 35, 27); therefore through a draw ad parallel to be, as the projection of the required straight line. Join ba, and through c draw cd parallel to ba; if D be the point whose projection is d, and whose index & is a+y-6, the straight line AD in space will be parallel to the given straight line BC. For AD and BC are in parallel vertical planes, and they make equal angles with horizontal lines in those planes, because ad is equal to be, and d-a is equal to y-ẞ. PROBLEM VI. A plane being known by sufficient data, to determine its scale of slope. Case 1.-The plane being given by three points in it, not in a straight line. Let a1 b47 C32 (fig. 15) be the projections of the three points A, B, C in the plane, with their indices. Join ab, and in it find the point d which is the projection of the point in AB, having the same index 320 as the third point C (Prob. I.); and join cd, which is therefore the projection of a horizontal straight line in the plane. Since cd is the projection of a horizontal in the plane, drawing af parallel to cd, it is also the projection of a horizontal in the plane. Drawing any line ef perpendicular to cd, this will be the projection of a line in the plane in the direction of the greatest inclination, or in the direction of the scale of slope (9, 10), and e, f will be the projections of points in the plane at the heights 32 and 11 from the zero plane. Dividing, therefore, the distance fe into 21 equal parts, the several points will be the projections of points in the plane at intervals of an unit in level: and thus the scale of slope of the given plane is constructed. Case 2.-The plane being given by two points in it, and its angle of inclination to the horizon; or, which is the same problem, to determine the scale of slope of a plane that shall pass through two given points, and have a given inclination to the horizon. Let the given points in the plane be a35, b20 (fig. 16), and LMN the given angle of inclination of the plane, which must not be less than the inclination of the straight line a35b20 to the horizon. Draw MP perpendicular to MN, and equal to fifteen on the scale of the plan, the difference of the indices of the points A and B; draw PL parallel to MN, and LN to MP. From the centre a at the distance MN describe a circle; and from b draw be touching it in c. The |