49. The inclination of a plane is sometimes expressed by a fraction, the numerator of which represents the difference of level between any two points, while the denominator represents the horizontal distance between the same points. In any right-angled triangle, the perpendicular represents the numerator, and the base the denominator of this fraction. Thus, in Fig. 24, if a b = b b', i.e., if the base equal the height, the fraction is, and the angle of inclination is 45°. Again, if a b 3, and bb' 1, the fraction is, that is to say, for every unit of level, the = = horizontals are 3 units apart. In the same way, a plane of will have its horizontals 3.5 units apart for every unit of level, and so on, The student will now understand the solution of the following problem. PROBLEM 36. Two pickets with their upper extremities on the same level and 7 feet apart, stand vertically out of the side of a hill, 6 and 8 feet respectively. Determine the slope of the hill, and give the fraction which represents it, scale -inch to 1 ft. с Fig. 59. 19 Construct the right-angled triangle a 3 c, making a b, 7 feet (7 times inch); and a c, 2 feet (2 times 3-inch); c b represents the slope of the hill. Now, since a b = 7, and a c = 2, the fraction is, the horizontals being 3.5 units apart for every unit of level. The drawing is completed by making c d, 6 times, and b g, 8 times 1-inch. 50. If the plane of a circle is horizontal, its plan will be K a plane inclined at 52° to contain a line inclined at 36°, and to be parallel to another line inclined at 32°. For the principle involved in the solution of this question, see Prob. 22. PROBLEM 29. Required the section of a road given by its contours. Def. In Geometrical Drawings, sections are employed to furnish those details which cannot be represented by a plan or elevation merely, and are designated sectional plans and sectional elevations. When an object is cut by a vertical plane, i.e., by a plane at right angles to the plane in which the object is situated, the section is termed a profile. Let the contours be a b, c d, ef, etc., taken at equal verti be, make cd, 1 inch (that being the length of 6 feet when drawn to the scale of), e ƒ', '2 inch, and so on to m n', 4 inch. Through the points n', h', f', d', and a, draw the curve line as shown in the figure. This line will represent the sectional elevation of the given road. In representing objects in section, it is usual to show the part sectioned by a series of equi-distant parallel lines inclined at 45° to the base line. The direction in which the section is taken, is indicated by a line (A B in the drawing), called the section line or section plane. Since, in the case before us, A B is at right angles to the horizontals, the section is a profile, and it may be remarked, that a section so taken is the only one which shows the real lengths of the horizontal distances. Obs. Supposing the plan to be drawn to a scale of 1 mile to 1 inch, the distances between the horizontals will be 352, 880, 704, and 528 yards respectively. PROBLEM 30. The observed angle B A C between two lines B A, a C, whose inclinations to the horizon are 32° and 37° respectively, is 40°; required the horizontal projection of the angle в A C. Obs. 1°. In the operations of surveying, it is usual to consider all prominent objects to be connected by right lines forming triangles, the problem then consisting mainly in measuring the angles of these triangles. In representing the survey upon paper, that is to say, in making a map of a country of which we have made a survey, we have to reproduce these triangles upon a reduced scale, in the same order as that in which they were observed. To do this, the angles of the triangles should be situated in a horizontal plane. If the plane of an angle, that is, if the plane containing the lines forming an angle, is inclined to the horizon, we have to find its horizontal projection, but not the angle itself (see Prob. 23, Fig. 42). Now, this projection can be found if we know the measured magnitude of the angle, and the inclinations of its two sides to the horizon. Solution. Let A, A', be the horizontal and vertical projec of the other side. If from A as a centre, and with radius A C, we describe an arc C D E, the side A'c must meet the horizontal plane at some point in this arc. To determine its position, we must find the distance between two points in the lines containing the observed angle. We must, therefore, construct this angle. To do this, draw from A', Aa, making with A' B an angle a A B of 40°. Next, make ▲'a equal to a' c, and join в a; в a is the distance sought. From centre B, and with radius в a, describe an arc cutting BDE in D, and join a D. The line A D is the horizontal projection of the second side, and the angle BAD the horizontal angle required. Obs. 2°. The operation performed in the present problem is known by the name of the "reduction of an angle to the horizon." PROBLEM 31. To determine a line which shall be perpendicular to two lines not parallel, nor lying in the same plane. Fig. 54. B me a L Let a b, c d be the plans of the given lines. Draw the plane mm, m m' to contain one of the lines as cd. Now, this plane must be parallel to the line a b. Therefore, draw a b', the elevation of a b, parallel to mm. In a b take any point g, and find its elevation, g. From g', draw go at right angles to m m. Find go, the plan of go. Since g'o is at right angles to the plane mm, m m', its plan go' will be at right angles to the horizontals of the plane. parallel to a b, meeting c d in n. Next, from o', draw on to go', meeting a bin p; n p is perpendicular to a b, c d. B a Fig. 55. (a) Let a b, c d, the plans of the given lines, be parallel. Find a b', cd, the elevations of a b, c d. The point h in which these lines intersect each other, is the elevation of the required line; its plan is h' i. |