C20 plane a20b2020 is a horizontal plane at the height 20 from the zero plane, and the point A, (5) being at the height 15, equal to LN, above this plane, any line joining A and the circumference of the circle will make an angle with the horizontal plane equal to the given angle LMN. The line joining A and c20, or a5 and and the horizontal line a2020 are both perpendicular to the horizontal line b2020, and consequently the angle a5c2020 is the angle of inclination of the plané a35b20c20 to the horizon, which is therefore equal to the given angle LMN. Dividing the horizontal line ca into 15 equal parts, we have the required scale of slope. Case 3.-The plane being given by a point in it, its inclination to the horizon and the direction of inclination. The inclination of the plane being given, to find its scale we have only to draw a right-angled triangle, having the angle at the base equal to the given inclination, and the perpendicular opposite equal to any number of units; the base being divided into the same number of units will give the divisions of the scale. Drawing, therefore, through the given point, a straight line in the given direction, and dividing it from that point at intervals equal to those in the base of the right-angled triangle, it will be the required scale of the plane. PROBLEM VII. The scale of slope of a plane being given, to find the index of a point in the plane when its projection is given. A straight line drawn through the projection of the given point, perpendicular to the given scale, will be the projection of a horizontal line passing through the given point in the plane; and therefore its intersection with the scale will give the index of the point. The converse of this problem, viz. to find the projection of a point in the plane when its index is given, is indeterminate, because every point in the horizontal line which is in the plane, and passes through the point in the scale denoted by the given index, has the same index. PROBLEM VIII. On a given plane, to draw a straight line to pass through a given point and to have a given inclination to the horizon, not greater than that of the given plane. From the given inclination, the ratio of the base to the perpendicular, in a right-angled triangle of which the hypothenuse has that inclination to the base, is given. Draw therefore on the given plane, whose scale of slope is 38, 41 (Plate X. fig. 17), a horizontal line having an index differing by one, or any number of units from that of the given point [387], and from the projection of that point as a centre, at the distance of the base of the corresponding right-angled triangle, describe a circle, cutting the projection of the horizontal line [39.7, 397] the straight line [38-7, 397] joining the projection of the given point and either of the intersections will be the projection of a straight line in the plane, and which has the given inclination. PROBLEM IX. To find the intersection of two given planes. The general solution of this problem has already been given (13); but we have now to make some remarks on the character of the dihedral angle formed by the surfaces of the planes, and also to notice some particular cases of the problem. The intersections of pairs of horizontals, having the same index in each of the planes, determine the line of intersection of the planes (13). When the angle contained by the pair of horizontals of the higher level is included by the angle contained by the pair of the lower level, or the angular point of the higher is turned towards that of the lower, the angle formed by the surfaces is salient, or they form a ridge: when the angle contained by the horizontals of the higher level includes the angle contained by those of the lower, or the angular point of the higher is turned from that of the lower, the angle formed by the surfaces is re-entering, or they form a channel. Or when the higher divisions of the scales of the two planes approximate, the planes form a ridge; when the lower divisions approximate, the planes form a channel. Particular cases.-1. When one of the planes is horizontal, we have only to find on the scale of the inclined plane a point which has the same index as the horizontal plane: the horizontal straight line passing through this point will be the projection of the planes' intersection. 2. When the horizontal straight lines in the two planes are parallel, the intersection of the planes will be horizontal and parallel to these; and its projection must be a perpendicular to both scales, and pass through points in each which have the same index. By placing the scales of the two planes contiguous to each other, the coincidence of the points which have the same index may be immediately seen, at least approximately, and the intersection thus determined. The intersection of the two planes may, however, be readily determined by finding their intersections with a third auxiliary plane, since the point of meeting of the two intersections must be a point in the intersection of the two planes; and the intersection will be the perpendicular from this point on either of the scales. Let ab, cd, as divided (fig. 18), be the scales of the two planes. It is evident from inspection that the index of the horizontal line in the two planes which has a common index must be nearly 28-5. To find this index by construction, through any two points, as 24 and 27 in each of the scales, draw perpendiculars to the scales; these will be horizontals at the respective heights in the two planes. Draw two parallels ef, gh, to represent the horizontals at the heights 24 and 27 in an auxiliary plane, intersecting the corresponding horizontals in the two planes in f, i and h, k; join fh, ik, and produce them to meet in 7: fl is the intersection of the auxiliary plane with the plane whose scale is ab, and il its intersection with the plane whose scale is cd; and consequently the intersection of fl and il, is a point in the intersection of the two planes; drawing therefore the horizontal Im, perpendicular to both ab and cd, it is the intersection of the planes of which ab and cd are the scales. When the construction is correct, lm will cut the two scales in points having the same index. PROBLEM X. To find the intersection of a given straight line with a given plane. Let b27.430.7 (fig. 19) be the given straight line, and cd the scale of slope of the given plane. Through a30-7 and b27.4 draw the parallels ae, bf as the horizontals in a plane passing through the given line; and through the points 307, 27'4 of the given scale, draw the horizontals in the given plane, intersecting the corresponding horizontals in the auxiliary plane in the points e30-7, f27-4; the line joining these points is the intersection of the auxiliary plane with the given plane. The point 929-3, in which the given line cuts this intersection, is therefore the intersection of that line with the given plane. PROBLEM XI. Through a given point, to draw a plane parallel to a given plane. This problem is immediately reduced to Case 3, Prob. VI.; for if through the given point a plane be drawn having the same inclination to the horizon as the given plane, and in the same direction, it will be parallel to that plane. The scale, however, of the required plane may be more readily determined from the consideration, that, since the planes are parallel, the difference between the indices of points in them which have the same projection must be everywhere the same; and consequently the scales of the two planes will be the same as regards the intervals of their divisions, but commencing at different indices, as in figure 20, where 22.4 is the index of the given point, and 19, 23 is the scale of the plane passing through it, parallel to the plane whose scale is 37, 40. PROBLEM XII. Through two given straight lines, to draw planes parallel to each other. 4°33'1 €33.7 If through a point in one of the given straight lines a straight line be drawn parallel to the other, this last line will be parallel to the plane passing through the intersecting lines (Prop. 21); and will therefore be wholly contained in a plane passing through any point in it, and parallel to the former plane. On these principles we have the construction in figure 21. Through the point a35-4 in one of the given straight lines d35-4031, a straight line a35-47 is drawn parallel to the other straight line c21.219.5 (Prob. V.), 35-4-337 being equal to 21.2-19.5, and the projections of the lines being equal. Finding the point f, in the straight line a95-40991, whose index is 33.7 (Prob. I.), the straight line f3-737 is a horizontal in the plane of the intersecting lines; and its scale of slope being perpendicular to this, is readily constructed. The scale of the plane parallel to this plane, and passing through a point in the line c21-2d19.5, and therefore through the line, is determined by Problem XI. PROBLEM XIII. From a given point, to let fall a perpendicular on a given plane. The horizontal line in the plane, passing through the foot of the perpendicular, being at right angles to the perpendicular, and also to the straight line of greatest inclination in the plane, the plane passing through the perpendicular and the line of greatest inclination is at right angles to the horizontal straight line (Prop. 4), that is, it is vertical; and it is therefore the vertical projecting plane both of the perpendicular to the given plane and of a straight line of greatest inclination in it. The projection of the perpendicular, therefore, coinciding with the direction of the scale of slope of the plane, if besides the given point, the index of a second point in it be found, the perpendicular will be determined. We will first consider the perpendicular to the given plane as it exists in the vertical plane passing through it, and cutting the given plane in a straight line of greatest inclination. Let A (fig. 22) be the given point, and BC the intersection of the vertical plane passing through the perpendicular from A to the given plane, with this plane. Let AH be a horizontal straight line cutting BC in H. Take HE equal to any number of units on the scale of the plan, and draw EF perpendicular to AH: make AG equal to EF, and draw GI perpendicular to AH. Then since the triangles AGI and FEH are equiangular, and AG is equal to FE, GI is equal to HE; that is, if AG be taken equal to the excess of the index of the point H or the point A above the index of the point F in the plane, the excess of the index of the given point A above that of the point I will be the number of units in HE on the scale of the plan. This principle gives the following very simple method of determining the index of a second point in the perpendicular to the given plane. Let a be the given point, 40, 80 the scale of slope of the given plane, and 0, 50 the scale of the plan. The projection of the perpendicular to the plane is al parallel to the scale of slope. In this scale find the point h whose index is 78, that of a78; take hf equal to any number of units, 20, on the scale of the plan; and find the index of the point f, 64-67, by means of the scale of slope. Take ai, on the projection of the perpendicular, equal to 78-64.67 or 13.33 on the scale of the plan: then from the foregoing principle, the index of the point i is 78-20=58, the index of a diminished by the number of units (on the scale of the plan) in hf. It is evident that it is unnecessary to find the point h, on the scale of slope, having the same index as the given point a, since the distance equal to the assumed number of units on the scale of the plan being measured from any other point in the scale of slope, the difference in the indices of the two points will be the same. Two points being found in the perpendicular, we may find any number of points in it (Prob. I.), and construct its scale. We may also find its intersection with the given plane (Prob. X.), and the true length of this perpendicular (Prob. II.), or the distance of the given point from the plane. When the given point is in the plane, a second point in the perpendicular is determined on precisely the same principle. When the scale of slope of the plane is not given, it is not necessary to construct it; all that is required for the determination of a second point in the perpendicular being the projection of two horizontals in the plane. Thus let 53, 53 and 65, 65 (fig. 23) be two given horizontals in the plane, and a4 the given point. The projection of the perpendicular to the plane is in the direction al perpendicular to the horizontals. The distance between these is 18 on the scale of the plan, and ai being taken on that scale equal to 12 (65—53), the index of i is 84-18 or 66. PROBLEM XIV. Through a given point, to draw a plane perpendicular to a given straight line. The given straight line being perpendicular to the required plane, the scale of slope of this plane must be in the direction of the projection of the line or parallel to it. Drawing, therefore, for the scale of slope of the required plane, a straight line parallel to the projection of the given straight line, and passing through the projection of the given point, and taking this point as a point deter |