B Let A and B be the two given points in the untraced line; P the given point through which a straight line is required to be drawn 4 parallel to the straight line joining A and B. P Join PB, B being the more distant from P of the two given points, and divide BP in any convenient ratio in the point C. Join AC, and produce it to D, so that AC CD :: BC: CP, and join PD. Then PD is the straight line required. : For, the triangles ACB, PCB have ‹ ACB = < PCD (31), and the sides about these equal angles proportionals, .. the triangles are similar (71 Cor. 1), and .. equiangular; .. < ABC= ‹ BPD, and PD is parallel to AB (34). Or, if it be more convenient to bisect PB in C, CD must be taken equal to AC, and PD will be parallel to AB, as before. ARCHITECTURAL MOULDINGS, ARCHES, &c. 197. The profile of all Architectural mouldings is made up of straight lines and arcs of circles*; and this profile traced upon paper is called the 'working drawing', because by it the workman executes the design of the Architect. It is important therefore that these working drawings' be constructed upon right principles, whatever those principles may be; and although it is no business of ours, as Geometricians, to discuss the fundamental principles of Architecture, yet there are two rules of construction so often observed, (with however many exceptions) that it seems desirable to point them out in the following examples both of Mouldings and Arches. These rules are, I. When an arc of a circle runs into a straight line, at a certain point, so that the straight line is, as it were, a continuation of the arc, that straight line should be a tangent to the circle at that point. And * For no other reason probably than the comparative ease with which straight lines and circles are put to use in constructive art. II. When an arc of one circle runs into an arc of another, the two arcs should have the same tangent at the point of junction. It follows as a consequence of Rule II, that the centres of the circles and the point of junction will be in the same straight line (143). As a simple instance of the former Rule either of the following mouldings is of common occurrence:— A B B D in both which ‹ BAC= ‹ ACD = a right angle, and upon AC as diameter a semi-circle is described; .. AB is a tangent to the circle at A, and CD at C. 198. There are other simple mouldings for which a quadrant is described instead of a semicircle, such as the following: The construction is obvious. E is the centre of the circle of which AEC is a quadrant. AB touches the quadrant at A; and the arc AC commences, or terminates, at C with its tangent at right angles to CD, thus distinctly repudiating, as it were, a continuation of the curve in that direction. Observe how the rule in question is carried out in the annexed compound moulding: The moulding begins and ends at right angles to the horizontal line, at A and C. It consists of several distinct parts- but observe how each of these parts is tangential with those next to it. May it not be, that this neighbourly union of the parts is that, which gives fitness and repose, and therefore, beauty to the whole? Let Architects decide. It is further to be observed that in this example the centres of the three circles are in one vertical straight line. 199. The next class of mouldings requires two quadrants, so joined together that they have a common tangent at the point of junction. The quadrants may be equal as in figs. (1) and (2), or unequal as in figs. (3) and (4). A is the point of junction of the arcs; and OAo is a straight line. Hence the tangents to the two arcs at A coincide, being at right angles to the same straight line OAo (143). The mode here shewn of joining the arcs of two different quadrants so as to produce one continuous curve is of frequent application in the Arts. The object is so to join them, that they may appear to belong to each other; and unless this be done by making them have the same tangent at the point of junction, the compound curve will appear broken at that point, and consequently offensive. It is true, that Mouldings are frequently met with in which the circular arcs meet the horizontal lines neither as tangents nor at right angles, each arc being less than a quadrant; but even in such cases Rule II. is always observed in good examples, that is, the arcs meet each other tangentially. The annexed diagram represents a case of this kind. BC is a straight line bisected in A. Upon AB, AC two equilateral triangles AOB, AoC are constructed. Then the arc AB is described from centre O, and arc AC from centre o. And because ▲ BAO = one-third of two right angles CAO, .. OAo is a straight line (31); and it passes through the two centres and the point of junction A, .. the arcs touch each other at A. ARCHES are of various kinds, and serve various purposes, which it does not fall within our province to explain. But we will proceed to describe the working drawings' from which the Arches in most common use are constructed. 200. PROP. XCVIII. To draw a semi-circular arch of given span. e E d D Take the straight line AB to represent the given span. Bisect it in C; and with centre C, and radius AC, describe a semi-circle ADB. Increase this radius by Aa, or Bb, the depth of the arch-stones, and describe another semi-circle adb from the same centre. Divide the inner semi-circumference into an odd* number of equal parts, according to the number of stones which are to form the arch; and join the several points of division with the centre C: produce these lines to the outer circumference, and they will determine the joints of all the arch-stones. Aa, Ee being two contiguous joints thus ад B b Odd, because on statical grounds it is not well to have any joint vertical, as would happen, if an even number were chosen. PART II. 6 determined, AEea is the model for each one of the archstones, which are equal and similar in all respects. OBS. In practice it is not necessary that the outer boundary, adb, be strictly semi-circular. Provided Aa, AE, and Ee are constructed as above, the actual boundary beyond ae may be of any form we choose. If this arch be placed so that AB is horizontal, and the piers on which it rests are vertical, it obeys Rule I, (197), the straight lines which continue the arch downwards from A and B being tangents to the circle at those points. 201. PROP. XCIX. To draw a segment-arch of given span and rise. e d D E A Let AB be the given span; and CD, drawn at right angles to AB from its middle point, the given rise. Find the centre of the circle which passes through the three given points A, B, D. With this centre, and radius OA, describe the arc ADB. Join OA, OB, and produce them to a, b, a making Aa = Bb = depth of arch-stones. With the same centre, and radius Oa, describe the arc adb. Divide ADB into an odd number of equal parts; and join the several points of division with the centre O. Produce each of these lines to meet the outer arc, as OE to e, and they will determine the joints of all the arch-stones. Aa, Ee, being two contiguous joints thus determined, AEea is the model for each one of the arch-stones, which are equal and similar in all respects. OBS. The first observation appended to Prop. XCVIII. applies here also; but not so the second. When this arch is connected with vertical piers or abutments, it will appear broken at A and B, because the vertical lines through those points will not be tangents to the circle. Its advantage over the semi-circular arch is obviously that of giving the same span with a much less rise, which is a great gain often in the case of bridges and other similar constructions having roads or canals over |