This table shows that, in practice, for surmounted arches, the limit x= √2p, or the thickness obtained for the construction by graphical means is more than sufficient, since it gives results greater than those that the experiments require, excepting only in the cassinoid; but even in the case of that curve the graphical construction comes nearer to experiment than the result of the first formula. 1405. It is moreover to be observed, that the pointed is the most advantageous form for surmounted arches composed of arcs of circles. We have had occasion to speak, in our First Book, of the boldness and elegance exhibited in this species of arches by the architects of the twelfth and thirteenth centuries; we shall merely add in this place that where roofs are required to be fire-proof, there is no form so advantageously capable of adoption as the pointed arch, nor one in which solidity and economy are so much united. 1406. Next to the pointed arch for such purpose comes the catenary (the graphical method of describing which will be found under its head, in the Glossary at the end of the work), and this is more especially useful when we consider that the voussoirs may all be of equal thickness. Application of the Method to surbased Arches, or those whose Rise is less than the H.'f Spun. F M 1407. For the purpose of arriving at just conclusions relative to surbased arches, three models were made of the same thicknesses and diameters, with a rise of 35 lines, and in form elliptical, cassinoidal, and cycloidal. We however do not think it necessary, from the similarity of application of the rules, to give more than one example, which is that of a semi-ellipse (fig. 577.), in which, as before, the formula is IIr E The lines described in the foregoing examples being drawn, we have TI, represented by d, being 24·84, we have 2pd b, which expresses the sum of the vertical efforts m + n (39.5 × 9) a, being always 120, 355'5 == is a 120 16543-44 131-94 2242-94 +8·76-2·96=25-22 lines, or a little less than 25 lines. 1408. In the model it was found that a thickness of 26 lines was necessary for the pier, and the lower voussoirs were connected with it by a cementing medium. Without which precaution the thickness of a pier required was little more than one tenth of the opening. Taking the square root of double the thrust, that is, of 666, we have 25.81, about the same dimension that the graphical construction gives. The experiments, as well as the application of the rules, require the following remarks for the use of the practical architect. 1409. I. The cassinoid, of the three curves just mentioned, is that which includes the greatest area, but it causes the greatest thrust. When the distance between the intrados and the extrados is equal in all parts, it will only stand, supposing the piers immoveable, as long as its thickness is less than one ninth part of the opening 1410. II. The cycloid, which includes the smallest area, exerts the least thrust, but it can be usefully employed only when the proportion of the width to the height is as 22 to 7 in surbased arches, and in surmounted arches as 14 to 11. The smallest thickness with which these arches can be executed, so as to be capable of standing of themselves, is a little more than one eighteenth of the opening, as in the case of semicircular arches. 1411. III. The ellipsis, whose curvature is a mean between the first and second, serves equally well for all conditions of height, though it exerts more thrust than the last-mentioned and less than the cassinoid. 1412. It is here necessary to remark, that too thin an arch, whose voussoirs are equal in depth, may fall, even supposing the abutments immoveable, and especially when surbased; because, when once the parts are displaced, the force of the superior parts may lift up the lower parts without disturbing the abutments. Raking Arches. 1413. Let ACA' (fig. 578.) be the model of a raking arch of the same diameter and thickness as the preceding example, the voussoirs of equal thickness, and the piers of different heights, the lowest being 10 inches or 120 lines in height, and the other 14 inches or 174 lines. The tangent at the summit is supposed parallel to the raking lines that connect the springing. 1414. This arch being composed of two different ones, the mean circumference on each must be traced, and each has its separate set of lines, as in the preceding examples; E the horizontal line KL of the smaller arch is produced to meet the mean circumference of the other in S, and the interior line of its pier in g. n 1415. The part KLS represents the horizontal force of the part of the arch KGS, common to the two semi-arches ; so that if a joint be supposed at S, the part LK represents the effort acting against the lower part of the smaller arch, and LS that against the lower part of the larger arch. These parts resist the respective efforts as follows: the small arch with the force represented by iK, and the greater one with the force represented by gS. But as gS Fig. 578. is greater than LS, transfer LS from g to f to obtain the difference fS, which will show how much LS must be increased to resist the effort of the larger half arch; that is, the effort of the smaller one should be equal to Lf; but as this last requires for sustaining itself that the larger one should act against it with an effort equal to KL, this will be the difference of the opposite effort, which causes the thrust against the lower part of the smaller arch and the pier from whence it springs. Hence, transferring fL from L to q, taking the half of iq and transferring it from L to h, the part hK multiplied by the thickness AB will be the expression for the thrust represented by p in the formula x= 2p+2pd-2mc + 62 b 42 Having found hK=30 and AB=9, we have for the value of p 30 × 9=2741, and for that of 2p-549, d which represents IT, being 291, 2pd=161954. In 2me; m, which represents MK x AB, will be 124 x 9=111, and 2m-222. e, which represents iK, being 8, we have 2mc=222 × 8=1776. The height of the pier represented by a being 174, we have 2pd-2mc 16195-1776 174 The vertical effort represented by b, or TF x AB, will be 41 × 9=375, Substituting these values in the formula, we have √549 +82-81 +4.64-2·16=23-08 for the thickness of the greater pier from which the smaller semi-arch springs. For the half of the greater arch, having produced the horizontal line IK'L', make K'r equal to VL', and bisect rL' in t; the line K't represents the effort of the smaller against the greater arch, which resists it with a force shown by i'K'; thus making K'q' equal to 'K, the effort of the thrust will be indicated by q't x AB, whose value p in the formula will be z=√360+118 65+51 48-7-175-15-855 lines for the thickness of the smaller pier. BB Taking the square root of double the thrust, we should have for the larger pier 23-44 line and for the smaller one 19 lines. In the geometrical operation, for the larger pier mal Bu equal to hK and Bn equal to 2AB; then upon un as a diameter describe a semicirc cutting the horizontal line BA produced in E. BE will be the thickness of the pier, an will be found to be 23 lines. For the smaller pier make B'u' equal to q't and B'n' equ to 2 A'B'. Then the semicircumference described upon un as a diameter will give 19 lin for the thickness. 1416. By the experiments on the model 22 lines was found to be the thickness necessar for the larger pier, and 18 lines for the smaller one. Arch with a level Extrados. 1417. The model of arch fig. 579. is of the same opening as the last, but with a level extrados, serving as the floor of an upper story. The thickness of the keystone is 9 lines. To find the place of fracture or of the greatest effort; having raised from the point B the vertical BF till it meets the line of the extrados, draw the secant FO cutting the interior circumference at the point K, and through this point draw the horizontal IKL and the vertical HKM The part CDKF will be that which causes the thrust, and its effort is represented by 18.26×22 PSR Fig. 579. - 207.46. The The height of the pier, represented by a in the formula, = 183 The area of the upper voussoir FKCD=667·44; but as the load of the haunches is borne by the inferior voussoir, we must subtract the triangle FKH= remainder 459.98 multiplied by KL and divided by the are KD, that is, 422.24, represents the effort of the upper part. FBKHXIK KB " is 651-07 × 18.86 459-98 × 35-14 38.28 which becomes 1418. That of the lower part, represented by 263.67. The difference of the two efforts=158.57 will express the thrust or p of the formula, and we have 2p 317.14. 1419. The piers being supposed to be continued up to the line EC of the extrados will be greater than the arm of the lever of the thrust which acts at the point K. Thus the expression of the arm of the lever, instead of being a + d, as in the preceding examples, will be 2pr a-d, and the sign of must be changed. In numbers, a 1420. In the preceding examples, 2mc, which represented double the vertical effort of the superior voussoir multiplied by the arm of its lever, becomes nothing, because it is comprised in the addition made to the lower voussoir; so that the formula now is b, then, which always expresses the vertical effort of the half arch, is therefore Substituting these values in the last formula, we shall have r = √31914 — 38·12 +20·25 −4·5=12.88 lines. = Experiment gives 14 lines as the least thickness that can be relied on. To find the thickness by the geometrical method, make Km equal to IK and Bh equal to mL, Bn to double CD, and upon nh as a diameter describe the semicircumference cutting the horizontal line OB produced in A: then BA=174 lines is the thickness sought. 1421. Rondelet proves the preceding results by using the centres of gravity, and makes the result of the operation 12-74 instead of 12.80, as first found. But the difficulty of finding the centres of gravity of the different parts is troublesome; and with such a concurrence of results we do not think it necessary to enter into the detail of the opera tion. A different Application of the preceding Example. 1422. The model (fig. 580.) is an arch similar to that of the preeeding example, having a story above it formed by two walls, whose height is 100, and the whole covered by a timber roof. The object of the investigation is to ascertain what change may be made in the thickness of the piers which are strengthened in their resistance by the additional weight upon them. 1423. The simplest method of proceeding is to consider the upper walls as prolongations of the piers. 1424. In the model the walls were made of plaster, and their weight was thus reduced to of what they would have been if of the stone used for the models hitherto described. The roof weighed 12 ounces. We shall therefore have that 100, which in stone would Lave represented the weight of the walls, from the difference in weight of the plaster, reduced to 75. In respect of the roof, which weighed 12 ounces, having found by experiment that it was equal to an area of 576 lines of the stone, both being reduced to equal thicknesses, we have 12 ounces, equal to an area of 13.82 whose half 6 91 must be added to that of the vertical efforts represented by b in and Changing these terms into and the formula becomes The height of the piers or a in the formula = 183 +75=258. p does not change its value, therefore 2p (as in the preceding example) = 265 86. d, the difference between the height of the pier and the arm of the lever, will=75. In the model a thickness of 11 lines was found sufficient to resist the thrust, and taking the root of double the thrust the result is 13 lines. 1425. By the geometrical method, given in the last, taking from the result 174 lines, there found, the value of, that is, 5·58, the remainder 113 lines is the thickness sought. 1426. It may be here observed, that in carrying up the walls above, if they are set back from the vertical BF in hf, the model required their thickness to be only 6 lines, because this species of false bearing, if indeed it can be so called, increases the resistance of the piers. This was a practice constantly resorted to in Gothic architecture, as well as that of springing pointed arches from corbels, for the purpose of avoiding extra thickness in the walls or piers. The area KFCL of the upper part of the arch will be 1223.10, from which subtracti that of the triangle FKG, which is 590-82, the remainder 832-28 being multiplied 1 30.73 and divided by 32.7 makes the effort of this part 782-44. 1429. The area of the lower part is 697 95, to which adding the triangle FKG=390-8 we have 1088 77, which multiplied by 23-27 and divided by 52-15, gives 485-82 for i effort. The expression of the thrust, represented by p in the formula, b, representing the sum of the efforts of the semi- arch, will be Substituting these values in the formula, we have the equation x=√593-24-100-64+79-21-8-9-15.01. By taking double the square root of the thrust the result is 23-91, a thickness evidentl too great, because the sum of the vertical efforts, which are therein neglected, is con siderable. 1430. The geometrical method gives 19 lines. The least thickness of the piers from actual experiment was 16 lines. 1431. Rondelet gives a proof of the method by means of the centres of gravity, as i some of the preceding examples, from which he obtains a result of only 13-26 for the thick ness of the piers. Consideration of an Arch whose Voussoirs increase towards the Springing. 1432. The model (fig. 582.) has an extrados of segmental form not concentric with its intrados, so that its thickness increases from the crown to the springing. The opening is the same as before, namely, 9 inches, or 108 lines. The thickness at the vertex is 4 lines, towards the middle of the haunches 7 lines, and at the springing 14 lines. The centre of the line of the extrados is one sixth part of the chord AO below the centre of the intrados; so that (See 1390, obs. 22, and 1441). The radius DN=68'05 The are BKKC=42-43 258 75 x 38.18 =232.47. 1433. The area KHDC of the upper part of the arch is 258.75, that of the lower part BAHK 486 5; hence the effort of the upper part is represented by the expression 1434. The half segment A Be being supposed to be united to the pier; BeHK, whose area is 178, is the only part that can balance the upper effort; its expression will be 66 24. The difference 178 x 15.82 E PSR Fig. 582. of the two efforts 166-23 will be the expression of the thrust represented by p in the formula |