sary to sustain the half-keystone and the two voussoirs N, O. The two halves of the arches united bore a weight of 5 lbs. 2 oz. before the first voussoirs gave way. 1380. To find the effect of each of these voussoirs when the arch is raised upon its piers, let fall from the centres of gravity N, O, S of these voussoirs the perpendiculars Nn, Oo, Ss, in order to obtain the arms of the levers of the powers P, Q, R, which keep them in their places, tending at the same time to overturn upon the fulcrum T the pier which carries the half arch, and we have their effort Px Nn+Qx Oo + R x Ss. The height of the pier being 195 lines, we have Nn=244.94 Oo=256.26 and Ss 260.50, whence we have The effort Px Nn = 83·4 x 244.94, which gives 20427.996 Total effort in respect of the fulcrum, 163898-804 1381. The pier resists this effort, 1st, by its weight or area multiplied by the arm of the lever determined by the distance Tu from the fulcrum T to the perpendicular let fall from the centre of gravity G upon the base of the pier. 2d. By the weight of the half arch multiplied by the arm of its lever VY determined by the vertical LY let fall from the centre of gravity L, and which becomes in respect of the common fulcrum T= Tt or VB-BY, in order to distinguish BY, which indicates the distance of the centre of gravity of the half arch (and which is supposed known because it may be found by the rules given in 1275. et seq.) from the width VB that the pier ought to have to resist the effort of the half arch sought. In order to find it, let P, the effort of the arch above found, be 163898.804. arr 1382. The area of the pier which represents its weight multiplied by the arm of the lever will be ar x That of the half arch multipied by its arm of lever will be where a + c will be bx + bc, whence the equation P: + bx + bc, shown by VB+ BY, which we have to solve. Now first we have 2 Multiplying all the terms by arr to eliminate xx, we have the second power; but as xx + bb xx+ 2br a 26 a a arr an expression in which a is raised to is not a perfect square, that is to say, it wants the square of half the known quantity which multiplies the second term; by adding this square, which is to each side of the equation, we have xx + + first member by this means having become a perfect square whose root is a + a' we shall which becomes, by transferring to the other side of the + a a 2p-2bc bb 2p-2bc bb b + a equation, x= in which a being only in the first member of the equation, its value is determined from the known quantities on the other side. Substituting, then, the values of the known quantities, we have which gives x=281 lines instead of 2 inches and 5 lines, which was assigned to the piers that they might a little exceed equilibrium in their stability. Proof of the above Method by another Method of 1383. A proof of the truth of the hypothesis in the preceding section is to be found in the method proposed by Bossut in his Treatise on Mechanics. Let the voussoir N (fig. 568.) standing on an inclined plane be sustained by a power Q acting horizontally. From the Fig. 568. 1 centre of gravity let fall the vertical Nn, which may be taken to express the weight of the voussoir. This weight may be resolved into two forces, whereof one, Ne, is parallel to the joint, and the other Na is perpendicular to it. In the same manner the power Q expressed by QN in its direction may be resolved into two forces, whereof Nf will be parallel to the joint and the other Nd perpendicular to it. Producing the line from the joint HG, drawing the horizontal line GI and letting fall the vertical HI, consider the line HG as an inclined plane whose height is HI and base IG. Then the force Ne with which the voussoir will descend will be to the weight as the height HI of the inclined plane is to its length HG. Calling p the weight of the voussoir, we then have Ne=px and the force Na which presses against the plane as the base of the plane IG is to its length, which gives the force Na=p × HG HG IH IG IG 1384. Considering, in the same way, the two forces of the power Q which retain the voussoir on the inclined plane, we shall find the parallel force Nf=Q × and the perpendicular force Nd=Qx HG The force resulting from the two forces Na, Nd, which press against the joint, will be expressed by p× HG Q× GH; and as the voussoir only begins to slide upon a plane whose inclination is greater than 30 degrees, the friction will be to the pressure as the sine of 30 degrees is to its cosine, or nearly as 500 is to 866, or of its expression. Calling this ratio n, we shall, to express the friction, have 100 IG IG IG (px + Q xTG) GH xn. II As the friction prevents the voussoir sliding on its joint, in a state of equilibrium, we shall have the force Nf equal to the force Ne, less the friction; from which results the equation All the terms of which equation having the common divisor HG, it becomes QxIG=px HI-(p × IG - Q × IH) × n ; and, bringing the quantities multiplied by Q to the same side of the equation, we have Q× IG+(Q× IH) x =px HI-(px IG) × n; which becomes 1385. Thus for the third voussoir N (fig. 567.) placed on an inclined plane of 40 degrees, HI which represents the sine of the inclination will be 643, and its cosine represented by IG, 766, the expression of the friction n will be 500, or nearly. The weight of the voussoir expressed by its area will be 473, which several values being substituted in the formula, we have which gives Q-836, the expression of the horizontal force P, which will keep the voussoir N in equilibrium on its joint instead of 83-4, which was the result of the operation in the preceding subsection. HI-nxIG 1386. The same formula Q=pG+XIH gives for the voussoir M on an inclined joint of 60 degrees, whose sine HI is 866 and cosine IG 500, Q=473 × X =273·4; 886-x500 500+1 × 866 instead of 273-3, which was the result of the operation in the preceding section. 1387. For the half-keystone, the sine HI, being of 80 degrees, will be expressed by 985, and its cosine IG by 174; the half-keystone by 2361, and the friction by The formula now will be Q=236 × which gives Q=282.2, instead of 281 found by the other method. These slight differences may arise from suppressing the two last figures of the sines, and some remainders of fractions which have been neglected. Multiplying these values of the powers which keep the voussoirs in equilibrium upon their beds by the several arms of the levers, as in the preceding calculations, their energy will be as follows: For the total force in respect of the fulcrum T=163851·56. Which is the value of p, and being substituted for it in the formula x = as well as the values of the other letters, which are the same as in the preceding example, we have x= /163851 56 × 2—2128 x 2x12 2128 2128 2128 + =28.16 lines for the thickness of the piers, instead of 284 lines found by the preceding operation. Application of the Principles in the Model of a straight Arch. 1388. The second model to which the application of the preceding methods was made was a straight arch of the same sort (fig. 569.), whose opening between the piers was 9 inches. The arch was 21 lines high and 18 lines thick. It was divided into 9 wedges, whose joints were concentric. To determine the section of the joints, the diagonal FG was drawn on the face of the half arch, and from its extremity F touching the pier, the perpendicular FO meeting O in the vertical, passing through the middle of the opening of the piers, all the sections meeting in this point O. Each of the sections of the piers which support the arch forms an angle of 21° 15' with the vertical, and of 68° 45' with the horizon. 1389. In considering each of the wedges of the half arch as in the preceding method, it will be found that in order to retain the voussoir A on the joint IF (of the pier) which forms with the horizontal line NF an angle of 68° 45', we have Fig 569. Fig. 570. The height of the piers being 195 lines to the underside of the arch, and 216 to the top of the extrados, it follows that the arm of the lever, which is the same for all the wedges, is 206, from which we derive for the thrust p of the formula, b which expresses the area of the half arch=1219; c which expresses the distance of its centre of gravity from the vertical Fn=24, and the height of the pier a=216. stituting these values in the formula, we shall have Now, sub Experiment gives 44 lines for the least width of the piers upon which the model will stand. But it is right to observe that from the impossibility of the joints being perpendicular to the intrados, the forces of the wedges press in a false direction on each other, as will be seen by the lines Fa, 1c, 2e, 3g, perpendicular to the joints against which the forces are directed, so that such an arch will only stand when the perpendicular FG does not fall within the thickness of the arch; and, indeed, this sort of arch is only secure when it comprises an arc whose width is equal to the section upon the piers IF, as shown in fig. 570. Observations on the Way in which Stones forming an Arch act to 1390. Let the semicircular arch AHCDNB (fig. 571.) consist of an infinite number of voussoirs acting without friction, and only kept in their places by their mutual forces acting on each other. It will follow 1. That the first voussoir, represented by the line AB, having its joints sensibly parallel and horizontal, will act with its whole weight in the vertical direction IE to strengthen the pier. F H Fig. 571. 2. That the vertical voussoir CD, which represents the keystone, having also its joints sensibly parallel, will act with its whole weight horizontally to overturn the semi-arches and piers which carry them. 3. That all the other voussoirs between these two extremes will act with the compound forces Gn, nm, ml, Kl, Kh, hg, gf, ƒT, which may each be resolved into two others, whereof one is vertical and the other horizontal: thus the compound force Kh is but the result of the vertical force 4h, and the horizontal force 4K. 4. That the vertical force of each voussoir diminishes from T to G, where, for the keystone CD, it becomes nothing, whilst the horizontal forces continually increase in an inverse ratio; so that the voussoir HN, which is in the middle, has its vertical and horizontal forces equal. 5. That in semi-circular arches whose extradosses are of equal height from their intradosses, the circumference passing through the centre of gravity of the voussoirs may represent the sum of all the compound forces with which the voussoirs act upon one another in sustaining themselves, acting only by their gravity. 6. That if from the points T and G the vertical TF and horizontal GF be drawn meeting in the point F, the line TF will represent the sum of the vertical forees which assist the stability of the pier, and FG the sum of the horizontal forces which tend to overthrow it. 7. That if through the point K the horizontal line IKL be drawn between the parallels FT and CO, the part IK will represent the sum of the horizontal forces of the lower part AHNB of the vault, and KL those of the upper part HCDN. 8. The lower voussoirs between T and K being counterpoised by their vertical forces, the part of the arch AHNB will have a tendency to fall inwards, turning on the point B, whilst the voussoirs between K and G being counterpoised by their horizontal forces, the part HCDN of the arch will re-act upon the lower part by its tendency to turn upon the point A. 9. The horizontal forces of the upper part of the arch shown by KL acting from L towards K, and those of the lower part shown by IK opposite in direction to the former, that is, from I to K, being directly opposed, would counterpoise each other if they were equal, and the arch would have no thrust; but as they are always unequal, it is the dif ference of the forces which occasions the thrust, and which acts in the direction of the strongest power. 10. If we imagine the width BO of a semi-arch constantly to diminish, its height remaining the same, the sum of the horizontal forces will diminish in the same ratio, so that when the points B and O are common, the horizontal force being annihilated, nothing remains but the vertical force, which would act only on the pier, and tend to its stability, thrust vanishing, because, instead of an arch, it would, in fact, be nothing more than a continued pier. 11. If, on the contrary, the height OD diminishes, the width BO remaining the same, the curve B and D would, at last, vanish into the right line BO, and the arch would become a straight one. In this case, the vertical forces which give stability to the pier being destroyed, all that remains for sustaining the arch are the horizontal forces which will act with the whole weight of the arch; whence this species of arches must be such as exert most thrust, and circular arches hold a middle place between those which have no thrust, and flat arches, whose thrust is infinite, if the stones whereof they are formed could slide freely on one another, and their joints were perpendicular to their lower surfaces, as in other arches. 12. The inconveniences which result from making the joints of flat arches concentric have been before noticed. If the stones could slide freely on one another, as they only act in a false direction, their forces could never either balance or destroy one another. 13. A vast number of experiments made by Rondelet, upon fifty-four models of arches of different forms and extradosses, divided into an equal and unequal number of voussoirs, showed that the voussoirs acted rather as levers than as wedges, or as bodies tending to slide upon one another. 14. As long as the piers are too weak to resist the thrust of the voussoirs, many of them unite as one mass, tending to overturn them on a point opposite to the parts where the joints open. 15. Arches whose voussoirs are of even number exert more thrust than those which are of unequal number, that is, which have a keystone. 16. In those divided into uneven numbers and of unequal size, the larger the keystone the less is their thrust, so that the case of the greatest thrust is when a joint is made at the vertex, as in the case of arches whose voussoirs are divided into equal numbers. 17. A semicircular arch divided into four equal parts has more thrust than one divided into nine equal voussoirs. 18. Arches including more than a semicircle have less thrust than those of a similar span, the intradosses and extradosses being of similar forms. 19. Thrust does not increase as the thickness of an arch increases; so that, cæteris paribus, an arch of double the thickness has not double the thrust. 20. A semicircular arch whose extrados is equally distant throughout from, or, in other words, concentric with, the intrados, when divided into four equal parts, will only stand when its depth is less than the eighteenth part of its diameter, even supposing the abutments immoveable. 21. Whenever, in an arch of voussoirs of equal depth, a right line can be drawn from its outer fulcrum to the centre of the extrados of the keystone (fig. 572.), fracture does not occur in the middle of the haunches if the piers are of the same thickness as the lower part of the arch. 22. Arches whose thickness or depth diminishes as they rise to the vertex have less thrust than those whose thickness is equal throughout. 23. Semicircular and segmental arches whose extrados is an horizontal line have less thrust than others. 24. As long as the piers in the models were too weak to resist the thrust, it was possible to keep them in their places by a weight equal to double the difference between the thrust and resistance of one pier, acting by a string suspended passing through the joints in the middle of the haunches, or by a weight equal to that difference placed above each middle joint of the arches, as in fig. 572. From these experiments and many others, a formula has been deduced to determine the thickness of piers of cylindrical arches of all species whose voussoirs are of equal depth, whatever their forms; and to this we shall now introduce the reader. Fig. 572. Method. ド M 1391. Having described the mean circumference GKT (figs. 573, 574.), from the points G and T draw the tangents to the curve meeting in the point F. From this point draw the perpendicular FO cutting it in the point K. This point is the place of the greatest effort, and of the consequent failure, if the thickness of the piers is too weak to resist the thrust. 1392. Through the point K, between the parallels TF and GO, draw the horizontal line IKL, which will represent the sum of the horizontal forces as will the vertical TF express the vertical forces; the mean circumference GKT will express the compound forces. E F M B h Fig. 573. R Fig. 574. 1393. The arches having an equal thickness throughout, the part IK of the horizontal line multiplied by the thickness of the arch will express the horizontal effect of the lower part of either, arch, and KL multiplied by the same thickness will express that of the upper part. These two forces acting in opposite directions will partly destroy each other; thus transferring IK from K to m, the difference mL multiplied by the thickness of the vault will be the expression of the thrust. This force acting at the point K in the horizontal direction KH, the arm of the lever is determined by the perpendicular PH raised from the fulcrum P of the lever to the direction of the thrust, so that its effort will be expressed by mL x AB × PH. This will be resisted 1. By its weight represented by the surface EP × PR multiplied by the arm of the lever PS, determined by a vertical let fall from the centre of gravity Q, which gives for the resistance of the pier the expression EP x PRx PS. 2. By the sum of the vertical efforts of the upper part of each arch, represented by MK x AB acting at the point K, the arm of their lever in respect of the fulcrum P of the pier being KH. 3. By the sum of the vertical efforts of the lower part represented by IT multiplied by AB acting on the point T has for the arm of its lever TE. Hence, if equilibrium exist, mL x AB × PH = PE × PR × PS + MK × AB × KH + IT × AB × TE. But as in this equation neither PR (= BE) nor PS nor KH nor TE is known, we must resort to an algebraic equation for greater convenience, in which |