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1373. If to this experiment the principles of mechanics be applied, considering the plane of 30 degrees inclination as a horizontal plane, the vertical faces ED FR may be considered a inclined planes of 60 degrees. On this hypothesis it may be demonstrated by mechanics, that to sustain a body between two planes forming an angle of 60 degrees ( fig. 566.), the resistaoce of each of these planes must be to half the weight sustained es HD is to DG, as the radius is to the sine of 30 degrees, or is 1 is to 2.
EQUILIBRIUM OF ARCHES. 1374. The resistance of each parallelopiped represented by the prism ABDE (fig. 565.) being equal to half their weight, it follows that the weight to be sustained by the two prisms should equal one quarter of the two parallelopipeds taken together, or the half of one, which is confirmed by the experiment. This agreement between theory and practice deter. mined Rondelet to apply the hypothesis to models of vaults composed of voussoirs and wedges disunited, made of freestone, with the utmost exactness, the joints and surfaces nicely wrought, as the parallelopipeds in the preceding example.
1375. The first model was of a semicircular arch 9 inches diameter, comprised between two concentric semi-circumferences of circles 21 lines apart. It was divided into 9 equal voussoirs. This arch was 17 lines deep, and was carried on piers 2 inches and 7 lines thick. It was found, by gradually diminishing the piers, which were at first 2 inches and 10 A lines thick, that the thickness first named was the least which could be assigned to resist the thrust of the voussoirs.
1376. The model in question is represented in fig. 567., whereon we have to observe, - 1st. That the first voussoir, I, being placed on a level joint, not only sustains itself, but is able to resist by friction an effort equal to one half of its weight. 2d. That the second voussoir, M, being upon a joint inclined 20 degrees, will also, through friction, sustain itself; and that, moreover, these two voussoirs would resist, previous to giving way on the joint AB, an horizontal effort equal to one half of their weight. 3d. That the third voussoir, N, standing on a joint inclined at 40 degrees, would slide if it were not retained by a power PN acting in an opposite direction. 4th. That taking, according to our hypothesis, an inclined plane of 30 degrees, whereon the stones would remain in equilibrium as an horizontal one, the inelined point of 40 degrees may be considered as an inclined plane of 10 degrees, supposing the surfaces infinitely smooth. 5th. That the effort of the horizontal power which holds this voussoir in equilibrium upon its joints will be to its weight as the sine of 10 degrees is to its cosine, as we have, in the section on Mechanics, previously shown. (1255 et seq.)
1377. The model of the vault whereon we are speaking being but 9 inches, or 108 lines in diameter, by 21 lines for the depth of the voussoirs, that is, the width between the two concentric circumferences, its entire superficies will be 4257 square lines, which, divided by 9, gives for each voussoir 473 square lines. Then, letting the weight of each voussoir be expressed by its superficies, and calling P the horizontal power, we have
P: 479::sin. 10°: cosin. 10;
Or, P : 473::17365 : 98481 ; which gives P=8360The fourth voussoir, being placed upon a bed inclined at 60 degrees, will be considered as standing on a plane inclined only at 30 degrees, which gives, calling Q the horizontal power which keeps it on its joint, —
Q:473::sin. 30° : cosin. 30.
Or, Q : 473::50000 : 86603 = 273 1978. The half-keystones, being placed on a joint inclined 80 degrees, are to be considered as standing on an inclined plane of 50, the area of the half key which represents its weight being 236. If we call R the horizontal power which sustains it on its joint, we shall have the proportion
R: 236}:: sin. 50 : cosin. 50;
or, R : 236]::76604 : 64279; which gives R=281% 1379. Wishing to ascertain if the sum of these horizontal efforts, which were necessary to keep on their joints the two voussoirs N, O, and the half-keystone, was capable of thrusting away the first voussoir upon its horizontal joint AB, the half arch was laid down upon a level plauc of the same stone without piers, and it was proved that to make it give vay an horizontal effort of more time 16 ounces was required, whilst only 10 were neces
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 Nc 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 Nc=px Hi, and the force Na which presses against the plane as the base of the plane IG is to its length, which gives the force Nu=p HG
1384. Considering, in the same way, the two forces of the power Q which retain the Foussoir on the inclined plane, we shall find the parallel force Nf=QxCH
and the perpendicular force Nd=Q HG The force resulting from the two forces Na, Nd, which press against the joint, will be expressed by px HG
+ Qx 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 13 of its expression. Calling this ratio n, we shall, to express the friction, have
(pxGG + Q x G), 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 Nc, less the friction; from which results the equation
HG=P * HG -(px - Q x C):
Q IG=p HI -(px IG-Q R IH) xn; and, bringing the quantities multiplied by Q to the same side of the equation, we have
Q x 1G+(Q IH)xn=p x HI -(p < IG)xn; which becomes
which is the formula for each voussoir, substituting the values for the expression. 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 seen or já nearly. The weight of the roussoir expressed by its area will be 473, which several values being substituted in the formula, we have
643 — * 766
766 + }} * 643 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.
HI1386. The same formula Q=px 1G7nxiu gives for the voussoir M on an inclined joint of 60 degrees, whose sine HI is 866 and cosine IG 500, Q=473
886 - 18 500
500 + 11 x 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 236.), and the friction by a
985 - 8 x 174 The formula now will be Q=236} * 174 + x 985, which gives Q=282-2, instead of 231% 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 beglected. 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 voussoir N, 83.6 x 244.94 = 20476.98
0, 273.4 x 256.26 = 70061 .48
S, 282.2 x 260.50= 73313.10 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 r =