EXAMPLE II. Required the furface of the fruftum of a spindle generated from an arc of the fame ellipfe as in the laft example; the height of the fruftum being 15, and the central diftance 10. Here the arc generating the fruftum, is equal to half the arc in the laft; and confequently that arc is 15426, and the fpheroidal fruftum = 1843'40985. = Then 1843-4098 1843 4098969.246 fruftum required. 15426 X 31416 X 20 = 874163 the furface of the PROBLEM XVII. To find the Solidity of an Elliptic Spindle. RULE I. 1. Divide the fquare of the perpendicular are DE by 3 times the fquare of the parallel axe AB, and multiply the quotient by the cube of FG the axe or length of the fpindle; and call the product P. 2. Find the area of the elliptic fegment FDG from which the fpindle is generated; multiply this area by 4 times CH the central distance, and call the product q 3. Multiply 1.57079 by the difference between P and Q, and the product will be the content of the fpindie FDGN. That aa That is, 2n × (773 — 4cs) = the spindle FDGn. 3bb Where n = 78539, a = DE, b = AB, l = FG, c = the fegment FDGF.* CH, and s aa-xx cc ddxx ¿√ec — xx — 2b2)=pddx × x -2pbyx; and the fluent is s = pddx × 2bp x area HIKD = the fruftum DKON. When x isa, the above theorem will become 2 pd d 3cc 366 × a32bp x area FDH DFN the half of the spindle. And if from the femi-fpindle be taken the fruftum, there will remain pddee × за -e 2bp × area FKI = the fegment FKO of the fpindle, e being the height FI of the fegment. Corol. 1. If d be fuppofed = c, the ellipfe will become a circle, and accordingly the theorems above given will become the fame with thofe before found for the circular fpindle. Carol. EXAMPLE. The axes of an ellipfe being 50 and 30, required the folidity of a fpindle generated from an arc of it, about Corol. 2. If H coincide with c, b will vanish, a will be = c, and the theorems will become the fame with thofe before found for the spheroid. Corol. 3. Putting DDN the greatcft, and d = Ko the least diameter of the fruftum, = H1 its height, c = CH the central distance, s the elliptic femi-fegment KDM, and n = 78539&c. Then, in the foregoing theorems, b = x, b = c, d = c + ¿D, = area HIKDs + 1⁄2 dh, c = b x DD + DC a = b√ C + 1⁄2 D √ DD + DC-dd-dc and DD+DC=dd-dc which values being substituted in the theorems above, give } nh × [2DD + dd−8c ( − D+d+3,5)] for the value of the fruftum, or half a cask in the form of the middle zone of an elliptic spindle; and ¦nl×[2DD—Sc(−D+2)] 3S for half the fpindle when s the arca DFH, and /= HF. Corol. 4. But in real practice, fuch as cafk gauging, none of thefe rules can be used, because we have not given either the axes of the ellipfe or the central diftance; and to accommodate rules to that purpose, we muft introduce another dimenfion of a fruftum, befides its length and greatest and leaft diameters. Thus, putting m Po the diameter through R the middle of the length HI, and the other letters as before. Then, by the property of the ellipfe, (c+D)2 - (c+α)2:4; (c+r)2 - (c+ 1)2: 1, hence 4(c + {m)2 - 3(c + D)2 = (c + d), and — 1⁄2 2 I 3D + d2 4722 = PQ C = X 4 And the last two theorems be 3D d + 4 m about its chord parallel to, and at the distance of, 9 from the tranfverfe axe. Here 159 6 DH the height of the generating fegment; and, by the property of the ellipfe, DC: AC :: 2√DH × HE : FG = 40, the base of the elliptic fegment, or length of the fpindle; alfo, by prob. 6, the area of the fegment is 167-7345. Divide 3 times the generating fegment by the length of the fpindle; from the quotient fubtract the greatest diameter of the fpindle; multiply the remainder by 4 times the central diftance, and fubtract the product from the fquare of the greatest diameter; then the difference multiplied by the length of the fpindle, and the product by 5236 will give the content of the fpindle. 2 That is, nl × [D2-4c(−D + 33)] = the spindle. 3 Where D DN the greatest diameter, and the reft of the fymbols as in the laft rule.-By corollary 3. EXAMPLE. Required the folidity of an elliptic fpindle, whofe length is 40, greateft diameter 12, and the central diftance 9. Here 26 the height of the generating fegment, which is therefore the fame as before, the area being 167.7345. Then Then 3*14159×40×[122—4×9 × (−12+3 × 167·7345)] 6 40 - 1·74024) = 2578.56 the = 3.14159 X 80 X (12 — 174024) content required. RULE III. From three times the fquare of the greatest diameter, take 4 times the fquare of the diameter in the middle between the greateft diameter and the end; and from 4 times the faid middle diameter, take 3 times the faid greateft diameter': Divide the former difference by the latter, and of the quotient 44 will be the central diftance. 3DD-4mm the central distance; D being the greateft diameter, and m the middle dia meter. Then proceed as in the last rule. EXAMPLE. If, as in the example to the laft rule, the greatest diameter be 12, and the length 40, required the content, fuppofing the diameter at 4 of the length to be 6 x (√213) or 9:49546. the central diflance, the fame as in the last ex ample; and therefore the content will be 2578.56, as beford. PRO |