volving arc. Subtract the latter product from the former; and the remainder being multiplied by 6.2832, will give the surface required. 2. The convex surface of any segment or zone of a circular spindle, cut off perdendicularly to the chord of the revolving arc, may be found in the same manner; using the length of the solid, and the part of the arc which describes it, instead of the length of the whole spindle and whole arc. EXAMPLES. 1. Required the solidity of the circular spindle AB CD, whose length AB is 30, and middle diameter CD 16 inches. D F 152 225 By Rule 2, Prob. 14, Part II., we have +8= 8 8 × ÷ (36.125 +8=28.125 + 8 = 36.125, the diameter CF; and 8 = 5.3333 ÷ 29.565 =.18039; then 1+.18039 × 30 = 1.18039 × 30 = 35.4117, the length of the revolving arc ACB; and by Rule I., Prob. 16, we have 35.4117 × 18.0625639.6238, half of which is 319.8119, the area of the sector AOBC. Again, CO · CE 18.0625 = and OE x AE 8 = 10.0625 = OE; 10.0625 × 15 = 150.9375, the area of the triangle AOB; then 319.8119-150.9375=168.8744, the area of the revolving segment ACB. Now, by the Rule, we have 3 --168.874410.0625) 2 × 12.5664 = 3375 168.8744x5.03125) × 12.5664 3 = (1125 849.6493) × 12.5664 = 275.3507 × 12.5664 3460.16703648 inches, the solidity required. 2. What is the solidity and superficial content of a circular spindle, whose length is 96, and middle diameter 72 inches? Ans. Solidity 138.51593 feet. PROBLEM XV. To find the solidity of the middle frustum or zone of a circular spindle. RULE. From the square of half the length of the whole spindle, take of the square of half the length of the frustum; and multiply the remainder by the said half length of the frustum. Multiply the revolving area which generates the frustum, by the central distance, and subtract this product from the former; then the remainder being multiplied by 6.2832, will give the solidity required. EXAMPLES. 1. What is the solidity of the frustum ABCD, whose length mn is 40, its greatest diameter EF 32, and its least diameter AD or BC 24 inches? Draw DG parallel to mn, then we have DG = mn =20. Also, EF - AD = 16 – 12 = 4 = EG; and DG2+EG2 = 201⁄2 +42 = 400+16=416, the square of the chord DE; then by Theo. 12, Part I., DEo÷EG 416 = = 104, the diameter of the generating circle, 4 half of which is 52, the radius OE or OH; consequently OË – EI = 52 — 16 = 36, the central distance OI. - O12 = 522 362 2704 · Again, OH 1296 = 1408, the square of HI, half the length of the whole spindle. 4 By Rule 3, Prob. 17, Part II., we have, = .03846, 104 the tabular height; and the corresponding Area Seg. is .00994; then .00994 × 104o = .00994 X 10816 = 107.51104, the area of the segment DEC. Also, mn x Dm 40 x 12 = 480, the area of the rectangle mDCn; and 107.51104 + 480 587.51104, the generating area mDECn. Now, by the Rule, we have (1408 - 20) -) × 20 = 133.33333 × 20 = 1274.66667 × 20 = =25493.33334, the first product; and 587.51104 x 36 = 21150.39744, the second product, which taken from the first, leaves4342.9359; then4342.9359 x 6.2832 = 27287.53484688 inches, the solidity required. 2. What is the solidity of the middle frustum of a circular spindle, whose length is 50, greatest diameter 40, and least diameter 30 inches? Ans. 30.84257 feet. PROBLEM XVI. To find the solidity of a cylindrical ring. RULE. To the thickness of the ring, add the inner diameter; and this sum being multiplied by the square of the thickness, and the product again by 2.4674, will give the solidity. Note, The surface of a cylindrical ring may be found by the following rule: To the thickness of the ring, add the inner diameter; and this sum being multiplied by the thickness, and the product again by 9.8696, will give the surface required. EXAMPLES. 1. What is the solidity of a cylindrical ring, whose thickness AB or CD is 6, and the inner diameter BC 20 inches? D Here 20+ 6 × 62 × 2.4674 = 26 × 36 × 2.4674 =936 x 2.4674-2309.4864 inches, the solidity required. 2. What is the solidity of an anchor-ring, the inner diameter of which is 24.6 inches, and its thickness 6.4 inches ? Ans. 3133.005824 inches. 3. Required the solid and superficial contents of a cylindrical ring, whose thickness is 9, and inner diameter 32 inches. Solidity 8194.2354 inches. Ans. Superficies 3641.8824 inches. PROBLEM XVII. To find the solidity or superficies of any regular body. RULES. 1. Multiply the tabular solidity, in the following table, by the cube of the linear edge of the body, and the product will be the solidity. 2. Multiply the tabular area, by the square of the linear edge, and the product will be the superficies. Note. The above table shows the solidity and superficies of the five Regular Bodies, when the linear edge of each is 1, or unity. EXAMPLES. 1. What is the solidity and superficies of the tetraedron ABCD, whose linear edge is 8 inches? Here.1178511 x 83.1178511 x 512-60.3397632 inches, the solidity; and 1.7320508 x 82 = 1.7320508 × 64 110.8512512 inches, the superficies. 2. What is the solidity of the hexaedron ABCD, whose side AB is 6 feet? |