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PROBLEM VI. To find the solidity of a frustum of a cone.

Example 1. .7854×4=3.1416, and .7854×21=1.7671, which added to 3.1416-4.9087-the sum of the areas of the two ends. Log. 3.1416=0.497078, and log. 1,7671= 0.247247, and the sum of these logarithms = 0.744325, which divided by 2=0.372162, whose natural number=2. 356=the square root of the product of the areas of the two ends. 4.9087+2.356-7.2647, which multiplied by 24= 174.3528 the number of cubic feet in the mast.

Example 2. .7854×16=12.5664, and .7854 × 9=7.0686, which added to 12.5664-19.6350-the sum of the areas of the two ends. Log. 12.5664=1.099920, and log. 7.0686= 0.849308, and the sum of these logarithms 1.948328, which divided by 2=0.974164 whose natural number=9.423=the square root of the product of the areas. 19.6350+9.423 29.058 which multiplied by 3=87.174- the capacity in cubic feet. 87.174×1728÷231=652.143 wine gallons.

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Example 3. .7854 X 49-38.4846, and .7854X36=28.2 744. Log. 38.4846=1.585288, and log. 28.2744=1.4513 96, and their sum=3.036684, which divided by 2=1.5183 42 whose natural number=32.98. 38.4846+28.2744= 66.7590 the sum of the areas of the two ends, and 32.98 =the square root of the product of these areas, which added together become 99.7390, which multiplied by 22=26 6.0372 the capacity in cubic feet. 266.0372x1728=45 9512.5983, which divided by 282=1630.0212=the number of gallons of ale which the vat will hold.

PROBLEM VII. To find the surface of a sphere.

Example 1. 3.14159×7930=24912.80870, which multiplied again by 7930-197.558, 572.99100=the number of square miles on the surface of the earth. Second method, by logarithms. Log.7930-3.899273, which multiplied by 2 7.798546, which added to 0.49714942, (log. 3.14159)= 8.29569542, whose natural number 197,558,473+.

Example 2. Log. 2,800,000=6.447158, and log. 3.141 59 0.49714942, which subtracted from 6.447158, leaves 5.95000868 the logarithm of the diameter, which added

to 6.44715 (log. circum.)=12.39766 whose natural number =2,495,547,600,423 square miles.

Example 3. 3.14159×13=40.84067=the circumference. Again, 40.84067 × 13 : = 530.92781 = the surface of the whole sphere, which divided by 2=265.46435=the number of square feet on the surface of the hemispherical dome.

PROBLEM VIII. To find the solidity of a sphere.

Example 1. Log. 7930 3.899273, which multiplied by 3=11.697819=log. cube diam. Log. 0.5236=1.719000, which added to 11.697819=11.416819, whose natural num. ber=261,107,228,915 the solidity of the earth.

Example 2. 43-64, pacity of the sphere. gallons.

and 64X.5236-33.5104=the ca 33.5104 × 1728 ÷ 231 = 250,667

Example 4. Log. 2180-3.338456, which multiplied by 3=10.015368, which added to 1.719000 (log. 0.5236)=9. 734368, whose natural number=5,424,600,000 miles.

Example 1. Art. 72. Log. 65.45-1.815910, which di minished by log. .5236=2.096910, which divided by 3=0. 698971 whose natural number-5 feet.

Example 2. 16755÷62.5=268.08 cubic feet, whose logarithm=2.428264, and 2.428264-1.719000 (log. .5236)= 2.709264, which divided by 3.903088 whose natural number=8 feet.

PROBLEM IX. To find the convex surface of a segment or zone of a sphere.

Example 1. By Trig. 134, (Fig. 15.) R: AC::cos ACD : CD. Substituting numbers, 1: 3965: cos 23° 28′; CD; and, by the tables, 10.0000000; 3.5982432::9.9625076 : 3.5607508=3637, and 3965-3637-328-the hight of the segment. 7930 × 3.14159 × 328-8,171,400.2536.

Example 2. By Trig. 134, (Fig. 15.) R: CG:: sin ECG ; GM. Substituting numbers, 13965: sin 23° 28': GM;

and, by the tables 10.0000000: 3.5982432::9.6001181: 3.1983613=1578.9. Log. 7930=3.8992732, and log. 3.1 4159 0.4971490. 3.1983613+3.8992732+0.4971490= 7.5947835, whose natural number 39334850.213 which multiplied by 2=78669700.426=the area of the zone.

Example 3. DN= (Fig. 15.) CP-CN-PD=39651578.9-320=2058.1=DN. 3.14159×7930=the circum. sphere, and 3.14159×7930×2058.1=51273000.132=the surface of one of the temperate zones, and 51273000.132× 2=102546000.264 the surface of the two temperate zones. 8171400.253x2=16342800.506 the surface of the two frigid zones, and the surface of the torrid zone=78,669,750. 426. By addition, the surface of the whole globe=197,558, 501.196.

PROBLEM X. To find the solidity of a spherical sector. Example 1. See Figure 15.

By the 1st example in PROBLEM IX, the spherical surface =8171400, which multiplied by 13213, the radius (=} diameter)=10,799,867,000 miles.

PROBLEM XI. To find the solidity of a spherical segment. Example 1. 252-625, 625X.7854-490.8750, 490.87 50 × 83.5236×4=2231.61=the solidity.

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Example 2. See Figure 15. R: AC::sin ACD: AD. Substituting numbers, 17930: :sin 23° 28' AD; and, by the tables, 10.0000000: 3.5982432::9.6001181 : 3.198 3613-1578.92, which multiplied by 2=3157.84 the diam. of the base of the frigid zone. By example 1 Art. 73, the hight 328, whose log. 2.5158738 which multiplied by 3=7.5 476214. Log. 0.5236=1.7189996, which added to 7.547 62147.2666210=18466123. The diameter of the base as found above 3157.84 whose log.=3.1983610 and this into 2=6.9987820, which added to 1.8950909 (log. 0.7854)= 6.8938729, and this added to 2.2148438 (log. 164 the hight) 9.1087167 1,284,534,242, which added to 18, 466,123=1,303,000,365 = the solidity of one of the frig

id zones.

PROBLEM XII. To find the solidity of a spherical zone or frustum.

Example 1. 122-144, 102-100,(4-3)2=5.3333, which added to 144+100-249. 3333, and this into 3 times the distance (12)=2991.9996, and this into .5236=1566.630990 56 the answer.

Example 2. By instituting the proportion as in the 2d example, Art. 75, we find the radius of the larger end to be 3637, and by the same example the radius of the smaller end=1578.9. 36372=13227769, and 1578.92=2492925. 21. The distance=(Art. 73.) 2058.1, whose square=43257 75.61, of which=1411925.2033 and 13227769+2499922 5.21 +1411925.2033 = 17132619.4133, which multiplied into 6174.3 (3 times the dist.)=105781932043.53819, and this into .5236=55,387,419,616.996596284 the solidity of one of the temperate zones.

Example 3. By the last example the solidity of the two temperate zones=55,387,419,616.996596284×2=110,774, 839,233.993192568; and by example 2d Art. 75, the solidity of the two frigid zones=1,303,000,365×2=2,606,000, 730; and by example 1st Art. 71, the solidity of the whole globe 261,107,228,915. The two temperate zones+the two frigid zones 113,380839,963.993192568, which subtracted from the solidity of the whole globe, leaves 147,726, 388,951.006807432=the solidity of the torrid zone.

Note. The answer given by Day to this 3d example may be obtained by adding the temperate and frigid zones as given by him, and subtracting the sum from the solidity of the globe as appended to the 1st example Art. 71. To obtain all the answers to the questions in this section precisely as given in the text book, is no easy task; and, if this could be done, little would be gained in point of accuracy.

Example 4. 3.14159 X 25=78.53975 the circumference, which multiplied by 4=314.15900=the answer.

Example 5. 402 ×.7854 1256.6400, and this into 9= 11309.7600. 183=5832, and this into .5236 = 3053.6352, which added 11309.7600=14363.3952=the answer.

PROMISCUOUS EXAMPLES OF SOLIDS.

Example 1. 33-27 ft.=46656 inches=capacity of vessel. No. of balls 2 inches in diameter=183-5832 No. of balls 4 inches in diameter 93 729 No. of balls 6 inches in diameter

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63= 216

No. of balls 4 inches in diameter= 43= 64
Solidity of a ball 2 inches in diameter=23×.5236= 4.1888
Solidity of a ball 4 inches in diameter 43 X.5236=33.5104
Solidity of a ball 6 inches in diam. =63×.5236= = 113.0976
Solidity of a ball 9 inches in diam. =93×.5236= 381.7044
Solidity of 5832 balls 2 inches in diam. = 5832 × 4.1888=

24429.0816

Solidity of 729 balls 4 inches in diam.

24429.0816

=

729X33.5104 =

Solidity of 216 balls 6 inches in diam. =216×113.0976 =

24429.0816

Solidity of 64 balls 9 inches in diam. = 64 X 381.7044 = 24429.0816

The solidity of the balls subtracted from the number of inches in the capacity of the vessel, leaves 22226.9184, which divided by 231 (the number of inches in a wine gallon)=96.22.

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Eaxmple 2. See Figure 17. NV3: RV3:: whole pyramid upper part; or, making x=RV, 27 : x3::3:1; and, 3x3=27, x3=9, and x=3/9=2.08 feet 24.96 inches. Again, let x=the hight of the two upper parts; then, as before, 27x3:3:2; and 3x3-54, x3-18, and x=3/18 =2.62 feet 31.449 inches. 31.449-24.96=6.48. 3631.449-4.551. Therefore, the hight of the several sections are 24.96, 6.49, and 4.551.

Example 3. See Figure 10. AC and CF each 15 inches. By Euc. I. 47, AC2+CF2=AF2=a square of a side of the prism, that is, 225+225=450 inches, or 31 feet the area of one end. 3156 175 cubic feet.

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Example 4. Log. 890,000-5.9493900, and this into 3= 17.8481700 log. 890,0003. Log. 0.52361.7190000, which added to 17.8481700-17.5671700. Log. 7930=3. 8992732, and this into 3 11.6978196, which added to

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