As centinel, laid at his feet, Poor Tray watch'd the flock on the plain; Suspended, his crook, on the tree, Hung ready his hand to receive; Ah! ne'er shall we hear it again! The shepherd shall traverse the plain. But though he to death is consign'd, And no more the lov'd bard shall we see, His song in a wreath is entwin'd, And that wreath forms a GARLAND for me! TO GRAY. Next see ethereal GRAY, While from her eyelids gush'd the soul-assuaging tear! Oft shall his numbers soothe me to repose, TO GOLDSMITH. Next hapless AUBURN's friend my bosom cheers, Of flatt'ring slaves, who spurn'd the harmless swain ; Poet belov'd! my vanquish'd heart is thine, TO BURNS. And whae is he that syngs sae weel, And should auld mokie sorrow freeten, And, while I breathe, whene'er Ise scant,TM Mathematical Department. MATHEMATICAL QUESTIONS IN NO. VI. ANSWERED. 1. Qu. (75) Answered by Mr. Brewer, Private 2nd R. L. Militia. THE area of the garden is readily found = 3136320 inches; and putting r for the depth of the pond, its diam. will be 8x, and by mensuration (48x + x2) ,5236x=25,6564x3 is its concavity; also 64 x2 x ,785450,2656 a2 is the surface of the pond. Hence, 3136320 - 50,2656 x2 25,6564 x3, from whence x is found 48,98515 inches = 4,0820954 feet, and 8x, or the diameter, 32,65676 feet. Again, by Mr. B. Brooke. Let 8 x and x denote the diam. and depth of the pond; then its capacity will be 25,6564x', and its surface= 50,2656 x2; but by the question 25,6564x3 + 50,2656 x2 3136320 inches the area of the garden; whence 48,9848, therefore the diam. is 10,8855, and the depth 1,3607 yards. In the same manner it was answered by Messrs. J. Baines, Reading; J. Bamford, Holthead; W. Bruster, Donington; J. Butterworth, Haggate; J. Cattrall, Plymouth; J. Cummins, Holbeck; W. Dunn, Broughton; J. Ely, Doveridge; J. Gawthorp, Leeds; W. Harrison, Burton; J. Hine, Plymouth; S. Jones, Liverpool; M. Macann, Lutton; R. Maffett, Plymouth; A. Nesbit, Farnley; Rylando; J. Smith, Alton Park; E. Webster, Armley Mills; and J. Winward, Plymouth. 2. Qu. (76) Answered by Messrs. Bamford, Cummins, and Ely. By adding the given surfaces together, we have 314,16 for the whole surface; from whence we find the diam. 10, and the circumference = 31,416; then, by mensuration, 6 is the depth, and 4,713 the radius of the segment immersed; therefore its solidity, or the quantity of water displaced, is 387,851; but, by the principles of hydrostatics, 1728 387,851 :: 1000: 14,028 lbs. the weight of the globe. Again, by Messrs. Brooke, Bruster, and Webster. The diameter of the globe is 209,44 + 104,72 3,1410 its solidity. Therefore, by hydrostatics 1728: 387,851 ::1000: 224.45 ounces the weight required. The same, by Messrs. Macann and J. Smith. Since the curve surface of the globe is given 314.16 inches, its diam. is easily found 10 in. and as the superficies of spherical segments are as their altitudes, we 10 x 2 have the height of the immersed segment = 3 and its solid content= 10472 27 cubic inches, is the con- tent of the water displaced by the globe; therefore, by hydrostatics, 1728: 1000 :: 10472 : 224,4513 ounces, is the weight, as before. It was answered also by Messrs. Armitage, Baines, Brewer, Cattrall, Gawthorp, Hine, Jones, Maffett, Nesbit, Rylando, Tomlinson, and Winward., E GL H F 3. Qu. (77) Answered by Mr. J. Smith, Alton Park. Let ABCD and ABK represent vertical sections of the conic frustum and the cone completed, EF the dividing section; and draw LC parallel to GK. Then by sim. triangles LB =2: LC 12 :: GB 5: GK = 30, therefore, IK GK-GI=18. D K Put ,7854n, theu, by mensuration, AB × GK × 1000n, is the content of the cone ABK, and DC' x IK X × 2=216n, is the solidity of DCK. Half the difference of these 392n is the solidity of the frustum EFCD, which, added to 216n, gives 608 n for the solidity of the cone EFK. Now, the cones ABK, EFK,, eing similar solids, we have 1000n, 608 n b :: GK 30: HK = 30 25.41494; therefore, HI distance required. 608 =36086/76= 1000 HK-IK=7,41494 is the Again, by Mr. Armitage, Rochdale Academy. First, 10-6:12:: 6, 18 what the cone wants in length to be complete, and 18+ 1230 length of the whole cone; then, by sim. solids, as 785,4 (= content of the whole cone): 30:: 477,5232 (= the solidity of the top cone and half frustum): 25,41494, and the distance from the bottom 25,41494 — 18 — 7,41494, as before. Otherwise, by Mr. J. Cattrall, Fifer 2nd R. L. Militia. By similar triangles LB: LC :: BG: GK = 30; consequently, 785,4 is the solidity of the whole cone, and 615,7536 is the solidity of the frustum, whence, 785,4615,7536 169,6464 solidity of DCK, 615,7536 therefore, 169,6464+ 2 =477,5232 is the so |