A complete and practical solution book for the common school teacher . 2 XttX KX) = cu. in. (12) The amount consumed is ++ cu. in. Rule for finding a Spherical Segment.—A spherical segment withone base is eqtiivalent to half a cylinder having the same base andaltitude, phis a sphere whose diameter is the altitude of the seg-ment. PROBLEM 462. By boring- through the center, what is the diameter of an aug-erthat will bore away one-half of a ball 6 inches in diameter? Solution. (1) Let O be the center of the ball. (2) OB=r=3, OF=CD= R, the radius of theauger bit, a
A complete and practical solution book for the common school teacher . 2 XttX KX) = cu. in. (12) The amount consumed is ++ cu. in. Rule for finding a Spherical Segment.—A spherical segment withone base is eqtiivalent to half a cylinder having the same base andaltitude, phis a sphere whose diameter is the altitude of the seg-ment. PROBLEM 462. By boring- through the center, what is the diameter of an aug-erthat will bore away one-half of a ball 6 inches in diameter? Solution. (1) Let O be the center of the ball. (2) OB=r=3, OF=CD= R, the radius of theauger bit, and CB=^. (3) ThenCT)2=R2 = (2r—x)x, and CO =r—x. (4) The volume removed equals a cylinder, ra-dius R, and height 2(r—x), together with two segmentsof the ball of height x, and radius R. (5) The volume of the cylinder is 2(2r—x) XTr(r—x). (6) The volume of the two segments is ^ttx2 (Qr—2x). (7) Adding and equating to half the volume of ball fr2*-, we have 4*3—12^2+12r2^2—4r3=—2r3 . . (1). (8) (l)-^4 = ;t:3—3r-*2+3r2;t2— r3= —\r* . . (2).. FIG. 121. MENSURA TION. 233 (9) &JZ)=x—r=rf% . . (3)._ (10) .-. R* = (2r—x)x=r*(l—jf2) . . (4). (11) 2R=2rV (1—iV2) =6V (1—if 2), diameter of auger bit, or in. PROBLEM blacksmith had an iron ball,Which he did a bullet call;He found out, by some nice trick,It was exactly one foot thick;And silvered oer with plate,And which he proved by scales and weight;The silvers weight he did pronounce,Was just exactly half an ounce;He took the ball and made it hot,And forged there out an iron rod;The rod was just three inches thick,Round and long, just like a stick;Now how much silver will it takeTo plate the rod that he did make,And put it on exact as thickAs that which on the ball did stick?Solution. (1) The volume of the ball is 128X^, or 288tt cu. in. (2) This is also the volume of the cylindrical rod. (3) The area of the end of the rod is (§) 27r=2^7r. (4) The length of the rod must be 288iH-2£*r, or 12
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
