=2; and by writing for 22 in the Value of HS, و we have y VHS, and y√2 x x √22yx = the ey Area of the Parallelogram. Again put v OP, and by the Property of the Curve xy::x : v2 •.• y2x2—y1v2= 2x2 and v2 and 422 = 4y2. fuppofed Square, but 4x yo=ay= 16x22 a the Area of the a y and y2= 4xc and by putting for y and its Value above, we have the Area of the Parallelogram, and 2 c the Area of the Square. Then by the Queftion 3:2:: 2c2 16x + c2+a2 c 16x+c2+a2° This Equation re duced gives x√3a+a√5 76,4204 the Transverse, and 2 y29,1917 jugate Diameters. W. W. D. The fame anfwer'd by Mr. Jof. Pilgrim. the Con Put b 1752 the given Area of the Ellipfis, whofe Diameters AD and RT are to be 'determined, c3.1459 the Area of a Circle whofe Radius is Unity, a=OR the SemiConjugate and ́e the Semi Tranfverfe. Then 1:c:: ae b ca cae the Area of the Ellipfis whence caeb, and—e. Again fuppofe the Radius of a Circle, then 0.5 aa-half b ca the Side of its infcribed Square. But a:::V0.5aa: OE half the Length of the greatest Parallelogram that can be inferibed in the Ellipfis (as may easily be demonstrated.) dba Let do. Then the fame half Length, and da caa RO the half Breadth of the Parallelogram. Confequently ddb c the Area of the Parallelogram. Again, put = ON half the Side of the greateft Square that can be in ;uu (per Conics) whence' bbcca a bbcc + ctat uu of the Area of that Square. But (per Quest.) 3:2:: 36 b c ca a bbcctct and confequently bb36caa-cca*. Whence by Divifion, Tranfpofition, and compleating the Square a a = 36 2 C V9bb 4cc bb 213 0147, and a CC 14,5950. Whence RT29,19, AD=76.4206, EF 54,0375. The Parallelograms Breadth 20,6404, and PN Hence the following RULES. 27,2685. ift As I: V0,5: the conjugate Diameter of any Ellipfis: the Breadth of the greatest infcribed Parallelogram. And Tran. Diameter: the Length of the greatest Parallelogram. 2d As 0,785397 the Area of any Ellipfis:: 0,5: the Area of the greatest Parallelogram that can be infcribed in it. Many other Perfons Names may be seen in the Catalogue, who truly folved this Question. (2) Queft. 112 by Mr. Cha. Smith. By the Question the Content of the Glafs is ,523598x3= I 0,523598 (1,909862) then x3— nxx, hence Log. x × 3 = Log. n + Log. xxx, or Log. ,2810019 × × × =Log. ××3 —Log. »'.' (x=3-Log.") 3- Log. x x, from this Equation x is greater than 2,15 and less than 2,16; whence (per Sherwin's Tables) x is easily determin'd 2,1596,523598x35,2737305, or xx5,27373 exact to 5 Places both Ways. Hence 52,7373 Cubic Inches or 14 Pints and 2,2068 Inches is the Quantity of Wine drank. The fame anfwer'd by Mr. Steph. Hartley. Let 6,7854, then will Log. xxx, and by a common Table of Logarithms (and your laft Year's Diary, p. 31.) x is found 2,1601 Inches, and the Wine drank 1,8274 Pints. The The fame answer'd by Mr. Rich.Gibbons. Put 60,5236, then by Queft. bxxx, now was b 0,5 then x2, but as it is a small matter more, fuppofex 2,1 its Log. is 0,3222193, and the Logms of bx3 -xx is equal 0,008997; and, by taking the Log" of 2,2— 0.3424227,then the Logarithms of bx3xx is 0,0071622, thefe two Logms (and not their natural Numbers) wrought by Trial and Error, give a new Log. of x 0,3334703 and by taking the Log. of 2,16 for a second Value of x, by Trial as before we get the Log. of x=0,3343813, by which Log. of bx3 is 0,7221434 Confequently x is 2,15964. and of -xx is 0,7221433S This Method of working converges fafter than any other Method I know of; and is evidently a further Improvement in this useful Doctrine. This Queftion was curiously folved by Mr. Jof. Orchard, Writing-mafter and Teacher of the Mathematics in Gofport, Mr. J. Holden, Mr. Geo. Godhelp, Mr. R. Butler, Mr. Rob. Peckham, Mr. Wm Spencer, Mr. Tho. Garrard, Mr. Tho. Mofs, Mr. Wm Kingfton, Mr. John Wigglefworth, Mr. John Williams, Mr. Wm Abbott, Mr. Paul Sharp, and many others, as may be seen in the Catalogue. B (3) Queft. 113 answered by Tabularius. I Given BODO OE -- ACb, and AO=c; ..OC= bcm; Let GCx, then GQ m+x, and, per 47 E. 1. GE GF=√a^m2-zmx-x2, whence x √ a2 m2+x2 or a2x2 — m2x2 -zmx3- -x+ is a Maximum: in Fluxions 2a2xx — zm2xx --b m x2 x 4 4x3 x=0... a2 — m2—3mx—2x2—0; whence by Reduct. x = a2 m 3 m 4 ; and thence the Area is known. The fame answer'd by Mr. Wm Enefer. It may easily be demonftrated that the Parallelogram required muft ftand at Right Angles with the Diameter that paffes through the Centres of both Circles. It is likewife evident, from the Construction of the Lune, that this Problem admits of three Cafes. For In the annex'd Scheme (fuited to the ift Cafe) let BEIFDCB be the given Lune, EFC the Parallelogram, then we have given by Queft. OBOE OF≈a, AC6 and OAc, put GC (the Width of the Parallelogram), then GIa-b +c-x and AG=a+b-c+x. Th. per 35 E. 3. AG x IG •GE)2 = a2b2 + 2 b c 2 b x c2 + 2cx - x2, .. - c2+20x `x√ a2-b2-2bc-2bx-c2+zex+x half the Area of the Parallelogram a Max, in Fluxions 242xx—2b2xx+4b cxx6b2xx — 2c2xx+6‹x2x-4x3x=0, which Equation being a2-b2+zbc-c2 properly ordered gives x2 + 36—3c 2 2 Xx= 2 whence by compleating the Square, extracting the Root, and 30-36 18a2+b2 — 2br+c2 +V proper Reduction, x= 4 16 a ge neral Theorem for the 1ft and 2d Cafts; and fince in the laft Cafe b muft of neceffity be equal to c, therefore x = a√ a Theorem for the last Cafe. The fame anfwer'd by Mr. Will. Spencer. Given OE=a, AC=b, AO=c, then is OC = b -C, which put equal d. Let x = CG, the Breadth of the Parallelogram, then is xd OG, and (per 47 E. 1.) √√a2x2-2dx— EGGF: hence the Area of the Parallelogram is 2x √√2x2- 2xdda Max. in Fluxions za2xx-4x3x-6dx2x —2d2x x=0..42, 2x2-3dx — d2 a2-d2 2 › x= whence xa+d2 16 - 3d, which is a general Theorem for a Parallelogram. Curious Answers to this Question were received alfo from Harland Widd, Mr. J. Holden, Mr. Charles Smith, Mr. Geo. Gedhelp, Rimfide, Mr. Rub. Butler, Mr. Stephen Hartley, Mr. John Watchorne, Mr. Philip George, Mr. Tho. Garrard, Mr. Wm. Kingstone, Mr. Richard Gibbons, Mr. John Wigglefworth, Mr. John Williams, Mr. Paul Sharp, and Mr. Abr. Lord, the Propofer. (4) Qaeft. (4) Question 114. answer'd by Mr. Tho. Mofs. Now as the Port is, by the Question, 250 Leagues due West off I, it is therefore manifest that it must lie in the great Circle EIF paffing thro' I at Right Angles to the Meridian PHI; therefore let IT be the given Distance, draw the Meridian PTD: Then in the Right angled Spherical Triangle PIT are given PI (the Comp. of Lat. of the Place of Obfervation) and TI, whence PT (the requir'd Comp. of Latitude) and alfo the Angle IPT (the Port's Difference of Longitude from the Meridian PHI) is eafily obtained: Moreover here is given Hr the Difference of Latitude, and HM the Distance fail'd; whence by Mercator's Sailing the Diff. of Longitude (or the Angle HPM) may be found, and confequently the requir'd Longitude of the Port T (or the Angle TPL) for the Angle MPL is given by the Queftion. 2.E.D. |