M F will ben2p2x a + 3nb the Distance fought. 1654320 Inches (26: 1, nearly) N. B. If the Curve AMB be confidered as the Involute of Archimedes's (or any other Kind of) Spiral, the Length thereof will be always found, by multiplying Nn and its Fluxion together, dividing the Product by the Radius of Curvature of the Evolute (at #), and then taking the Fluent of the Quotient. XII. QUESTION 432 answered by Mr. O'Cavanah. Y Divifion, and extracting of the Square Root on both Sides of the BY ? propofed Equation, we have X -I is further tranf น : whence, fuppofing A and B to denote the two circular Ares whose Tangents are y ( and u n 2 we fhall, by taking the Fluent, have A+ C = X B; C being a conftant Arch, ferving to correct the Fluent. Pq From this Equation, when either of the Quantities z or x is given, the other may be determined (from a Table of Tangents) in all Cafes, except when imaginary Quantities enter into the Confideration: Thus, ≈ being fuppofed given, the Tangent u I will also be given; and from thence (by the Table) the Arch B correfponding; whereby the Arch A X B- C), and its Tangent y, will be known, and When 7, p and q are such, that‡ √. then may the Relation of ≈ and ≈ be algebraically expreffed, by a finite x Equation: For, here the Fraction 12 is a rational Quantity, where and s are Integers, may be affumed = ; and we fhall have s X A+C=rB: But, if e be taken to denote the Tangent of the Arch C, That of A † C will be= Whence (by the Theorem at p. 41. in Ladies Diary for 1755) the Tangent of s xA+C will be had s X Icy 2 &c. which must neceffarily be equal to the Tangent of B = r being Integers, the Series's will terminate, and the Relation of y (= Mr. Rollinfon (the Propoler) answers this Question, in almoft the very fame Manner. Peter Walton has brought out an elegant Solution to it, from p. 66 of Landen's Mathematical Lucubrations; and, in a Way not greatly different, the Solution is alfo given by Mr. R. Butler, and Mr. L. Charlton. XIII. QUESTION 433 anfwered by Peter Walton. L ET A denote the Force which, acting at the End n, at right Angles to the Rod, would accelerate the Velocity of that End about (G) the Center of Gravity of the Rod, at the fame Rate as the faid Velocity is accelerated by the Action of the Rod against AB; and let A" denote the Force which, acting at the End m, at right Angles to the Rod, would retard the Velocity of that End about G, at the fame Rate as the faid Velocity is retarded by the Action of the Rod again AC: Alfo let & be the Velocity of m, or n, about G; v, the Velocity of G towards the Horizon; w, its Velocity in a Direction parallel to the Horizon; t, the Time the Rod has been in Motion, while it touches AC; T, the Time from the Rod's ceafing to touch AC; d, the Distance of G from A B, at the Commencement of the Motion; x, the Distance of G from A B, at any Time after the Motion has commenced ; 24, the Length of the Rod; s= 16 Feet; and y=√ a2 — x2. Then will the abfolute Weight of the Rod, the Preffure against A B, and the Preffure against AC, be as 25, a A. a A" and respectively: 3y 3 x By Means of which Equations, we find 3 w ww+u u = while the Rod continues to touch A C. Hence, by taking the correct Fluents and expunging and w, we have v = √3ds-3sxxy = 3 ds-3sx x Van. 2 ; whofe Fluxion is = o, and 2 d a confequently x== when A" o, i. e. when the Rod caufes to 3 ax touch A C. Moreover t is correct Fluent, when x is 2 d is the Time required in the Question. After the Rod has ceafed to touch AC, (A" being =0, and w inva avy. From which Equations, we get u u- S x Hence, by taking the correct Fluents, expunging u, and putting d— d3 ; whereby the Time from the Rod's ceafing to touch A C, to its Coincidence with AB, will be found; from whence and the given (invariable) Velocity of the Center of Gravity from the PlaneA C,during that Time, the required Distance of the End of the Rod from A C will be obtained. Mr. O'Cavanab has alfo obliged us with a Solution to this, very difficult, Problem, which agrees in every Particular with that exhibited above. the Distance of the Center of Ofcillation (P) from the End = (con<fidered as the Point of Sufpenfion), it will appear, in like Manner, "that the Celerity with which the faid End recedes from the Plane “AC, -b-dxyy When this Quan "tity is a Maximum, the End (m) will quit the Plane A C; after which "the relative Celerity wherewith n recedes from M H, will be 2 y X "df x T — y dd k by y wherein r is a conftant Quantity, to be deter "mined (like m above) from the Value of y, when the End m ceases 66 to touch the Plane A C." *Note, That, in this Author's Solution, d= Mn, y = MH, c = the first Value of y, and ƒ= 16 Feet. 12 The PRIZE-QUESTION answered by Mr. E. Rollinson. LET ABCD reprefent theCafk,orVel- A mum, when that of becomes equal to D KG L I H B the given Quantity IE. Hence, by the general Rule for the Refolution of Ifoperimetrical Problems (Vid. p. 101 of Simpson's Miscellaneous Vyy - aa yx ; and confequently xa x Hyp. Log. y+ Vyy aa a anfwering to the Catenaria; in which a IG (because when y = a, x will be o.) = The Equation of the Curve being thus known, the Fluxion of the generated Solid, 3,1416 × y y x, will be found = 3,1416 a X aaax; the Double of which, when x, and y= b, will be the whole Content of the Veffel. But now to find a (which is yet unknown), the general Equation, b, and xc, will give ca X Hyp. Log. b+Vbb-aa H. L. come Hyp. Log. + √v v — I = =0, 5 v (in the present Cafe) whence is found = 1,1788; and from thence a = 31,388. Hence the Content will come out 16121 Cubic Inches, or 69,79 WineGallons. And the Curve-Surface will be lefs than that of the circumfcribing Cylinder, in the Proportion of 95,54 to 100. The Solution by Mr. Rollinson, here given, being the only true one that came to hand before Candlemas-Day, that Gentleman is therefore intitled to a Prize of twelve Diaries, without the Chance of drawing Lots. New MATHEMATICAL QUESTIONS to be answered in the next Year's DIARY. I. QUESTION 434 by Mifs T. S―e; Addreffed to Mr. U.T -r, who took the Liberty to afk ber the following Questions, viz. What Age? What Fortune? And what Height fhe was ? MY Height, Sir, in Inches, is three Times my Years ; My Fortune their Squares will both fhew; Put all thefe together, there then, Sir, appears From which, Sir, determine the Things you requir'd'; As Lovers of Science I always admir'd, Thofe Favours, perhaps, I may grant. II. QUESTION 435 by Mr. Patrick O'Cavanah. A Tortoise once (or top lies) (*4494% But half the Ground fhe run before ; Run with a Hare, and won the Prize;And in the third was only reckon'd The Hare, in the first Minute's Space, One half the Space the went the fecond'; In Ratio Geometrical: Mean Time the Tortoife ftill crept on, III. QUESTION 436, by Mr. John Brickland, Teacher of the Mathematicks in Oxford. WH THAT is the Side of that Equilateral Triangle, whofe Area coft as much paving at 8 d. a Frot, as the pallifadoing the three Sides did at a Guinea a Yard? IV. QUESTION 437, by Mr. Thomas Mofs. FROM the vertical Angle of a Triangle (whofe Bafe is 70, and its two Sides 40 and 50) to draw a Line, terminating in the Bafe, fo as to be a Mean-propertional between the two Segments of the Bafe made thereby. Ti V. QUESTION 438, by Mr. Peter Walton. O place three Circles (whofe Kadii are 10, 15, and 20) fo that three RightLines may be drawn each, to touch all the three Circles. VI. QUESTION 439, by Mr. W. Bevil. TH SW, two Ports do bear, Whofe Distance, if I rightly gu fs, Is fifty Leagues, nor more nor lefs; Between which Ports a Current goes, SE, three Knots, as I fuppofe : The Quere is, how muft I fail From northern Port, in five-knot Gale, |