ftract this from the Time of the Sun's Entrance in Come, there m mains the Time answering to the 25th of April near NooL. WHEE up Sun's Declination is 16° 43′ 55′′, being nearly the Compremer e Latitude. The Sun being not in Apogee 'till 19th of Jung u. tion, after paffing the Tropic, will be flower that for be adjusted by taking the Difference of Declination at the refpeftiv Times for a Day, and proportioning his Longitude, the Latnude i 73° 16'. Mr. Alexander Naughley fays, the Latitude is st the longest Day from April 24, to July 26, being one Fourth Ferg the Year. Mr. W. Piron, of Petworth, in Suffex, directs tan tenterat the one Eighth of the Year to or from the Time of the Wanker Summer's Solftice, and find the correfponding Declination for ter refulting, and the Complement thereof will be the Late requise ·73°. 15'. 42". N. or S. confirming the Truth at = Solution. Mr. Thomas Allen, of Gosberton, Lincolnshire, makes on Lights the Year (365 d. 5 h. 48'. 54") = 45 d. 15h com Sun's Recefs from either Tropic in that Time = 63 47 35. Thesefore 66°. 31′.+6o. 41′. 30′′.=73°. 12. 30°. He required titude. IX. QUESTION 50, answered by Mr. F. HOLD: CALL the four Hands, H, M, S, and T, wie Wietsen. 1,12,720,43200, refpectively in the fame Time, au to fres of thofe Motions will reprefent the Proportion of ther Welove each other in the fame Time: By comparing which m that in one Fourth of a Revolution of M from H, I will now 1. 2 B 11 12 volution before M; that is, when M has movet íron è 4 Revolution. In the firft of thele Time a'r near Conjunction; in the fecond, H and M are in Cage, Mr. the third and fourth Times only are to be cases t irft, Mr. ir. Gibfon, cant Quarter at the Time of the exact Quadrature of H and S. Now I 2876 Rev. paft the Quadrature, S is found by Proportion to be with M in the former Case, and as much short of it in the latter; and because the Motion of T from S is 60 Times that of S from M, it follows, that S and T are always conjoin'd in the Quadrature of S 60 and M; and therefore T is Rev. before S in the former Cafe ; 2870 and as much behind it in the latter; but the vacant Quarter is the second before S in the former Cafe, and the first behind S in the latter; therefore the latter is to be chofen, at which Time in the exact Quadrature of H and T, the whole Circle being divided into 518388 Parts, H and M will be 129447 diftant; M and S 131724; S and T 127620; and T and H 129597.-The Time being 5 h. 10'. 55". 40′′. . 3༠34༠ from their Conjunction, when all the Hands are first and fartheft .43199 afunder. X. QUESTION 51, anfwered by Nichol. Dixon of Blackwell. LET ABb, s= Time of the Ciftern's fillb ing in Seconds = 1500"; then s: h : : 1′′ : ~ the Velocity of uniformly filling, when the eva uniformal Velocity of Defcent, evacuating Cock being open, and supplying the Cock ftopt; then B Velocity of the Surface upwards at C, both Cocks being open. Again, Mr. Emerfon's excellent Doctrine of Fluxions, and corrected by the Rules given in Problem 12. p. 102. of the fame Book) is b x 2 L × Log. X Log. ; therefore the whole Fluent is 25 L X Log. Mr. F. HOLDEN'S Solution. Let a the whole Depth of the Ciftern, and x any Depth, in Inches, required; then a is the Velocity with which the Water 즐 would be raised by the supplying Cock; x2 the Velocity with which 플 it would be depreffed by the evacuating Cock; whence a2 -x2will will be the Velocity with which the Water rifes, when both Cocks are opened; the Time (being as the Quantity divided by the Velocity) Fluxion of the Time, whofe Fluent is 2px2+2pa ---x (that the Flu Now, because the fupplying Cock ftract 10 x from 50 I 5 Log. will always be negative) the Remainder will be the Time in Minutes of the Ciftern filling to any Height x, whence Times Hyp. Log. of -- x (because the 1.1565 cant Quarter at the Time of the exact Quadrature of H and S. Now S is found by Proportion to be I 2876 Rev. paft the Quadrature, with M in the former Cafe, and as much fhort of it in the latter; and because the Motion of T from S is 60 Times that of S from M, it follows, that S and T are always conjoin'd in the Quadrature of S 60 and M; and therefore T is Rev. before S in the former Cafe ; 2870 and as much behind it in the latter; but the vacant Quarter is the second before S in the former Cafe, and the firft behind S in the latter ; therefore the latter is to be chofen, at which Time in the exact Quadrature of H and T, the whole Circle being divided into 518388 Parts, H and M will be 129447 diftant; M and S 131724; S and T 127620; and T and H 129597.-The Time being 5 h. 10. 55". 40′′. 3༠34༠ from their Conjunction, when all the Hands are first and fartheft 43199 afunder. X. QUESTION 51, anfwered by Nichol. Dixon of Black-well. LET AB=b, s Time of the Cistern's fillb ing in Seconds = 1500"; then s:b:: 1": b-bx being open, and supplying the Cock stopt; then = abfolute Velocity of the Surface upwards at C, both Cocks being open. Again, Mr. Emerfon's excellent Doctrine of Fluxions, and corrected by the Rules given in Problem 12. p. 102. of the fame Book) is b2 x 2 L X Log. min. ; therefore the whole Fluent is 25 L X Log. 25x t. 1.15717 5.54246 Mr. F. HOLDEN'S Solution. Let a the whole Depth of the Cistern, and x any Depth, in Inches, required; then a2 is the Velocity with which the Water would be raised by the supplying Cock; x2 플 the Velocity with which it would be depreffed by the evacuating Cock; whence a' -x2will be the Velocity with which the Water rifes, when both Cocks are opened; the Time (being as the Quantity divided by the Velocity) Fluxion of the Time, whofe Fluent is 2px2+2pa ¿1-x2, or 2px2+2pa1× Hyp. Log. a a Now, because the fupplying Cock raifes the Water one Inch in one Minute p musta 2 must = 1, i. e, a 25 =5; fo that the Height x being Inches, fub I ftract 10 x from 50 Times Hyp. Log. of 1- x (because the 5 Log. will always be negative) the Remainder will be the Time in Minutes of the Ciftern filling to any Height x, whence |