lities, learning and patriotifm. 2. Men of letters receive no patronage from the great in this ifland. This is a difhonour to them, not to those who stand in need of their countenance and protection. If learning, as Ovid remarks-cmolfit mores nec finit effe feros-foftens the ferocity of rude nature, and polishes our manners, we may eafily determine the precife ftate in which the minds of thofe great are, who fuffer learned merit to pine in obfcurity and penury. Alexander looked to letters for immortality and not to his victories. The paffage in the Archia poeta oficero deferves notice. "Alexander, cum in Sigao ad Achillis tumulum adftitiffet, O fortunate, inquit, adolefcens, qui tuæ virtutis Homerum præconem inveneris.-Et verè, nam nifi Ilias extitiffet, idem tumulus qui corpus ejus contexerat, nomen etiam obruiffet." 3. The Fellows of our University, from whom much is expected, have not that -otium cum dignitate-which they fo highly merit. Confined for many years to a small pittance which we hope to fee enlarged from the ample revenues of the foundation or engaged in the inftruction of youth and collegiate duties, they have neither the means or the opportunities of rendering themfelves eminent in the walks of literature. When they arrive to the rank of Seniors, they are fo far advanced in life, that their habits are formed, and compofitions of great extent or importance cannot be expected from them. Not one publication worth notice has appeared this month, yet the literary genius of Ireland is ftill alive, its fcintillations are visible in the Anthologia, which has hitherto, and continues still to be honoured by the productions of men of taste and erudition. The originality of its materials, and the beauty of its engravings will be found equal to those of the belt Magazine in Europe. does the Editor defire or expect the favourable and ample patronage he has as yet experienced, longer than he is able to promote the caufe of literature, virtue, and the arts in Ireland. Nor MATHEMATICS. Solutions to Queftions in the Anthologia Hibernica, for the Month of February 1793, by Cornelius Kennedy. Solution to the 2d Queftion. Ivide 37440 by 10.4X60 we get 60 whofe fquare 3600 & 3600+603660 the greatest impetus and Vzo +60 2 b the diftance of the object on the top of the perpendicular, then b: 60 :: rad. to fine of an angle which put = d and b: 3660x2-60% 60 to m also 3660Xm 2873 (p) then 3660: p: rad'. : fine 51.43 (C) then c-d-double compl. elevation and 90 levation let e its fine, then R-e: R:: 3660×2: y the greatest randon R—e : yey then 3660+ey 10.999.19. Impetus at the object on the defcending plane confequently. 10. 4×60× 3660xey=65145.6 the force in pounds as required. Solution Solution to Question 3. CONSTRUCTION, DEMONSTRATION. Because the altitude DABG=AB= the LGAB=45, and from the nature of the circle 2 V2 r x-xx=QM and xX V2 r x-x xarea of the parallelogram Qm which let be multiplied by PSPA and × × × 2 V2 r x-x x=2 x x the fluxion of the ungula AGE. T Solution to Queflion the 4th. rx-xx O inveftigate this question from first principles would take up too much room and time. I have inveftigated it by the method used by Emerson in his fluxions for finding the centre of gravity, &c. and found the greatest horizontal force 18 X 16 2000lb. 2250 and the greatest perpendicular force =2000XI! =5000lb. Solutions to the Queflions in the Anthologia Hibernica, for February 1793, by Solution to Question 3d Draw DB=2r, and upon it defcribe a Square, and a Semicircle, then will the Square reprefent the Section C thro' the Axis of the Cylinder, and the Semicircle half the Bafe. Now put DG=x then will GB=2r-x and per nature of the Circle 2 GK-2 \jzrx -x 2 which multiplied by gives 2 x rx-x2 - the double Area of the fpace G Kar and becaufe DGK is right and the angle GDL half a right angle GL= GD hence 22rx-x2X LG=DG gives 2xx D KL n S B 2xx Varx-x2 or quired. for the fluxion of the Ungula DAB as re 2rx-x2 .18 16 Solution to Question 4th. By the nature of the Center of Sufpenfion, Ofcillation and Gravity, we find = x 2000 2250lb. the greatest Horizontal force, and by the pofition of the Axis of the Ofcillating Body when its force on the Center of Sufpenfion in a perpendicular direction is the greatest, we get greatest Perpendicular Force as required. I + 12 X 2000 = 5000lb. the Scho. From this ingenious prob. it appears that the force of Bells when rung in peal, must be great on their Axis of Motion, and on the frame of Walls that refifts. Mr. Anderson's Solution of the Queflion, proposed by him, Athologia Hibernica, p. 136. SOLUTION. By the method of finding the Centres of Gravity and Ofcillation of Bodies, it appears that the diftance of the Center of Gravity from the point of Sufpenfion, is 2 feet, and that of Oscillation 2, and by an easy procefs in Mechanics we get 3×2 = 6×3 18 = X 2000 = 2250lb. the greatest Horizontal force; alfo 2x2 2000 x 1+ 812 16 12 = 2000 × = 5000lb. the greatest perpendicular force. Mr. Mooney's Solution of the two Queflions propofed by him, Anthologia Hibernica, page 131. Question 1. CONSTRUCTION. L ET any given right line A B be fo cut in C (per 11. E. 2.) that the rectangle ABC may be equal to the fquare of AC; then from the centre A, with the radius AB defcribe the circle BDE, in which infcribe BD equal to AD, and join DC; then the triangle ACD is that which was required. E A D B DEMONSTRATION, DEMONSTRATION. From the 10. E. 4. AC-DC, and the angle CDB CAD-CDA; likewife the angle DBC BCD but per 32 E. 1, angle DC B⇒CAD+CDA, confequently angle DBC, is double CAD or CDA, and angle ACD=CBD+BDC therefore the angle ACD CAD+CDA+BDC, but fince CAD=CDA= BDC, the angle ACD is treble of either of them; that is done, therefore, which was required. 37440 10.4x60 SOLUTION OF QUESTION 2. 2 2 = 60 whose square=3600 feet the impetus at the top of the castle then 3600-460—3660 the greatest impetus of the piece and (200)+60) 208. 8= the distance of the object on the top of faid caftle then 208.8: 60:: Rad'. to fine 3660X2086 270.6′ and 208.8: 3660X2-60 ::60: 2086 also 2873 then 3660:2873: Rad3. to fine 51°. 43′ now 51°. 43'-27°. 6'=24°. 37′=double the Comp. of the elevation hence 90° 24°37′=770. 41'=the elevation on the plane of the horizon the depreffion of the plane then Rads.-fine 77°. 41 Rads. :: 3660x2: 319130 feet the greatest randon on the defcending plane, laftly Rad. fine 77°. 41 319130 7339.99 then 3660+7339. 9=10999. 99=the impetus at the object on the defcending plane, confequently 10.4x6cX 10999.99-65145. 6lb. the abfolute force with which the object on the defcending plane is ftruck, as required. 2 Answer to the Geometrical Queftion which requires the Conftruation of an Ifofceles Triangle whofe vertical Angle is treble either of the Angles at the Bafe. LE ET the vertical angle be that of a regular pentagon, which is, three fifths of two right ones, and therefore treble either of the remaining angles of the Ifofceles. SEMAJ NOTLIMAH. Question 5 G IVEN the difference of the fquares of the fides and the area of a rectangle to find the fides. Dublin, March 25th, 1793. SEMAJ NOTLIMAH. Question 6, by Mr. Michael Mooney, near Dublin. Quære, on what day of the prefent Year, will the Sun's afcenfional difference and midnight depreffion, be the fame in the City of Dublin? I SIR, To the Editor of the Anthologia. SEND you the following new queftion, and hope if it meet with your approbation you'll give it room in your useful Magazine for April next. Ο Question 7, by Cornelius Kennedy. Na certain part of the earth I obferved the latitude and longitude of a star when on the meridian to be 4°. 36' 8" South, and 65°. 5'. 36" refpectively, and at the fame time the refraction was found to be 1'. 8' 2". Quære the latitude of the place of observation. ERRATA TO THE THEORY or COLUMNS. Page 209 1. 10 from the bottom, for o6) read o 1. 5 from the bottom, for 2 cb2a — 2 cn2 b2 a— 3 e b2 a read 2 cb2a — 2 cu3 b1a — 3 c b2 a 3 Page 210 5 from bottom for 426Xb2 read 42b+b2. |