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was called to the British throne - as a We have already given a very full member of that Society, which flour account of this discovery*, which has rishes by the distinguished liberality of justly attracted the notice of all Euits royal patron - and, last of all, rope. We have now to inform our as a person now more immediately readers, that Mr. Herschel, by his imunder the protection of this excellent provements in telescopes, has discovermonarch, and owing every thing to ed a volcano in the moon. The cirhis unlimited bounty, I cannot but cumstance has been thus related: wish to take this opportunity of ex- A few weeks ago, as a lady was presling my sense of gratitude, by looking through his telescope, she giving the name Georgium Sidus, told Mr. Herschel that she saw a “ Georgium Sidus,

light like a star in the moon; upon -jam nunc, affuejce vocari.

which Mr. H. said that he was glad

Virg. Georg. she had taken notice of it, as he had to a ftar, which (with respect to us) long observed it; and was convinced, first began to shine under his auspicious by his observations, that it was no ftar, reign.

but a VOLCANO. * By addressing this letter to you, Such is the story which we have Sir, as President of the Royal Society, heard, and from very good authority. I take the most effectual method of At some future period, if any further communicating that name to the Li- particulars of this discovery transpire, terati of Europe, which I hope they or if the relation should have been will receive with pleasure. I have the falsely stated to us, we shall lay the honour to be, with the greatest respect, whole before our readers. Sir, your moft humble And most obedient servant,

* See our Magazine for July last. W. HERSCHEL."

MATHEMATICS. ANSWERS TO MATHEMATICAL QUESTIONS. . 12. QUESTION (IV. Auguft) answered by the Rev. Mr. HELLINS, Teacher of the

Mathematics and Philosophy,

CONSTRUCTION. ET EFRQ. be the given rectangle, and

K

c the points E and F describe a circle, so that the upper segment of it, ECF, shall contain the given vertical angle. Draw the diameter KG perpendicular to EF, and produce it to meet QR in H. Then, if the seg. EGF falls within ine rect. divide ab in c (by Prob. 17.

E B. v. Simpjon's Geom.) so that ac x cb thall be = KG X GH, and from G in the

G semi-circle GCK, apply GC=ac, also from C through the points E and F, draw lines to

B b cilt QR, produced both ways in A and B,

А. O DH R and ABC will be the triangle required.

DEMONSTRATION. The angle ACB is = the given 'vertical angle by the const. and since the arc EG=FG, it is evident that CD bileets the vertical angle. Join the points KC; then will the triangles GCK and GHD be equiangular and by Theor. 24, B. III. Simpson's Géom. KGXGH=GCxGD. Now GC = ac and acxcb=KGxGH by conit. It, therefore, is evident, that ac xcb-aix GD, and GD=ch, and conkquently that CD=ab the given bisecting line,

It

3 T2

It is obvious that the circle may either touch or cut the line OR, which are the two other cases of the problem; when it touches the line QR, the construction istoo easy to need a descriptions and when it cuts QR, ab must be produced by Prob. 18, B. v. Simpson's Geam. so that cb * ca = GHXGK. The reft of the construction the faine as in the first cate.

This Queftion was also answered by Mr. Isaac Dalby, Mr. Hampfhire, and Mr. J. Meritt.

e

13. QUESTION (V. Auguft) answered by Mr. WiLLIAM Kay,

CONSTRUCTION. Make DA equal the sum of a line drawn to make any given angle with the bale, and the greater segment of the bale, cut off by it; and draw DC indefinitely making the angle ADC =half the angle that the line drawn from the vertical angle is to forin with the base. From A, with a radius equal to the longer side of the triangle describe an arc cutting DC in c and C. Join AC, and from C draw CB to make A

E B the angle ACB = the giren vertical angle, and meeting AD in B: then will ABC be itre triangle required.

DEMONSTRATION. Draw CE, making the angle DCE=the angle EDC. It is manifeft from Eue. 1. 6. that CE = DF, and contequently that CE + EA = DA the line given. Moreover the angle EDC + ECD=CEA=2EDC.

Mr. Robbins, the proposer, Mr. Dalby, Mr. Hampshire, Mr. Merrit, and Mr. Webb, answered this Question. 14. QUESTION (VI. Auguft) answered by Mr. JEREMIAH AINSWORTH, the

Propofer. Bifeet the arch AC in D, draw DEL cutting the circle ABC again in b, and join Ab, bC; allo join DA, DC, EC; and let fall the perpendicu

; lars EG, EM, EF. Then (Euc. II. 13.) DE2 -D02+E02-2DOXFO, but, by conliruction, EO2-KO2 I KOX NH=DO2 F 2DO X FÍ; therefore DE2 — 2DO2 F 2DO X FI-2DOXFO = 2DOX DO FFI-FO=2DO X DI-DLXDI

K DA?, or DC2: consequently DE = DA, = A

1 DC; and therefore a circle described from D, as a center with the radius DC, or DA will pass through the point E. But (Fig. 1.) by Euc. III. 21. the anglebCA=bDA=2ACE (Euc. III. 20.) or (Fig. 2) the 4 ATM=ADE (being fupplements of the equal angles BCA, DA) = 2ACE (Epc. III, 20.) and consequently CE bifects the angle ACB in the first cale, or ACM in the ad cale; also bE bisects the angle ABC, because D is, by construction, the middle of the arc AC: consequently the circle GHM touches the sides of

GI the triangle AbC, and therefore Ab coincides wiih AB, b with B, and consequently BC tonclres

M the circle NMK.

Q. E. D.

Bb

OFM

G

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The Same demonstrated by Mr. ISAAC DALBY.

P

Let a circle circumscribe, and another be in{sribed in the triangle ABC (Fig. 1.) also let a

N

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circle touch CA and the fides BA, BC produced (Fig. 2.) draw BD to bifect the vertical angle, through the centers draw OK, and through O draw DL. Let fall the perpendicular EF and join DA, DC. Then will DE=DC=DA; that is, the locus of the centers of the two circles, one touching ihe fides internally (Fig. 1.) and the other externally (Fig. 2 ) will be in a circle described with the radius DE, DC, or DA: the former (Fig. 1.) is demonstrated, Prob. 67, page 251, Britith Oracle; in the latter (Fig. 2.) draw EC, the external angle ECM-ECA= ECD+ACD=BEC + CBE =BEC + ACD, therefore ECD-DEC; hence DC= DE.

This being determined, we have by Euc. VI. 8. Corol. 2DO (DL) X DI = DC2=DE2=E02 +DO? $ 2DO XOF (hy Euc. II. 13.) the last factor being pofitive or negative as the center E is above or below the center 0; hence EO2 = 2 DI-DO420FDO: but EF, IA, are pendicular to DO, therefore, IFONE the semi-diameter; that is, DO-DIEOF=IF in Fig. 1. Of being positive or negative as E is above or below 0; hence 2D0—2D1E20F=2IFNH, therefore 2DO-NH=2DIE 20F, hence in Fig. 1. EO? DO-NHXDO=OK-NH XOK, because OK=DO; but in Fig. 2. 2DI + 2F0 = 2DO + NH, therefore E02-10+NHxDO-OK+NAXOK,

Corol. If EP be drawn at right angles to EO, then a circle described about P (with radius PE=EN) will touch the circle whose center is 0.

Another Demonstration by Mr. GEORGE SANDERSON.
From any point B, in the circle BKCA, draw BA, and BC tu touch the circle
MNH (see the fig. to the first solution) then AC will touch it in the point G.

Through the points B and E draw BE to meet the circle BKCA, in D, Fig. 1. or to cut it in Fig. 2. draw the diameter DL cutting CA in I, join CD, AD, and AL, allo draw EG perpendicular to AC. Then, because AB and BC touch the circle, and BD palies through the center, it is ma-telt that the angle ABD= CBD-ALD, and that the right angled triangle BME, LAD are fimilar :: BE : EM :: LD (20K): AD (CD) hence BEXCD=20K XEM; but in Fig. 1. ED XEB=KK XOF+OK (Euc. 111.34) =OK2-OEP (Simp. Geom. B. II. Theo. 7) but OE2 - OK-OKHN (2EM) by construction : ED EB - 20K XEM, therefore, DE = DC; whence by Prob. lxvii. page 251, British Oracle, E is the center and EM the radius of a circle inscribed in the triangle ABC.

EDXEB=EK XEO+OK (Euc. III. 36) OE? — OK2 .: DE X EB=20KXEM, therefore ED=CD, hence the angle CDB = 2CEB, moreover, because CD CA, DO is perpendicular to CA, and confequently parailel to EG, but the angle DBC-DCA ; the angle CDO = BEM= CDB+BDO=2CED + DEG, take from both the equals (BEM and 2CED-DEG) the common angle CED, there remains the angle MEC - CEG, but M and G are right angles :: ÉG =EM=| HN as was to be deinonstrated.

In Fig. 1. the lines EF, EG are omitted.
15. QUESTION (VII. Auguft) answered by Mr. ISAAC DALBY.

CONSTRUCTION.
Take AG = 30 m. the

P
ML KI H G

R
diff, of lat. of the ships at
the beginning of the chace,

inake the < GAR= 45°
draw GR +GA, join AR, W
and make RP=2AR; then

N D
GP is the distance run by
the prize, and AR, RP the

S

B
whole distances run by the
Mhip on the larboard and

A
ftai board tacks relpectively.
For A and G being the places of the ship and prize, at the commencement of the .1,
chace, and G right N. from A, and the wind at N. N. W. and our Thip being
clofe hauled, it runs from A, N. E. and W. on her Jarboard and ftarboard tacks
respectively; hence, if A, B, C, D, E, F, &c. and G, H, I, K, L, M, &s, be

contemporary

In Fig. 20

E

contemporary places of the ships respectively, while that from A is running to B, that from G is got to H, and HB is + AB, and while that from B runs to C, the other is arrived at I, and IC is + CB, and so of the other tacks; hence it is evi. dent from the parallelism of the lines that AB+CD+EF, &c. is = AR; and for the same reason BC+DE+FW, &c, = RP: whence this conttru&tion.

To determine AB the run on the first tackBecause 3AR (AR+RP) and GP are the whole distances run, AB and GH, and their equals BO, GO, must be in the fame ratio, viz. 3AR : GP :: AB (BO): GH (GO) therefore take any point n in GA, and in nb whole direction is + AR take nb so that GP: 3AR :: Gn:nb, through b draw GB, and AB is the run on the first tack; then draw BHLAR, and make BC || RP and =2AB, draw CI+CB, and A,B,C, G,H,I, are contemporary places of the tips; now, if AP be drawn, and BP supposed to be joined, it is evident the thip will be in those lines at her respective tacks: this needs no demonstration. For the diftances run on the following tacks, we hare by lim. triang. GA: A3

GSX AB :: IC (GS):

CD, the run on the second larboard tack; and GA:
GA
GS?

GS2

GSP X AB GS (IC) :: GS (IC): =IN, hence GA : AB :: (IN): GA

GA

GA2 = EF the run on the third larboard tack : hence, if we conlider the ships as points, that from A will make an infinite number of tacks before it arrives at the point P; for the runs on the Jarboard racks are the terms of the descending geometrical

GSXAB GS2X AB series, AB,

&c. continued ad infinitum ; where the first GA

GA2

GS term is AB, common ratio and sum = AR, and the terms of the series

GA' doubled are evidently the ruins on the starboard tacks.

C-LCULATION: ✓ ZxAG = 42,426m. = AR the run on the larboard tacks, this doubled is 84,852 =RP, on the Itarboard, which is 4,895 knots per hour; and RP-GR= 54,852 the dift. run by the prize, which is 2,1 knots per hour : and drawing va ll BA, by sim. triang. Ga (Gn+V 2 xbn2) : ab :: GA : AB, that is ✓ 2X3AR?

GS +GP: 3AR :: GA : 16,257 the run on the firt tack, or AB ; and = 0,62414

GA the common ratio of the series.

N.B. We are informed by O'Bryen, in his translation of L'Hofte, that this me. thod of chasing to windward was first put in practice by Adm. Barnet in the Me. diterranean,

This Question was answered very ingeniously by Mr. G. G. and Mr. G. San. derson.

MATHEMATICAL QUESTIONS.

30. QUESTION I. A Mathematical Ænigma. THE difference between the indices of the 4th letter of the first word, and 4th letter of the 2d word, being multiplied into the difference of their squares, will produce 1323: and their sum, being multiplied into the sum of their squares, will produce 10503, the index of the said 4th letter of the 2d word

The sum of the indices of the 5th letter of the first word, and 2d letter of 2d word being added to their product, makes 47; and the difference of their squares is 528; the index of the said zd letter being leaft.

The zd letter of the 2d word, the 2d letter of the first word, the first letter of the 2d word, the 3d letter of the first word, and the 4th letter of the first word are in arithmetical progresion.

The

being leaft.

1

2 3

The sum of the squares of the indices of the 5th and 6th letters of the 2d word is 520; and their product being added to the square of the index of the 6th letter, the sum will be 448; the said 6th letter being leaft.

The sum of the indices of the first letter of first word, and 3d letter of ad word being subtracted from the sum of their squares, will leave 62; and if their product be added to their sum, it will make 35; the said 3d letter being least.

The 7th, or laft letter, of 2d word is the same as the third letter of the first word. Note. If figures be put over the twenty-four letters of the alphabet in this

4 5 manner a, b, c, d, e, &c. I call them indices; because, if the index be known, the letter it belongs to is known; and these letters being disposed as is here directed, they will form the name of the propofer.

31. QUESTION II. by Mr. GEORGE SANDERSON. July the 5th, 1775, the distance of the moon's farthest limb from Antares was observed to be 55° 41' 17.", when the altitude of the moon's lower limb 42° 2' 15", and the altitude of Antares 29° 12' 20"; required the latitude and longitude of the ship, the dip of the horizon being 4' 20".

32. QUESTION III, by STEREOGRAPHICUS. I have a stereographical projection of the sphere on the plane of the meridian, in which the parallels of delineation are drawn to every 5 degrees; but suspecting that the parallels near the equator are not truely arcs of circles, I measured the distance from the center to the intersection of the hour circle of 3, and parallel of 5 degrees of declination; and found it just 7,52 inches the diameter of the primitive being a yard. Froin hence it is required to determine the error in declination (if any) of the point of intersection, supposing the hour circle is correctly drawn.

33. Question IV. by Irk. Three points being given, it is required, by the intersection of right lines only, to find a 4th, in the fame plane, where lines drawn from the former three thall make given angles with each other.

QUESTION V. by Mr. DUFFAUT, of the Academy at Greenwich. A fhip fails in the N. W. quarter from lat. 55o. N. and 5° 30' E. longitude into lat. 58° 20' making her difference of longitude and diltance equal; the then alters her course, and runs into lat. 60°, making the difference of longitude thereby half as much again as the distance; she thifts her course a second time into lat. 63° 24', so that her difference of longitude is double the distance. Required the three courses, the longitude in, and her bearing and distance from the place of her departure.

35. QUESTION VI. by y DRACONIS. Supposing two sides of a spherical triangle to remain constant, it is required to find the relation of its angles when the area is a maximum.

The answers to these questions are required to be fent, poft-paid, to Mr. Baldwin, in Paternoster-row, London, before the ift of March, 1784.

1

NATURAL HISTORY.
TH
THE following defcription of the Boron UPAS, or Poison-Tree, which

grows in the island of Java, and renders it unwholesome by its noxious vapours, has been procured for the London Magazine, from Mr. Heydin

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