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ASTRONOM Y.

WE have received the following letter from the ingenious and worthy Mr. Magellan. If any of our readers fhould have feen the comet which Mr. Pigott mentions, we fhould be much obliged to them if they would favour us with their obfervations. We are happy to be the first to announce this appearance to the aftronomical world.

FOR THE LONDON MAGAZINE.

EXTRACT OF A LETTER FROM EDWARD PIGOTT, ESQ OF
YORK, TO MR. DE MAGELLAN, F. R. S.
Dated November 22, 1783.

"SIR,

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WE fhall now lay before our readers Mr. Herfchel's letter to Sir Jofeph Banks, refpecting the name of the ftar which he difcovered fome time fince: A LETTER FROM WILLIAM HERSCHEL, ESQ. F. R. S. TO SIR JOSEPH BANKS, BART. P. R. S.

FROM THE PHILOSOPHICAL TRANSACTIONS.

"SIR, "BY the obfervations of the most eminent astronomers in Europe it appears that the new ftar, which I had the honour of pointing out to them in March, 1781, is a primary planet of our folar fyftem. A body fo nearly related to us by its fimilar condition and fituation, in the unbounded expanfe of the ftarry heavens, must often be the fubject of the converfation, not only of astronomers, but of every lover of fcience in general. This confideration then makes it neceffary to give it a name, whereby it may be diftinguished from the rest of the planets and fixed ftars.

"In the fabulous ages of ancient times, the appellations of Mercury, Venus, Mars, Jupiter, and Saturn, were given to the planets, as being the names of their principal heroes and divinities*. In the prefent more

. In

philofophical era, it would hardly
be allowable to have recourse to the
fame method, and call on Juno, Pallas,
Apollo, or Minerva, for a name to our
new heavenly body. The firft con-
fideration in any particular event, or
remarkable incident, feems to be its
chronology: if in any future age it
fhould be afked, when this laft-found
planet was difcovered? It would be a
very fatisfactory answer to fay,
the reign of King George the third.”
As a philofopher then, the name of
GEORGIUM SIDUS prefents itself
to me, as an appellation which will
conveniently convey the information of
the time and country where and when
it was. brought to view. But as a sub-
ject of the best of Kings, who is the
liberal protector of every art and
fcience-as a native of the country
from whence this illuftrious family

* M. DE LA LANDE'S Aft. § 639

was

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We have already given a very full account of this difcovery*, which has juftly attracted the notice of all Europe. We have now to inform our readers, that Mr. Herfchel, by his improvements in telescopes, has discovered a volcano in the moon. The circumftance has been thus related:

A few weeks ago, as a lady was looking through his telescope, the told Mr. Herfchel that the faw a light like a far in the moon; upon which Mr. H. faid that he was glad she had taken notice of it, as he had long obferved it; and was convinced, by his obfervations, that it was no ftar, but a VOLCANO.

Such is the story which we have heard, and from very good authority. At fome future period, if any further particulars of this difcovery transpire. or if the relation fhould have been falfely ftated to us, we fhall lay the whole before our readers.

* See our Magazine for July laft.

W. HERSCHEL."

MATHEMATICS.

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

L

Mathematics and Philofophy.

CONSTRUCTION,

ET EFRQ be the given rectangle, and

ab the given bifecting line. Through the points E and F defcribe a circle, fo that the upper fegment of it, ECF, fhall contain the given vertical angle. Draw the diameter KG perpendicular to EF, and produce it to meet QR in H. Then, if the feg. EGF falls within the rect. divide ab in c (by Prob. 17. B. v. Simpson's Geam.) fo that ac × cb fhall be KG × GH, and from G in the femi-circle GCK, apply GC ac, alfo from C through the points E and F, draw lines to cut QR, produced both ways in A and B, and ABC will be the triangle required.

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DEMONSTRATION.

The angle ACB is the given vertical angle by the conft. and fince the arc EG=FG, it is evident that CD bifects 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. KG×GH=GC×GD. Now GC = ac and acxcb=KGxGH by conit. It, therefore, is evident, that acxcbac×GD, and GD=cb, and confequently that CD ab the given bifecting line. 3 T2

It

It is obvious that the circle may either touch or cut the line QR, which are the two other cafes of the problem; when it touches the line QR, the construction is too easy to need a defeription; and when it cuts QR, ab must be produced by Prob. 18, B. v. Simpson's Geom. fo that cb x ca GHXGK. The rest of the conftruction the fame as in the firft cafe.

This Queftion was also answered by Mr. Ifaac Dalby, Mr. Hampshire, and Mr. J. Meritt.

13. QUESTION (V. Auguft) answered by Mr. WILLIAM KAY,
CONSTRUCTION.

Make DA equal the fum of a line drawn to make any given angle with the bale, and the greater fegment 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 form with the bafe. From A, with a radius equal to the longer fide of the triangle defcribe an arc cutting DC in C. Join AC, and from C draw CB to make the angle ACB the given vertical angle, and

and

A

meeting AD in B: then will ABC be the triangle required.

DEMONSTRATION.

E B

Draw CE, making the angle DCE the angle EDC. It is manifeft from Eue, I. 6. that CEDE, and confequently that CE + EA DA the line given. Moreover the angle EDC +ECD CEA=2EDC.

Mr. Robbins, the propofer, Mr. Dalby, Mr. Hampfhire, Mr. Merrit, and Mr. Webb, answered this Question.

14. QUESTION (VI. Auguft) anfwered by Mr. JEREMIAH AINSWORTH, the

Propofer.

Bifect the arch AC in D, draw DEL cutting the circle ABC again in b, and join Ab, bC; also join DA, DC, EC; and let fall the perpendicu lars EG, EM, EF. Then (Euc. II. 13.) DE2

DO2+EO2—2DOXFO, but, by contruction, E02 K02 KO × NHDO2 2DO × FI; therefore DE2=2DO2 2DO × FI-2DOXFO =2DO×DO FI-FO=2DO × DI=DLXDI = DA2, or DC: confequently DE = DA, = DC; and therefore a circle defcribed 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 angle bCA≈bDA=2ACE (Euc. III. 20.) or (Fig. 2) the ACMADE (being fupplements of the equal angles ¿CA, ¿DA) 2ACE (Enc. III. 20.) and confequently CE bifects the angle ACB in the firft cale, or ACM in the zd cale; alfo bE bifects the angle ABC, because D is, by conftruction, the middle of the are AC: confequently the circle GHM touches the fides of the triangle AbC, and therefore Ab coincides with AB, b with B, and confequently BC touches the circle NMK. Q. E. D.

The Same demonftrated by Mr. ISAAC DALBY.

Let a circle circumferibe, and another be infcribed in the triangle ABC (Fig. 1.) alfo let a

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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 the fides internally (Fig. 1.) and the other externally (Fig. 2) will be in a circle defcribed with the radius DE, DC, or DA: the former (Fig. 1.) is demonstrated, Prob. 67, page 251, British 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) × DI =, DC DEEO2+DO2±2DOX OF (by Euc. II. 13.) the laft factor being pofitive or negative as the center E is above or below the center O; hence EO2 = 2 DI➡DO±10FXDO: but EF, IA, are pendicular to DO, therefore, IF NE the femi-diameter; that is, DO-DI±OF÷IF in Fig. 1. OF being pofitive on negative as E is above or below O; hence 2DO-2Dα2OF=2IF_NH, therefore 2DO-NH2DI±20F, hence in Fig. 1. EO2=DO-NHXDO=OK-NH XOK, because OK-DO; but in Fig. 2. 2DI + 2FO = 2DO + NH, therefore EO-DO+NHxD0=OK+NHXOK,

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 Demonftiation by Mr. GEORGE SANDERSON.

From any point B, in the circle BKCA, draw BA, and BC to touch the circle MNH (lee the fig. to the firft folution) 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, alfo draw EG perpendicular to AC. Then, becaufe AB and BC touch the circle, and BD pafles through the center, it is made that the angle ABD= CBD-ALD, and that the right angled triangle BME, LAD are fimilar EM :: LD (20K) : AD (CD) hence BEX CD=20K×EM; but in Fig. 1. ED XEB=FKXUE+OK (Eue. III. 3¢) —CK2—OE2 (Simp. Geom. B. II. Theo. 7) but ОE2 ≈ OK2—OK×HN (2EM) by construction · ED × EB - 20K × EM, therefore, DE DC; whence by Prob. Ixvii. page 251, British Oracle. E is the center and EM the radius of a circle infcribed in the triangle ABC.

BE:

In Fig. 2. EDXEB=EK×EO+OK (Euc. III. 36) – OE2 — OK2 ·: DE × EB=20KXEM, therefore ED=CD, hence the angle CDB 2CEB, moreover, because CD CA, DO is perpendicular to CA, and confequently parallel to EG, but the angle DBC=DCA. the angle CDO BEM = CDB +BD0=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. EG EM HN as was to be demonftrated.

In Fig. 1. the lines EF, EG are omitted.

15. QUESTION (VII. Auguft) answered by Mr. ISAAC DALBY.

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For A and G being the places of the fhip and prize, at the commencement of the chace, and G right N. from A, and the wind at N. N. W. and our ship being clofe hauled, it runs from A, N. E. and W. on her larboard and starboard tacks refpectively; hence, if A, B, C, D, E, F, &c. and G, H, I, K, L, M, &c, be

contemporary

contemporary places of the fhips refpectively, 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 fo of the other tacks; hence it is evident from the parallelifm of the lines that AB+CD+EF, &c. is = AR; and for the fame reafon BC+DE+FW, &c, = RP: whence this conftruction.

To determine AB the run on the firft tack- -Because 3AR (AR+RP) and GP are the whole diftances 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 z in GA, and in nb whose 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 BH+AR, and make BC || RP and 2AB, draw CI+CB, and A,B,C, G,H,I, are contemporary places of the fhips; now, if AP be drawn, and BP fuppofed to be joined, it is evident the fhip will be in thofe lines at her refpective tacks: this needs no demonstration.

For the distances run on the following tacks, we have by fim. triang. GA: AB CD, the run on the fecond larboard tack; and GA :

:: IC (GS):

GSXAB

GA

GS (IC): GS (IC):

-IN, hence GA: AB ::

GS2
GA

GS2 × AB

(IN) :

GA2

GSZ GA EF the run on the third larboard tack; hence, if we consider 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 larboard tacks are the terms of the defcending geometrical GSX AB GS2XAB feries, AB, &c. continued ad infinitum; where the first

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GA

GA2
GS
2
GA

term is AB, common ratio

and fum AR, and the terms of the feries

doubled are evidently the runs on the ftarboard tacks.

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LCULATION.

√2XAG2 42,426m. AR the run on the larboard tacks, this doubled is 84,852 RP, on the starboard, 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 ba || BA, by fim. triang. Ga (Gn+ √2 × bn2): ab :: GA: AB, that is √2×3AR2 GS +GP: 3AR:: GA: 16,257 the run on the first tack, or AB; and =0,62414 GA

the common ratio of the feries.

N. B. We are informed by O'Bryen, in his tranflation of L'Hofte, that this method of chafing to windward was first put in practice by Adm. Barnet in the Mediterranean.

This Question was answered very ingeniously by Mr. G. G. and Mr. G. Sanderfon.

MATHEMATICAL QUESTIONS.
30. QUESTION I. A Mathematical Enigma.

THE difference between the indices of the 4th letter of the firft word, and 4th letter of the 2d word, being multiplied into the difference of their fquares, will produce 1323: and their fum, being multiplied into the fum of their fquares, will produce 10503, the index of the faid 4th letter of the 2d word being leaft.

The fum 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 fquares is 528; the index of the faid 2d 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 progreffion.

The

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