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The fame by Mr John Dalton, Teacher of the Mathe

matics, Kendal.

The weights of the parts of the fruftum on each fide of the prop will be thus, the greater 17367 lb, the lefs 0.862 lb. And from the directions of the writers on fluxions, the distance of the fulcrum from the center of gravity of the greater part is 160415 inches, and from that of the lefs part 1'4524 inches; which diftances, multiplied by the refpective weights, give the momenta of the two parts, the difference of which is the momentum of the required weight, which divided by 3, its diftance from the fulerum, gives the weight 92 4461b. The fame by Mr John Afpland, of Soham.

GIGHEF ABDCAB='3. Put EF6, DC = b = 3.3, HI =6=27; 53a + 2 a b + b2 2 × 4 b = G then willa+ab+b2 S16*041375 be the diftance of the center of gravity of

B

the н. Again, put ABS 3.c2 +2cb + b2 2 × ÷ 7 = DE from point - 1-3, then will {63+66 +614523 be

the diftance of the center of gravity of DB from the point H. Now the folidity of the part DE is 600'21 inc. and its weight 17.3671b; and the folidity of DB is 2979 inc. and its weight 861971b. Put now 16 04137, 14523, W17 367, and w8197, alfo the weight fought; then rw+lx dw, and hence x dw 92 4471b, the answer.

Ingenious anfwers were also given by Meffrs Adams, Allenfis, Ball, Birch, Baden, Bretherick, Burrow, Cansfield, Cullyer, Cunliffe, Emes Evans, Hornby, Howard, Jackfon, King, Lowry, Mudge, Rowe, Saul, Scholefield, Surtees, Taylor, Terry, TTodd, Waugh, White, Whiteboufe, Williams, Woolfion, and Young.

V QUESTION 867 anfvered by Mr Matt Terry,

Land Surveyor.

109

Since the fteelyard accurately weighs 60cwt or 6720lb when 2431b is fufpended on the 29th divifion, theref. 6720243 :: 29:1 2240 d the length of the fhort arm of the fteelyard; and d: 118 112 5245, inftead of 112. Again, d: 1 :: 243: 231 7242 lb, inftead of 2 cwt or 224lb. Hence, in weighing with the fmall weight, or 118 lb, the error on the firft divifion is 5245, on the 2d twice as much, on the 3d thrice as much, and fo on to the 29th divifion, where the error is 15 21 lb. And, in weighing with the greater weight, or 243 lb, from the 29th to the 59th divifion, the error on the 30th divifion is 77242 lb, on the 31ft twice as much, on the 32d thrice as much, and fo on to the laft or 59th divifion, where the error is 30 times 7'7242, or 231 7261b, or 2 cwt 7 lb.

This question was answered nearly in the fame manner by Mers Adams, Allenfis, Amicus, Afpland, Ball, Birch, Boden, Bretherick, Burrow, Cavill, Cullyer, Dalton, Evans, Griffith, Howard, Jackson, Mudge, Rowe, Saul, Scholefield, Smith, Stevenfon, Surtees, Taylor, Waugh, White, Whiteboufe, Williams, Woolfton, and Young.

VI QUESTION 868 anfwered.

There are two ways of confidering this problem, viz. either as the femiparabola is a right one, or an oblique one. According to the common acceptation, it is taken as, and ufually understood to be, a right parabola, But fome of our correfpondents have taken it as oblique, which makes the conclufion different, though each solution be right in its own way. We fhall infert an example or two of each.

1. For the Oblique Parabola, by Amicus.

Circumfcribe the given circle with an equilateral triangle CH, touching it in A, B, C; bifect AF in T, and to FH apply TDAT=T}; join AD, and draw T E parallel to F H cutting A D in E; bilect re in v; then with the vertex v, axis v E, and ordinate rightly applied AD, deIcribe a parabola D VAL cutting G H in 1, fo is DALHD the femiparabola required.

F

E B

G L

C P H

For fince TV = VE, TA and TD are tangents to the parabola; and fince TDTFFD, TD is parallel to GH; confeq. the ordinate L is parallel to the tangent TD, and the abfciffa DH to the axis VE; wherefore HD A L is a femiparabola; and because the tangent GF is bifected in A, it is manifeftly that required.

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By fim. triangles, as HDH :: 2 : 3 :: BO; AD =
BH3, or BH BF BO3 AB; and DB: DH A Ƒa
LH, or LH3B0; and DP (perp. to GH)

the area of the femiparabola = 2B02.

FC BO; and

2. For the Oblique Parabola, by Mr T Todd, of

Darlington.

It is proved in art. 29 Simpfon's Flux. that the equilateral triangle GBA is the leaft that can circumfcribe the given circle DEP, where BA, perp. to the radius or, is a tangent to both curves in the point E; PE parallel to GA; and BD perp. to The OB 2r, and GB 2√3. Alfo fince BE EA, and, by the parabola rv v B

GA.

P

B

FA

GB, it appears by Simpfon's Geom. p. 201, that the required femiparabola G VEF is now given, being the greatest that can be infcribed in the faid triangle, and the leaft that can circumfcribe the given circle;

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hence then VP: VG:: 1:3: PE :GF2-972, or GF37Br, and the area of the femiparabola GVEF is GV X GF X fine

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3. For the Right Parabola, by Mr Ifaac Saul, Holland near Wigan.

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By theor. 8 Simp. Max. and Min. the least rightangled triangle circumfcribing the circle, is when the legs BC, BA are equal. Alfo the greatest parallelogram that can be infcribed in any curve is when the fub-tangent CD DECB. Theref. circum. the given circle with the right-angled triangle ABC, making AB BC; bife B c in D, and DC in F; then to the vertex F, axis FD, and or- B

dinate p, defcribe the parabola F EG required. Then, putting a for the radius of the circle, B = a + a√2, and Bc — a √ 6 + 4/2 2a+a√21, and BBC, and DE2DFBC; hence D FB: DE2 BG2, or I3BC2: BG2 BC2, and BG BC3. Confeq. the area of the femiparabola, or BF X BG, is 3 X 2 BCX B C √3 = B c2 √3 = a2 × 3 √ 3 + √ 6 = 5·047.a2, as required.

2

Solutions were alfo given by Meffrs Bretherick, Cunliffe, Dalton, Evans, Hornby, Howard, Jackson, Mudge, Rowe, Sanderfon, Taylor, Waugh, White, Williams, and Young.

VII QUESTION 869 anfwered by Mr John Howard, Teacher of the Mathematics, Carlifle.

Let z be the zenith, p the pole, and o the fun's place; then we have given z P 40° the co-latitude, alfo zo being the co-altitude, and Po the co-declin. the altitude is 90 zo, and the delin. РО - go, the fum of which is po -zo 34° 40' PD, by taking oD oz. Then in the triangle z DP, are given P z, PD, and P30°, 0

to find the fide D'z and PDZ, the fupplement of which is the LODE = LOUD, Then in the ifofceles triangle oz D are given the Bale z D, and the angles at the bafé, to find oz or op 67° 31′ the co-altitude; confeq. POPD + DO = 102° 11', and then PO-90 12° 11′ the declination fouth, answering to October 25th.

The fame by Mr Rd Waugh, Bufhblades, Durham. Let a and b fine and cofine of 50°, c and d the fine and cofine of 34° 40', m cofine of 30° the LP, and x and y the fine and cofine of the declination. Then cy-dx: cos, zo; and, by a well-known

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Solutions were also given by Meffrs Adams, Amicus, Afpland, Clement, Cock, Cullyer, Dalton, Evans, Hornby, Jackfon, Mudge, Rowe, Taylor, White, Williams, and Young.

VIII QUESTION 870 anfvered by the Rev L Evans, of Hungerford.

The times in which pendulums make an equal number of vibrations being as the roots of their lengths, we have 39: ✓ 12 :: 20: 11.06 the time in which a pendulum of 12 inches makes 20 vibrations. Now let depth of the well, a 16 feet the space through which a heavy body falls in the first second of time, b = 1142 feet the fpace through which found paffes in the fame time, c11" 06 the time given from the first defcent of the ftone to the hearing of the found. x7 the time of the ftone's delcent; and b:::′′:

Then by the law of fall-S ing bodies ax: 12 :2

= { = time of thefound's }

bafcent;

afcent; wherefore

B2 + abc — b√ b2 + 4abc

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It was alfo ingeniously anfwered by Melis Adams, Allenfi, Amicus, Afpland, Ball, Mrs Baufor, Birch, Boden, Bretherick, Burman, WC, Canfield, Cavill. Clement, Cock, Cooper, Corving, Cullyer, Dalton, Dening, Dixon, Hornby, Howard, Jackson, King, Lowry, Mafon, Mudge, Roope, Rowe, Saul, Scholefield, Smith, Stevenson, Surtees, Taylor, Terry, Todd, Waugh, White, Whitehouse, Williams, Wood, Youle, and Young.

IX QUESTION 871 anfwered by Amicus.

If three tangents be drawn to the fame given circle, their interfections will form a triangle, and if the interfection fartheft from the centre be called the vertical angle, the tangent between it and the point of contact will be equal to half the perimeter of the triangle, if the triangle does not circumfcribe the circle; and to half the diff. c between the fum of the fides and bafe if it does. Which property holds equally in fpherical trian- Al

gles as in plane ones, and is too evident to need a particular demonftration here. This premifed,

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Let the great circles PA, PB, forming the given vertical angle at p be drawn, fo that APBP half the given perimeter, and the lefs circle A T B touching them in A and B; and round the pole P at a given diftance therefromthe given perp. of the triangle, draw the reprefentation of another leffer circle; then the reprefentation of a great

circle c being drawn to touch thofe of the two leffer ones, will, from' what is premifed, form the reprefentation CPD of the triangle required. Now the diftances of the poles of the great circle cTtn from each of the points of contact T and t muft 900 or a quadrant, the distance then from Pa quadr. - Pt, and from o the pole of the other leffer circle a quadr. To; whence, by common fpherical geometry, the reprefentation of this pole, and confequently of the great circle c TTD, are readily drawn. And that whether the projection be ftereographic or orthographic.

The fame by Mr Ifaac Dalby.

Lemma. If two great circles touch a leffer circle, the fegments of the great circles between the points of contact and points of interfection on the fame fide of E the leffer circle are equal. This is too evident to need a demonftration.

S

A

D

R

W

Projection. Make the DNC the G given vertical angle; take NC, ND each the perim. defcribe the arcs CP, DP perp. to NC, ND; about P as a pole defcribe a leffer circle at the dift. of PD or PC, which, it is evident, will touch N D, NC in D and c; alfo about N as a pole defcribe a leffer circle AR B at a diftance the given perp. and defcribe the great circles s w, ELF parallel to the leffer circles AR B, DTC refpectively; then by prob. 21 Emerfon's Stereog. Proj. defcribe a great circle G IOL to cut the great circle s w fo that the

LIW comp. of NB or NA to 90°, and the great circle ELF fo that the GLE Comp. of PD or PC to 90°; and GNO will be the triangle required. For it is well known, that if a great circle be defcribed to touch two leffer ones, it will cut their parallel great circles in angles the complements of the leffer circles diftances from their roles, therefore the great circle G10L will touch the leffer circles in T and R hence by the foregoing lemma, the arc TG GD, and Tooc, therefore 'GT+TO+GN + NO = ND + NC the perim, and defcribing the arc NR to the point of contact R, it will be perp. to the bafe Go, and the given perp. by conftruction.

Mr John Howard and Mr Wm White alfo answered this queflion.

X QUESTION 872 anf. by Mr Fas Young, of Pruddos.

The figure being drawn as per question, &c. and LGHK a tang. to the curves at H; put the radius CA or C Mr, the ordinate HE

coy; then is CEOH

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0 C

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