s; alfo let x and x be the two remaining terms of the feries; then by the question, ≈2 a2+b2 + c2 + d2 ---- + x2 = 5+x2; hence, to have z rational, it is evident that s+x2 must be a fquare; affume it s+x2—s— x12 = s2 — 25x+x2; hence x = SI here s is evidently an odd number; it therefore - ; * ! +b2+c2+d2 -

2

that appears

a

- - - - to n — 2 terms must be fome integral fquares whofe fum is an odd number greater than 1, and the two remaining

Ex. If n be 3, or there be three terms, the leaft of them must be greater than 2; if it be taken 3; thens 39; hence

2

4, and == 5; therefore the three may be 3, 4, 5

2

If n=4, or 4 terms; let the first two terms be 2, 3, the sum

S—I
2

-6, and

2.

of whose squares is 13=s; theref. =7, the other two terms; the four being 2, 3, 6, 7. And fo on, for any number of terms.

This question was also ingeniously answered by Meffieurs Amicus, Barnes, Bengal Officer, Bournley, Buchanan, Cavill, Creggs, Cullyer, Cunliffe, Dalton, Haycock, Holt, Howard, Kinnebrook, Lawton. Leybourn, Liddell, Mooney, Park, Pearfon. Rowe, Ryley, Sampson, Skewes, Stevenfon, Taylor, Tomlinson, White, and Whiting.

X QUESTION 933 anfwered by Mr Alex. Rowe, of

Reginnis.

A pendulum 39 inches long vibrates feconds, and the lengths of different pendulums are reciprocally proportional to the fquare of the number of their vibrations made in one and the fame time; therefore as 39 ;; n ; n √. the time or number of feconds in which

39

the pendulum / performs n vibrations. Then, by Dr. Hutton's Compendious Measurer (a fmall book particularly ufeful in Schools), the velocity of found is about 1142 feet in a fecond of time, or a mile in 4 feconds; whence as 1": 1142 :: V. 11420V.

39首

39 nd, a general rule for the diffance fought in feet.

182

&

Nearly in the fame manner was the solution alfo given by Meffieurs Amicus, Barnes, Bengal Officer, Buchanan, Bulmer, Burdon, Burton, Cavill, Cock, Cox, Craggs, Cullyer, Cunliffe, Deaden, Elfob Evans, Ewbank, Garnet, Glendenning, Greaves, Griffb, T. H, Harrifon, Hartley, Haycock, Hickinbotham, Holt, Kinnebrook, Law son, Leybourn, Liddell, Mifs Mafon, Mooney, Mulcafier, Nichols,

Nicholfen, Park, Pearfon, Pritty, Roberts, Ryley, Sampson, SanderJon, Simpfen, Skewes, Stevenfon, Swarn, Taylor, Tomlinfon, White, Whiteboufe, Whiting, Wilkinson, Williams, Woolcott, and Yeuart.

XI QUESTION 934 anfwered by the Propofer, John: Ryley, of Leeds.

Put 24 the area of the cistern's bafe, m32 feet the velocity acquired by a falling body in one fecond of time, the area of the aperture at the bottom, 9 the quantity run in by the filling cock in a fecond, b5 the height, t the time required, and the variable height of the water in the cistern.

Then, by Art. 1. of Dr. Hutton's Mathematical Mifcellany, the time of exhauftion will be defined by ✔=452700" by the

queftion; whence n will be had =

2 r

n

m

I

142 7

foot 10091 fquare inch

Alfo, the velocity of the iffuing fluid being

mx, the quantity run out in one fecond will be expreffed by n ✔ mx; and conf.quently q-nmx will be the rate of the vefiel's filling per fecond; alfo, t-nt mx the fluxion of the quantity in the veffelrš. =(putting v2x, and sn✔m}

qt.

Therefore =

rx
9-n√mx

the time required.

66 min. 14 fec. or 1 hour 6 min. 14 fec

"N. B. This queft was published, in the Leeds Mercury, by a pupil of mine, and two anfwers given to it, the former of which was intirely falie, making the answer 36′ instead of 66′ 14′′, the exact anfwer, and the latter quite unintelligible."

Nearly in the fame manner were true folutions given by Meffieurs Amicus, Barnes, Buurnley, Buchanan, Cowing, Cullyer, Cunliffe, Dowden, Garnett, Haycock, Holt, Howard, Kinnebrok, Lawton, Leybourn, Rowe, Sampfen, Stevenfen, Taylor, Vertigo, White, Wilkinson, and Woolcott.

The other anfwers that were received were wrong, owing to the fuppofing the efflux of the water at the bottom to be conftant, whereas it is continually variable, owing to the varying height of the rising fluid in the vessel. XII QUESTION 935 anfwered by Mr Wm. Lawton. Let DCE be the boat, which it is evident will be the largest when it is infcribed in the femicircle A DEB, whofe centre is c, in which cafe it is the radius 2CE 2. = 100 =

CF2 + FE2x2 + y2 = x2+

169x2 3x2+5x+7

the root of which equation is x7*4494946.

A

D.

Again, the fluxion of the area is p

13

13

correct fluent of which is 33√32+5x+7-33 √7 +

673

7.449 &c.) 19.2175, the double o which is 78 435 feet, the area of the whole ection D C E of the boat.

Ingenious anfwers were also given by Melleurs Amicus, Barns, Bournley, Craggs. Cullyer, Cunliffe, Dalton, Dowden, Garnett, Haycock, Howard, Rowe, Ryley, Stevenfon, Taylor, Tomlinfun, Vertigo,

White, and Woolcott.

XIII QUESTION 936 answered by Amicus.

In Mr Flamsteed's, and other graphical methods of computing folar eclipfes, the fhadow is fuppofed to fall upon the plane of the imaginary circle bounding light and darkness; and its fize is made the fame as if it really fell upon fuch a plane; whereas it ought to be the orthographic 1 projection on fuch plane of the fhadow as it really is upon the furtace of the carth, or of the portion of the furface covered thereby, which, being nearer the moon, must be greater than the imaginary fhadow on the imaginary plane. Confequently the horizontal diameter ought to be augmented according to the altitude. The fame anfwered by the Propofer, Mr Geo. Sanderson.

To determine which of the diameters is to be made ufe of in folar eclipfes, let us fuppofe the centres of the earth, moon, and fun to be in the fame right line where T is the centre of the earth, c the mcon's, s the fun's, and м the vertex of the penumbral cone, whofe flant fide touches the moon in the point o: Alfo let BEZ reprefent a part of the earth's difk, ED the femidiameter of the penumbra, or fection of the penumbral cone (which in this cafe is a circle); ao is the plane of projection, which in Flamsteed's method touches the moon's orbit, and is perdendicular to Ts, the line connecting the centres of the earth and fun: Join c E, and draw E A parallel to p c meeting a bin A.

B

E

Mr Flamfteed, Mr Keill, and others, make the angle E MD equal to the apparent femidiameter of the fun, equal to ECD OEC AED-OEC; but ECD is the apparent fernidiameter of the penumbra (ED) as feen from the moon; which by reafon of the great diftance of the fun (TS) in refpect of E D and D c*, is projected on the plane (ab) into a line equal to itself, which by parallel lines is

equal to A c. But the angle oc is equal to the apparent femidiameter o c of the moon as seen from E; which angle is manifestly equal to the apparent horizontal femidiamiter augmented according to its altitude; therefore the fem penumbra Ac in Flamsteed's projection, must be made equal to the angles AEO+OECEMD+OEC; or to the apparent femidiamete of the fun, plus the horizontal femidiameter of the moon augmented according to its altitude.

* Mr. Flamfeed fuppofes the eye to be in the centre of the fun: whence all the points on the earth's difc are orthographically projected on the plane a b.

XIV QUESTION 937, anfwered by Amicus, the Propofer. From the centre o, along the tranverse

c and conjugate o A, of the given ellipfe, fet ffoF: OD: fide of a fquare: its diagonal, and with radius o A defcribe a circle; join D F, perpend. to which draw OE, which continue to the circle in G ; through G para el to DP draw the tangent RI meeting the tranfverfe in R, and conjugate produced in 1; parallel to co and through G draw the ordinate of the ellipfe KQ; join IK, which produce to L till KLIK; through L parallel to the tranf

I

verfe draw LP, and it will cut off the fegment required.

10

For IK, which cuts the tranfverfe produced in s, by a known property of the ellipfe, is a tangent thereto at K; and fince, by conftr. Fo2: DOI:2: FEDE, DE2FE, therefore GI=2GR, and KI = 2 KS; but, by conftr. KILK, therefore K SSL, and, Kм being joined, QKMP is a parallelogram, and because KI= KL, it is the greateft n the fegment r M C K A, and its double the greatest in the fegment of the whole ellipfe, and its length 2KQ is the greateft; because it confifts of the whole chord 2 MP bounding the elliptic fegment.

Since, by conftr. 2001Q, and Go2=A02 = 20'10 = 3002, OAO√, GQ=AO√, and KQ=co✓ now the length of the greatest parallelogram that can be infcribed in the whole ellipfe is known to be co√2, and co√2:2 KQ :: √3:2. True answers were also given by Mers. Barnes, Cullyer, Garnett, Howard, Kinnebrook, Leybourn, Mooney, Pritty, Ryley, Vertigo, White, and Woslcott.

XV or PRIZE QUESTION 938 anfwered by Lieut. Mudge.

Let A B be the furface of the water, and CDE

the cylinder. Put n

the specific gravity of oak, that of water being 1, a 7854, 2g32% the force of gravity, CD any space afcended at any time t, and the velocity.

Then ad2bx is the weight of water difplaced, or force of the water acting against the oak, and adnb weight of the cylinder, or force of the eak against the water; therefore a d2 × b. -X- nb

or ad2 x mb.

is the motive force urging the cylinder upwards, putting m1n; and confequently, dividing by ad2 nb the matter

moved, gives

mb
-x
nb

f the accelerative force. Hence, by forces

(fee Dr. Hutton's Conic Sect. and Select Exercifes, theor. 10, pa. 169), v v = 3 g ƒ x=25 xbxxx, the fluents of which give 2 =

nb

X 2mbx x2, and when xp, we have.

the velocity required.

2g b n

Alfo for the time, by theor. 11 of the above, ;=-=

2bn

above) are t√2

g

the fluents of which (by form. 10, pa. 171 as

XA or the arc whofe verfed fine is

diameter I j which gives the time required.

Thus far the folution, as required by the queftion; but it may not be improper to make a few obfervations on this problem.

OBSERV. I. When vo, then x2 bm, the greatest length that can rife out of the water. And when is a max. by making the flux. of the value of vo, there refults xbm for the height rifen when the velocity is the greateft; and which therefore is juft half the whole height to which the cylinder will rife before it stops. OBSERV. 2. When the cylinder is in the act of finking, after x the fame notation remaining, and v denoting now the defcending being when x2bm, the correction vanishes, and

2bm;

velocity,

then v2=

28 b n

X 2 bmx

2 the fame expreffion as before, and the cylinder will fink juft to its first pofition, or till its upper end be level with the furface of the water. It therefore appears that the cylinder will continue to leap up and fall down, if tenacity, friction, or fome fuch force did not take place between the furface of the cylinder and the water. But as experience fhews there is fuch a force, we must have recourfe to experiment for the true folution in this cafe.

Let then s denote the fum of the friction, attraction, &c. upon the

Therefore, by obferving the height afcended out of the water when