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Otherwise, by Messrs. Butterworth and Kay. ANALYSIS. Suppose the thing done. and that ACB is the triangle required; then, by the question, the ratio of(A C + BC) to the area of the triangle ACB is given, but the angle ACB being given, the ratio of (AC + BC) - A B’ to the area of the given triangle is given also; (Eu. Datu, 76,) therefore, the ratio of (AC+CB) to (A C + BC)'- A Ba is given, and, by division, the ratio of A B’ : (AC+BC) is given; but

A B is given, therefore (AC + B C) is given, consequently the area is given : hence we have the base, vertical angle, and the area to construct the plane triangle, which is a well-known problem.

Other answers were sent by Messrs. Eyres, Gawthorp, Macunn, "esbit, Rylando, and Whitley. 15. Qu. (89) Answered by Messrs. Hine, Maffett,

Rylanda, and Winward. Analysis. Imagine it done; ac and that the point C is found; take BP an nth part of the given line, join P.C, also draw C D perp. to A B, and A x parallel to CD, meeting PC produced in x; then, n.DB-M.DC=nPB= n.DB-n.DP, and by equal additions and subtractions, n. DP=m.DC, orm:n::DP:DC; and by parallel lines PD:DC::PA:Ar; therefore m:n::PA: A.x is a given ratio, and PA is given, therefore A x is given, and x is a given point; hence this construction-Take B P an nth part of the given line, and draw A x perp. to AB, take Ar a fourth proportional to m, n, and PA, join P, which will cut AC in C, the point required; the truth of which is evident from the Analysis, Again, by Mr. S. Jones, Richmond Academy, Liverpool.

CONSTRUCTION. Take HB=an nth part of the given line, and draw BI at right angles to A B, and such that I B: H B : :n :m, join I, H, which produce to meet A C'in C, the point required.




pendicular C D, then the triangles HBI, HDC are similar, and CD:HD::IB:HB::n: m, by construction; therefore, m. CD=n. HD In.DB.R.HB, addn.HB-m. D C to both sides, and n.DB- -m.DC En.HB= the given line by the construction. 2.E.D.

And in the same manner it was answered by Messrs. Butterworth, Eyres, Gawthorp, Kay, Nesbit, and Whitley

16. Qu. (90) Answered by Rylando. COMPOSITION. Let A D be the given right line, and A, B, C, D, the given points in it; make the angles B AF, BCF, each equal to half the given vertical angle, and through the points A, F, C, A describe a circle; through B draw F E, to meet the circle

F in E, join A E, CE, and AEC will be the triangle required.

DEMONSTRATION. For the LA CF=LA EF= ¿ CAF=<CEF, being in equal segments of the same circle; therefore, E F bisects the 2 AEC, and passes through the point B; also, by construction, the sides A E, C E, and AC, pass through the points A, C, and D, respectively. 2.E.D.

Almost exactly in the same manner were the answers given by Messrs. Butterworth, Gawthorp, Jones, Kay, and Whitley. And other elegant answers were sent by Messrs. Eyres, Hine, Maffett, Nesbit, and Win. ward.

17. Qu. (91) Answered by Messrs. Eyres, Hine, Maffett,

and Winward.
Analysis. Suppose it done,
and DR drawn as required
through the given point P;
join KP, SP, and produce

KP to A, making KP. PA
-the given space, that is KP.
PA = DP.PR, or K P:


PD::PA:PR; and since the KP DLRPA, the triangles KP D, PAR, are similar, and 4 PRA= Z PDK = a given L, and P A is a given line; hence, the locus of R is 'a given circle, but PS is a given lide, and P R'S a given 2, therefore the locus of R is also a given circle; the intersection of these loci, determines the point R, through which the required line may be drawn as required. Ergo Solutum.

The same answered by Messrs. Butterworth, Kay, Rylando, and Whitley.

COMPOSITION. Let PKS, be the given points; join them, and upon PK, PS, describe seginents of circles to contain the given angles ; in PS take the point L, so, that the rectangle PS X PL may be equal to the

K given rectangle PRX PD; make the LPLD = the given

PRS, and draw L D meeting the circle in D; through D, draw PDR, and the thing is done. For, as the 2 PL D is equal to the PRS, by construction, the triangles PRS, PLD are similar, and PR:PL:: PS: PD, and consequently PRXPD=PLX PS = the given rectangle.

It was answered also by Messrs. Eyres, Gawthorp, Jones, and Nesbit. 18. Qu. (92) Answered by Mr. A. Hirst, the Proposer.

Let C A and ca be the radii of two wheels, Bb a common tangent, Cc a linė joining their centres, i the intersection of Bb, Cc; then, when the teeth have the epicycloidal form, which is now well known to be the best, they will always act upon each other in the common tangent Bb, and they evidently Tact with the greatest advantage at the point of intersection i-; but that they may move with the least friction, each tooth must move with equal velocities at i, in order to which, C i must have to ci, the direct ratio of their radii, their number of teeth, &c.

That is, because the triangles C Bi, cbi are similar, we have CB=CA:cb = ca:: Ci:ci; whence it appears that the radii, or diameters of the wheels, ex. cluding the length of the teeth, are directly as the number of teeth.

In order to apply the numbers given in the question, as an example, we have 60 feet = 720 inches, and twice the depth of the working point is 2,6 inches, also 720 2,6 = 717,4; then, 1000 : 717,4 :: 100 : 71,74 :: 10; 7,174, consequently 71,74 + 2,6 = 74,34 inches, is the diameter of the greater wheel, and 7,174 + 2,6 = 9,774 inches is the diameter of the less.

Note. When the great wheel is much larger than the small one into which it works, the teeth of the small wheel will come off to an edge, and, consequently, cannot work the full depth ; in this case we must proportion the diameter of the large wheel to the depth at which it works, to the whole depth of the small wheel, as above.

A pair of wheels made according to the above method, under my direction, were lately put up by a friend of mine, and they are now working, the small one at the rate of 400 revolutions in a minute, and with the least noise of any I have ever yet heard.

It is well-known that the difficulty of pitching wheels increases with the difference of their diameters.

Other answers were sent by Mr. Gawthorp and Mr. Nesbit.— Rylando refers to art. 163, Marrat's Mechanics, for a true solution to this question, and observes that the original solution in the Gentleman's Diary for 1795, from whence that article was taken, was invented by the late Mr. John Banks, Lecturer in Natural Philosophy at Manchester; and that he, Rylando, had this information from Mr. Saul, the proposer of the question.

As this is a question of great utility in practice, we trust our able correspondents will some of them give us their further opinions on this difficult subject.

19. Qu. (93) Answered by Rylando. ANALYSIS. Let ABC be the required triangle, CD the line dividing the base A B in the given ratio. Make DI: DC equal the given ratio, and join I B; then CI and the angle C B I are given ; also the ratio of CA:BI is given, and the locus of B is a given circle. Hence we have given the base, vertical angle, and sum or difference of one side, and a line to which the other side has a given ratio, to determine the triangle. Take BP:BI in the given ratio, and join I P; the CBI, and ratio BI: BP being given, the LC PI will also be given, and the locus of P will be a given circle passing through C, I, to contain the CPI:and since BP is the line to "which BI has the given ratio, the line CP will obviously be given, and of course the locus of P is likewise a circle to center C and given radius CP: the intersection of these two known circular loci determines P, and, consequently, the triangle ABC. The point P will fall on the same, or the contrary side of B, with regard to C, according as the second datum is the difference or the sum of B C, CA.

Again, by Mr. Whitley, Masbro'. Let A B C represent the triangle required, and B G the given line dividing the base AC into the segments AG, GC having a given

E ratio. Through C draw E F parallel to B G, meeting AB in E, and AF a parallel to B C in F; A and let BG produced meet A F in H. Then by similar As, GC: H AG::GB:GH = a given line, because G B and the ratio of GC

VF to AG are given. By parallel lines AG: A C::GH:


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