126 



♦ KNOWLEDGE 



[Feb. 13, 1885. 



M. They are usually convenient ; but in truth you can 

 use any form you please, so only that you assure yourself 

 that each rectilinear element diflers eventually from the 

 corresponding curvilinear element, by an area which even- 

 tually bears an evanescent ratio to either. 



A. Give me au example where some other elements than 

 rectangles are employed. You have already used triangle?, 

 but I should be glad of another example. 



J/. Let us for a change deal with the epicycloid ; — 



Fig. 2. 



Let A (Fig. 2) be the vertex of the epicycloid A P D, 

 A./} B the generating circle (radius a) ; B D the circular 

 base (radius?)); AT the enclosing circle (radius 6 + 2«) ; 

 <) D T radial. 



Through two neighbouring points P, Q, describe arcs 

 Py. Q '/ about 0. Join A.p, A q, and draw tangents P L, 

 Q M to the epicycloid. Then we know that if ^ L, O m M 

 1 >e drawn cutting B D in / and m, I P and ut Q are .square 

 to PL, Q M respectively. Thus regarding L Q M as a 

 triangle (which it is, in the limit) and comparing its area 

 with that of the triangle j) A q, we see that since the sides 

 L Q, Q M are equal to ;; A, A.q in the limit, the areas 

 )' Aq, L Q M, must be proportional to the angles p A q, 

 L Q M. Now it is obvious that if m M were parallel to 

 / L the angle L Q M would be equal to the angle p A q (for 

 M Q, L Q would make the same angles with these parallels 

 that A q, Ap make respectively with A B). But since in M is 

 inclined to Jjl, we must add this inclination; viz. the 

 angle I O M, to the angle ;) A q, to get the angle L Q M. 



A. How are we to do this ? 



M. From the way in which the generating circle rolls in 

 marking out the arc PC,), we know of course that / »i is 

 equal to p q. Hence, joining G p, q, we have 

 angle 1 m : p C q :: G p : O I :: a : b 

 angle 10 m:pAq::2 a:b (for angle p Cq=i2p Aq) 

 and angle LQ M ( = 1 7)i + p A q) : p A Q :: i a-i-h:b 

 wherefore area L Q M : area p Aq y.i a + b -.b. 

 This being true of all such elements of the re.spective areas 

 A Q D T and ApqB,it follows that 



area A Q D T : semicircle A;;(/B:: 2 a + b -.b 



or area A Q D T=^^^— . generating circle. 



A. What is the area of A Q D B 1 



J/. That is easily determined. We know the area 

 ABDT. 

 A. I do not. 

 If. Pardon me ; I thought you did. 



A. Well, I suppose the area is easily obtained. Area 

 B D is equal to .V the rectangle under B and the 



semi-oircumferenoe ofOBD= — — Then 



area A T=area B D . ( 1 = — ( , ) 



[ b- / 2\ b J 



w a b 



wherefore area A B D T=0 A T-0 B D = 2 tt a-Y'lti'') 



J/. That is right ; but there is a better way of obtaining 

 this result. Imagine A B carried round to D T. Its 

 centre C describes the arc C c, and by a useful property, 

 easily established, the area equals rectangle under arc C c 

 and diameter A B 



or area A B DT=7rff . — — . 2 a=2i^a-| Jl-\as before. 

 b \ b ) 



A. We have now, then, only to take area A Q D T from 



area ABDT. 



31. Precisely ; giving us 



area AQDB = 2.«=(:i+i) _ .,r[^) 



_Tra-/2a+3b\ 

 ~^[~b~) 



Thus the whole area enclosed between one arc of an epi- 

 cycloid and its circular base 



(^') 



generating circle. 



When the base is straight, or l> is iuGnite, this result 

 I since — then vanishes ) becomes, 



V ^ / 



area between a 03 cloid and its base = 3 . generating circle, 

 as already shown. 



.1. How about the hypocycloid 1 



M. You will find no difficulty in dealing with that 

 curve by the same method. It will be a useful exercise. 

 You will obtain the following results, — 



Area between ciiiej'cloid and base = ( -^ + -— ) generating circle. 



Area between cycloid and base = 3 times generating circle. 

 Area between hypocycloid and base = [3 — r^ J generating circle. 



A . So that if you have three generating circles of radius 

 a, one rolling outside the second rolling inside a circle of 

 radius b, while the third rolls along a straight line, then 



Epicycloidal area -|- hypocycloidal area = 2 cyoloidal 

 area. 



M. Yes ; willi many other relations of a similar kind 

 noted among the problems to my " Geometry of Cycloids." 



The whole of the new rolling stock of the Mei'3ey Railway and 

 Tunnel Companj- is, says the Engineer, to be fitted by the Pintsch's 

 Patent Lighting Company with the Pintsch gas-light. Two lights 

 will be fitted in each compartment, and will cost then less than one 

 oil-lamp. The system is gradually finding its way into the car- 

 riages on oijr leading railway systems. The Midland llailway 

 Comp.any has 121 coaches fitted on this system, the Great ^Vestern 

 38; the South-Eastcrn, 151; the Metropolitan District, 301 ; the 

 District, 350; the London and South-Western, 553; the Great 

 Eastern, 592; the Caledonian, 208; the Glasgow and South- Western, 

 250 : the North British, 36, and the Mersey Railway, 5G. This 

 gives a total of 2,059 coaches fitted on this system on British 

 lines; but the number thus lighted in all countries is over 17,500, 

 besides which there are locomotiyes fitted with head-lights burning 

 Pintsch's gas. 



