June 5, 1885.] 



♦ KNOWLEDGE ♦ 



479 



CHATS ON 

 GEOMETRICAL MEASUREMENT. 



By Richard A, Proctou. 



(^Continued from page 432.) 

 THE COXE. 



A. Is the cone as easily dealt with as the cylinder 1 



if. Quite as easily. 



A. Will you take first the curved surface of a conel 



M. I can only deal with the curved surfuce of a right 

 circular cone — the curved surfaces of oblique cones and 

 cones on elliptical bases requii-e the higher analyses. 



A. What is the curved surface of a right circular cone 1 



M. It is equal to a triangle having the circumference of 

 the circular base of the cone for base, and the height of the 

 cone for height 



A. How do you prove this 1 



M. Let C D B, Fig. 1, be a right cone on the circular base 

 DAB; CA, CB, OD being straight lines from the vertex 

 to diflFerent points on the circumference of the base. Then 

 we might divide the whole curved surface of the cone into 

 small areas like C Ao, and show that each of these ap- 

 proaches in area to a triangle Laving Aa for base, and C A 

 for height Adding all such triangles, we should obtain a 

 triangle having the circumference of the circle DAB for 

 base, and C A for height. But it is simpler to suppose the 



It 



Fiff. 2. 



cone rolled on a plane, till its whole surface has touched 

 the plane. If the line C A, Fig. 1, touches the plane at 

 the beginning of the rolling, as in C A, Fig. "J, it is obvious 

 that the surface rolled over will be such a sector as 

 C A B A', where A B A' is the arc of a circle, round C as 

 centre, rolled over by the circle DAB, and C A, CV«, C B, 

 C D in Fig. 2 are the lines touched by the correspondingly 

 lettered lines of the conical surface C D B in Fig. 1, — C A' 

 being where C A of Fig. 1 again reaches the plane. 



A. As in the rolling of the cylinder, this result of the 

 rolling of the cone seems obvious. But how is it proved 

 that the circular edge of the cone's base will trace out a 

 circular arc \ 



M. As you begin to roll the cone from contact with 

 C A, it is evident that, as A T ',the tangent at A to the 

 circle DAB, is square to C A, the beginning of the 

 rolling starts the trace A «, square to C A ; and this square- 

 ness is maintained throughout. Hence as a circular arc 

 round C as centre, and at distance C A, is the only curve 

 which can pass through A, and be everywhere square to 

 each successive position of the equal generating lines of the 

 cone, as they successively touch the plane, this arc is the 

 trace left by the rolling edge of the circle D A B of Fig. 1. 



A. Do you not here assume that the vertex C of the cone 

 will always remain at the same point in the plane 1 



A[. Xot actually ; for the proof only requires in reality 

 that all the lines in which the rolling cone touches the 

 plane shall be equal, which of course they must be. But, 

 one might assume as self-e\'ident that will not move on 

 the plane ; for C is the trace of a rolling circle of infinitely 

 small dimensions, and such a circle — i.e., a point, cannot in 



any finite number of rollings traverse a finite distance. 

 will therefore remain always at the same point C, Fig. -. 



A. We can, of course, now tleterniine the area of a 

 frustum of a cone, such as D a B, between two parallel 

 circular sections DAB and d ah (Fig. 3). 



FiR. 3. 



M. Yes ; the curved area D a B is equal to the dilierence 

 between two triangles, the larger having circumference 

 DAB and height C D, the smaller having circumference 

 dab for base, and height Cd. 



A. Will it not be convenient to express the areas of 

 cones and frusta of cones in terms of the height, radius of 

 base, and the like I 



M. It will ; for we must remember that the purely 

 geometrical character of our inquiry will not l^e afiecterl by 

 the use, now, of letters to represent lines, Ac. ; or by the 

 use of the trigonometrical symbols : — 



Suppo.se fir.st, radius of cone's base to be r, and side of 

 cone to be d. 



Then curved area of cone = h ■ l-nt-d = -rd. 



If height of cone is A, C D = ■v/r- -f li\ and 

 curved area of cone = txt v r- + /(-. 



If half the vertical angle D C B of cone is «, i.e., Z D C O 

 =a, where is the centre of the base, and height=/t, then 

 D 0=A tan a, and C D— A sec a ; 



.-. curved area of cone = 7r/i' tan n sec i\. 



If we have a frustum, as D a B, such that D O =r, 



7 \ 



CO = h and Go = />' ; then do = ^ . r ; and 



/t 



curved area of frustum = -rh — -r ~. /<' 



// 



h ^ 



If half vertical angle D C B = <., C O =/*, and C o = h', 



then , 



curved area of frustum = 7r //- tau « sec u — Kh- 



tan a sec a 



= - tan a sec n (/(.- — Ji!~) 



A. Can we express the area conveniently in terms of its 

 slant side Dd 1 



M. Let us see. Suppose as before V>0=t; C D=(/, and 

 D d=s. Then 



, T-. ^d C Id - S) 



/.curved area =-rd 



.^r^Az.').{d-.) 

 d 



^{d--id-sf] 



='^(2d-s)s 



d ^ ' 



Now, it is easily seen that the circle K L M obtained bjf 



