of Electric Oscillations along Parallel Wires. o'l5 



the two distances of opposite wires divided by the distance of the 

 similar wires. 



§ 10. 2n Wires at corners of Regular Polygon, 

 consecutive ivires being opposite. 

 It is clear from symmetry thatD] = — D 2 = i) 3 = — D 4 = <£c; 

 therefore equation (14) gives at once 

 f 



'if — A — Bi2 + B 13 — . . . Bi( 2Mt _2) + B l (2»-l) — Ai2 w 



, 012 &U . . . Oj 2n ,_„. 



= log— (28) 



° a bis Old • • • 0i (2»-i) 



Let r be the radius of the circumscribing circle, then 



012— 2r sin — , o l3 =2r sin — — , &c: 

 "In "In 



it . Sir . (2n— 1 )tt 



■f 9 » Sln s~ sm s - • • • sm ^ — 



/_ 1 „„ 2r , i„„ 2/i 2rc 2n 



.\ -^ = log h log 



<r ° a . 2tt . 4vr . (2n-2)7r 



sin ,r— sin ~ — . . . sin ■ 



2n 2-n ' 2n 



= log — + log - 



2r 

 = log — , (20) 



Therefore the speed and attenuation with the given arrange- 



2r 

 ment are the same as for a pair of leads at distance — . 



n 



§ 11. n Wires at corners of Regular Polygon, 



all carrying similar currents. 



Here Di = D 2 = D 3 , &c, and we have 



■-V=A + B 12 + B 13 + .. . Bi„ 

 c 



i 2i , , 1V1 . i . T . 2w . (/i-l)7r 



= /tloo- loga — (n— -1) log zy— log sm -sua — ,. .sill 1 - — 



°jc ° J ° ° n n n 



m ., p . 7T . 2-7T. . (» — l)lF 



Write a tor sin -sm — ... sm , 



17 ?t w n 



2U-1) 2 



w , / <y 2 r n q n 2 \ ,„, N 



(30) 



2* 



