SIXTH ORDINARY MEETING. 43 



tind the larger 619,000, or 433,000 4- 186,000 miles. The volume of 

 this space is — 



— (5280) 3 (10) 9 | (619) 3 - (433) 3 I cub. ft. 



Also, the surface of the sun is 4tt (433) 2 (10) 6 (5280) 2 sq. ft. 

 Therefore 1 cubic foot of ether is agitated by — 



4 7t(433) 2 (10) 6 (5280) 2 x 5500000 



L* (5280) 3 (10) 9 { (619) 3 - (433) 3 1 



5500000 - . , - 



loot-pounds oi energy. 



5280 x 279000 



Let m represent the mass of each ether-particle, or the average 

 mass if the ether-particles are not uniform, and n the number of 

 such particles in a cubic foot, so that nm = M •will be the number 

 of pounds of ether in a cubic foot. 



Using the ordinary equation of the harmonic curve — 



• / 2?rz \ 



y = asm I — — + a\ 



it will be seen by differentiating twice that the maximum velocity of 



. 2 7r a 

 any particle owing to any single wave is — - — V , where a is the 



amplitude, I the wave length and V the velocity of propagation. 

 Hence the energy of a particle whose mass is in, under such 

 circumstances is — 



7Tb '' 7C Ou 



— — - — V 2 foot-pounds. 

 9 X 



Therefore the energy of a cubic foot of ether is — 



,m 2r 2 a 2 Y 2 M 2^a 2 Y 2 



I 



9 & 9 ^ 2 



Equating these two expressions for the same quantity of energy 

 ■we get as the mass of a cubic of ether 



g^ 5500000 lbs 



M = 2ttV (186000) 2 (279000) (5,280/ 

 It will be seen that the only assumption involved in this calcula- 

 tion is thac the average velocity of the ether particles may be taken 

 ito be equal to the maximum velocity in consequence of a single wave 

 ,motion. 



