SIXTH ORDINARY MEETING. 43 
and the larger 619,000, or 433,000 + 186,000 miles. The volume of 
this space is— 
af (5280) (10° | (619) — (433) \eub. ft, 
Also, the surface of the sun is 4 (433)? (10)® (5280)? sq. ft. 
Therefore 1 cubic foot of ether is agitated by — 
4 m (433)? (10)° (5280)? x 5500000 
a (5280) (10) | (619)? - (433)° | 
_ 6500000 
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 mm = M will be the number 
of pounds of ether in a cubic foot. 
Using the ordinary equation of the harmonic curve— 
5 ed aad 
y = asin ( 7 + a) 
foot-pounds of energy. 
it will be seen by differentiating twice that the maximum velocity of 
: : ‘ ee : 
any particle owing to any single wave is arate where a is the 
amplitude, A the wave length and V the velocity of propagation. 
Hence the energy of a particle whose mass is m, under such 
circumstances is— 
V? foot-pounds. 
Therefore the energy of a cubic foot of ether is— 
pam 2atatV™ Mant aN? 
ha | 
g R g x 
Equating these two expressions for the same quantity of energy 
we get as the mass of a cubic of ether 
gi? = 5500000 ies 
M = 9,3q2 (186000)? (279000) (5,280 ,° 
Tt will be seen that the only assumption involved in this calcula- 
tion is that the average velocity of the ether particles may be taken 
to be equal to the maximum velocity in consequence of a single wave 
motion. 
