250 



Professors W. E. Ayrton and J. Perry on 



which it did at a point corresponding with 20 o, l C. This 

 subtracted from 69°*62, the initial internal temperature, was 

 taken to represent v ; so that 



i> = 49°-52: 

 the values of v were calculated by subtracting from each value 

 of T (the internal temperature shown by the curve A A A) 

 the corresponding value of x (the external temperature shown 

 by the curve a a a). From the various values of t, the time, 

 and of v the corresponding values of m and log N were calcu- 

 lated; and the series of numbers obtained is shown in the fol- 

 lowing Table : — 



t. 



T. 



X. 



v. 



m. 



log N. 



1050 



447 



19 



25-7 



000110 



1-90014 



1200 



4037 



18 69 



21-68 



113 



1-80629 



1350 



3676 



18-38 



18-38 



no 



1-89462 



1500 



33-68 



1808 



15-60 



109 



1-89341 



1650 



30-98 



1776 



13-22 



110 



1-89155 



1H00 



2875 



1746 



11-29 



105 



1 -89304 



1950 



26-8 



1755 



9-65 



104 



1-89491 



2100 



25-21 



169 



831 



099 



1-90002 



2250 



23-88 



1674 



71 4 



101 



1-90415 



2400 



2273 



166 



613 



101 



1-90793 



2550 



21-7 



1H-5 



5-2 



109 



1 -90650 



2700 



20-75 



16-4 



4-35 



119 



1-89902 



Hence 



Mean m 



= 0-001075 



Mean log N 



= 1-89846 



Mean N" 



= 79-15. 



sin a — a cos a N 



a— sin a cos a 2v Q 





79-15 





99-04 





= 0-7999; 





a = 128°-62, 





or 2-2445 radians 





K = 0-00583, 



consequently 



whence 



and 



E = 0-00236. 



Second method. — As it is still an unsolved problem to find 

 the general solution for loss of temperature in a globe when 

 the external conditions are varying, the form of the functions 

 being different from those employed in Fourier's solution, we 



