96 PEOCEEDINGS OF THE AMERICAN ACADEMY 



We have again — 



m.m. 



(6716.4) 681 X 0.000 6716.4 = .45739 = — sin t s + sin (r 2 — x. 2 ). 

 B 681 X 0.000 6866.8 = .46763 = — sin i a -j- sin (r a -f x 2 ). 

 i a = 15° 02' 30" r 2 = 46° 13' 27" 



rev. rev. 



whence 2x 2 = 50' 54" = 14.490 ,\ 1. = 3' 30".8. 



The mean of these values, 3' 29".7 is taken as the value of one 

 revolution of the micrometer head in arc, at the time the above 



rev. 



measures were made on B X2 and A w and B to B i2 = 6.087 = 21' 16".4, 



rev. 



A to A 12 = 7.727 = 27' 00".4. 



For B , ® 681 X 6866.8 = .46763 = — sin i-\- sin (r — 10'38".2) 



„ B 12 , (3) 681 X I B 12 = = — sin i + sin (r -|- 10' 38".2) 



© i-\-r =61° 16' t = 14° 47' 38" 



Substituting in (|) the values of % and r found by ©, we have 



m.m. 



X B l2 = 0.000 6928.0. The value of the corresponding line in 



m.m. 



Angstrom is 0.000 6927.8. 



The result of this example showing a sufficiently near approach to 

 accuracy for our purpose, the same method was used for determining 

 the wave-length of A i2 , starting with the known value of A . Measure- 

 ments at this part of the spectrum are more difficult than in the 

 more brightly illuminated portion. Three attempts, on as many 

 different days, gave — 



m.m. 



X A 12 = 0.000 7678.4 

 0.000 7677.8 

 0.000 7679.4 



The mean of these = 0.000 7678.5 was taken as the value of 

 XA, r 



The stray light of higher refrangibility and greater illuminating 

 power than the red, and which is reflected from the lens, overpowers 

 the red to such an extent that some means of removing it is necessary. 

 For this purpose, a small direct-vision prism, placed over the slit, 

 allowing only the red to pass, seems best ; but red glasses have been 

 found to answer nearly as well. 



