22 KNUT LUNDMARK, GLOBULAR CLUSTÉRS AND SPIRAL NEBUL^E. 



K + 0,601 X — 0,789 Y + 0,124 Z = - 0,410 

 K + 0,764 —0,612 +0.203 —0,095 



K + 0,789 —0,614 —0,023 —0,010 



which give the normal equations: 



1 0,000 K — 6,698 Y — 0,757 X + 1,699 Z = — 1,025 



— 6,698 +4,893 +0,035 —0,768 +0,603 



— 0,757 +0,035 +3,606 —0,730 

 + 1,699 —0,768 —0,730 + 1,501 



0,016 

 0,545 



from which follow the elimination equations: 



1 ; 0000 K — 0,6698 Y — 0,0757 X + 0, 1699 Z = — 0,1035 



1,0000 Y — 1.1597 X + 0,9091 Z = — 0,2211 



1 ,0000 X — 0,0573 Z = — 0,0660 



1 ,0000 Z = — 0,3440 



which give the following values for the unknown quantities: 



X = — 0,086 Z = — 0,344 F = 0,3547 

 7 = -0,008 Ä = -0,057 4= +5° 



D = + 76° 

 If we put K = 0, we obtain: 



1 ,0000 F + 0,0071 X — 0,1570 ^ = + 0,1232 



1 ,0000 X — 0,2011 Z = — 0,0055 



1,0000 Z = — 0,3673 



X = — 0,079 V = — 0,381 



7 = + 0,066 A = - 40° 

 Z= -0,367 £= + 74° 



£"., = ± 0,080 E v =± 0,154 (m. e.) 

 E y = ± 0,069 ^ = ± 59° 

 ^=±0,130 ^=±44° 



Residuals. 



v 



K = — 57 



K = 



V 



K= — 57 



/f = 



v 



v 



v 



v 



-170 km 



+ 84 km 



+ 16 km 



-160 km 



-114km 



- 128 km 



-125 



- 25 



- 14 







+ 102 



+ 89 



+ 10 



- 18 



- 22 



-410 



+ 265 



+205 



-300 



- 68 



+ 52 



- 95 



- 92 



- 80 



+ 225 



-148 



-155 



- 10 



-101 



- 84 



