December 1, 1898.] 



KNOWLEDGE. 



287 



on Tuesday, the 27th, at 11.39 p.m. On the 27th she will 

 be totally eclipsed, and the following data may be useful: — 



First contact with shadow 



Beginning of totality 



Middle of totality 



End of totality 



Last contact with shadow 



li. 



9 47-8 G.M.T. 



10 57-4 



11 42-1 



12 2G-8 



13 36-4 



Diagram showing the path of the Moon througli tlie Eartli's Shadow, 

 I)ecember 27th. 



The first contact on the Moon's limb is at an angle of 

 112° from the north point towards the east, and the last 

 at 95" towards the west. The magnitude of the eclipse, 

 that is, the distance at mid-totality from the Moon's 

 most immersed limb to the boundary of the shadow 

 nearest to the opposite limb, divided by the Moon's 

 diameter, will be 1-883. The eclipse is illustrated in 

 the above diagram, showing the part of the shadow 

 traversed by the Moon, and indicating also the points 

 of contact and the times of occurrence of the principal 

 phases. During the eclipse there will be occultations of 

 sixteen stars, ranging in magnitude from 8-7 to 9'5, for 

 which particulars will be found in the " Companion to the 

 Observatory." 



On the 29th there will be an occultation of i Canori, the 

 disappearance talsing place at 10.1 p.m., 92° from the north 

 point (180° from the vertex) ; and the reappearance at 

 11.16 P.M., 803" from the north point (834° from the 

 vertex) , reckoned through east. 



Correct Solutions of No. 1 have been received from 

 H. S. Brandreth, G. G. Beazley, W. de P. Crousaz, D. R. 

 Fotheringham, H. Le Jeune, -J. M'Robert, A. E. White- 

 house, W. Ciugston. 



,/. M'Robert. — Received solution too late to acknowledge 

 last month. 



^4/^;,„._The reply to 1. Q to B6 is Kt to Kt7, a cleverly 



provided defence. The Indian puzzle which you send is 



very pretty, but it has become evident that our solvers will 



I not attempt anything longer than a three-move problem. 



I G. F. /■., anil Fl'. B. Stead.— U 1. Q to B3, KxR, 



and there is no mate. 



X. E. Meaiex. — We insert your 3-move problem below. 



Mr. Bolton's 11-move mate is too long fjr this column. 



[With regard to problem No. 2, our worst fears have 

 been realized. It is evident that the readers of Knowledge 

 are not attracted by sui-mates, and they will not be pub- 

 lished in future. At the same time we may quote the 

 opinion of the best problem judges to the effect that there 

 is more scope for originality in the sui-mate than in the 

 direct mate. Every possible phase of the latter has now 

 been exhausted.] 



PROBLEMS. 



No. 1. 



By F, W. Andrew. 



Black (V). 



(J^^tss Column. 



By 0. D. LooooK, b.a. 



Communications for this column should be addressed to 

 C. D. LooocK, Netherfield, Camberley, and posted on or 

 before the 10th of each month. 



Solutions of November Problems. 



No. 1. 



(By A. C. Challenger.) 



. Q to Et7, and mates next move. 



No. 2. 



(By P. H. Williams. I 



1. B to K5, P to Q3. 



2. Q to QB7, PxB. 



3. Q to B3, P to KiJ. 



4. B toQ3, PxB (or A). 



5. Q to Bsqch, R to KtS. 



6. Kt toB6, RxQ mate. 



(A.) 



4. P to K6. 



5. Q to Bsqch, R to KtS. 



6. R to K2, RxQ mate. 



'n Hi 



mm ^ mm = wm. 



i ^ ^ ^*# 

 '^£^ mm mm. 



'm% 



^m, A '^'i-j^ 'mi 



km 



White ,1I) 



White mates in two moves. 



No. 2. 

 By N. E. Meares. 



BLiCI (10). 



White (10). 



White mates in three moves. 

 [We consider it advisable to state that Castling is 

 allowed in this problem ; also that the key-move is a check. 

 In spite of this the problem is well worth trying, some of 



