335 
dD, 0D, 1 1 A 
Ket Say = Gey, BUN ar COS els —5) nn ae) 
As B>b this expression is + for a S30°. For the total field 
we finally have 
2(F 1 A ls 
0 (01 +s) = 8} — | sin? B sin 28 + sin? a cos Zal 1 —— EE) 
dz? 6° a 
Equalizing the bracketed terms to zero gives a relation between 
a and @; neglecting 57/45? we- find 
45° 48° 50°46’ | 54°44’ | 60° 
Sha Td oo | Gr O° 14 40 | GOs. 
as corresponding sets. For «a —= 60° we obtain the same value for £, 
- i.e. non-protruding frontal rectangles. 
In excentric axial points at a distance z from the centre A the 
value of the first term is 
e.g. for C= al 
the value: 8 = 90° 
2ab 
9 (a) = 43 tal « 5 
That of the second term for one pair of inclined planes 
B? — 2Bzx sin a cos a 
N,(z) = 25 sin a cos a pj da es 
b? — 2ba sin a cos a + x’ sin? a 5 
b— à sin a cos at B — « sin a cos a GE 
+ 2tgal tand — tan—1—________ 
esn’ a © sin? a 
By means of (4%) and (5%) the total field may be calculated for 
any axial point, whether the 4 inclined planes intersect in one line 
or not; only in the former case do they form an “optimum-surface” 
with regard to A. 
For excentric points on an equatorial axis of y we find as the 
irste term, for ¢ == 0 
pipe Re nes 
i) ar =DE ey ; 
and as the second term for two pairs of inclined planes 
B* + 2By sin’ a 4- y? sin? a \ 
Dy) = 235 sin a cos a | log ——+—___—\— 
b* + 2by sin? a + y' sin? a 
B? — 2Bysin’? a + y’ sin’ a ‚ysn'atb 
OE. + 2 tg a tan A + Ci) 
b 2by sin? a + y? sin’? a y sin a cos a 
4 ‚gsm a—b ‚yen ad B ; y sin? a— B 
an tan tan 
a ysinacosa ” y sin acos a y sin a cos « 
The distribution of the field is thereby completely determined ; in 
