of Eclinhurgh, Session 1883-84. 
745 
This solution fails if (Sa2a3)2 - , or TVa 2 a 3 = 0; for then 
the three points A,B,C, are in one line at starting. But this, and 
similar cases of failure (when they are really cases of failure) are 
due to an improper selection of three of the points. We need not 
further discuss them. 
But it is interesting to consider how the vectors P can he found 
when one position of the reference frame has been obtained. Keep- 
ing, for simplicity, to the system of three points, we have by the 
solution of the equations above the following data : — 
= ^ Sa^3 = e', + =f , = = 
where e, e\ /, y, g\ li are known numbers ; which, as the equations 
from which they were derived were not linear, have in general more 
than one system of values. The second, third, and sixth of these 
equations give 
^38 . a2tt3^2 “ 'b (/“ b ^ ^ (^ 2^2 * 
Provided coplanar with a 2 ,a 3 , this equation gives, by the 
help of the fifth above, a surface of the 4th order of which is a 
vector. But is also a vector of the plane ^a.^P 2 = e, and of the 
sphere T ^2 =" 9^ Hence it is determined by the intersections of those 
three surfaces. 
But if S . a 2 a^P 2 vanishes, the equation above gives (by operating 
with S . Va 2 tt 3 ) 
0 = h(Ya^a^)^ - (/- S^2«3)S . P 2 Y. e'S . a2Va2tt3 , 
which gives a surface of the second order (a hyperbolic cylinder) in 
place of the surface of the fourth order above mentioned. This 
may, however, be dispensed with : — for P^ is in this case deter- 
mined by the planes Sag/?^ = e and S . agag/Jg = 0, together with the 
sphere T/Sg = g. 
3. Note on the Occurrence of Drifted Trees in Beds of Sand 
and Gravel at Musselburgh. By James Geikie, LL.D., 
F.E.S. 
1 am indebted to Mr William Robertson for calling my attention 
to the interesting phenomena which form the subject of this 
