1877.] The Rev. S. Haughton on Physical Geology. 537 



If (p be the angle of separation between the axes of rotation and figure, 

 or semiangle of the cone, we have 



dd = sm du. 



The solid earth revolves with an angular velocity, w, round Obt (today's 

 axis), while the ocean revolves with the same rotation round Oay (yester- 

 day's axis). 



Let F be a coefficient depending on the friction between the ocean and 

 its bed, and such that, if the ocean be moving relatively to the earth with 

 any angular velocity, 



E X angular velocity = couple produced (5) 



Resolving the angular velocity of the ocean round Oy (yesterday's 

 axis) into its components, we find the frictional couples 



(1) Round OC =Fw cos 0, 

 Round Oy'=F(jj sin 0. 



Resolving the latter into its components, we find 



(2) Round Ot" = Yio sin du, 



(3) Round Of =Fw sin <p 



1. The component of the ocean rotation round OC being the same as 

 that of the solid earth (for yesterday's and today's axes are at the same 

 distance from the axis of figure), there is no frictional couple developed 

 round OC. 



2. The frictional couple round 0*" produces its full effect, for there is 

 no rotation of the solid earth round this axis. 



3. The rotation of the solid earth round Of is w sin 0, and the compo- 

 nent of the ocean rotation falls short of this by the quantity 



du 2 



— (i) sin — . 



Hence we have a frictional couple round this axis, retarding the equa- 

 torial rotation and equal to 



. du 2 



— _b w sin — — 



The frictional component round the axis Of is much greater than that 

 round 0^' ; wherefore, neglecting the latter for the present, and resolv- 

 ing the couple round Of round the axes x and y, we have, writing 

 p=oj sin 0, 



Jjdt = — ~Fp sin u du, 

 Wdt= — Fp cos u du 9 



