54 The Rev. S. Haughton on Physical Geology. [Mar. 8, 



In calculating the motion of the pole caused by the ocean excavations, 

 the weight of the sea-water must be considered, and, by chance, it bap- 

 pens that the weight of the sea-water somewbat more than counter- 

 balances the weight of the surface-rock excavated ; so that the depression 

 of the ocean-surfaces of the earth beneath the zero plane have had little 

 or no effect in shifting the position of the pole. 



Assuming 1*026 and 2*75 as the densities of sea-water and surface- 

 rock, we have for the excess of weight of water added above that of rock 

 excavated, expressed in depth of rock, in miles, 



2 x 1-026-0-58 x 2-75 



275 =(>1 ' m3le * 



The introduction of the weight of the sea will thus give us (raising 

 the zero plane by 0*17 of a mile) 



t= +1-45 mile (continent), 

 t= 0-00 „ (ocean). 



The formulae (8) may be brought into a shape fit for calculation in the 

 following way multiplying both sides by r we have 



rd= =935-6^^(l-cos3\), 



which gives the displacement of the pole in English miles. If we assume, 

 for convenience of the quadrature, 



we have 2r=7916 miles, 

 rcll=34o 

 2=1-45 mile, 



M=|"crr 3 cubic miles of surface-rock, which has half the 

 mean density of the entire earth. 



We may calculate from these data 



M= 519440 million cubic miles of surface-rock, 

 r Hdl= 7836-6 „ 



Hence, finally, 



r0= -14-11(1- cos 3 \) (10) 



This equation expresses that a semilune of continent 5° in width, 

 elevated from the pole to the equator, being 345 miles in width at the 

 equator and zero at the pole, will push the earth's axis away from it 

 though a distance of 14-11 miles. 



If we imagine a continent occupying 90° of longitude of a semilune, 

 and extending from the equator to the pole, we find, if I denote the hour- 

 angle from the meridian bisecting the continent, — 



