of a Liquid Substratum beneath the Earth's Crust. 223 



reliance can be placed upon the above estimates. It is, how- 

 ever, necessary to remark that all the accidents to which I 

 have referred would tend to diminish the earth-tide, and to 

 lessen the consequent diminution of the ocean-tide, and so to 

 weaken any argument against the theory of a liquid sub- 

 stratum derived from the circumstance that a diminution of 

 the semidiurnal tide has never been detected. 



(11) It appears, however, to be chiefly upon the fortnightly 

 tide that reliance is placed for answering the question whether 

 the earth is rigid or not, because friction and the interference 

 of continents will have a much smaller effect upon that slow 

 tide. Now the fortnightly tide in an equatorial canal would 

 have a velocity of about 54 feet per second. If we substitute 

 this value for a, neglecting the effect of friction, and putting 

 k=6Q miles, we obtain 



H = cxOG. 



If we use a larger value for k, the result will be but little 

 affected. 



(12) The effect of friction would not altogether disappear 

 even in the case of the fortnightly tide ; and the supposed 

 mountain-roots, which appear to be a necessary accompani- 

 ment of a liquid substratum, would still interfere with the for- 

 mation of the tide, and diminish the value of E, and lessen the 

 diminution of the ocean-tide. 



Suppose, then, that if there were no friction in the sub- 

 stratum, the ocean-tide would, as shown above, be diminished 

 to 06 of its undiminished value. Let us inquire what the 

 coefficient of friction in the substratum would need to be to 

 bring the ocean-tide up to 0*7 of its undiminished value. Now 



H = c{l- §t? cos 2 2S(2—f v ^ nearly. 



Hence the diminution varies as cos 2 28. By our supposition 



1-0-7 2os 



or 



28=30°. 



fax' . 



Now (§ 4) 



tan 28 = 



2(^-* /2 ) 



and cc r , the equatorial velocity of the fortnightly tide, — 54"28 

 feet per second; k= 60x5280 feet; and «=209xl0 5 feet; 

 g = 32*2 feet. From this equation we find 



/=0'0178tan28. 



