ATTRACTION ON TIDAL PROMINENCES. 27 



Now, in order to estimate the greatest effect possible, 

 let us suppose that the greatest elevation is in the middle 

 of the quadrant, i.e. 45 before quadratures ; and further 

 that the elevation is not diminished by friction. 

 Let H = the moon's greatest horizontal force. 

 w = angle from the moon. 

 e = greatest elevation. 



Then the tangential force at any point = ZTsin 2o>, and the 

 elevation = e sin 2<o. For it is proportional to the velocity, 

 which as we have seen (p. 14) = Fcos 2 (w - ), and if 

 8 = 45 this = Fsin 2w. Multiplying by the element of 

 the equator, and the unit of width, we get the moment 

 = He sin 2 2o> x rdw x r. Now sin 2 2w is always positive, 

 varying from to 1 and from 1 to 0. Substituting its 

 equivalent ^(1 - cos 4w) we divide the moment into a 

 periodical part ^ Her* cos 4wtfw which varies from - %Het*du> 

 to + -^ Her*(fo, and therefore produces no permanent effect, 

 and a constant part which is = -J- Hei*d&. 



Summing this round the circle, and multiplying by the 



coefficient of friction, we have for the whole moment Heirr z f. 



Taking the density of the earth as 5, the moment of 



inertia of the equatorial section of the earth is - irr*. 



Dividing the former by the latter, we have for the angular 

 tion ~T-^- 

 1 depth of sea 



(negative) acceleration ~T-^- 



XT rr 



NOW ' H 365000 6440 



If we assume the depth of the sea to be 3 miles, the angu 



s 



lar (negative) acceleration becomes nearly = .. . .... - . 



93 billions x r 



