TIDAL FORCE. 3 



component of xn is equal and parallel to nh, the 

 perpendicular on the radius, and is, therefore, pro- 

 portional to Ik, which is one- third of nh. Note that 

 Ik = El sin HEi<= Ex cos If^sin MEfr= Jr sin 2 Mft 

 that is to say, the tangential force is proportional to sin 2 

 (angle from moon). The same reasoning applies to the 

 dotted letters in the further hemisphere. 



The vertical component being in the same line as gravity 

 (either in the same or an opposite direction) cannot directly 

 produce any motion. In fact, it could not do so unless it 

 actually exceeded the force of gravity. And, as will be 

 seen by-and-by, it is too minute to produce any indirect 

 effect. 



The tide-producing force then always acts towards EM 

 (in the direction of the arrows in fig. 2). From this we 

 can deduce theorems relating to the place of high and low 

 water, &c., without requiring to determine the magnitude 

 of the force, which will be hereafter taken into account. 

 At present we need only observe that it is very small 

 compared with gravity. 



First, then, let us consider the case of water limited to 

 an equatorial canal. The moon being supposed in the 

 equator, we shall establish the following theorems : 



i. If there were no friction it would be low water 

 under the moon, and high water in quadratures. 



n. Friction accelerates the times of high and low water. 



in. In addition to the oscillatory motion of the water 

 there is a constant current produced by the action of the 

 moon. 



iv. The effect of friction on this is to increase the 

 length of the day. 



B2 



