170 NEWTON S riuNCiriA. 



pyramid diminish without limit, the pressures on the two 

 sides C O A and B C A will be normal and respectively 

 equal to 



p x area C A, q x area B C A. 



Resolving these, parallel to B, we have 

 p area C O A area C O A 

 q area B C A area B C A 5 



because this latter ratio expresses the cosine of the incli 

 nation of the two planes ; hence 



p = q. 



Similar equations hold by symmetry for the other sides. 

 And therefore the pressure is equal in all directions. 

 This includes Prop. xix. of Section V. 



3&amp;lt; \ Note I. 



4. J 



5. Newton proceeds to consider the equilibrium of a sphe 

 rical mass of fluid, like our atmosphere, resting upon a 

 spherical concentric bottom, and gravitating towards the 

 centre of the whole. The object is to determine the pres 

 sure on any point A of the bottom. Divide the fluid into 

 concentric orbs of equal thickness dx. Now any part of 

 a fluid at rest may be supposed to become rigid ; for it 

 will then resist and be resisted by the remainder of the 

 fluid in exactly the same manner as before. Draw, then, 

 any cylindrical canal from the point A to any point B 

 in the surface of the fluid, and suppose its superficies to 

 become rigid. This canal will be divided into elements by 

 the concentric orbs. Let d s be the length of any one 

 of these elements, and F the force of gravity ; then the 

 weight of that element is Fds. This acts directly to 

 wards the centre, that is, along dx. Resolving along the 

 canal, the force with which this element tends to press 

 the bottom of the canal is Fdx. The same is true for 



