ORIGINATING THE CONSTRUCTION OF A HYDROSTATIC QUADRANT. 189 



consequence of certain assumed spaces of the inner surface, being in 

 contact with each of the contained fluids. 



If 0:zzhalf a right angle, that is, if each fluid cover a space of 45 

 degrees; then 20 zz 90, and consequently, sin.20 zr 1 and cos.20:zrO, 

 while sin. 0:=! ^/ 2, and cos.0:=r | ^2 ; therefore, by substitution, 

 equation (142) becomes 



l(s' 



- J(s ' 



Now, let the fluids be mercury and rectified alcohol, as in the pre- 

 ceding example, then we shall have 



829 V2- 



i (14000829)^2 -f 829 



which answers to the natural tangent of 28 55' nearly. 

 Again, if 0zza right angle, that is, if each fluid cover a space of 

 90 degrees on the inner surface of the tube; then, 20zzl80, of 

 which the sine and cosine are respectively and 1, while the sine 

 and cosine of 0, are respectively 1 and ; consequently, by substitu- 

 tion, equation (142) becomes 



s' -4- s 



tan.zzz-7-- . /IAA\ 



s 1 s (!44). 



203. This is a very neat and obvious expression, and the practical 

 rule derived from it, may be enunciated in the following manner. 



RULE. Divide the sum of the specific gravities of the two 

 fluids by their difference, and the quotient will give the 

 natural tangent of an arc, which being estimated from the 

 lowest point of the tube, will indicate the highest point of the 

 heavier fluid. 



If therefore, the contained fluids be mercury and rectified alcohol, 

 as in the preceding cases, we shall have 

 14000 -|- 829 



ta "-*=14000-829 =U2587 - 

 which answers to the natural tangent of 48 23' 19''. 

 We might assume other particular values of the spaces in contact 

 with the fluids, and thereby deduce corresponding forms of the equa- 

 tion ; but what we have already done on this subject is quite 

 sufficient. 



