MR. POWER ON THE ABSORPTION OF THE SOLAR RAYS, ETC. 
19 
leads to the same results, and gives a geometrical meaning to the language employed. 
By a well-known theorem in statics, if one force be equivalent to two others, and 
lines be drawn making any given angle with the directions of the three forces, the 
sides of the triangle intercepted by these lines will respectively represent the forces 
in magnitude and sign. 
Let us agree to regard as positive those motions which tend to the right hand of a 
person supposed to be swimming on the plane of incidence in the direction in which 
any one of the three rays proceeds. Now the directions of the three rays themselves 
being those of the three excursions hh'h p each turned from right to left through a right 
angle, it follows from the theorem just enunciated, that the sides of a triangle which 
are respectively parallel to the three rays, will properly represent the transverse velo- 
cities in magnitude and sign. 
Let PO, OP' and OP, be the directions of the incident reflected and refracted rays ; 
take m any point in the latter, and draw mn parallel to OP', meeting PO produced in n. 
By the rules of composition, On is equivalent to Q/n and mn, and since all three are 
measured in the direction in which the rays proceed, their signs must all be regarded 
as positive. Consequently h, h! and h t have the same sign and are severally pro- 
portional to On, O m, mn, and therefore to sin O mn, sin O tun, sin nOm ; that is, to 
sin (0+0), sin 2$, and sin (0—0). 
h' sin (3 — 8,) 
Ji- 
ll, 
We have, therefore, 
sin (0 + 0,) 
sin 20 
h sin (0 + 0 y ) 
15. These equations, combined with the former, 
serve to determine h' h, 0 r 
Substituting the above values in the equation 
of No. (13), put under the form 
a cos ( 
we get 
Now 
* 3 
a cos 0 
A' 2 ] a,cos0, A, 2 
A io r — o To 
ar l Id f u f Id 
TT- { sin 2 (0+6) — sin 2 (6 -0)} = 4 sin 2 0 cos 2 0. 
sin 2 (0+0) — sin 2 (0—0) 
= {sin(4-M,)+sin (d— ^)} {sin (0+0 )- sin (0-0)} 
—2 sin 0. cos 0 r 2 cos 0.sin 0 r 
Therefore, substituting and dividing by the common factors, we get 
a sin 0, a : sin 0 
or 
sin Og^—dh sin 
This equation, combined with h'—h sm ^ ^ 
sin (0 + 0,) 
h t = h - 
sin 20 
sin(0 + 0,)’ 
