C. 8. Lyman’s new form of Wave apparatus. 389 
circle and that of the particle’s orbit ; that is, putting R andr 
for these radii respectively, v for the orbital velocity, (= ="), 
and D for the difference in question, 
a got ae 
EF 
When r equals R, then D=r, or half the height of the wave. 
represented by the transverse wires. ‘The elevation in question 
is equal to the height due to the particle’s orbital velocity; or, 
is a third proportional to the diameter of the rolling circle and 
the radius of the orbit; or, is equal to the area of the orbit 
divided by the length of the wave; that is, putting H for this 
elevation, J for the wave’s length, and the other symbols as 
y2  -y2 = mp? 
When ¢ equals R, H = = or one-fourth the height of the 
wave. To this elevation is due one half the mechanical 
energy of a wave—the other half to the motion of its parti- 
cles. That energy is, in other words, half potential, half actual. 
_ 15. The decreasing diameter of the orbits with depth. This 
is seen in the shorter crank-arms below, and the decreasing 
amplitude of sway of the upright elastic wires, down to their 
points of rest, which mark the depth of still water. The de- 
crease of the orbits in diameter takes place in a geometrical 
ratio, and is approximately one half for each increase of depth 
equal to one ninth of a wave length; or, more exactly, putting 
rand r' for the radii, respectively, of a surface orbit and of 
kK 
one whose middle depth ish, itis r’-=re*®, 
radius of the rolling circle, and e the 
nally rectangular, or rectangular ng at rest. 
