Centrifugal Force as Geotropic Stimuli . 
Total period (/ + T) = 7 m. 50 s. 
t 
-j, — 4 * 5 - 
^ = 4-086. 
Point of equilibrium is 0-5 cm. below middle of box. 
r — 6-5 cm. 
Cent, force = 4-36 mg. 
C - 4 - 5 1 mg. 
Thus ^ = 4.5, £= 4 . Si . 
735 
CT 
mg.t 
= i-oo. 
Experiment 4. Sept. 15, 1912, 10.14 p.m., to Sept. 16, 9.25 a.m. 
Temp. = 20 0 C. 
Radius of rotation at middle of box = 7 cm. 
Total period (/ + T) = 7 m. 30 s. 
/ 
Y = 5 > 6 5 - 
7 ? = 4 - 37 - 
Point of equilibrium is at middle of box. 
.-. r = 7 cm. 
Cent, force = 5*38 mg. 
C — 5-45 mg. 
Thus ~ = 5 ‘ 65 , — = 5 - 45 - 
T mg 
CT 
mg . t 
= 0-96. 
Experiment 5. Sept. 19, 1912. 9.50 a.m. to 5.25 p.m. 
Temp. = 20 0 C. 
Radius of rotation at middle of box = 10 cm. 
<9 = 105°. 
Total period (/+ T) = 7 m. 30 s. 
t 
Y = 5 - 30 - 
R = 3-85. 
Point of equilibrium is 1 cm. below middle of box. 
r = 9 cm. 
Cent, force = 5-40 mg. 
C = 5-47 mg. 
t C 
Thus -7= = 5-30, —=5.47. 
T 
CT 
mg. t 
mg 
— 1-03. 
