i86 
Journal of Agricultural Research 
Vol. VII, No. 4 
Tabi.3 XVII.— A comparison of the energy received by direct radiation with'the energy 
dissipated by transpiration during midday 
Period and crop. 
July 7-16, 1914. 
Wheat, Kubanka. 
Wheat, Galgalos. 
Oat, Swedish Select. 
Oat, Burt. 
Barley. 
Millet, Kursk. 
Millet, Siberian. 
Cowpea. 
Aug. 2—11, 1914. 
Alfalfa, Grimm, E23-20-52. 
Alfalfa, Grimm, E23. 
Alfalfa, Grimm, E23 (in open).. 
Alfalfa, Grimm, 162-98A.. 
Hourly 
transpiration. 
Radiation (cal. 
per sq. cm. 
per hour in 
shelter). 
Cross section of 
plant normal 
to sun’s rays 
(area of plant 
shadow). 
Ratio of radia¬ 
tion energy 
received to 
energy dissi¬ 
pated through 
transpiration. 
Gm. 
400 
58 
Sq. cm. 
2 , OOO 
0. 54 
469 
58 
I, 920 
• 44 
434 
58 
2, 040 
•51 
4 °s 
58 
1,940 
I, 480 
• 55 
220 
58 
• 73 
239 
58 
1,920 
I, 980 
•87 
257 
58 
•83 
3 i 7 
58 
2,230 
.76 
283 
64 
I, 840 
.78 
332 
64 
2,230 
. 80 
326 
80 
2,200 
2,870 
1. 01 
529 
64 
•6S 
The area of the plant shadow in the case of the grain crops has been 
determined by considering the plants in a pot to have the form of a 
Fig. 4.—Determination of the area, on a plane 
normal to the sun’s rays, of the shadow of a 
cylinder of diameter d and height h in terms of 
the angular departure (2) of the sun from the 
vertical. 
right cylinder, the diameter and 
height of which were determined 
from direct measurements and from 
photographs. The method of com¬ 
puting this area is readily seen from 
figure 4. Let x be the angle made by 
the sun’s rays with the vertical at 
midday (zenith distance). The pro¬ 
jection of the right cylinder of height 
h and diameter d on a plane normal 
to the sun’s rays would give a rec¬ 
tangular figure with elliptical ends 
whose total length is d cos x + h sin x . 
In this expression^ cos x represents 
the minor semidiameter of each el¬ 
liptical portion, whose major semi¬ 
diameter is -. The area of the el- 
2 
• * 7 T 
liptical portion is consequently ~-<P cos x, while the area of the rectang¬ 
ular part of the figure is h d sin x. The total area (a) is therefore— 
a — — d 2 cos x 4 - h d sin x. 
4 
