eee 
On Sun-heat and Radiation. 93 
But this would not answer for calculation, as H passes through 
90°, at the time of the equinoxes, when tanH becomes infinite and 
sind vanishes. I therefore expand in terms of cosH as follows: 
L cos? 1°3) cos®H 2\3°5 cost 
Tv 
H=7- | cos H+ 5. aod Be Oe GT + &e. (9) 
and, 
VY 1—cos?*H 
tan? A =) ee 
a cos Hf 
us 1 cos*H 1 costH 1:3 cos6H 1:3°5 cos8® be. (6 
acral TTR TLS) oF Lea? am) Toa oara eee 
Therefore, 
H—tnH=4——~,—5e0s H— 7, cost H— 00st H 1H & 
—tan TOM Come Dae SOT cos _ cos? =o cos Cc. 
(7) 
But, 
tan A sin 6 
COS aE a ae 
fe sin A sin 6 sin?d 1:3 sintd | 1:3: sin€g 1:3-5-7 sin86 & 
MIG. con j 2 12° 2 *123° 8 To34° 98 of 
(8) 
and 
1 ge he C08 x sin? 6 x aa sin4 6 a Tao ing 1:3:5 
cos H CTE MS © ere ae br a mcs eae 
sin’ 0 
Tey (9) 
Hence, finally, 
The heat received in one day = 24 sin \ sin 6 (H—tan H : 
= 2A 
wv, ; 
3 sin A sin 6 
sin?6 1 sin! un) 1:3 sin§ o 
ia aac a 
+ cos {1 — 
sin? X sin? 6 ( sin?§ 1-3 sintd 
Sweax. | Borg has as + &e. as 
J 
sin4 dX sin‘ dé ( Bon 
ee esi Ore we, 
24 cos® ru (| 2 
sin® \ sin® 0 ¢ 
8&0 cos? X Ve Keng: 
