May. 1911. 



KNOWLEDGE. 



183 



If the ratio of the length of a stick to that of its shadow- 

 is i, the tangent of the angle of elevation is -5. and a = 26 34'. 

 Now, on Mav 1st the Snn's declination is 14' 49' N.. or 

 A = 90 - 14° 49' = 75' 11 '. 



h'lGUKE 1. 



0, the Latitnde of London is 51° 31' iiearl\ 

 Making these snbstitntions we find that 



2 



■5436 



— = 32° 56' nearlv. 

 2 



/( = 65° 52' 

 = 65- -866 

 Hence, the number of honrs before or after the Sini has 

 crossed the Meridian is 



65-866 „ , ,„ 

 — - _ 4-39 



or 4". 23"\ 

 Hence the time is about 7.37 a.m. or 4.23 p.m. 



The Sun will attain nearly the same declination on 

 August 13th, so that the same ratio between the length of a 

 stick and that of its shadow would hold at the above times on 

 this date. 



To solve the second part of the (|uestion, finding the dates 

 when the length of a stick and th.it of its shadow are eipial at 

 noon in Latitude London, we have the well known equation 



Colat. + Dec. = Meridian Altitude. 



Where Colat. = 90 - Lat. = 90 - 51 31' 

 = 38° 29' 



In this case M.A. = 45°, since the tangent of the angle of 

 elevation is 1. 



Hence Dec. = 45' - 38' 29' North. 

 = 6° 31' N. 



The Sun attains this Declination about April 7th and 

 September 7th. On these two dates, therefore, the length of 

 the shadow will be equal to that of the stick at nomi. in 

 Latitude of London. 



To find the Latitude, assuming we know the ratio between 

 the length of a stick and that of its shadow at noon on 

 April 20th. 



Suppose a stick 7-ft. 1-in. high cast a shadow (i-ft. long at 



noon. 



85 

 If a be the angle of elevation, tan a = - — 



72 



= 1-1805. .■. a = 49= 44' 



Now. on .April 20th, the Declination of the Sun is IL 13' N. 

 Colat. + Dec = Meridian Altitude. 

 .-. Colat. = 49° 44' - !!' 13' 

 = 38= 31 

 Hence Latitude is 90° - 38° 31' 



= 51 29 

 or, about the Latitude of London. 



On August 24th, the Sun attains nearly the sauie declination, 

 so that this ratio would hold on that date also. It should be 

 noted that the time found in the first part of the question will 

 be apparent time. The mean time, or clock time, is found 

 from it by the equation. Mean Time — Apparent Time = 

 Equation of Time. As, however, the equation of time is never 

 more than sixteen minutes, there is no considerable error in 

 taking the two as coinciding. On none of the dates given 

 above does the equation of time amount to five mimites. 



IRhV.I M. D.winsnN. 



30. In answer to (Question 30, of "Interested" (March issue of 

 2 " Knowledge ■'), may I suggest 



^ I the following aspossiblesolutions 



to the queries therein mentioned? 

 (1) How can the time of day 

 be ascertained by measuring the 

 ratio of the length of a stick 

 to that of its sh.'idow on a given 

 date ? 



In Figure 1, OT represents 

 the stick. O A the length of 

 the shadow of the stick cast by the Sun, OZ the zenith. 

 Then 



tan-' OA 

 OT 

 In Figure 2, E H W is the horizon, l: 

 equator, P the celestial pole, / the zenith. 



S Z = Sun's zenith distance (found 

 by means of the stick). 



ZF = Colatitude -= (90' - 51° 30') 

 = 38° 30'. (If London is the place 

 of observation). 



S P = Sun's North Polar distance 

 = 90° — Sun's Decl. (found from 

 almanac on given date). 



< S Z M = Sun's Azimuth, i.e.. 

 Sun's angular distance from Meridian 

 measured from the north point, 

 through E S H W". 



Then from Spherical tri 

 Sin^PZ ^ Sin SZP 

 SP 



Figure 1. 



Sun': 



zenith distance at instant of 

 observation. 



Sin SZ 



Sin 



^leSZ P. 

 Sin S P Z 



Figure 3. 



Sin SZP 

 Sin S P 



and this gives the Sun's Eastern 



or Western hour angle according 



as the observation is before or 



after noon. If the observation is 



before noon, 12 — Sun's hour angle 



gives the approximate time. If 



after noon the time is gi\'en by 



Sun's hour angle alone. 



(2) .At what times in the year will 



same as that of its shadow at noon ? 



the length of stick be the 

 (Latitude of London.) 



When the length of shadow equals length of stick the Sun's 

 zenith distance is then 45° and 51" 30' - 45°, gives the Sun's 

 declination at the required times .'. Sun's Declination = 6" 30 . 



I'rom the almanac we find that the Sun's declination has 

 this value on the 7th April, and the 7th September. 



(3) Can the latitude of a place be determined by comparing 

 the ratio between the length of a stick and that of its shadow 

 at noon ? 



Sun's zenith distance at noon 



length of shadow 

 length of stick 



Sun's Declination - < O O S, (E Q W represents the equator! 

 (P the pole) and < OOZ = Latitude required. 



Latitude=< SOQ+<SOZ. L. O. G. P. 



-= tan 



<zos 



