394 BAROMETRICAL MEASUREMENT OF HEIGHTS. 



Example 1. 



Suppose the height of the barometer, reduced to the freezing point, to be h' = 

 295.39 Paris lines; the temperature of the air t' = 11°. 8 Reaumur, and the latitude 

 <^ = 51° 48'; the increase of heat downwards being 1° Reaumur for 100 toises. 

 What is the height of the barometer, reduced to the freezing point, at a station 

 lower by h = 498.2 toises ? 



Then 



In th' case t := t' -\- 4°.98 = 16°.78, and t -\- t' = 28°.58. 



log k = 2.69740 

 Table I. for 28°. 58 gives a = 5.99538 



Table II. for 51° 48' gives c = -f- 0.00026 

 Table III. for 498 toises gives c' = — 0.00007 



log M = 8.69297 — 10 



u = 0.04931 

 log b' = 2.47040 



log b = 2 51971 

 Barometer at the lower station b r=; 330.90 Paris lines. 



Example 2. 



Suppose the reduced barometer b' = 598.6 millimetres ; the temperature of the 

 air t' = 18°. Centigrade ^ 14°. 4 Reaumur; the difference of elevation h = 2217 

 metres , <^ = 3°. The temperature of the air at the lower station t z=i 27°. 5 Cen- 

 tigrade =; 22°.0 Reaumur, and t -\- t' = 36°. 4 Reaumur. 



^, 1,5 log 2217 = 3.34577 



Then log/* = | ^ _^ g^^jQj^ 



3.05595 V = 3.06 

 a = 5.98750 

 c = — 0.00112 

 c' = —0.00015 



log u = 9.04218— 10 



u = 11020 

 log b' = 9.7^14 



log b = 9.88734 

 Barometer at the lower station b :^ 771.5 millimetres. 



2. For Computing Differences of Elevation from Barometrical Observations. 



Given the unreduced height of the barometer at the lower and upper station, 

 B and B'; the temperatures of the attached thermometers, T and T'; the temperatures 

 of the air, t and t' ; and the latitude, ^. 



To find /i, or the difference of elevation between the two stations. 



Subtract (log B' — 10 T') from (log B — 10 T), paying due attention to the 

 nature of the signs of T and T', and taking the numbers 10 T and 10 T' as units of 

 the fifth decimal. Calling then (log B — 10 T) — (log B' — 10 T') = «, or if the 

 heights of the Barometers are reduced to the freezing point, log b — log b' =^ u, 

 take. 



In Table I , A with the argument t -\- t\ and make r = log u -\- A. 

 In Table II., with the argument (/>, take c reversing the sign. 

 D 54 



