BAROMETRICAL MEASUREMENT OF HEIGHTS. 



Example 1. 



Suppose the height of the barometer, reduced to the freezing point, to be I 1 = 

 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 ? 



In th' case * = t' + 4.98 = 16.78, and t + t' = 28.58. 

 Then 



log h = 2.69740 



Table I. for 28.58 gives a = 5.99538 



Table II. for 51 48' gives c = + 0.00026 

 Table III. for 498 toises gives c' = 0.00007 



log u = 8.69297 10 



u = 0.04931 

 log V = 2.47040 



log I = 2.51971 

 Barometer at the lower station b = 330.90 Paris lines. 



Example 2. 



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

 air t' = 18.0 Centigrade = 14. 4 Reaumur; the difference of elevation h = 2217 

 metres ; <f> = 3. The temperature of the air at the lower station t = 27. 5 Cen- 

 tigrade = 22.0 Reaumur, and t + t' = 36 .4 Reaumur. 



, * flog 2217= 3.34577 



log h = | + 971Q18 



3.05595 v = 3.06 

 a = 5.98750 

 c = 0.00112 

 c' = 0.00015 



log u = 9.04218 10 



u = 0.11020 

 log V = 9.77714 



logJ = 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 ^, 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') M, or if the 

 heights of the Barometers are reduced to the freezing point, log b log b' = w, 

 take, 



In Table I., A with the argument t + *', and make v = log u + A. 

 In Table II., with the argument <, take c reversing the sign. 



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