172 
Er Strømfladens Hojde over Overfladens Niveauflade 
Ah Meter og det tilsvarende Tryk i Atmosfærer 2\p, saa 
har man (da i Overfladen h=0 og p=0), naar Tætheden 
er S: 
ao S (1 — 6 cos 2 g) 1 +06. h) A 
Ap Atm. = loa 
Størrelsen af S kan variere, som Tversnittene X VII, 
Pl. XL, og XX VIII, Pl. XLI, vise, mellem 1.0265 (73° N. Br.) 
og 1.0284 (61° N. Br.). Factoren for Ah bliver i første 
Tilfælde 0.09957, 1 sidste 0.09967. Den største Værdi af 
Ah er 1.4 Meter. I dette extreme Tilfælde vilde den 
første Factor give Ap = 0.1394 Atm., i sidste 0.1395 Atm. 
Om man regnede med Factoren 0.1 fik man 0.1400 Atm. 
Forskjellen i Resultatet, 0.0006 eller 0.0005 Atm., svarer 
til Kviksølvtryk af resp. 0.46 og 0.39 Millimeter. Jeg 
sætter derfor 
Ap Atm. = 0.1 Ah Meter. 
I Niveaufladen H= 300 Favne have vi 
p=54.6438 + 53.23 (S — 1.02783) Atm. 
for 800 Favnes Vandsojle, 
10.027 
Tæthedsfladen Ah = — > : 
1 — BP cos20 
53.23 (1.02783 — 3) 
Meter, 
1.0027 
= ——_—____ . 93.23 (1.02783 — I 
altsaa Ap l= pene 53.23 (1.02783 ) 
Atm. for Tæthedsfladen, 
altsaa ptAp= 
; 1.0027 | 
94.6438 + 53.23 (S — 1.0278 — —— 5} 
54.6438 + 53.23 ( 1.02783) |1 T= pas a 
Sættes gm = 70°, faar man 
1.0027 
1 — Bcos 29 
Den højeste Verdi af IN er 1.02804. Med denne 
bliver 3 — 1.02783 = 0.00021 og 53.23 (3— 1.02783) x 
00% EE Å å Å 
å ENE Å = 53.23 x 0.00021 «-0.00321 =0.000034 
| — Bcos2@ 
Atm., hvad der svarer til et Kviksolvtryk af 0.026 Millimeter, 
der kan sættes ud af Betragtning. 
Det hele Tryk i 300 Favnes Niveauflade bliver følgelig 
1.0027 Mp 
= =0:00321. 
= 1.00321; 1 — ——=>=== 
LOOSE 1 — Bcos2p 
J == 
P35) = 54.6438 + 7 Atm, hvor W er Vindfladens Hojde. 
I Niveaufladen H = 300 Favne have vi saaledes fol- 
gende Tryk i Atmosfærer i de forskjellige Stationer: 
h Favne (Fms.) = 
sure in atmospheres, we shall get the static pressure at 
the point of the surface of level. 
If the height of the current-surface above the sur- 
face of level of the top-surface is AA h metre, and the cor- 
responding pressure in atmospheres is Ap, we have then 
(since at the surface h = 0 and p = 0), the density being S, 
a, S (I — 6 cos 2 g) 
1.89877 Ah Meter. 
The value of S can vary, as shown by the trans- 
verse sections XVII, Pl. XL, and XXVIII, Pl. XLI, be- 
tween 1.0265 (lat. 73° N) and 1.0284 (lat. 61° N). The 
factor for Ah will in the former case be 0.09957, in the 
latter 0.09967. The greatest value of Ah is 1.4 metres. In 
this extreme case the former factor would give Ap = 0.1394 
atm., the latter 0.1395 atm. Assuming computation with 
the factor 0.1, we should get 0.1400 atm. The difference 
in the result, 0.0006 or 0.0005 atm., corresponds to a 
mercury-pressure of respectively 0.46 and 0.39 millimetres. 
Hence I take 
Ap atm. =0.1 Ah metre. 
At the surface of level H= 300 fathoms, we have 
p = 54.6438 + 53.23 (I — 1.02783) 
atm. fora column of water 
300 fath. in height, 
vr 
102053 93 (102183 >) 
the surf. of density Ah = 3 — 
% 1 — 6 cos2 @ 
metres ; 
TG 3 
= pas (102708 — 5) 
1 —pPcos2y 
atm. for the surf. of density; 
Ap 
hence 
and therefore p + A\ p= 
Mews dl as tones) la —_ 
1— 6 cos 29 
Putting p= 70°, we get 
pe Se EE 000358 
1 —fcos29 1 — B cos 2 
The highest value of X is 1.02804. With this value 
+ — 1.02783 = 0.00021. and 53.23 (3 — 1.02783) x 
1.0027 
DT =fews2e 
atm., which corresponds to a mercury-pressure of 0.026 
millimetres, and may be neglected. 
Hence the whole pressure at a surface of level at 
300 fathoms will be 
= 1.00321; 1 
l = 53.23 x 0.00021 x0.003821 = 0.000034 
P300 = 54.6438 + kid atm., W denoting the height of 
the wind-surface. 
At the surface of level H= 300 fathoms, we have 
therefore the following pressures, in atmospheres, for the 
various Stations. 
