Full resolution (JPEG) - On this page / på denna sida - Sidor ...
<< prev. page << föreg. sida << >> nästa sida >> next page >>
Below is the raw OCR text
from the above scanned image.
Do you see an error? Proofread the page now!
Här nedan syns maskintolkade texten från faksimilbilden ovan.
Ser du något fel? Korrekturläs sidan nu!
This page has never been proofread. / Denna sida har aldrig korrekturlästs.
172
Er Strømfladens Højde over Overfladens Niveauflade
A h Meter og det tilsvarende Tryk i Atmosfærer A]), saa
har man (da i Overfladen h = 0 og p — 0), naar Tætheden
er S:
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
surface of level of the top-surface is A h metre, and the
corresponding pressure in atmospheres is Aj), we have then
(since at the surface h — 0 and]? = 0), the density being .s’,
Ap Atm. = "" «ff -.",«» 2 THI + t.h) Ah Favno (Fm } =o. St 1 -<i eos 2 ,,) Ah
1 — 1,1) 1.82877
Størrelsen :if S kan variere, soul Tversnittene XVII,
Pl.XL, og XXVIII, Pl.XIjI, vise, mellem 1.0265 (73°N. Br.)
og 1.0284 (til" N.Br.). Factoren for Ah bliver i forsto
Tilfælde 0.09957, i siilste 0.09907. Dell storste Værdi af
Ah er 1.4 Meter. I dette extreme Tilfælde vilde den
forsto Factor give Ap = 0.1394 Atm., i sidste 0.1395 Atm.
Om mail 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
Aj) Atm. =0.1 Ah Meter.
I Niveaufladen H = 300 Favne have vi
p = 54.6438 -f 53.23 (2 — 1.02783) Atm.
for 300 Favnes Vandsøjle,
TiethedsHaden Ah — , "’"^V -53.23 (1.02783 —5)
1 — [i eos 2 <f
Meter,
altsaa Ap= , 53.23 (1.02783 - 2)
1 — ß eos 2 cp
Atm. for Tæthedsfladen,
altsaa p-{-ZSp =
54.6438 + 53.23 (2— 1.02783) jl —
- ß eos 2 (pl
Sættes (f- 70°, faar man
, ’f87, =1.00821, »-T ’f7, =
1 — p eos i <j 1 — I eos 2 If
Den højeste Værdi af 2 er 1.02804. Med denne
bliver 2— 1.02783 = 0.00021 og 53.23 (2— 1.02783) x
- 0.00321.
1-
1.0027
= 53.23x 0.00021 X0.00321 =0.000034
I — ft eos 2 fl
Atm., hvad der svarer til et Kviksølvtryk af 0.026 Millimeter,
der kan sættos ud af Betragtning.
Det hele Tryk i 300 Favnes Niveauflade bliver følgelig
W
P......= 54.6438 •+ Atm., hvor II7 er Vindfladens Højde.
I Niveanfladen // — 300 Favne have vi saaledes
følgende Tryk i Atmosfærer i de forskjellige Stationer:
The value of S can vary, as shown by the
transverse sections XVII. Pl. XL, and XXVIII. Pl. XLI.
between 1.0265 (lat. 73» N) and 1.0284 (lat. 61° N). The
factor for Ah will in the former case ho 0.09957, in the
latter 0.09967. The greatest value of A h 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 A h metre.
At the surface ot level H— 300 fathoms, we have
p = 54.6438 -+ 53.23 (2 — 1.02783)
atm. for a column of water
300 fatli, in height,
10.027
53.23 (1.02783 — 2)
metres;
53.23 (1.02783 — 2)
the surf, of density Ah —
j-hence Ap = -—
ß eos 2 (,
1.(1027
• ß eos 2 (p
atm. for the surf, of density;
and therefore p -(- A p =
54.643a 53.23 (2— 1.02783) 11 —
1.0027 \
1 — ß eos 2 (fj
Putting rp= 70°, we get
1.0027
= 1.00321; 1-
1 -
1 — ß eos 2 (p
The highest value of i
2 — 1.02783 = 0.00021 and
1.0027 1
1 — ß eos 2 (p,
atm., which corresponds to a mercury-pressure of
-<).()( 1321.
1-.
1.0027
— ß eos 2 (p’
1.02804. With this value
3.23 (2— 1.02783) X
= 53.23x0.00021 X 0.00321 = 0.000034
0.026
millimetres, and may be neglected.
Hence the whole pressure at
300 fathoms will be
W
surface of level at
IF denoting the height of
the wind-surface.
At the surface of level // = 300 fathoms, we have
therefore the following pressures, in atmospheres, for the
various Stations.
<< prev. page << föreg. sida << >> nästa sida >> next page >>
