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.
175
Ståt. No.
Ståt. No.
1.02785
374.6538
0.0200
1.02774
274.6243
0.0712
:!0r>
1.02787
274-6592
0.0209
274.680
1.02761
274-5893
0.0874
v 274.696 274.677
1.02785
274.6538
0.0176
0.0592
274.602
:!us
1.02785
274.6538
0.0192
274-673
1.02781
274-6431
00453
274.688
0.0302
374.684
H,
1.02746
274.5489
0.0977
1.027S4
274.6512
0.0343
1.02741
274-5355
0.1258
274.661
1.22782
274.6458
0.0409
1.02759
274-5839
0.0829
Værdierne for Pm, P1(I0I, og P,WM ere afsatte i
Karterne Pl. XLV, XLVI og XLVII, og derefter ere
Isobarerne trukne.
For at finde det System af horizontal»1 Bevægelser,
altsaa Bevægelser i Niveaufladen, der svarer til
Isobar-Systemet, kunde man gaa frem ligesom i Meteorologien,
og anvende Formlerne tor retliniedo æquidistante Isobarer,
under Forudsætning af at Frictionen var proportional med
Hastiglieden. Kaldes Gradientkraften (,</ G i Meteorologien)
G, Massen af en Kubikmeter Vand q, Afbojningsvinkelen
mellem Gradientens Retning (lodret paa Isobaren, fra det
højere mod det lavere Tryk) og Bevægelsens Retning a,
Jordens Omdrejningshastighed io, Bredden tf,
Frictionscoeffi-cienten k og Hastigheden i Meter per Secund u, saa har
man i saa Fald:
G
sin a = z (1
The values for P„„,, Puml. and P1M0 are set off on the
maps, Pl. XLV, XLVI. and XLVII, and the isobars
drawn accordingly.
I11 order to find the system of horizontal motions, or
the motions at the surface of level corresponding to the
isobar-system, one might proceed as in meteorology, and
apply the formulæ for straight equidistant isobars,
assuming the friction proportional to the velocity. Now, if
we call the force of the gradient (it G in meteorology)
G. the mass of a cubic metre of water q, the angle of
deviation between the direction of the gradient
(perpendicular to the isobar, from the higher to the lower pressure)
and the direction of the motion u. the angular velocity of
the rotation of the earth to, the latitude rr, the
friction-coefficient k. and the velocity in metres per second it, then
G ■ o
sin a = 2 01 sin <r . u
G
eos a — k . ic
1 = k. H
Kaldes Afstanden (i Meter) langs Gradienten Aa-, og
den dertil svarende Trvkforskjel i Kilogram (henført til
den virkelige Tyngde paa Stedet)
saa har man
G =
zv*
Da en Kubikmeter Kviksølv vejer 13595.9 Kilogram,
og en Atmosfære er lig Trykket paa en Kvadratmeter af
en Kviksølvsøjle af 0.70 Meters Hojde (ved 0° og
Normal-tyngden), bliver, naar Trykket regnes i Atmosfærer,
Ajr-’’ = 13595.9x0.7« Ap" = 10333 Ap".
Kaldes Søvandets virkelige, af Sammentrykningen
paavirkede, Tæthed S, vejer en Kubikmeter Søvand 1000 S
Kilogram.
Er f/u, Normaltyngden, har man altsaa
1000 ti
0= *. ’
Indsættes disse Værdier i Bevægelsesligningerne,
faar man
Now, if we call the distance, in metres, along the
gradient A.r, and the difference in pressure corresponding
to it, in kilogrammes (referred to the true gravity at the
place), AjJ,
we have
G =
A p^
Aa"» ’
As a cubic metre of mercury weighs 13595.9
kilogrammes, and an atmosphere is equal to the pressure 011 a
square metre of a column of mercury, 0.76 metre in
height (at 0° and standard gravity), then, assuming the
pressure to be computed in atmospheres,
Ap’* = 13595.9 X 0.7G Ap" = 10333 Ap"’.
Calling the actual density of sea-water, acted on
by compression, S, a cubic metre of sea-water weighs
1000 S kilogrammes.
With ffi5 as the standard gravity, we have therefore
1(100 S
" a. ’
If these values be substituted into the equations of
motion, we get
<< prev. page << föreg. sida << >> nästa sida >> next page >>
