<html> <head> <meta http-equiv="content-type" content="text/html; charset=ISO 8859-1"> <title>Ross Ice Shelf could break off swiftly</title> </head>
<body>
<p>
By James T. Conklin on Wed Mar 21 13:16:22
-0700 2007 (in
reply
to Ross Ice Shelf could break off
swiftly )
<br />
<br />
Original article: <a href="http://technocrat.net/d/2006/11/30/11603";>Ice
Shelf could break
off swiftly</a>
<br />
<br />
Inconvenient Facts regarding rising
ocean
level:
<br />
<br />
1. The ocean level is unchanged when
floating ice
melts.
<br />
<br />
2. The ocean is a spherical body of
water. The ocean
volume
varies as the cube of the ocean radius.
Therefore, for the
ocean
radius to increase 20 feet, the ocean
volume must increase
8,000
times more than for a 1-foot radius
increase. For the ocean
radius to increase 40 feet, the ocean
volume must increase
64,000
times more than for a 1-foot radius
increase.
<br />
<br />
<br />
===========================
<br />
<br />
<br />
<strong>James, your observations are not
true at
all.</strong>
<br />
<br />
<br />
=================================================================
<br />
<br />
<strong>1. The ocean level REALLY IS
unchanged when floating
ice
melts.</strong>
<br />
<br />
We will neglect salinity here, because it
makes no signifant
change in formulas.
<br />
<br />
For more detailed information see
explanation in this thread
(By
Wayne Gramlich on Fri Dec 01 09:28:29 -0800
2006) and google
"Archimedes law" or
"Archimedes
principle".
<br />
(e.g. http://en.wikipedia.org/wiki/
Archimedes_principle)
<br />
<br />
It's a 6th grade physics here in Serbia
(kids 11-12 years
old) and I guess elswhere in the world
(including USA), so
you
shouldn't find it difficult to
understand. :-)
<br />
<br />
<br />
<strong>2. The other observation is also
false.</strong>
<br />
<br />
=================================================================
<br />
<br />
<br />
There are several ways to proove your
"mathematics" is
false. The first goes...
<br />
<br />
<strong>Conventions used in text:</strong>
<strong>-------------------------</strong>
Digit grouping symbol: <strong>.</strong>
Decimal symbol: <strong>,</strong>
Digit grouping example: <strong>one million = 1.000.000 = 1.000.000,00</
strong>
<strong>Data:</strong>
<strong>-----</strong>
<br />
Average Earth (as a geoid, oblate spheroid,
or even a
sphere)
diameter is R = 12.742 km (the average
radius is half that
value,
r = 6.371 km) (source 1 or source 2).
<br />
The average depth of the World Ocean is 3
711 m = 3,711 km
(source 2).
<br />
The volume of the sphere is V = (4/3)*(Pi)*
(r^3).
<br />
Pi ~ 3,14; (Pi is just a number)
<br />
Surface of World Ocean is about 71% of
Earth's surface.
(source 2)
<br />
The volume of all the oceans would be
something like:
<br />
V(oceans) = (V1 - V2) * 0,71 , where V1 =
(4/3)*(Pi)*(r^3),
V2 =
(4/3)*(Pi)*((r-3,711)^3).
<br />
<br />
<br />
<strong>Math:</strong>
<br />
V(oceans) = 0,71 * (4/3) * (Pi) *
[ (6371^3) - (6371 -
3.711)^3 ]
<br />
V(oceans) = 0,71 * (4/3) * (Pi) * 451,62 *
10^6 = 1,342 *
10^9
km^3
<br />
<br />
(or ~ 1.340,7 million km^3 as UN says; we
got 0,1% smaller
result
because of several approximations: r, Pi,
geoid ->
sphere,
etc.)
<br />
<br />
So let's assume: V(oceans) ~ 1,340.7
million km^3 (source 2)
<br />
<br />
"The World Ocean (Oceanosphere)
contains on the order
of
1,340.7 million km3 of water, making up
1/800th of the total
volume of the Earth (1,083.3 billion
km3)." (source 2)
<br />
<br />
<br />
<br />
Now, back to your <em>false</em> math...
<br />
<br />
For the ocean to increse 20 m, 10 m, and 1
m, respectively,
the
ocean volume must increse:<strong><br />
<br />
V(oceans + 20 m) - V(oceans) = V(oceans) - V
(oceans - 20 m)
=</strong>
<br />
<br />
(...Sorry, I should have calculated using
the first formula,
but
I used the second; it doesn't matter, the
result is
the
<br />
<br />
same; however, just to let you know so you
don't get
confused; I'm too lazy to write (and
calculate) it all over
again.)
<br />
<br />
= {0,71 * (4/3) * (Pi) * [ (6371^3) - (6371
- 3,731)^3 ]} -
{0,71
* (4/3) * (Pi) * [ (6371^3) - (6371 -
3,711)^3 ]} =
<br />
<br />
= {0,71 * (4/3) * (Pi) * 454,05 * 10^6} -
{0,71 * (4/3) *
(Pi) *
451,62 * 10^6} =
<br />
<br />
= 1,3497 * 10^9 km^3 - 1,3425 * 10^9 km^3 =
<br />
<br />
<strong>= 0,0072 * 10^9 km^3</strong>
<br />
<br />
<br />
<strong>V(oceans + 10 m) - V(oceans) =</
strong>
<br />
<br />
= {0,71 * (4/3) * (Pi) * [ (6371^3) - (6371
- 3,721)^3 ]} -
{0,71
* (4/3) * (Pi) * [ (6371^3) - (6371 -
3,711)^3 ]} =
<br />
<br />
= {0,71 * (4/3) * (Pi) * 452,84 * 10^6} -
{0,71 * (4/3) *
(Pi) *
451,62 * 10^6} =
<br />
<br />
= 1,3461 * 10^9 km^3 - 1,3425 * 10^9 km^3 =
<br />
<br />
<strong>= 0,0036 * 10^9 km^3</strong>
<br />
<br />
<br />
<strong>V(oceans + 1 m) - V(oceans) =</
strong>
<br />
<br />
= {0,71 * (4/3) * (Pi) * [ (6371^3) - (6371
- 3,712)^3 ]} -
{0,71
* (4/3) * (Pi) * [ (6371^3) - (6371 -
3,711)^3 ]} =
<br />
<br />
= {0,71 * (4/3) * (Pi) * 451,74 * 10^6} -
{0,71 * (4/3) *
(Pi) *
451,62 * 10^6} =
<br />
<br />
= 1,3428122 * 10^9 km^3 - 1,3424555 * 10^9
km^3 =
<br />
<br />
<strong>= 0,00036 * 10^9 km^3</strong>
<br />
<br />
<br />
Thus we see that <strong>for the ocean
radius to increase 20
m,
the ocean volume must increase around 20
times more than for
a 1
m radius increase.</strong>
<br />
(Not "8.000 times more" as you
claim.)
<br />
<br />
<br />
<strong>For the ocean radius to increase N m,
the ocean volume must increase around N
times more than for
a 1
m radius increase.</strong>
<br />
<br />
<strong>For the ocean radius to increase 40 m,
the ocean volume must increase around 40
times more than for
a 1
m radius increase.</strong>
<br />
(Not "64.000 times more" as you
claim.)
<br />
<br />
<br />
<br />
<span style="text-decoration:
underline;"><strong>Note one
more
important fact:</strong></span>
<br />
- <strong>volume of the World Ocean (all
oceans
together)</strong> is about 1,3424555 *
10^9 km^3.
<br />
- <strong>1 m increse equals</strong>
0,00036 * 10^9 km^3.
<br />
- <strong>ratio is:</strong> [0,00036 *
10^9 km^3] /
[1,3424555 *
10^9 km^3] = <strong>0,0002682 ~ 0,027%</
strong>
<br />
<br />
<strong><span style="text-decoration:
underline;">Consequence:</span></strong>
<strong>in order to have ocean level rise
<span
style="text-decoration: underline;">1 m</span>, we need ocean volume
(or mass) <span style="text-decoration:
underline;">increase
only
0,027%</span> of its current volume/mass.</
strong>
<br />
<br />
<br />
=================================================================
<br />
<br />
Or, <span style="text-decoration:
underline;"><strong>more
simplified with one more math
approximation:</strong></span>
<br />
<br />
<br />
<strong>(a+b)^3 = a^3 + 3*(a^2)*b + 3*a*
(b^2) + b^3</strong>
<br />
<strong>(a-b)^3 = a^3 - 3*(a^2)*b + 3*a*
(b^2) - b^3</strong>
<br />
<br />
<br />
Thus:
<br />
(r+x)^3 = r^3 + 3*(r^2)*x + 3*r*(x^2) + x^3,
<br />
(r-x)^3 = r^3 - 3*(r^2)*x + 3*r*(x^2) - x^3,
<br />
<br />
where:
<br />
r = 6371 km,
<br />
and x is ocean radius increse (e.g. 1 m, 10 m, 20 m, 100 m).
<br />
<br />
Since <strong>"r" and
"x" are not of the
same
order of magnitude</strong> and since
<strong>"r"
is 60.000-6.000.000 times larger than "x"</strong>, we can say
<br />
<br />
<br />
<strong>3*r*(x^2) and (x^3) are negligible</
strong>, so we
take
only first 2 addends:
<br />
<br />
<br />
<strong>(r+x)^3 ~ (r^3) + 3*(r^2)*x
(very close
approximation)
<br />
(r-x)^3 ~ (r^3) - 3*(r^2)*x &nbsp;
(very close
approximation)</strong>
<br />
<br />
<br />
<em>Then we have:</em>
<br />
<br />
<strong>V(oceans + x m) - V(oceans) ~ V
(oceans) - V(oceans -
x m)
= </strong>
<br />
{0,71 * (4/3) * (Pi) * (r-x)^3} - {0,71 *
(4/3) * (Pi) *
(r^3)} =
<br />
{0,71 * (4/3) * (Pi) * [(r^3) - 3*(r^2)*x]}
- {0,71 * (4/3)
*
(Pi) * (r^3)} =
<br />
{0,71 * (4/3) * (Pi) * 3*(r^2)*x} ~
<br />
<br />
~ 2,97 * 3*(r^2)*x ~
<br />
<br />
~ 8,92 * (r^2)*x ~
<br />
<br />
~ 120.18 * (10^6) * 1 km * x =
<br />
<br />
<strong>= k * x,</strong>
<br />
<br />
where <strong>k = const</strong>,
<br />
<strong>k ~</strong> 120.18 * (10^6) * 1 km
=
<strong>120,180,000
km</strong>
<br />
<br />
<em><strong>It's linear function,</strong>
so you could say
that <strong>for
the ocean radius to increase N times 1
meter, the ocean
volume
must increase N times more than for a 1 m
radius increase.
<br />
<br /></strong></em>...or, as we've seen
earlier,
<strong>ocean need N times 0,00027 of its
volume/mass, i.e.
N times 0,027 % of
its volume/mass, for N meters rise.</strong>
<br />
<br />
=================================================================
<br />
<br />
<br />
Sources (about Earth diameter, Ocean
surface, Earth
surface):
<br />
<br />
1. Wikipedia, http://en.wikipedia.org/wiki/
Earth#Shape
<br />
2. United Nations Atlas of the Oceans,
DISTRIBUTION OF LAND
AND
WATER ON THE PLANE:
<br />
<br />
http://www.oceansatlas.com/unatlas/about/physicalandchemicalproperties/
background/seemore1.h
tml
<br />
<br />
=================================================================
<br />
<br />
<strong>Regards,
<br />
Miloš T. Kojašević
<br />
(Milos T. Kojasevic)
<br />
<a href="http://www.ff.bg.ac.yu";>Faculty of
Physics</a>
<br />
<a href="http://www.bg.ac.yu";>Univerzitet u
Beogradu /
University of Belgrade / Belgrade University</a>
<br />
Belgrade, Serbia, Yugoslavia
<br />
<br /></strong>
</p>
</body>
</html>
