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Science Forum Index » Space Forum » Rounded Rocks in Gusev Crater
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| Dick Morris |
 Posted: Thu Jan 08, 2004 12:04 pm |
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drdoody wrote:
Quote:
"Blurrt" <nsaulr@optushome.com.au> wrote in message
news:3ffb432f$0$18746$afc38c87@news.optusnet.com.au...
Just looking at the rocks on the floor of Gusev crater - they are rounded.
This implies water (or some other fluid) erosion.
Or wind.
Doc
We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the rocks
I've seen down in the deserts of Utah and Arizona. |
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| Marvin |
| Posted: Thu Jan 08, 2004 1:02 pm |
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Dick Morris <richard.a.morris@boeing.com> wrote in
news:3FFD8DAC.54C3FC1B@boeing.com:
Quote: We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the
rocks I've seen down in the deserts of Utah and Arizona.
You have an easy way of identifying earth rocks that have been exposed to
the surface for 2billion+ years? wow!
Get a clue please.. The earth surface is *active*. There are frequent(by
geological scales) glaciers marching past, lots of airborne stuff (dust,
sand, *rain*, *ice* , birds!, oxygen) that simply are not found in the same
form on Mars. Extrapolating your backyard-commonsense observations *will*
*not* *work* in an inherently foreign environment like Mars. |
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| Marvin |
| Posted: Thu Jan 08, 2004 1:02 pm |
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Dick Morris <richard.a.morris@boeing.com> wrote in
news:3FFD8DAC.54C3FC1B@boeing.com:
Quote: We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the
rocks I've seen down in the deserts of Utah and Arizona.
You have an easy way of identifying earth rocks that have been exposed to
the surface for 2billion+ years? wow!
Get a clue please.. The earth surface is *active*. There are frequent(by
geological scales) glaciers marching past, lots of airborne stuff (dust,
sand, *rain*, *ice* , birds!, oxygen) that simply are not found in the same
form on Mars. Extrapolating your backyard-commonsense observations *will*
*not* *work* in an inherently foreign environment like Mars. |
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| Dick Morris |
| Posted: Thu Jan 08, 2004 1:37 pm |
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Marvin wrote:
Quote:
Dick Morris <richard.a.morris@boeing.com> wrote in
news:3FFD8DAC.54C3FC1B@boeing.com:
We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the
rocks I've seen down in the deserts of Utah and Arizona.
You have an easy way of identifying earth rocks that have been exposed to
the surface for 2billion+ years? wow!
Get a clue please.. The earth surface is *active*. There are frequent(by
geological scales) glaciers marching past, lots of airborne stuff (dust,
sand, *rain*, *ice* , birds!, oxygen) that simply are not found in the same
form on Mars. Extrapolating your backyard-commonsense observations *will*
*not* *work* in an inherently foreign environment like Mars.
Take a few deep breaths and settle down. We don't know how long those
rocks have been exposed, and if you have some examples of smooth,
rounded rock surfaces that have indisputably been produced by wind
erosion I would like to see them. |
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| Dick Morris |
| Posted: Thu Jan 08, 2004 1:37 pm |
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Marvin wrote:
Quote:
Dick Morris <richard.a.morris@boeing.com> wrote in
news:3FFD8DAC.54C3FC1B@boeing.com:
We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the
rocks I've seen down in the deserts of Utah and Arizona.
You have an easy way of identifying earth rocks that have been exposed to
the surface for 2billion+ years? wow!
Get a clue please.. The earth surface is *active*. There are frequent(by
geological scales) glaciers marching past, lots of airborne stuff (dust,
sand, *rain*, *ice* , birds!, oxygen) that simply are not found in the same
form on Mars. Extrapolating your backyard-commonsense observations *will*
*not* *work* in an inherently foreign environment like Mars.
Take a few deep breaths and settle down. We don't know how long those
rocks have been exposed, and if you have some examples of smooth,
rounded rock surfaces that have indisputably been produced by wind
erosion I would like to see them. |
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| Chris Hall |
| Posted: Thu Jan 08, 2004 1:38 pm |
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In <3ffd9b28.0@news1.mweb.co.za>, on 01/08/04
at 08:02 PM, Marvin <kitfox@gettalife.you.dumb.ass> said:
Quote: Dick Morris <richard.a.morris@boeing.com> wrote in
news:3FFD8DAC.54C3FC1B@boeing.com:
We have a lot of rocks here in Washington State, and such smooth,
rounded surfaces I've seen only on river rocks. Ditto for all the
rocks I've seen down in the deserts of Utah and Arizona.
You have an easy way of identifying earth rocks that have been exposed to
the surface for 2billion+ years? wow!
Get a clue please.. The earth surface is *active*. There are frequent(by
geological scales) glaciers marching past, lots of airborne stuff (dust,
sand, *rain*, *ice* , birds!, oxygen) that simply are not found in the
same form on Mars. Extrapolating your backyard-commonsense observations
*will* *not* *work* in an inherently foreign environment like Mars.
Neither will the assumption that all or even most of the rocks you see on
the surface of Mars today have been sitting on or even near the surface
for 2 Ga. That sort of assumption doesn't even work on the moon, which is
far less active than the surface of Mars. Just look over to "sleepy
hollow" to find a candidate for excavation. The only way we'll rigorously
know the surface age of Mars rocks is to look at the cosmic ray exposure
ages (some short lived isotopes, some stables produced by spallation
reactions). That will take a sample return mission, I'm afraid (for good
field control). Having said all that, they DO look like water rounded
rocks, don't they?
--
Chris M. Hall, Associate Research Scientist
Dept. of Geological Sciences, University of Michigan
Specialization is for insects |
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| Olaf van der Zalm |
| Posted: Thu Jan 08, 2004 4:21 pm |
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Quote:
I'm pretty sure I'm talking out of my ass here, but here's my question.
Is
the Reynolds number for Martian atmospheric flow (wind) in any way
comparable to Earth-normal water flow?
It is a fairly straitforward process calculating the Reynolds number, but
what comes after isn't. There you have to make a lot of assumptions and to
worsen things you can't compare with water that easily since Co2 gas is a
compresible fluid and water is incomressible (or nearly).
If you however want to work out the friction factor of the Martian wind I
might have some ideas on how to takle the problem. Start with a turbilent
flow across a flat plate (assuming turbular flow is save since we can
calculate the Reynolds number and see it's large). Asume the wind is an
ideal gas (not true, but if we haven't got any better data we have to use
this). Finally assume a zero pressure gradient. That will allow you to use
equations for turbular pipe flow, which are better understood and apply very
well in this case. An empirical power-law velocity provile with an exponent
of 1/7 is typically used.
Now for the reynolds number:
Re=U*L/nu=U*L*rho/mu (since nu = mu/rho)
where
U = stream velocity [m/s]
L = characteristic length of the flow geometry (i.e. distance from tube
entrance or hydraulic diameter for tubes) [m]
nu = kinematic viscosity [m^2/s]
mu = viscosty [kg/(s*m)]
rho = density [kg/m^3]
so for water at say 283 K at atmospheric pressure:
Re = U*L/1.304E-6
And let's say the water speed is (related to the wind speeds):
U1 = 10 m/s (5 beaufort)
U2 = 20 m/s (8 beaufort)
U3 = 30 m/s (11 beaufort)
I don't know what real martian values for the wind speed are so I gave a
few.
We get:
Re1 = 7.7E6*L
Re2 = 1.5E7*L
Re3 = 2.3E7*L
Now it largely depends on the characteristic lengths, but we can say that
for L>4mm (for these water speeds) the flow must be turbulent since
Re<2.5E3 : Laminar flow
Re>4E3 : Turbulent Flow
now I'll look at Mars once again starting with the equation for the Reynolds
number:
Re=U*L/nu=U*L*rho/mu (since nu = mu/rho)
We don't know the viscosity and density (or kinematic viscosity) so we have
to calculate it. Without giving the exact way of calculating it, you can
calculate the (absolute/dynamical) viscosity of a 100% carbon dioxide fluid
using the emirical Sutherland correlation and arive at 1E-5 kg/(s*m). This
not perfect for these low pressures but comes close.
To get a density we have to assume an ideal gas (not true). The equation of
state:
p = rho*R*T
and
R = Ru/Mm
where
T = temperature [K]
p = pressure [N/m^2]
Ru = universal gas constant = 8314 N*m/(kmol*K)
Mu = molecular mass
so:
Re = U*L*p*Mm/(Ru*T*nu)
Re = U*L*1E3*44/(8314*283*1E-5)
Re = U*L*1870
For the same speeds we get
Re = 1.87E4*L @ U=10 m/s
Re = 3.74E4*L @ U=20 m/s
Re = 5.61E4*L @ U=30 m/s
So there's a big difference between water and wind on mars..
Olaf |
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| Olaf van der Zalm |
| Posted: Thu Jan 08, 2004 4:21 pm |
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Quote:
I'm pretty sure I'm talking out of my ass here, but here's my question.
Is
the Reynolds number for Martian atmospheric flow (wind) in any way
comparable to Earth-normal water flow?
It is a fairly straitforward process calculating the Reynolds number, but
what comes after isn't. There you have to make a lot of assumptions and to
worsen things you can't compare with water that easily since Co2 gas is a
compresible fluid and water is incomressible (or nearly).
If you however want to work out the friction factor of the Martian wind I
might have some ideas on how to takle the problem. Start with a turbilent
flow across a flat plate (assuming turbular flow is save since we can
calculate the Reynolds number and see it's large). Asume the wind is an
ideal gas (not true, but if we haven't got any better data we have to use
this). Finally assume a zero pressure gradient. That will allow you to use
equations for turbular pipe flow, which are better understood and apply very
well in this case. An empirical power-law velocity provile with an exponent
of 1/7 is typically used.
Now for the reynolds number:
Re=U*L/nu=U*L*rho/mu (since nu = mu/rho)
where
U = stream velocity [m/s]
L = characteristic length of the flow geometry (i.e. distance from tube
entrance or hydraulic diameter for tubes) [m]
nu = kinematic viscosity [m^2/s]
mu = viscosty [kg/(s*m)]
rho = density [kg/m^3]
so for water at say 283 K at atmospheric pressure:
Re = U*L/1.304E-6
And let's say the water speed is (related to the wind speeds):
U1 = 10 m/s (5 beaufort)
U2 = 20 m/s (8 beaufort)
U3 = 30 m/s (11 beaufort)
I don't know what real martian values for the wind speed are so I gave a
few.
We get:
Re1 = 7.7E6*L
Re2 = 1.5E7*L
Re3 = 2.3E7*L
Now it largely depends on the characteristic lengths, but we can say that
for L>4mm (for these water speeds) the flow must be turbulent since
Re<2.5E3 : Laminar flow
Re>4E3 : Turbulent Flow
now I'll look at Mars once again starting with the equation for the Reynolds
number:
Re=U*L/nu=U*L*rho/mu (since nu = mu/rho)
We don't know the viscosity and density (or kinematic viscosity) so we have
to calculate it. Without giving the exact way of calculating it, you can
calculate the (absolute/dynamical) viscosity of a 100% carbon dioxide fluid
using the emirical Sutherland correlation and arive at 1E-5 kg/(s*m). This
not perfect for these low pressures but comes close.
To get a density we have to assume an ideal gas (not true). The equation of
state:
p = rho*R*T
and
R = Ru/Mm
where
T = temperature [K]
p = pressure [N/m^2]
Ru = universal gas constant = 8314 N*m/(kmol*K)
Mu = molecular mass
so:
Re = U*L*p*Mm/(Ru*T*nu)
Re = U*L*1E3*44/(8314*283*1E-5)
Re = U*L*1870
For the same speeds we get
Re = 1.87E4*L @ U=10 m/s
Re = 3.74E4*L @ U=20 m/s
Re = 5.61E4*L @ U=30 m/s
So there's a big difference between water and wind on mars..
Olaf |
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| Christopher P. Winter |
| Posted: Sat Jan 10, 2004 10:30 pm |
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On 08 Jan 2004 04:28:15 GMT, patrick23sull@aol.com (Patrick) wrote:
Quote: I think you guys are forgetting the time scale. The water
probably flowed 3 billion years ago. The rocks were deposited
by crater impacts later. A couple of billion years is plenty of
time for dust in the atmosphere to round the rocks.
It should be easy to tell the difference once the rover gets moving.
River-eroded rocks would tumble, and be eroded on all sides. Wind-eroded
rocks (above a certain size) would be more eroded on the side facing the
prevailing winds. Also, those embedded in the soil should not be much eroded
on the buried portions. |
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