Penetrating Radar?

Without knowing the shape of one of the surfaces of a solid, it is very difficult to calculate a volume. If you make some assumptions about the non-visible surface, and model those, you can then generate a model of the other (visible) surfaces and overlay it on your hypothetical modeled surface. There are quite a few software packages that will do the modeling of the known surface and generate an STL or IGES model. If you have photogrammetric photographs, you can use 123D-Catch from Autodesk (bonus: it's free!), and Photomodeler (http://www.photomodeler.com/index.html, not free), as well as others. The unknown surface can be modeled in Sketchup (free, but has a learning curve), Blender (ditto), or any CAD package such as AutoCAD. If you can get that far with all of this, generating the volume will be trivial. There are other packages for creating voxels (volume pixel) that might work for you too.

As you are under water, you may have to use bathymetry to generate a 3D model instead of the photogrammetric method, in which case Sketchup may be the tool of choice for a basic model.

I think the algorithm used by most of these packages for calculating volumes is based on Green's Theorem, modified for a third dimension - see:

Computing areas using Green’sTheorem and a Software Planimeter

Paul Davis and Serban Raianu

TEACHING MATHEMATICS AND ITS APPLICATIONS, Volume 26, No. 2, 2007

I hope this helps, Neil.

I think you need some kind of x ray or gamma scanner, similar to use for see fossils structures (like dinosaurs) inside of the rocks. It's not my speciallity but i hope help you.

The volume of solid can be determined by sampling the depth at several

locations and averaging. The greater the number of samples, the better the

accuracy.

Sampling in a grid pattern gives the best results.

Dear Pablo,

I read up the details on penetrating radars, but for an underwater situation, especially in the ocean, I am not sure if it will work very well... Do you have any information on this?

Best wishes,

Dominique.

Dear Neil,

Assuming the unknown surface structure was considered initially, but modelling it seems to work well too because then I can demonstrate situations for various base structures... I am looking into the details of the software and Green's Theorem for volume calculations, but finding out about the unknown surface is still on my mind, more for the reason of curiosity...

I have taught myself to use AutoCAD, so I hope learning to use Sketchup should not be too difficult and time consuming...

I will let you know how the modelling work helps the study...

Thanks!

Dominique.

Dear Michael,

Thank you for the explanation on the measurement of visible surface dimensions...

Best wishes,

Dominique.

I am going to attempt to explain a concept I have been working on, and it may prove

useful at some future date.

If you locate an irregular area of "Continental" material protruding from the surface of the Ocean floor, you can determine its vertical extent (thickness) solely by

knowing the density of the "Continental" material in comparison to the

density of the Ocean Floor Material.

This is almost exactly the same as an Ice berg floating in an ocean of water.

The Ice berg, being made of ice has a lower density than the salt water of the ocean.

The variation in the case of the "Continental" material is that there are three

materials to consider. You must also include the Height and density of the

salt water above the Ocean floor as a contributing downward force on the

oceans, and you must also include the height and density of the saltwater

above the raised continental material in your calculations.

Because the area of the ocean floor is vast, and the gravity of the earth is

far more than large enough to plastic-aly deform the Earth into a spheroid,

the surface area ratios do no come into play at a small scale.

At Hydro-static equilibrium, The Downward force on the magma under the

oceans is H 1 ( Oceans ) x Density ( Oceans ) + H 2 ( Basalt ) x Density

(basalt) = H 3 (oceans) x Density (oceans ) + H 4 ( continental ) x Density

( Continental ) + H 5 (Basalt) x Density ( Basalt ).

Since the actual continents and the total of the Oceans should be in or nearing

a static ( Hydro-static ) equilibrium, then the thickness of the continents

should be the maximum equilibrium height . The average is about

38.825 km thick, so this would be the maximum number used to calculate

the value of H 4 ( Floating "Continental" Materials ).

Then H 1 + H 2 < = 38.825 km, and similarly

H 3 + H 4 + H 5 < = 38.825 km.

A couple of simultaneous equations will provide the maximum thickness

of the Floating " Continental " Materials.

H 1 and H 3 are both salt water. H 1 > H 3. Density near 1.0

H 2 and H 5 are both basalt. H 2 > H 5 Density near 3.0

H 3 is " Continental " materials. Density < 3.0

At this point all you need is GOOGLE EARTH to find the

water depths to the top of the ocean floor, and the water depth to

the top of the " floating continental materials ", and you need

a single core sample to determine the density of the " continental "

materials, graph paper, and a calculator.

You could start by testing the value of 38.825 km against other peoples

work , and then reduce the number a little bit at a time until it

matches the thickness they got. for the floating continental materials.

As an aside, I have been working on a Rocky Planet density equation,

that seems to work fairly well if the Planet is mostly Rock, but misses

the mark quite a bit for small moons covered with ice, and it also

misses planets like Mercury that have a significant percent of the mass

as iron.

The " Average" Rocky Planet density equation: Den = Density

Den = (1+Pi) x 10^-9*R^3 + (1+sqrt 2) x 10^-1*R + 2904.697056 Kg/m^3

R must be in kilometers.

It seems to work fairly well in the range of sizes from the Moon Radius up to

Earth radius. The 2904.7 might be a good test density for a Planetary

surface average density. In the case of the Earth I would try the following

0.7096 ( density of basalt ) + 0.2904 ( density of continents ) = 2904.7 kg/m^3

If Basalt was 3.0 then: 0.7096 ( 3000 kg/m^3 ) = 2,128.8

2904.7 - 2,128.8 = 775.9

775.9 / 0.2904 = 2671.832 Rounds to 2,671.8 kg / m^3 ( continental )

Since we already know that surface rock is between 2665 and 2670 kg/m^3

prior to compression, 2,671.8 is not a bad number to start with when testing.

If you can get a better number for Basalt under the ocean floors, then

recalculate. The 0.7096 and 0.2904 are Ocean and Land fractions of the

of the Earth's surface area.

As a first try : 0.7096 ( 3000 ) + 0.2904 ( 2671.832 ) = 2904.7 ( a constant ).

If you lower the 3,000, the 2671.832 will rise in value.

if you raise the 3,000, the 2671.832 will decrease in value.

Getting the correct Density for Basalt (under compression)

is a Key to making this method work.

The Constant for the Density Equation is derived from:

Mu = G*M ( Earth ) = 398600.4418(9) where G = G (CODATA)

and Mu = G*M ( Moon) = 4902.8000 where G = G (CODATA)

G (CODATA) = 6.67384(80) E-11

The two radius are 6371.000 Km and 1737.11 Km

Dear Michael,

Although at first glance I would rather run away from such calculations, on the other hand they are extremely interesting to me and I wouldn't mind working them through one step at a time... These will definitely be a lot useful to me in the long run! It will also help me to improve my mathematical skills which need some development. :-)

Best wishes,

Dominique.

As it happens, I spent much of Saturday and Sunday working out the Math

for Ocean Waters, Oceanic Basalt, Density, Depth to equilibrium pressure,

Continents, Continental density, and depth to equilibrium pressure.

The Numbers changed a little.

Earth surface area ratios ( 0.7071 , and 0.2929 )

Earth water density ( 1.000 )

Water depth ( 3.682 km )

Basalt Density ( 3.007 )

Basalt depth ( 24.110 km )

Depth to equilibrium pressure ( 27.292 km ) for uniform density only

Earth continent density ( 2.6579 )

Earth Continent height ( 28.662 km )

Depth to equilibrium pressure ( 27.292 km ) for uniform density only

Projection of the continent above sea level ( 0.870 km )

Average Planet surface density equation :

Target average density = 2904.696056 kg / cubic meter

(0.7071 x 3007 ) + ( 0.2929 x 2657.9 ) = 2904.748 610

Both round to 2904.7 ( 2.9047 )

Oceanic pressure at equilibrium depth below sea level :

( 3,682 m ) ( 1,000 ) + ( 24,110 m ) ( 3,007 ) = 76.180 77 E 6 kg / m^2

Depth below sea level = 27.992 km

Continental Pressure @ equilibrium depth ( constant density )

( 28,662 m ) ( 2657.9 ) = 76.180 73 E 6 kg / m^2

Projection above sea level = 0.870 km

Depth below sea level = 27.992 km

Average vertical pressure at equilibrium depth = 76.180 75(2) E 6 kg / m^2

This would be the minimum depth to equilibrium pressure if the

pressure remained constant. Since we know it can not remain

constant, but the density climbs with increasing depth as an exponential

function this should be about 73 % ( e / (1+e)) = 0.731058578, we can invert

this number to find out exactly what the depth of continents is currently.

1.367879441 ( 27.992 ) = 38.289 681 32 km ( Oceanic )

Round to 38.290 km this is close to other methods I used to determine

the average depth of continents which gave values between

38.825 and 38.834 km.

As a check : ( continental ) 1.367879441 ( 28.662 ) = 39.206 km

Average the two ( 38.290 + 39.206 ) / 2 = 38.748 km ( Close enough )

I just usually round to 38.8 km but since I was showing the method

to find the depth of equilibrium pressure , I did not do much rounding.

Since the method checks out, then the Rocky Planet density Equation

with a density constant of 2904.696056 works for two

different things, Continental Thickness, and Density of the Earth and the Moon.

Den ( R.P. ) = ( 1+ Pi) x 10^-9 * R^3 + ( 1 + sqrt 2) x 10^-1 * R + 2904.696056

R must be in kilometers.

I do not round the numbers in the density equation because a very

small change in density makes a fairly large change in a Rocky Planets

Mass. The Product Mu = G * M is the only number we know about

planets to a significant number of digits. We do not need to know

G, we just need to know the ratio of G * M / G * M for numerous

planets to determine a Density Equation, then select a reasonable value of

G to determine a planet mass. I used G (codata) = 6.67384(80) E -11.

I noticed that the JPL ( NASA ) uses 6.67259(30) E -11.

No I am not starting all over again to calculate a new mass and a new

C = 2904.xxx xxx .

You don't.

Dear Michael,

I will look into these details while I work with the modelling aspect during this week... I will try a few practice runs with the equations and hopefully by then I should get a hold on the calculations...

Speaking of equilibrium pressures, is there a way to calculate for example the pressure exerted by a continental land mass over the basaltic crust tending towards isostatic adjustment? (an example of this would be in the event of sliding of a mass of continental material onto the ocean basin during tectonic events)

Best wishes,

Dominique.

@ Gavril : :-)

I ran some iterations on the depth of equilibrium pressure using a reduced

equilibrium height planet radius of 6369.5 km rather than 6371 km.

Basalt Plus the water 30.567 km ( 3.007 ) + 3.682 km ( 1.000 )

Basalt Plus water equivalency as all basalt 28.111 km ( 2.99228 )

Continental 31.446 km ( 2.6756 )

If all of the continental material, Basalt and water were mixed together

to a uniform density:

Uniform mixture : 29.0927712 km ( 2.909711854 )

Multiplier to convert the uniform density to gradually increasing density

( e+1 ) / e = 1.367879441

Actual depth to Uniform density : 39.7954 km ( below the equilibrium

radius )

Planet radius of equilibrium pressure :

6369.5 - 39.7954 = 6329.7046 km.

Continental height from the equilibrium pressure

up to the average surface height.

6371.870 - 6329.7046 = 42.1654 km ( generally accepted value is 40 )

Height to sea level :

6371.0 -6329.7046 = 41.2954 km ( up to sea level elevation )

Height to the ocean floor: 37.6134 km (ocean is 3.682 km deep )

Changing the Earth's Radius from 6371.0 to 6369.5

changed the value of the constant from

2904.696 056 to 2909.711 854

and increased the planets density from

5513.794307 to over 5517.690676. ( smaller radius increases the density )

The Mass must remain the same,

so if the Volume gets reduced, the Density must go up,

which in turn effects the surface constant

of the density equation, it must go up,

which in turn changes the depth to

equilibrium pressures , it must increase.

This is an iterative process to find the average density of the

basalt, the continents, the equivalent uniform density,

and the various depths of the materials to reach

a point of uniform pressure.

At these depths, the pressures are millions of kilograms

per square meter on any face of a cube, so differential

pressures immediately create plastic movement of rock

to minimize or eliminate differential pressures.

movement can be horizontal or vertical, or toroidal where

layers slide over other layers ( creep and deformation ).