Scifi science question

[garrowolf]

I have a setting where I have inertial compensators/contra-gravity systems that allow a ship to have up to 100gs of acceleration. I was thinking about mining Gas Giants and I was wondering how much pressure a system designed to negate/redirect 100gs could take. Basically I want to figure out how deep a ship could travel into a Gas Giant. I figured it would also make an awesome submarine warfare setting as well.

Any ideas?

 

[MarkB]

Compensating for acceleration and compensating for pressure differentials are really two different applications. Without knowing exactly how the technology is supposed to function, it's hard to say whether it would have any effectiveness in the latter application.

If you can describe the function of your theoretical system in more detail, that may help determine how it might be effective in this situation.

 

[garrowolf]

Well in most settings artificial gravity seems to be some sort of force field where anything within the field is pushed in one direction. Inertial compensation would be just another application of that but directed the opposite of the force of acceleration. This is the application I was thinking could be adapted to pressure resistance. Perhaps they are applying the field effect opposite of each other in that one is adding energy to a system (artificial gravity) and the other is taking energy away within the confines of the ship. I would also see this technology magnifying the force of a thruster or even replacing it altogether at higher tech levels.

 

[Deset Gled]

Mathematically:

1 g = acceleration due to gravity = 9.8 meters / second^2
Force = mass * acceleration
Pressure = force / area

So, with a very rough and hand waved explanation, you could say that the 100g factor of your inertial dampeners translates into being able to withstand 100x the pressure normally possible.

 

[garrowolf]

sounds good
thanks!

 

[tomBitonti]

Some numbers ...

Note: Piecing together a pressure/temperature profile for Jupiter took a bit of searching, and the results and terminology seemed to differ enough that I have only a small confidence in this one link:

From Jupiter

The Interior of Jupiter
This picture illustrates the internal structure of Jupiter. The outer layer is primarily composed of molecular hydrogen. At greater depths the hydrogen starts resembling a liquid. At 10,000 kilometers below Jupiter's cloud top liquid hydrogen reaches a pressure of 1,000,000 bar with a temperature of 6,000° K. At this state hydrogen changes into a phase of liquid metallic hydrogen. In this state, the hydrogen atoms break down yeilding ionized protons and electrons similar to the Sun's interior. Below this is a layer dominated by ice where "ice" denotes a soupy liquid mixture of water, methane, and ammonia under high temperatures and pressures. Finally at the center is a rocky or rocky-ice core of up to 10 Earth masses. (Copyright Calvin J. Hamilton)

And from:

Orders of magnitude (pressure) - Wikipedia, the free encyclopedia

100 kPa 14.5038 psi 1 bar

101.325 kPa 14.696 psi Standard atmospheric pressure for Earth sea level

100 MPa 14500 psi Pressure at bottom of Mariana Trench, about 10 km below ocean surface (1000 bar)

With:

1kPa == 1x10^3 Pa
1MPa == 1x10^6 Pa
1GPa == 1x10^9 PA

If the vessel can withstand 100x the pressure at the depth of the Mariana Trench, that would be 100 * 100 MPa == 10 * 1000 MPa = 10 GPa

Then also:

At 10,000 kilometers below Jupiter's cloud top liquid hydrogen reaches a pressure of 1,000,000 bar with a temperature of 6,000° K.

1x10^6 bar * 100 kPa / bar == 1x10^11 Pa == 10 x 10 x 10^9 Pa == 10 x 10 GPa

That would get you what seems to be a smallish percentage (10%, say 1000 km?) into the atmosphere. (Both density and temperature change with depth, so I don't think that would be 10%, but I can't say what the correct % would be.)

All very rough.

Also, to say, temperature might be just as big of a problem as pressure and gravity ...

Thx!

Tom Bitonti

 

[Umbran]

So, with a very rough and hand waved explanation, you could say that the 100g factor of your inertial dampeners translates into being able to withstand 100x the pressure normally possible.

That's some pretty frantic hand-waving to get to that conclusion. For one thing, your typical space ship is designed to withstand vacuum, not a positive external pressure - so 100x normal is 100 * 0, which isn't getting you much. Plus, there's a scaling issue to consider....

An inertial damping system would be designed to work on the ship and its contents, so they don't get squished when Wesley Crusher stomps on the gas. So, if the system provides protection against a 100g acceleration, maximum, that means the maximum force it can apply is

F = 100g * MassOfShip.

Now, say you direct that system outwards, over the skin of the ship, the entire surface area, to resist pressure. The pressure you can resist is then

P = F/AreaOfShip = 100g * MassOfShip / AreaOfShip.

P = 100g * DensityOfShip * VolumeOfShip/AreaOfShip.

So, if we figure that Density of ships is fairly constant - small ships or big ships, they're all kind of like modern naval vessels, with some metal with large volumes for people to move around in - then the determining factor is the size of the ship (Volume/Area is often thought of as "characteristic length"). Not all ships will be able to take 100x normal pressure. Really large ships with 100g inertial dampers will be able to take much less external pressure than small ships with the same kind of damping systems.

 

[tomBitonti]

Hi,

A prior post adjusted the question to say that the "100g" suppression field should be taken to increase the ships capacity to withstand pressures by a factor of 100. I haven't myself tried to map "acceleration tolerance" to "pressure tolerance".

I took the Mariana trench as an example of an extreme location which can be reached using current technology. Not sure how well you could use a bathysphere as a space ship, but it's a place to start. Nor do I know if we could actually go deeper.

Net, withstanding pressures equal to about 100x the deepest ocean trench on earth takes you 1000km (ish) into Jupiter.

Thx!

Tom Bitonti

 

[the Jester]

P = F/AreaOfShip = 100g * MassOfShip /AreaOfShip.

P = 100g * DensityOfShip * VolumeOfShip/AreaOfShip.

Just for clarity- you mean Surface Area here, right?

 

[Umbran]

Just for clarity- you mean Surface Area here, right?

Correct.

I should note that my analysis is also handwaving - I'm just trying to make that handwaving in a logical pattern

I am making a boatload of implicit assumptions about the inertial damping system. It would be just as reasonable (if not more reasonable) to say that having an inertial damping system does *nothing* to help you survive at pressure. "Intertial damping system" does not necessarily mean "general purpose gravity generator that you can tune to do what you want".