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#1 2011-11-20 14:14:49

JoshNH4H
Member
From: Pullman, WA
Registered: 2007-07-15
Posts: 2,546
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Space Towers

Space towers have long been considered to be impractical given the materials available.  For that reason, the focus in terms of static structures has largely been on tethers which are hung from orbit.  The basic reasons for having these structures are fairly simple, given that they could bring about tenfold (probably not more, but a 90% cost reduction is nothing to sneeze at) reductions in launch cost beyond the best that chemical rockets can do.

However, it seems to me that Space towers are too often neglected.  Especially given recent advances in nanostructured materials (I'm talking specifically about the new ultralight metallic microlattice developed at UC Irvine, though I do also mean the leaps and bounds currently being take in the field of materials science), I think that the age in which a space tower is a practical aim is getting to be very close to the present.

In order to evaluate if a material is capable of serving as a tower to GEO (or really any arbitrary height) it is necessary to compute the strength it needs as a function of density.  It would seem that the new board software does not contain the spoiler tag, so you can just skip over the derivation if you don't really care about it.

So, the necessary compressive strength can be written as:

S=∫(dP/dr)dr

Where P is pressure and r is radius.  Given from fluids that P=dgr (where d is density of the fluid and g is the local acceleration to gravity), dP/dr=dg.  Density is assumed to be constant, while the effective acceleration due to gravity changes according to the equation:

g(r)=GM/r²-w²r

G is Newton's Gravitational Constant, M is the mass of the planet, and w (I apologize to the gods of physics for not using an omega; I didn't know and couldn't find the alt-code) is the angular velocity, equal to 2π/T, for the case of Earth equal to about 7.2722e-5 rad/s.  In other words, the acceleration due to gravity is equal to the acceleration due to gravitational force minus the centripetal acceleration towards the center of the body.  Therefore,

S=d∫(GM/r²-w²r)dr

So the necessary strength is:

S=d(GM/r-w²r²)

Evaluated from the lowest r to the highest one; in this case from 6,378,100 m (earths radius) to GM/r=wSpace towers have long been considered to be impractical given the materials available.  For that reason, the focus in terms of static structures has largely been on tethers which are hung from orbit.  The basic reasons for having these structures are fairly simple, given that they could bring about tenfold (probably not more, but a 90% cost reduction is nothing to sneeze at) reductions in launch cost beyond the best that chemical rockets can do.

However, it seems to me that Space towers are too often neglected.  Especially given recent advances in nanostructured materials (I'm talking specifically about the new ultralight metallic microlattice developed at UC Irvine, though I do also mean the leaps and bounds currently being take in the field of materials science), I think that the age in which a space tower is a practical aim is getting to be very close to the present.

In order to evaluate if a material is capable of serving as a tower to GEO (or really any arbitrary height) it is necessary to compute the strength it needs as a function of density.  It would seem that the new board software does not contain the spoiler tag, so you can just skip over the derivation if you don't really care about it.

So, the necessary compressive strength can be written as:

S=∫(dP/dr)dr

Where P is pressure and r is radius.  Given from fluids that P=-dgr (where d is density of the fluid and g is the local acceleration to gravity; negative because pressure increases as you go down), dP/dr=-dg.  Density is assumed to be constant, while the effective acceleration due to gravity changes according to the equation:

g(r)=GM/r²-w²r

G is Newton's Gravitational Constant, M is the mass of the planet, and w (I apologize to the gods of physics for not using an omega; I didn't know and couldn't find the alt-code) is the angular velocity, equal to 2π/T, for the case of Earth equal to about 7.2722e-5 rad/s.  In other words, the acceleration due to gravity is equal to the acceleration due to gravitational force minus the centripetal acceleration towards the center of the body.  Therefore,

S=-d∫(GM/r²-w²r)dr

So the necessary strength is:

S=d(w²r²/2-GM/r)

Evaluated from the lowest r to the highest one; in this case from r=6,378,100 m to GM/r=w²r², or r=42,226,900 m (My calculated altitude for GEO).  Dividing each side by d, and plugging in, I get that S/d=62.2 MPa/kg/m³.  Note that this is compressive strength; also note that this same number applies to cables in tension without a taper. 

Now, many of us have probably heard of the new material that was just discovered at UC Irvine:  an Ultralight Metallic Microlattice material, which is ordered down to the nanometer region.  According to the sciencemag article, its density is .9 milligrams per cubic centimeter (.9 kg/m^3).  Disappointingly, its compressive strength is only a few kPa(see figure 3).

Given that reasonable densities lie in the region of 1000-5000 kg/m^3, we would need a compressive strength in the region of 60 GPa to 300 GPa.  Is that reasonable/achievable?  The other option is to allow the space tower to reach less than GEO, and simply do a small rocket launch from the top; for example, if your tower goes to 26,169,000 m (radius; that is to say, ~20,000 km above Earth's surface), your rocket can have a piddly 2 km/s delta V.  The compressive strength requirement in this case would be 49.0 MPa/kg/m^3.  Not much better, really.

Any ideas?  Or is the tether really more practical and near term?


-Josh

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#2 2011-11-20 16:28:05

Terraformer
Member
From: Ceres
Registered: 2007-08-27
Posts: 3,816
Website

Re: Space Towers

Your almost double post in confusing...

What's the figures like for a set of towers reaching to say 200km (I'm too tired right now to focus on even putting in the values)? I seem to recall a suggestion for a set of 100km tall space towers with a maglev track on top. Even without using magnetic acceleration on it, using a simple magnetic levitation track means you can cut out an awful lot of drag. Perhaps they could also be used to beam power, ala the proposal for airships previously. To accelerate to 7800m/s at a constant 30m/s^2 gives a time of 260 seconds and a distance of 1014km, of which only a fraction may have to be done on the track. Put it at maybe 150km altitude and coast upwards while using a rocket motor. Maybe 500km length will suffice? Certainly, if we get the total delta-V required from rocket down to approx. 5km/s, we get cheap reusable vehicles easily enough, with a mass ratio of 4 - and that's with comparatively low performing (at least Kerosene/LOX, if not Propane/H2O2) fuels.

One minor problem with only getting the stuff up to 2008 back is that people reading my old posts will think I don't know anything about rocketry or space colonisation...


"I'm gonna die surrounded by the biggest idiots in the galaxy." - If this forum was a Mars Colony

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#3 2011-11-20 17:32:11

JoshNH4H
Member
From: Pullman, WA
Registered: 2007-07-15
Posts: 2,546
Website

Re: Space Towers

Well, first of all, with regards to the idea that you seem to be talking about, I believe that is called a Launch Loop.  I learned recently in physics how eddy currents work, and in my opinion it is simply a phenomenal idea to use them to accelerate a vehicle.  I would love to do the calculations as to what kind of force you could expect, etc., and I plan to when I have time. 

In the case of a 200 km tower, you don't even really need to bother with the changing gravity.  You can more simply just assume that gravity is constant at 9.8 m/s² and find that the S/d ratio will have to be greater than gr, which is to say 9.8 m/s²*200,000 m.  That's a S/d ratio of ~2 MPa/kg/m³- Meaning a compressive strength between 2 GPa and 10 GPa for reasonable materials, with no taper.  That said, it can be almost guaranteed that a taper will be used in this circumstance, and I've heard that this kind of tower would be more than doable with steel.

Assuming that the materials are there, I like the idea of a space tower better than the idea of a launch loop because there need not be any parts which move very quickly, and this will enhance lifetime while at the same time reducing cost.  I believe that a launch loop involves a 2000 km long iron wire moving at 30 km/s and having a temperature of 500 C, which is not exactly conducive to long lifetime of the system.

And with regard to post history: Looking at the posts I made immediately before this backup was created, it would appear that I did not know that "increase" only has one c in it.  I am quite scared to think of what my posts look like way back at the beginning... I was just out of eighth grade, and I fear I knew very little about science or the English language.


-Josh

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#4 2011-11-20 23:12:45

JoshNH4H
Member
From: Pullman, WA
Registered: 2007-07-15
Posts: 2,546
Website

Re: Space Towers

Just for the record, I was at one point under the impression that it would be possible to build a space tower out of the newly discovered nanomaterial.  That misconception was based on this essay by Arthur C. Clarke.  In it, he states that if a tether with no taper can hang 4,960 km at 1 g (this is mathematically equivalent to a structure with no taper standing up), then it can serve as a space elevator.  Given a density of .9 kg/m³ and (okay, well, this isn't the actual strength of the material, but it is what I misread the graphs and thought it was at the time) a compressive strength of 154 kPa, it would be able to stand up to 17,500 km in 1 g. 

This is quite a discrepancy, actually.  I can't see any flaws with my math, yet at the same time the result that I'm getting is rather ludicrous.  As a BOTE, assuming that gravity were constant at 1 g from the Earth's surface to GEO, the S/d of a material for a compressive space tower would have to be 350 MPa/kg/m^3.  My calculation of 62.2 MPa/kg/m^3 is equivalent to a tower/tether that is 6,350 km tall/long.  Is Clarke wrong or am I?  Given that he is no longer with us, there is no way to compare notes, even if he had the time.


-Josh

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#5 2011-11-21 10:44:48

Terraformer
Member
From: Ceres
Registered: 2007-08-27
Posts: 3,816
Website

Re: Space Towers

A 200km tall tower... I know most of it would be empty space (or hyperlight solid, which is simply unusable empty space), excluding the elevator shafts and stations, but I have to wonder if the upper 10km could be used, and possibly rented out (high grade vacuum can be very useful). Stack spacecraft launching stations towards the top, and use the very top as an observatory...


"I'm gonna die surrounded by the biggest idiots in the galaxy." - If this forum was a Mars Colony

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#6 2011-11-21 10:51:22

JoshNH4H
Member
From: Pullman, WA
Registered: 2007-07-15
Posts: 2,546
Website

Re: Space Towers

Hmmm... Seems reasonable to me.  Dont forget about penthouses.  It could also serve as a stratellite, or a ground anchored satellite, so to speak, as well as a great source of data on the upper atmosphere.  Dropping something for 200 km will get a pretty significant time in freefall.  I guess it really depends on cost.  I think you would want to have revenue sources other than using it as a rocket platform.


-Josh

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