## Wednesday, April 24, 2013

### How Many Asteroids Would it take to House all of Humanity?

Arguing that humans should live inside asteroids naturally leads to the question of quantity - would there be enough resources in asteroids to be home to the entire population of humans?  What are the limits exactly?  The question is mostly unanswerable when the reference concept is rotating space stations, as this depends on the state of technology.  In this blog, I'm limited to a specific mechanism, which is containing air with the self-gravitation of an asteroid.  Therefore, statements about quantity can be conclusively made.  Of course, there is still ambiguity, which I will get into.

To begin with, I must return to the basic state equation that governs a gravity balloon.

$$P = \frac{2}{3} \pi \rho^2 G t^2 \frac{\left(3R^2 + 3 R t + t^2\right)}{(R+t)^2}$$

This relates 3 variables, pressure, density, and radius.  You need to draw the parallel between volume and the inner radius.  Plus, there's somewhat of a 4th variable, which is the total mass of the structure, which can be substituted in and out (numerically usually) with the shell thickness, t.  In the process of inflating an asteroid to make a habit the rock mass is constant.  Really, you could say this equation is dual to the ideal gas law, in that combining the two fully determines the inner radius (and all other parameters).

Now, given that we have this relationship and methods for solving it, the number of humans that an asteroid or collection of asteroids can house depends on two factors:

• how much space you require per person
• how low of a pressure is okay
I've dealt with the latter issue in previous posts.  Since using in-situ resources would be an absolute requirement, it would make sense to use as much atmospheric pressure as sea-level on Earth or less, where the lower limit is dictated by biological tolerability of low Nitrogen content.  Because of that, on this topic of availability throughout the entire solar system, I'll reference those values of sea-level pressure and 34% of sea-level pressure.

For the other limit, the amount of space to comfortably house a human, the numbers are much more ambiguous.  On the one hand, we can compare this to other space stations, so I looked up the crew and volume of the International Space Station (ISS), getting a volume per person in cubic meters.  We would want more volume than this, but it is an instructive lower bound.  For the upper bound, we can look to Earth itself.  How much atmosphere do we have per person?  That's not a straight-forward question because we all live at different pressures, but with a reductionist goal, I simply divide the mass of Earth's atmosphere by the sea-level density.  That gives the volume you would expect if you made a giant balloon of with all Earth's atmosphere and compressed it to sea-level pressure.  Obviously this is an overshoot, but that's good, because we now have a clear overshoot and undershoot.  These are different by roughly a factor of a million, so I introduced a middle-range number that's the square root of those two numbers multiplied.  It's fairly arbitrary, but as far as I can guess, it's probably the best shot at a comfortable volume requirement for people.  A 62 meter cube is awfully big, but we should consider that plants and other life would be important as well.

Reference Cases for Volume Requirement
ISSGeometric MeanEarth
V (km3) per person0.00000010.0002378470.565714286
V (km3) for all humans7001.7E+064.0E+09
Side length (m)4.662.0827.1
(side length is the side of a cube if the volume
was divided up to a cube for each person)

Unfortunately, I don't expect that I'll do better than a factor of a million.  I can get all kinds of (also subjective) numbers for population density per area, but there's no clear idea of the ideal population density of humans per unit volume.  In order to appreciate the full weight of the problem, it's useful to consider both extremes.

The one piece of information still missing is a list of existing asteroids.  This would be complicated (though not impossible) to do one-by-one, so I'm using a size distribution function instead.  The data to do this is very widely published, and I opted to simply use the size distribution list from Wikipedia.  The table there gives a cumulative distribution.  That is, the number of asteroids above so-and-so diameter.  Effectively, a cumulative distribution function (CDF).  I obtained a simple power distribution for the data.

Asteroid Size CDF

Now that we have the CDF, we can also obtain the PDF.  Of course, it is the derivative of the former, but since the CDF counts from largest (starting with Ceres) to the Nth asteroid, it's the PDF is the negative of the derivative.  This is simply an artifact of how the counting is done.

Distributions in km

$$\text{CDF} = N(>D) = \left( 1.77 \times 10^6 \right) D^{-2.12} \\ \text{PDF} = \frac{dN}{dD} = \left( 3.76 \times 10^6 \right) D^{-3.12}$$

The data was limited to sizes that I already know would be usable as gravity balloons to get the best possible numbers.  The exponent is somewhat consistent with other studies on the subject, which found an exponent of -4 (for the PDF) for the large asteroids.  I will still use my above number since I know for a fact that it reproduces the numbers I want in this narrow range.  With this, all of the information is there to find how many asteroids it would take to house all the human population given some pressure and volume requirement.

The form of the proposal is then to start with the smallest asteroids that is large enough to contain the desired internal pressure, "inflate" that asteroid to a very small volume, then move onto the next one.  This next one is larger and thus, we will be able to inflate to a larger inner volume.  This continues up the size scale of asteroids until the sum of the internal volumes of all the inflated asteroids comes out to the volume requirement.  Density is required for these calculations.  In previous posts I assumed a density of 1300 kg/m3, but here I'm changing that to 2000 kg/m3.  Reason being that larger asteroids tend to have more volume, and the 1300 number was only a conservatively low number for fairly small asteroids.

On the subject of density, we frankly have no good idea of how dense these asteroids are on the whole.  Generally speaking, the measurements of size come from back-calculations from the amount of light we get from the objects.  This light does fairly faithfully represent the surface it exposes to space, and that says nothing at all about density.  Because of this, we know diameters with much better certainty than what we know mass.  For much larger bodies, there is another method for calculating mass, which is to observe their gravitational interactions.  This method is completely impractical for the 5,000-some asteroids contained in the set that I'm interested in.  It probably wouldn't be possible to employ the method even if the resources were there to do it on one, and doing it for a large number of the asteroids is even more impossible  Given that, asteroid density will remain a very major source of uncertainty in all this writing.  There isn't a clear basis to employ the figure of 2000 kg/m3 for the density of the set, but the densities are sporadic in the first place.  I would prefer to just avoid multivariate PDF integration for the time being.  There is certainly room for someone to write an academic paper about reasonable expectations for the full size and density distribution among asteroids.  I've found myself at the practical limit of this research so that's my justification for using this generally representative number.  It's possible to test the method with different densities and it changes within possibly an order of magnitude with a reasonable density range.

Cumulative Volume from Inflating Asteroids

A fun way to give the answer to this query is to say "to get a pressure of 1 atmosphere you need to use asteroids A to Z", meaning that you would need to use all asteroids with sizes between those two candidates.  Because of that, I'm quoting a reference asteroid to get the central pressure with no inflated volume (that is the "A"), and then an example asteroid for every volume case (that is the "Z").  Here are the numbers for the two different pressures.

Asteroids Necessary to Obtain the Volumes at 1 Atmosphere
CaseCumulative
Volume
Goal
D (km)M (kg)Inflated
D (km)
Number
Used
Nth
Asteroid
Uninflated
Pressure
(atm)
starting size026.92.04E+160.0- 277 Elvira1.00
ISS70028.32.37E+162.6162 625 Xenia1.10
Geometric Mean1.66E+0642.88.21E+1626.4997 132 Aethra2.53
Earth3.96E+09241.41.47E+19537.41,639 65 Cybele80.38

Asteroids Necessary to Obtain the Volumes at 0.34 Atmospheres
CaseCumulative
Volume Goal
D (km)M (kg)Inflated
D (km)
Number
Used
Nth
Asteroid
Uninflated
Pressure
(atm)
starting size015.74.05E+150.0- 518 Halawe0.34
ISS70016.54.74E+151.6508 433 Eros0.38
Geometric Mean1.66E+0626.71.99E+1618.23,329 274 Philagoria0.98
Earth3.96E+09169.85.13E+18417.65,152 54 Alexandra39.79

For example, in order to get an atmosphere volume equal to that of Earth at sea-level pressure, you would need to inflate all the asteroids larger than Elvira and smaller than Cybele (and possibly include those two as well).  The value of this observation is that it shows that there is plenty of availability of asteroid material with which to make gravity balloon habitats.  In fact, it's not hard to see at all how this approach could easily house more people than all the terrestrial environments in the solar system combined.  I mean, provided that the middle-range value is a sufficient volumetric population density, then we have the potential for over 1000 times that volume, and that's not even touching the largest class of asteroids.

But how constructable would these things be?  As I will write about later, there are some valid concerns of stability although they should be manageable.  It follows that you work harder against the force of gravitational differentiation when you've "overextended" yourself by making an inner radius a great deal larger than the shell thickness.  By looking at the inflated diameter number, you see that that only needs to be true if you require a volume on the scale of what Earth has, which would be a mega-scale engineering problem by its very definition.  If you're looking at the middle-range volume requirement, the inner bubble is still relatively small compared with the overall dimension.

Regarding construct-ability, there's also the matter of the pressure before any inflation takes place.  This could be a problem by making the center uninhabitable while the atmosphere is produced and supporting infrastructure is built.  What is the maximum habitable pressure?  I searched around for a few numbers.  The compressed air reference points are fairly useless, because the atmosphere would be alien anyway.  So given that, the maximum safe SCUBA pressure might be pretty good guidance.  Comparing to the above table, it's clear that in order to make Earth volume scales, you will have to learn to deal with impractically high pressure, so it might be an uninhabitable environment to start out with.  Thankfully, the pressure declines fairly quickly once the inflation is started, which I wrote about in a previous post.
• trimix suggested                 3-4 atm
• compressed air limit           6.4 atm
• recommended diving limit   9.7 atm
• scuba record                     31 atm
The number of bodies in the above calculation is also of interest.  To some extent, requiring a huge number of space habitats to house your population defeats my previous arguments about the value of a large zero-gravity habitat, which is very easy movement between places.  If the interesting places you want to travel to are all in other habitats, you still are forced to traverse the less-ideal environment of space.  Plus, objects in the asteroid belt tend to have a year-length of 4 Earth years or more.  If we're following the natural flow of gravitational slingshots and such, it would take extremely long times to get to the other asteroids since they're distributed almost completely 360 degrees around the sun (although there is some distortion from Ceres).  Thus, the citification argument would dictate a minimization of the number of habitats which is the opposite of the calculation I've done - which is to use the smallest workable asteroids.  Obviously doing it that way wouldn't be attractive.  It's more realistic that someone knows how much total migration they want to see, then create only a single or a handful of gravity balloons that can handle that amount.  There is almost perfect flexibility to make those decisions.  Consider that the asteroid Cybele, by itself, would have enough volume (when inflated) to satisfy the middle-range requirement.  So you could build that gravity balloon and have 7 billion people move in there.  Those are just the kinds of numbers we're working with here.

Ceres is the largest thing in the asteroid belt, and you can see that the distribution function expects just about 1 object at the large diameter values of about 900 km.  The fact that my math predicts a diameter of only about 530 km is necessary for creation of Earth atmosphere volume shows that the idea is viable, because that doesn't exhaust all of the available mass.  In fact, since Ceres comprises the majority of matter in the asteroid belt, my numbers leave the vast majority of the mass on the table.  Ceres would be pretty impossible to break apart, for what it's worth.  I calculate its internal pressure to be on the order of 900 atmospheres, which is a higher pressure than our power plant turbines use.  This is a very high pressure.

Given some asteroid mass, then provided that it's large, the amount of volume you can obtain by making it into a gravity balloon scales as a 2/3rds power of the diameter.  That exponent comes from the fact that a gravity balloon's functional mass-efficiency is a result of the surface area (2) to volume (3) ratio.  One consequence of this is that you could actually produce a larger volume from the smaller asteroids by first smashing them into each other and then inflating the resultant body.  Ideally, of course, you would just start with a larger asteroid.

What about the moons of the solar system?  I've left those out of consideration so far.  But to get an ideas of the general abundance of things suitable for a gravity balloon, I looked at what internal pressures you can find in the moons of various planets.

Number of Moons of Planets with
Internal Pressures in Given Ranges
0.34 - 0.8 atm0.8-30 atmhigher
Earth001
Mars100
Jupiter1075
Saturn226
Uranus1425
Neptune193
Total282020

(Pluto not included because it's not a planet)

There are two major observations here.  One, the asteroid belt has a vastly superior population small bodies (it also helps that it's closer to Earth than most planets) that would be good for small gravity balloons.  If you wanted Ceres-sized things, however, there are a huge number of moons with absurdly huge internal pressures.  It would be extremely difficult to do this in practice, but it's obvious that the matter of moons is easier to get to than that of planets themselves.  The volume you could create with these objects goes through the roof.

If you inflated the moon in the simple way that I've written about so far, then you would end up with an object several times larger than the Earth itself (not the atmosphere, the whole thing).

## Saturday, April 20, 2013

### The Curious World of Artificial Gravity

Writing about space stations with artificial gravity, its fairly common to see people restrict themselves to a diameter of 500 meters or larger.  This follows mathematically from a commonly cited figure that Coriolis forces can be hard to tolerate at more than 2 rotations per minute.  If you require the production of 1 Earth gravity, then the required size is about 225 meter radius, and 450 meter diameter.  That is probably fairly reasonable.  It would be cheaper in terms of material to make stations even smaller, but there are a few reasons that very small habitats could be awkward, to say the least.  Let's look at the basic parameters of this size:
• Radius set to exactly 250 meters
• Edge velocity would be about 50 m/s, or 110 mph
• The change in pressure from the center to the surface is about 1.6% of an Earth atmosphere

For the pressure change from the center to surface, it has to be assumed that the air rotates along with the cylinder.  This is yet another good reason to illustrate these tubes in the way I have in the introduction, that is, a tapered end so that the edge itself isn't directly exposed to the atmosphere.  If the ends are open the the air partially doesn't rotate with the tube, and my view is that these are hurricane-like conditions.  The hydrostatic head from a few 100 meters is a lot, which is effectively what would drive currents.

The really interesting metric is the edge velocity.  Imagine that you hop in a train that is moving in the direction opposite of rotation in this world.  If the train accelerates to 110 mph, it will become weightless.  It still has to work against the drag force from the air (air rotates with the tube).  This phenomenon would exist for any moving thing to some extent.  Your "weight" decreases the faster you move in the direction opposite of rotation, because after all, you're closer to being stationary than the rest of the tube.

Car or Plane Example

It's hard to imagine exactly exactly how this would play out, but that's what we have physics for!  For an electric car, the friction forces include aerodynamic drag, drag on the tires, and some internal losses.  The drag force is fairly predictable as proportional to the density of air times the velocity squared.  The force friction force from the tires, however, is roughly constant.  This isn't a huge surprise when you consider that the concept of coefficient of friction supposes a constant force irrelevant of speed.  But that concept also supposes the force is proportional to the normal force ("weight" in other words) - and this is where things can get interesting.  Let's look at the case of a car driving in the opposite direction of rotation.  In this situation, we can formalize a few things:
• Weight of the car is equal to its acceleration, which is the speed of the rotating tube, minus its speed, squared times the radius.  R (V-v)^2
• The drag from the wheels is proportional to the weight
• The aerodynamic force is proportional to the square of velocity

I listed the constants for a typical car modeled with this perspective.  You can look at things in terms of force or power, but I prefer power because it reflect directly the amount of throttle you have to put into a car to keep it going at some given speed.

We also need to distinguish between a plane and a car, as well as address the issues with traction.  Driving a car is pushing it forward with the force of wheels on the ground.  This would stop working entirely as one approached 110 mph, the speed of the tube, because they would be near zero-gravity.  Tires only world with gravity.  The wheels losing grip and spinning isn't the only problem, you would also lose the ability to steer without slipping (and maybe tipping).  The opposite thing would happen as a car moving in the direction of rotation, the functional weight would increase, the grip of the wheels on the ground would increase, the weight the people feel in the car would increase, and you could steer tighter.  These cases are interesting to me, so I plotted them together here.

Power dissipation when moving against the direction of spin, versus with it
red - moving in direction of spin
blue - moving against direction of spin

A plane is different because it takes off while exerting force on the air.  The forward thrust is thus not dependent on the traction from the wheels.  Of course this is necessary because the plane plans to leave the ground.  You can notice in the above graph that the power required to drive is always increasing with increasing speed.  That's fairly "normal", and it's also what we would expect for a typical car driving in a 500 meter tube.  But to imagine a very curious case, I took the above car model and reduced the air friction by a factor of 10.  The effect this has is to make the tire friction dominate through more of the speed range.  But the tire friction decreases with increasing speed moving against the direction of ground motion.

Power dissipation when moving against the direction of spin
using tire friction + air friction, more tire friction than last graph

If you imagine that this model fits a plane, you discover that something truly strange happens.  At a certain speed you obtain a local maximum in power.  If you were accelerating, slowly increasing the throttle up to that point, then once you reached the point you would experience increasing speed with no increase in throttle.  By the time the system equalized, you would nearly be flying.  This brings up another good point, in an artificial gravity tube you can have wingless flight.  In fact, most flying things we're familiar with can hit 110 mph.  Basically, in this world, planes wouldn't have wings.  It's disputable whether it would have any "cars" in the first place.

Cross-Habitat Structures

You need to consider that for any artificial gravity environment, it will likely make sense to have structures that span the diameter of it.  There are a few reasons why, and the structural motivation is compelling.  Imagine that the artificial gravity tube was built with a uniform strength, designed for a given mass-thickness of load.  If you wanted to expand operations in a location further, it could be difficult.  In fact, putting any large amount of matter in a singular place would cause strange kinds of sagging in the structure.  The mass distribution really has to be uniform along the circumference, but structures that span the diameter could be the exception to that.  By building a tall building on both sides, they could be each others counterweight.  Considering that these would be held by tensile (and not compressive) strength, they could actually be cheaper to build than their equivalent building on Earth.

There is an obvious limitation to this, which is that the gravity decreases as you go higher.  If you are co-moving with the structure, then gravity changes linearly with radius.  So if you move up half the radius to the center of the tube, the gravity will decrease by half.  A 500 meter diameter tube, thus, could host a building 500 meters "tall" but the floors would have gravity evenly distributed along the spectrum of zero gravity to full gravity.  On Earth, we have a handful of occupied buildings that rise over 500 meters at the top, but the rest of mankind's buildings are shorter than that.  Examples:
• Burj Khalifa                       829.8 m
• Mecca hotel                      601 m
• One World Trade Center   541 m
• Petronas Twin Towers       458 m

If you imagine that gravity as low as 0.9g is tolerable, then you can only have 50 meters of usable floors in a cross-habitat structure, and only 25 meters on each end.  Most typical city centers with a decent skyline have a few buildings higher than 100 meters.  So in short, if an artificial gravity tube was to have the economic density of something like Manhattan, it's likely necessary to have a diameter larger than 500 meters, just to fit the buildings in.

Others have written about staircases that transition to zero gravity.  This idea actually occurs in several sci-fi ideas as well as serious discussions about space habitats.  As for a 500 meter diameter habitat, climbing ropes typically have lengths 30 to 80 meters.  While 500 meters is a lot, it's not unthinkable.  It's also fun to imagine how you could string such things ad-hock.  Anchor one end, then walk around the circumference holding the rest of the rope.  With those tied in sufficiently, you could climb straight across the habitat right away.

Spooky Vertical Motion from False Fields

A good read on the funny behavior of things in artificial gravity can be found here.  One of the strangest thing about living in an artificial gravity environment would be that when you drop something, it deflects some amount.  This reference illustrated this for two cases with 1g of gravity.  The dots show the location every 1/10th of a second.  It illustrates something falling from head-level, and the movement of a body's center of mass when jumping up straight.

Movement of drop from head-level and feet movement with jump
both are for 1g of gravity, taken from spacefuture.com

There are a number of ways that you could go about establishing the physics of such a place.  Logically, you might just want to look at some object, falling or moving, and then consider how the ground would move relative to it.  Another way is a full transformation of reference frames, which is a daunting task in many ways.  The ground both rotates and accelerates.  This nature is shared with orbiting reference frames that are tidally locked.  I tried my best at the problem on Physics SE.  It was surprising how difficult it is to do the full transformation of reference frames.  I want to keep this simple, so I solved a simplified case, to exclusively answer the question of how much an object dropped from head-level would deflect.

Diagram and calculations for deflection of dropped object

Walking through the math, you can observe that the dropped object follows the tangent line from the point where it was dropped.  Until it hits the floor it maintains its original velocity.  The distances that the object and the ground (the point that starts out directly below the object) travel before the hit are illustrated in red.  You can qualitatively observe that the dropped object travels the same total distance that it would have otherwise had it been held at its original elevation.  That allows the establishment of the needed relationships with simple trigonometry relationships.  These are more-or-less summarized in that image.  Then using these, as well as some other substitutions from the fact that the tube has a known gravity, I put the expression in two different forms.

Expression for the deflection distance for an object dropped at delta above the ground
Second expression is the deflection in terms of a fraction of the original height

Plugging in the numbers, this comes out to be quite significant.  If you use a drop height of 1.5 meters (about the shoulders of a human), here are the deflection values for a tube rotating at various rpm values, in order to get correspondence with my reference case and the above 1 and 4 rpm examples.

Given: one Earth gravity, 1.5 meter release height
• 4 rpm, 50 m radius, 25 cm deflection or 16%
• 2 rpm, 225 m radius, 11 cm deflection or 7.6%
• 1 rpm, 900 m radius, 6 cm deflection or 3.8 %
An important observation here is diminishing marginal returns.  The size you have to build the structure goes up very fast just to slow the angular speed down a little.  Plus, the material requirements grow proportionally to radius.  That means that for increasing the material requirements by about a factor of 4 you only halve the deflection effect.  Clearly this would be difficult to do away with entirely.  I think it suffices to say that people living there would always perceive the direction of motion.

The two examples discussed here of a car and a falling object are useful mathematically.  The reason being that this part of the fictitious forces comes from a cross-product of the object's velocity with the tube's angular momentum.  The angular momentum goes in the direction of the tube's axis.  That implies that you don't experience this type of force when moving in the direction of the tube's axis, and this is correct.  If you were to throw a ball, there is "distortion" in its path if thrown up or in a tangential direction, but not really when moving down the length of the tube.

Maybe this isn't all that bad of a thing.  I am reminded of a particular experiment which attached a special belt to study participants.  The belt was lined with vibrators.  Only the vibrator on the north side would vibrate, but it would do this all of the time.  It was found that the participants adjusted to life with the belt to an extent that was downright creepy.  Their brain had found a way to use the new sensation to help orient themselves and navigate as a classic example of sensory substitution.  I think the most amazing detail is that they reported nausea after they took the belt off.  It's not too far of a jump to imagine that this same thing may happen to people living in an artificial gravity environment.  Their minds would constantly be searching for the direction of rotation.  In fact, it may be stranger for them to consider our world than for us to consider their world.  On Earth, we seem to have nothing to distinguish between directions on the horizon other than a weak magnetic field from the center of Earth.  In artificial gravity you have a natural compass!

These kind of questions led others to propose the idea of Planetary chauvinism.  This is mostly proposed because of the difficulty of climbing in and out of gravity wells.  But it's worth mentioning that the "weirdness" of artificial gravity probably disappears depending on the amount of time you spend there.  We only speak about building things similar to Earth because Earth is all we know.  We can't judge a strange new world without ever have lived in it.  Being weird may be a good thing.  That leads me to the next topic - a mixed gravity environment, which I hope to write about in a future energy.  The gravitational balloon would be a truly mixed environment where one could go from gravity to no gravity and back again... even multiple times during a workday.  People would not only be perfectly comfortable in artificial gravity, but with zero gravity as well, not to mention everything in-between.

## Thursday, April 18, 2013

### What Would a Giant Space Habitat Look Like?

The concept of Rayleigh scattering is really neat.  I discussed the relevant attenuation coefficients for a large space habitat in a previous post.  It still stands to reason that the atmosphere would JUST reduce the light getting to places inside, but it would change the look and feel of it as well.  In a simplistic sense, the shorter wavelength light is reflected (thus a blue sky) and the longer wavelength passes through (thus, a red sunset).  This is illustrated beautifully by Wikipedia, and the basic mechanism goes as follows:

Example of Red/Blue Light Separation

It's a wild idea to consider how this kind of effect would manifest itself in an extremely large space habitat with lots of atmosphere like a gravity balloon.  Getting light in, in the first place, would be difficult, but once you accomplish that, there's also the need to distribute it.  I imagine that this would be accomplished with something like a tree of mirrors.

The seperation of short and long wavelength light would apply to every section of light delivery.  The mass-thickness that we may be dealing with are actually much larger than what you see on Earth during the daytime, since that only loses something like 4% of the visible light.  What we would be looking at is something much more akin to the behavior of Earth's sky near sunset, where brilliant colors are scattered all throughout the atmosphere due to an extremely long path length.  These are only present when the bulk of the light is hitting the atmosphere at a very shallow angle on Earth.  In a gravity balloon, everywhere it went, it would have a large distance of atmosphere to travel through.  Because of that, I imagine (quite inexactly) separation of light in the following ways:

The basic dichotomy is the difference between light that makes it through and the light that gets scattered.  The scattered light is what you see represent the beam's path itself.  The combination of the two would still produce fairly white light on things that it shines on.  However, if the mirrors don't directly reflect the direct light from the sun to walls, then those walls should appear reddish or orangeish depending how far the light beams have traveled.

How this light would be practically used is a more difficult problem.  Artificial gravity tubes at self-blocking.  No one wants to get their lighting to come up through the floor.  I'm not even quite sure how that would work.  The friction buffers complicate the issue even further.  The edges can't be open to atmosphere because of the speeds they're traveling at and the power dissipation that would happen because of that.  That leaves only two fairly difficult options to get the light to the people:
• A complicated system of mirrors to reflect light through the tube ends
• Have transparent surfaces throughout parts of the ground and friction buffers

I don't have good answers for these issues.  Obviously one solution is to just not use sunlight in the first place and use entirely artificial lighting.  It is somewhat depressing that the best illuminated ares would probably be the zero gravity regions.  It's hard to see a way around that.  For either approach, going through the tube ends or the tube sides, some amount of additional light collection would be a necessity.  Also, if the tubes are spinning at 2 rotations per minute, that could be difficult to deal with psychologically.  Still, if you imagine that the windows are circumferential, and that light is reflected into it from all sides, then it could be maintained relatively constant, and the same argument applies for the ends.  Nonetheless, each of these proposals winds up demanding reflecting the sunlight at minimum three times.  One alternative would be to simply locate the artificial gravity tubes in the direct line of sunlight propagation to begin with.  That may be more reasonable than it sounds if there are multiple access tunnels that have sunlight concentrated on them from space.  In fact, this is probably the best approach - the eliminate the internal mirrors in the above illustration.  The approaches for distributing the sunlight around the rotating tubes themselves is something I have not done, although it could be neat, I'm a little short on clarity on the exact design issues that would be at play.

### Price of Floor Space in a Artificial Gravity Habitat

In this post I will go into math needed to get a value for the structural mass needed to build a rotating structure that creates artificial gravity.  I'll then put the case into terms of dollar cost, and go over what kind of inputs and assumptions you need to get those numbers.  For the question to be answerable, you basically need area, mass, and engineering margins.

Strength Requirement

To start with you have to formalize your requirement.  With basic physics you can relate the minimum weight and strength of structural material for some load.  I did that on Physics Stack Exchange.  The equation for the case of supporting it by circumferential material comes out to something very simple.  Repackaging this to give the required amount of structural material, you can say the minimum structural material is the following.  Here, R is the radius of the spinning structure, g is the gravity you create, sigma is the material strength, and with rho, density, it becomes specific strength.

$$M_{struct} = \frac{ M_{load} }{ \left( \frac{( \sigma / \rho ) }{ R g } - 1 \right) }$$

I also obtained an equation for using cross-beams.  This was more difficult.  To put in terms of an equation that returns a minimum structural mass:

$$M_{struct} = M_{load} \frac{ R g \sqrt{ \pi }}{ 2 (\sigma/\rho) } \text{erf} \left( \sqrt{\frac{ R g }{ 2 (\sigma/\rho) }} \right) \exp{ \left( \frac{ R g }{ 2 (\sigma/\rho) } \right) }$$

As far as I'm concerned, both of these will reduce to the following, equation which is an approximation that works for both above equations under the assumption the material strength is much greater than the minimum for a zero-load structure.

$$M_{struct} = M_{load} \frac{ R g }{ ( \sigma / \rho ) }$$

Speaking of the zero-load structure, that is the case of a rotating thing that doesn't have to support anything.  You can get that from the above equation by setting the two masses equal and then you find a single value for material specific strength given the radius and gravity.  If you consider an artificial gravity habitat with a diameter of 500 meters, the material specific strength lower bound is 2,450 N-m/kg.  That number needs some context.  The strength I'm quoting of 2,450 N-m/kg is the absolute minimum strength, meaning that if you spun a hula-hoop of this size at this rate in zero gravity, then it would break apart if any part of it had strength less than that.  There are two multipliers you need to work against, which is the fraction of the mass that is load, and the engineering margin.  Practically, if these are both on the order of 10, then you will ideally want some material with a strength 100 times what that number is.  To give a reference, here are some specific strength values of material.

materialspecific strength
Nylon69,000
Aluminum214,000
Silicon carbide1088000
Carbon fiber (AS4)2457000
Carbon nanotube46268000

From this, we see that a 500 meter diameter tube isn't unreasonable from a materials standpoint.  Any reasonable assumptions for margin could probably be satisfied with fairly ordinary materials.  I'm particularly interested in Carbon steels, because the materials needed to make it is extremely abundant throughout the solar system.  A36 steel has a specific strength of 36,764 N-m/kg, which could be problematic under certain assumptions.

Pricing the Material

We can't know that much about the constraints of space manufacturing of structural materials in the future, but like other concepts I write about, it's reasonable to think that general engineering trends won't be unbroken.  In particular, the relationship between the cost to produce a material and its strength is a reasonably strong one.  Here is a graph showing the range of marketed materials on Earth today.  I notated some points on it so that I can employ it mathematically.

(reference)

Obviously there is a great deal of variation, but it still stands to reason that the general relationship may hold long into the future, and possibly indefinitely into the future.  We can use that line above to write an equation for the relationship.  It is a log chart with both axes, so the equation needs to reflect that.  The points go as follows:
• (10 £/kg, 1000 MPa)
• (1 £/kg, 100 MPa)
• (0.1 £/kg, 10 MPa)

These points have a direct proportionality, so the line can be summed up in a simple conversion constant.  That constant is 100 MPa for each pound/kg, or 100 (MPa-kg/£).  This is saying that the strength is equal to the specific cost times that constant.  The price is equal to the specific cost times the weight, by definition.  Combine these two, and use the density to convert the strength to specific strength, which leaves an extra density term.  That density combines with the cost coefficient to get a new type of coefficient which is the density dividedby the other figure.  Then, the previous expression for structural mass given a material and load value can be plugged into the price equation.  The specific strength cancels out (very interestingly), and we find that the cost only depends on the load and that new coefficient.

$$\text{strength} = \alpha (\text{specific cost}) \\ \text{price} = (\text{specific cost}) M_{struct} \\ \text{price} = (\rho / \alpha) R g M_{load}$$

This is a very interesting result, because it demonstrates that as long as the material is sufficiently strong, the price doesn't really depend on its strength.  Given this is true, you would generally opt for a material higher up on the above line.  You would prefer carbon fiber over aluminium just because it's easier to work with.

Let's get some values for this now.  I'm going to plug in a generic density to be representative of steels, use the above coefficient for price, and also the current conversion rate between the Great Britan Pound and US Dollars.  With this, we can put the price in terms of the radius of the structure and load that needs to be held up.

$$g (\rho / \alpha) = \frac{ 5,000 kg/m^2 }{ \left( \frac{ 100 MPa-kg }{ 1 \mbox{\pounds} } \right) \left( \frac{1 \mbox{\pounds}}{ 1.53 USD } \right) } \\ \text{ price } = \left( 3.2 \times 10^{-4} \frac{1}{m kg} \right) R M_{load}$$

The units are helpful, as they clearly reflect what's going into the equation.  The price for the same amount of load is greater for larger artifical gravity environments, this is pretty hard to argue with.  Also, this equation can be exersized on the basis of mass itself, or with mass-thickness.  The only difference for the latter is that different units will be given out.  If you use it with a kg/m2 metric, then the ultimate price you will get out of it is dollars / m2.

Expectations for Mass Thickness

Mass thickness is the amount of matter per unit area.  It's relevant for attenuation problems as a combination of distance and density.  It's also the metric that directly matters for the structural needs of a rotating artifical gravity habitat.  In order to competently talk about such a habitat we need some expectations of how much matter per unit density we'll want inside it.  One approach I have is to look at buildings.  A good way to get a sense of the mass per unit area for an office environment is to take the mass of a large building and divide it by the floor space.  I'm going to do that here for the (former) Sears tower.
• Tower mass: 222,500 tons
• Foundation footprint: 3,922 m2
• Floor space: 423,658 m2
• Mass-thickness by the footprint: 56,720 kg/m2
• Mass-thickness by the floor space: 525 kg/m2

For reference, the mass-thickness of Earth's atmosphere is about 10,000 kg/m2.  This is the same kind of discussion I wrote about in the introduction, although what I illustrated there was an assumed mass-thickness of 2,000 kg/m2.  Here, we see that office space can reasonably have a mass-thickness of 1/4th that, and only 1/20th that of supporting the atmospheric pressure itself.

It's also important to mention the water-depth equivelant.  The figure of 500 kg/m2 corresponds to a mass thickness that is equivelant to water 50 cm high.  If you think about it, that's pretty high.  That's a lot of stuff.  If you imagine a typical house, and pretend that everything turned into water and puddled, then how deep would one expect that to be?  The reason this is an important question is because these artifical gravity habitats in the gravity balloon would have a material cost directly proportional to the mass in them, which for planning a terresterial-esque habitat, is basically a matter of the mass thickness.  So with the office space example as a baseline, I want to compare what different size tubes would cost in terms of materials given different mass thickness values.  I'll start off with the 500 m diameter tube due to the reasons of Corolis forces, and then go up from there.

Price for 1 Square Meter of Floor Space
D (km)mu=250kg/m2mu=500kg/m2mu=1000kg/m2mu=2000kg/m2
0.5$20.00$40.00$80.00$160.00
1$40.00$80.00$160.00$320.00
2$80.00$160.00$320.00$640.00
4$160.00$320.00$640.00$1,280.00
8$320.00$640.00$1,280.00$2,560.00
16$640.00$1,280.00$2,560.00$5,120.00
32$1,280.00$2,560.00$5,120.00$10,240.00
64$2,560.00$5,120.00$10,240.00$20,480.00

This shows that if you have a normal office density in the smallest possible tube, you're looking at something like $40 per square meter. How does that compare to the alternatives on Earth? • Farm real estate, low:$ 0.12 / m2
• Farm real estate, high: $3.36 / m2 • Average US building:$ 896.49 / m2 (reference)

The farm comparision is at least a little bit unfair.  No one would think that you could farm with conventional methods in a gravity balloon, that just wouldn't make much sense.  So the building price is the best comparision point.  Still, I haven't worked in margin to the equation.  Let's say that, generously, you only need a factor of 3 lower than the yield strength.  That means that our low case is now $120 / m2. The probelm is, this isn't even all the costs. It's only the cost to get the ground itself rolled out there. You will still want a ceiling over your head, possible multiple floors, pluming, waste treatment, all other forms of infrastructure, sidewalks, real homes, I could go on. The price for floor space in our reference case of the US today includes all of these things in addition to the price of land itself. I should also mention, this is only for the gravity balloon. If you want to talk about a contained, rigid, artifical gravity habitat, the price per square meter would be 20 times that,$2,400 / m2, and this is frankly the best case scenario.  Living in space isn't cheap.  But then again, this is a reason the gravity balloon is a reasonable idea.  You avoid all the cost of making materials to contain the atmosphere.  That's managed by the larger engineering project of hollowing out space, controlling pressure, and worrying about the global dynamics.

Of course, all of this is based on the absurd assumption that materials would cost the same to make in space that they do on Earth, and that's clearly not true at all.  The zero gravity environment may be better for industrial production, and many people think that space could lead to the age of abundance.  The only point I want to make here is that there's a fairly strong upward sloping curve that you have to work against with those technology advancements.  Plus, I've not seen any numbers like this anywhere else.  If we want to take space settlement seriously, we need numbers.

## Tuesday, April 16, 2013

### Candidate Bodies to use as a Gravity Balloon

You could formulate the concept of the gravity balloon as a completely man-made construction.  Quite simply, once a space station grows to the scale of a Death Star, it make sense to stop building rigid walls as a pressure boundary, and instead design around the self-gravitation.  To fit into a realistic picture of the future of humanity, however, we need to make direct references to objects that we could use to hold an Earth-like atmosphere with minimal processing.  I touched on the availability of candidate objects in the introduction.  Since then I've delved into the knowledge base of asteroids to give a much more comprehensive picture and I hope to communicate that here.

The Obvious Candidates

Suitable bodies for the creation of a gravity balloon are extremely numerous in our solar system, however, they're not at the doorstep of Earth.  This is logical since the evolution of life required that Earth was not regularly hit by such bodies.  For what it's worth, the KT extinction event is thought to have been caused by an asteroid 10-15 km in diameter.  The factoid is interesting because it demonstrates that the minimum size needed to create a livable internal pressure in an asteroid is greater than the size that can cause mass damage on Earth, should it fall.

433 Eros

Predictably, the inner solar system is mostly devoid of the sweet spot of asteroids between about 20 and 60 km in diameter, the size necessary to contain a breathable atmosphere.  One notable exception is Eros, which is a fairly large asteroid that currently hangs around the orbit of Mars and could one day impact Earth.  Naturally, it's received a good deal of attention due to its unique status.  It is large enough to turn into a gravity balloon with an initial internal pressure of 70% of Earth's sea-level pressure, if it were spherical (density is reported to be 2.67 g/cm3, and is used in the calculations).  However, the shape clearly shows that it's not perfectly differentiated.  This isn't exactly a problem for a gravity balloon, since it just demonstrates that its material has tensile strength comparable to the gravitational forces.  This could be used to the advantage of people working with it, and could also make excavation somewhat more difficult.  It would definitely be one of the top targets of future asteroid hunters.

Eros, asteroid at about 1.46 AU, weighing 6.7 x 1015 kg

I also want to include a second scale of about 60 to 80 km (diameter).  These wouldn't be an ideal place to build a habitat right away, but with specific combinations of gases, humans can tolerate the pressures.  If such an object was inflated significantly, the pressure would decrease anyway.  Of course, development of that class would be relevant much further out in the future.

Phobos (moon of Mars)

Of the moons in the inner solar system, Phobos is the only suitable one.  Mercury and Venus are thought to not posses any natural satellites.  We could still discover some, but our efforts up to this point would have caught a mass lager than a 20 km diameter for either of these, so we're out of luck there.  Earth's moon is vastly too large by comparison (and Earth's second moon is vastly too small).  Development of infrastructure on the surface of the moon could, of course, be vital to space development but it couldn't hope to approach the flexibility that you could have in the inside of a gravity balloon.  Ultimately, you could consider the possibility of breaking up larger bodies to turn them into multiple gravity balloons, but this would require monumental capability, probably on the scale of what's needed for terraforming.  So for now we're just left with Phobos out of all the moons in the inner solar system, and I predict a central pressure of about 60% Earth's sea-level pressure.  The density is 1.88 g/cm3, and it is at least somewhat close to spherical.

Phobos, moon of Mars, about 1016 kg

This candidate has lots of positives.  For one, it's the only one that's currently being considered for manned missions by NASA.  While it's not likely that we'll be drilling over a kilometer into it anytime soon, this offers a connection with current buildup of space efforts.  Also, since there are significant public figures who are interested to see a million settlers on Mars, why not a million inside Phobos?  After all, the latter could turn out to be a better idea - less difficulty in mining and moving material, radiation protection without claustrophobic caves, and a full Earth gravity.  One of the major drawbacks is the fact that Phobos lies in a gravity well, in close orbit of Mars.  The orbital period is only 7.5 hours, as opposed to a full month for Earth's moon.  That makes it difficult to travel to and from other destinations in the solar system.  Because of that, Eros could have an advantage in energy budget for travel.  Tidal forces are also particularly strong, which would be an interesting technical complication.

Inner Edge of the Asteroid Belt

Now we're past Mars' orbit (1.66 AU) and we only have two workable candidates.  It's true that there are others, both comets and asteroids, that occasionally swoop into the inner solar system at the closest point in their orbit, but these would be energetically difficult to reach, just like far-flung candidates, so I'm excluding them.  In more scientific terms, I'm only interested in the location of the semi-major axis, which is a reasonable proxy for the energy required to reach them.  The area with a great abundance of suitable gravity balloon candidates starts where the asteroid belt begins.  To make the point, I gathered data on all asteroids within my size criteria (about 10 to 80 km diameter) and a semi-major axis closer than 2.5 AU.  This was done using the JPL small-body database search engine.

As far as I know, this data doesn't include any mass, and with good reason - mass is difficult to calculate, and often inferred from brightness.  They do give the calculated GM form of mass, calculated from orbital behavior, but this is only feasible for extremely large asteroids and almost none in my search criteria had that data.  That means all I had was diameter, but this isn't a huge problem.  I assumed a density of 1.3 g/cm3, which is largely thought to be representative of asteroids, or at least conservatively low for the majority of them.  With this density, I find mass, and use the methods I've previously described to find their internal pressure at the center.  I graphed this internal pressure over the semi-major axis values, which gives a good picture of the abundance of candidates throughout the inner solar system, up to 2.5 AU, covering some of the inner edge of the asteroid belt.

For the line that represents the limit of habitability, I'm using about 0.34 atmospheres, which is taken from Skylab.  In that mission, they didn't use Oxygen gas exclusively, but the Nitrogen content was much lower than on Earth.  That worked for them, although there are lingering concerns about using that type of atmosphere.  For a gravity balloon, the reasons likely be different, because the gasses that you can get access to are fundamentally limited since you're only interested in in-situ resources.  Oxygen gas would be comparitavely easier to make, so for early stages of self-sufficient space development it is thinkable we would use such a low atmosphere pressure.  Importantly, doing so also broadens the types of candidates that can be used.

The density assumption obviously causes errors, but for the majority of these, we don't actually have better data.  Give it another decade and that may change - this is one of the things I find most exciting about the subject right now.  A good example of the innacuracy is Eros, which is shown to be below 34% of sea-level pressure in the graph, which is obviously not the case.  Its density is very high, and that's why its in the wrong place.  Otherwise, the spead of pressures observed here reflects the general abundance curve of asteroids.  When I did the search, I included objects as low as 10 km diameter, just for illustrative purposes.  I then removed all those below the habitable pressure line from the set.

Even though these are all the same semi-major axis, they're not all the same since some orbits are more elliptical than others.  In order to illustrate this, I ordered the candidates from the reduced set (not all shown on that graph, about 50 in total) by the perihelion and plotted the perihelion and aphelion.  This shows that many of the candidates in the set have a nearly circular orbit, but on the other extreme, some come as close as about 1.7 AU.

Spread of Orbit Characteristics within the Selection Criteria

It's not immediately clear to me whether a more highly elliptic orbit is more or less favorable as a destination.  On the one hand, uniformity throughout its year would be nice, but it's possible that more highly elliptical orbits could offer opportunistic trips.  Either way, for the moment I selected the 9 candidates with the closest perihelion, just to have a small list to cite.

Asteroids within the Set with Closest Passes
(done with Tableizer)
AsteroidPerihelion (AU)Aphelion (AU)Inclination (deg)D (km)M (kg)P (atm)
323 Brucia1.673.1024.235.83.1E+160.75
220 Stephania1.742.957.631.12.1E+160.56
234 Barbara1.802.9715.443.85.7E+161.12
1108 Demeter1.803.0524.925.61.1E+160.38
916 America1.812.9211.133.22.5E+160.64
783 Nora1.812.889.340.04.4E+160.93
584 Semiramis1.822.9310.754.01.1E+171.70
219 Thusnelda1.832.8810.840.64.5E+160.96
284 Amalia1.832.888.148.77.8E+161.38

This list is somewhat arbitrary, so I'll also give the entire list.  If you want, you can search for the number and get information in public databases.  These are all the candidates that met the selection criteria:  67, 585, 198, 142, 584, 284, 261, 270, 186, 248, 432, 306, 113, 138, 126, 1963, 161, 623, 234, 182, 118, 435, 219, 131, 136, 783, 282, 495, 302, 877, 556, 189, 732, 474, 930, 323, 178, 376, 249, 169, 916, 757, 220, 1244, 1159, 572, 273, 1650, 917, 364, 565, 853, 443, 470, 1108, 1296, 908, 3345

If you continue to go further out from 2.5 AU, there are many many more.  Right now I'm not interested in those because, for one, they're more difficult to access from Earth, and two, they don't receive as much light.  Even at a distance of 2.0 AU, you receive 1/4th the sunlight as Earth.  Being able to continuously generate solar power without interruption from night is a benefit, but it may only break even the the common isolation values of about 208 W/m2 at 2.5 AU.  Everything in this post is intended to answer the question "where would you build a gravity balloon".  Phobos, Eros, and the above group are an obvious starting place, but beyond those it is less obvious.

It's hard to move something as large as the asteroids I'm talking about here.  NASA is currently working on a plan where they'll bring a near-Earth asteroid into orbit around the Earth-moon system.  But this plan calls for moving an object 7 meters in diameter that is already near-Earth.  For a gravity balloon, we're talking about objects 20 km in diameter, that are not near Earth.  The scope of moving such a thing is orders of magnitude beyond reasonability, and I place in the same fesiability category as terraforming - scale being in the millions of years.  So we're stuck with working with them in-place.

In order to give a good technical argument, I wanted to give the Delta v budget associated with accessing these places.  A complication, however, is that there isn't a single value associated with this.  To get from Earth to the surface of Phobos, for instance, one would have to travel out of Earth's gravity well, work against the gravity of the sun to get to Mars, descend to Phobos orbit, and then climb down Phobos' gravity well.  These are 4 components but the first one is identical for all possible locations.  Because of that I'm only going to focus on the latter 3, for purposes of comparison between the candidates.
• Delta-V needed to get to its orbit around the sun
• Delta-V needed to get to its orbit around its planet (in the case of a moon)
• Delta-V needed to descend to its surface
The first one is the real difficult one to calculate.  For that, I'm using the Hohmann transfer orbit equations from Wikipedia.  I'm combining the terms for the initial and final thrusts, giving the following expression.

$$\Delta v_{solar} = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+r_2}} - 1 \right) + \sqrt{\frac{\mu}{r_2}} \left( 1 - \sqrt{\frac{2 r_1}{r_1+r_2}}\,\! \right)$$
In this calculation, r1 will be the orbiting radius of Earth, because we're interested in missions starting from Earth, and r2 will be the semi-major axis of the destination object, which I've already tabulated.

For the case of reaching a planet's moon, we need a variation on the above formula.  To do this, I'm equating the r2 value to infinity and r1 to the orbit of the moon.  This isn't very accurate because it ignores everything about gravity assists and probably aerobraking, so it should be taken with a grain of salt.

$$\Delta v_{planet} = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{ 2 } - 1 \right)$$
To get the delta v for reaching or leaving the surface of an asteroid (with no atmosphere), that's just the formula for escape velocity.

$$v_e = \sqrt{ \frac{ 2 G M }{ r } }$$

With all these, we can get a simple table of the Delta-Vs.  In a simplistic sense, the values are addative, if you consider that a spaceship has to get to one location, stop, then go to the next.  Real life isn't exactly so simple, but this is the best I will do for now.  We also need to qualify that significant work is needed to get away from the Earth in the first place.  Low Earth Orbit costs about 10 km/s to begin with, and then from there to escape is on the order of 4 km/s.  Due to the rocket equation, however, accelerating 2 kg to 4 km/s is much much easier than accelerating 1 kg to 8 km/s.

Delta-V in (km/s) for sun, planet, and the object
ObjectR (km)M (kg)amoon R (km)sunplanetself
Phobos11.11.07E+161.529,377 5.5890.8850.011
433 Eros8.46.69E+151.46-5.07200.010
323 Brucia17.93.13E+162.38-10.02700.015
220 Stephania15.62.05E+162.35-9.90400.013
234 Barbara21.95.70E+162.39-10.04100.019
1108 Demeter12.81.14E+162.43-10.17800.011
916 America16.62.50E+162.37-9.96700.014
783 Nora20.04.36E+162.34-9.89400.017
584 Semiramis27.01.07E+172.37-9.99500.023
219 Thusnelda20.34.54E+162.35-9.92800.017
284 Amalia24.37.84E+162.47-10.32000.021

The main takeaway from this chart is that the propulsion needed to travel from Earth to the object dominates.  The value for the stellar Delta V for Phobos is just that of Mars.  Compare to a more accurate Delta-V, which includes some of the more harry real aspects of the real solar system.  The sum of all the transfers to get to Mars, I find comes out to 0.7+0.6+0.9 = 2.2.  This is much lower than the 5.5 I found, which is unsurprising considering the factors I didn't consider.  This also begs an interesting question, as to whether travel to Eros would actually be easier than travel to Phobos, and it looks like not, unless you are possibly considering the effort to get back to Earth (which should be more for Phobos).

Inflatability

So far I've only considered the challenge of getting to an asteroid, drilling to the center, and setting up shop there no matter how big of a volume.  This wouldn't be very useful in the long run unless the favorable scaling factors over rigid pressure boundaries could be exploited at some point.  Thus, it's a good question to ask how large of a volume could you get out of these objects before your pressure limits stopped you.  I've already covered the mathematics of doing this in previous posts.

The requirement of the maximum size will be that it has a pressure equal to what the Skylab space station had, about 34 kPa.  Then with some root finding routines, the inner radius that gives this pressure (given the object's mass and density) is found.  That then implies the internal volume.  This is fairly straightforward, and here are the numbers I obtained.

ObjectM (kg)rho (kg/m3)R inner (km)V (km3)
Phobos1.07E+1618764.74447
433 Eros6.69E+1526704.33340
323 Brucia3.13E+1613009.843993
220 Stephania2.05E+1613006.13966
234 Barbara5.70E+16130016.1417620
1108 Demeter1.14E+1613001.4312
916 America2.50E+1613007.811994
783 Nora4.36E+16130013.159525
584 Semiramis1.07E+17130024.7863773
219 Thusnelda4.54E+16130013.5810489
284 Amalia7.84E+16130020.2034507

I wanted a good way to illustrate this, since a large inner volume is the entire point of the gravitational balloon.  In the following illustration, I made the size of a dot proportional to the inner volume in a graph of the mass versus inner radius.  This means that the volume follows directly from the radius, as per 4/3 Pi R3, but the inner radius isn't a direct function of the mass because the densities of the objects can be different.  Indeed, this is why Phobos and Eros are outliers.  I had actual densities for those two while the inner-belt asteroids were assumed to be the same, so they lie on the same (imaginary) curve.  You can even see how Eros maintains almost the same volume (dot size) as Phobos, in spite of having much less mass.  This is because the effects of self-gravitation are greater for Eros, because it has a higher density.

The takeaway from this graph staring you in the face is that the potential size of a habitat grows very fast with size of the asteroid you use.  About half of the potential volume comes from Semiramis alone.  That's not even the largest in the group, out of the larger group of 50 or so asteroids in the inner edge of the asteroid belt, there are candidates much much larger than that.  If you've read my previous posts, you may be wondering how Semiramis can give such a large volume compared to my original large reference case in the introduction.  That's because I used a pressure of 0.8 atm in that case, and 0.34 atm here.  The latter might be somewhat uncomfortable, it's hard to say, but for the most near-term prospects of in-situ resource utilization, it's probably the most reasonable assumption.  It would be much smaller if you required 1 atmosphere, and Phobos and Eros would be entirely unusable.

In closing, I discovered a few things that I wasn't entirely certain of before.  Firstly, Phobos and Eros are quite inflatable if you use an atmosphere with reduced Nitrogen concentration.  In fact, at around 9 km inner diameter that would be quite spacious, larger than any other space habitat would could possibly hope to build, and large enough to build artificial gravity tubes.  Also, I was surprised by the sudden increase in density of large asteroids at around 2.2 AU.  Obviously this is because of the clearing effect of Mars, but it's still interesting.  There are lots and lots of options around there.  Finally, the difficulty to travel to those asteroid belt candidates is much more burdensome than the Eros or Phobos options.  Those two are probably the best we've got for anything near-term in the absence of other information.  But that other information could be substantial.  You would obviously have your pick of compositions if you went as far as the asteroid belt and that could also reduce the amount of equipment that has to be hauled in the journey, as well as the difficulty of actually manufacturing the atmosphere.

## Sunday, April 14, 2013

### Turbulent Friction Buffers

In the last post on the friction buffers, I outlined a case for a rotating structure with a diameter of 500 meters where 4 W/m2 of power input was needed and 100 friction buffers were needed. The figure of 4 W/m^2 certainly sounds reasonable, but perhaps the 100 sheets doesn't.  These are no more complicated than a simple sheet, but since the stability criteria isn't understood it's not quite understood how demanding their construction would be, and that many sheets sounds burdensome.  So far I've assumed that the flow must be kept laminar.  In order to be realistic we need to imagine turbulent movement of air between the buffers, thus improving efficiency by using space more liberally.

We will want a small number of stages.  We can use space up to a scale of a similar size to the tube radius itself.  That leads to a mathematical problem I would like to define.  If we extended the region of the friction buffer to a width of 100 meters or more, and kept the power dissipation the same, how many stages would be needed?

Formalizing the Problem

The fluid problem isn't very difficult to set up, as it's simply a form of the law of the wall.  The selection of the reference frame, however, needs to be made clear.  I opted for an approach of taking the average speed between two stages (call them n and n+1) as the reference frame.  Then, the fluid halfway between the two stages is roughly stationary.  This is then identical to the case of open flow within a channel up to that height.  This is how I look at it:

To get the friction factor, and thus the power dissipated, we apply the Colebrook-White formula, for the specific case of free flow over a wall.  This is why I was posting routines for numeric solution of this in the last two posts.

$$\frac{1}{\sqrt{f_{D}}} = -2 \log \left(\frac{\varepsilon}{12R_\mathrm{h}} + \frac{2.51}{\mathrm{Re}\sqrt{f_{D}}}\right)$$

The important thing to note here is that this applies for half of the channel between two sheets, as illustrated above.  From that information and the notations here, it's simple to find the Reynolds number.  For purposes of this post, I'll take the roughness to be zero.  Then the friction factor follows from the Colebrook-White formula.  Once you have that, you can find the sheer pressure.  Then the power for the given stage.  The total power dissipation is the sheer pressure times the velocity of the rotating structure itself.  This is because the retarding torque for each torque is exactly the same.  Summing up the equations we have:

(these equations are under construction as they are fully repaired)

Going From Total Tube Dimensions to Layers

$$d = \frac{ \text{dist} }{ 2 N }$$

$$U = \frac{ U_{total} }{ 2 N }$$

Friction Equations for Single Layer

$$Re = \frac{ U d }{ \nu }$$

$$4 f_{D} = f$$

Shear Pressure

$$\tau = f \rho U^2 \frac{1}{2} = 4 f_{D} \rho \left( \frac{U_{total}}{2N} \right)^2 \frac{1}{2}$$

Power per Area

$$\text{Power} = U_{total} \tau$$

This method allows use to relate the total power dissipation with all the other parameters.  Now we can ask questions such as "how many friction buffers do you need to keep power under so-many Watts using so-much space?"

Realistic Design of Friction Buffers

I'll continue to use the reference design of a rotating habitat with a diameter of 500 meters.  Reason being that it's the most straightforward way to establish feasibility.  Then we need to suppose some number of friction buffer stages and distance between them.  To start with, I imagined an area for the friction buffers equal to the total area inside of the tube.  This sound somewhat reasonable from a packing efficiency perspective.  That means that the outer radius of the structure, including the friction buffers, will be the square root of 2 times 250 meter inner radius.  I still haven't specified the number of friction buffers, so here is the power dissipation with different numbers of stages.

You can see that the power dissipated is very sensitive to the number of stages used.  According to my previous arguments, some small number of Watts per square meter is probably reasonable due to economic arguments, and comparison to a lightbulb.  This shows that around 15 or 16 stages would probably be reasonable toward this goal.

In terms of proportions, how would this look?  I made this image to illustrate the relative sizes.

The idea here is that the inner area is the habitable part of the structure and all the black circles around it rotate and successively lower speeds, which reduces energy consumption to just a few watts per square meter to keep it turning.  This does sound at least somewhat reasonable.  That's a lot of area to manufacture of the friction buffers, but it seems workable.  A person in the above image would be roughly 1.4 pixels tall.  So this is a pretty substantial size, and it would be fairly energetically cheap.

But let's say that we want to push the limits even further?  What if we use more space for the friction buffers?  I set the number of friction buffers as constant at 10, and then looked at the impact.

Here, you can see that using more area is much less effective than adding more friction buffer stages (although it would be easier).  Of course, at some point, increasing the outer radius more will increase the circumference so much that it won't actually save you any more material.  It's difficult to imagine the friction buffers would make sense at any significant multiple of the radius.  So what I've described here is definitely in the neighborhood of the optimal configuration that a space-faring civilization would pick.

These calculations have really helped me zero in on what this type of civilization would look like.  For what it's worth, the solar panels on the International Space Station produce about 120 kW.  This amount of power would sustain the rotation of about 30,000 square meters of land in a gravitational balloon.  Even at a (somewhat) rural population density of 250 people per square km, this would support about 8 people.

Even in the most extreme circumstances, the power needed to keep these artificial gravity structures rotating wouldn't be that much compared to the other energy needs of the people.  At least that's the case if you used the above 16 friction buffer layout.  Inhabitants might opt to use fewer at the gain of more energy.  Either way, the point I'm making is that these calculations clearly show that the energy wouldn't be a deal breaker.