Saturday, March 8, 2014

Murphy's reply

In general, Murphy's Do The Math crowd don't do math. Rather they appeal to Murphy's authority and hurl insults. Mike Stasse's forum was no exception:

Mike Stasse and his friend Maponos

Is Murphy an authority? Does his PhD means he's qualified?

I responded with The Most Common Delta V error. High school seniors typically mispatch conics the same way Murphy does. Murphy's level of expertise is somewhere below Orbital Mechanics 101 for liberal arts majors.

Stasse passed this on to Tom Murphy himself. And got a reply. Stasse quotes Murphy:
I don't dispute the more careful approach used on hopsblog. I put pieces together very simply, which may not represent more clever ways to manage interplanetary trajectories. That said, I stated clearly what I was doing, so that it's an easy job to pick it apart. I'm fine with that. I hope I never appealed to my authority as an orbital mechanics expert, because I am not
Did you get that, Stasse? By his own admission, Murphy's no expert. Perhaps Murphy hasn't appealed to his authority. But Stasse and his friends certainly have.

Murphy's quote goes on:
I just try to put scales on things and sort out roughly how hard things are. At the pace of a post a week (during that time)--on top of a busy job--I could not spend time polishing. 15 km/s (allowing a bit of rounding) is still frikin' hard, so my main point is barely scratched.
Murphy and his apologists typically claim his approximations aren't that far off and his points remain intact.

Murphy's point in this particular example doesn't stand. 15 km/s is about what it takes to put a geosynchronous communication satellite in place. This is doable as demonstrated by the large number of geosynchronous sats. Murphy's 20 km/s is about what it takes to land on the moon's surface and come back. 15 km/s and 20 km/s are vastly different delta V budgets.

But 15 km/s vs 20 km/s is isn't the worst Murphy error. He's done much worse. From Grab That Asteroid! in Murphy's Stranded Resources post:

The asteroid belt is over 20 km/s away in terms of velocity impulse. If the goal is to use the raw materials for production on Earth or in Earth orbit, we have to supply about 10 km/s of impulse. We would probably try to get lucky and find a nickel-metal asteroid in an unusual orbit requiring substantially less energy to reel it in. So let's say we can find something requiring only 5 km/s of delta-v.  . . .

To get this asteroid moving at 5 km/s with conventional rocket fuel (or any "fuel" that involves spitting the mass elements/ions out at high speed) would require a mass of fuel approximately twice that of the asteroid. As an example, using methane and oxygen,  . . .

Does fetching an asteroid take twice the rock's mass in propellent?

Ratio of propellent to dry mass can be found with Tsiolkovsky's rocket equation:
(Mass propellent)/(dry mass) = e(delta V/exhaust velocity) - 1

Let's see -- in Murphy's example delta V is 5 km/s. Exhaust velocity of oxygen and methane is about 3.4 km/s.

e(5 km/s / 3.4 km/s) - 1 = 3.35. So for every ton of asteroid, we'd need more than 3 tons of propellent. At first glance it looks like Murphy is being kind and even under estimating propellent needed.

But methane and oxygen isn't the only propellent. Xenon from an ion engine has an exhaust velocity of around 30 km/s.

e(5 km/s / 30 km/s) - 1 = .18

So about .2 tonnes (or 200 kilograms) of propellent to park a tonne of asteroid. 2/10 is not a "rough approximation" of 2.

But Murphy's error gets worse.

Murphy thinks we'd be lucky to find an asteroid outside of the Main Belt that takes 5 km/s to retrieve. Evidently he hasn't heard of Near Earth Asteroids. There are many asteroids that take much less.

The Keck study for retrieving an asteroid notes some asteroids take as little as .17 km/s. Let's plug in .17 km/s delta V:

e(.17 km/s / 30 km/s) - 1 = .006

So 6 kilograms of propellent to park a tonne of Asteroid. Now Murphy's guesstimate of twice the asteroid's mass is off by a factor of about 350.

Six kilograms is about the mass of two chihuahuas. Two tons is about the mass of two large horses, big horses as in Budweiser clydesdales.

Tom Murphy is a busy guy. So he uses furious handwaving to excuse info that's completely wrong.

Murphy's figures are often off by several orders of magnitude his PhD notwithstanding.

Neither Stasse's appeal to authority nor Murphy's "rough approximation" defense salvage Murphy's arguments.

Monday, February 24, 2014

The most common delta V error

Patching Conics
Time and time again I've watched people patch conics by using straight addition and ignoring the Oberth benefit. It is a very easy mistake to make. I'll give an example of this error for an earth to Mars delta V budget.

At perihelion an earth to Mars Hohmann orbit is moving about 33 km/s. 3 km/s is needed to leave earth's 30 km/s heliocentric orbit and enter this transfer orbit. In a similar fashion it takes 2.5 km/s to leave Hohmann transfer and match velocities with Mars.

But before we go from one heliocentric orbit to another, we need to escape the planet's gravity well. Earth's surface escape velocity is about 11 km/s. Mars' surface escape velocity is about 5 km/s.

The novice will look at these 4 quantities and simply add them. 11 + 3 + 2.5 + 5 is 20.5. They'll tell you it takes about 20.5 km/s to get from earth's surface to Mars surface.

But to accurately patch conics you need to use the hyperbolic orbit that takes you out of the planet's sphere of influence to a heliocentric orbit. Hyperbolic orbit speed is sqrt(Vescape2 + Vinfinity2). But what's Vinfinity? In this example it's the 3 km/s needed to go from earth's 30 km/s orbit to a Hohmann's 33 km/s. In Mars' neighborhood Vinfinity is the 2.5 km/s needed to exit Hohmann and match velocities with Mars.

If you remember high school math, sqrt( a2 +  b) should look familiar. It's the hypotenuse in the good old Pythagorean theorem! And that's what I use to visualize hyperbola speed:

The novice will tell once you've achieved the 11 km/s to escape earth's gravity well, you need another 3 km/s. The informed will tell you only another .4 km/s is needed.

For Mars to Earth the naive will tell you after you've reached 5 km/s to escape Mars then you need another 2.5 km/s to send the ship earthward. The savvy will tell you an additional .6 km/s is needed.

.4 vs 3 and .6 vs 2.5. In this case the novice method results in a 4.5 km/s overestimation of the delta V budget.

Erik Max Francis
I used to call this the Erik Max Francis Error. Delta V budgets from one planet to another would often come up in space usenet groups. Erik would use his Python BOTEC and give an answer to ten decimal places. People would ooooh and ahhhhh. Wow! Accurate to 10 significant figures! In reality Erik's answers were accurate to zero significant figures. I finally prodded him to correct his error. Now his BOTEC is accurate to 1 or 2 significant figures. So far as I know, he still gives answers to 10 decimal places.

Later I called it the Rune error. The total Vinf can be roughly estimated by subtracting Mars 24 km/s from earth's 30 km/s. And this is what Rune uses on the New Mars forum when Louis asks how much delta V is needed after you get out of earth's gravity well.
Brute force" trajectories would take about as much delta-v as is the difference between the orbital speeds of mars and earth, so about 29.8km/s (for earth) - 24km/s (for mars) = 5.8km/s
I have explained to the New Mars Forums many times that the speed of a hyperbola is sqrt(Vescape2 + Vinfinity2). Will Rune ever learn it? I doubt it.

But no matter, Rune's a member of Zubrin's cult. People expect Zubrinistas to be innumerate, nobody takes  them seriously. But sadly they are loud and high profile. John Q Public can be misled into thinking they speak for all space advocates.

It's more damaging when someone in authority commits this error.  Now I'm talking about Dr. Tom Murphy.

Professor Tom Murphy
This is from a graphic from Tom Murphy's Stranded Resources:

Murphy writes:
For instance, we travel around the Sun at a velocity of 30 km/s, while Mars sails at a more sedate 24 km/s. So to meet up with Mars, we have 6 km/s of extra velocity to burn, helping us up the hill. We speak of this as a Δv (delta-vee) adjustment to trajectory.
Same method as Rune for getting the total Vinf: Subtracting Mars' 24 km/s from earth's 30 km/s to get 6 km/s. An over estimation but not wildly inaccurate. And Murphy correctly shows earth's escape as 11 km/s and Mars escape as about 5 km/s.

But then Murphy straight up adds 11+6+5:

Crudely speaking, we must have the means to accomplish all vertical traverses in order to make a trip. For instance, landing on Mars from Earth requires about 17 km/s of climb, followed by a controlled 5 km/s of deceleration for the descent. Thus it takes something like 20 km/s of capability to land on Mars
Sounds like he's being generous to the poor deluded space cadets by rounding 22 down to 20. But the distance from earth's C3=0 to Mars C3=0 is not 6 km/s. It's about 1 km/s. Here's Murphy graph corrected for the Oberth benefit:

For comparison, Murphy's erroneous graph is left in but a shade lighter. From surface of Earth to Surface of Mars is about 16 km/s.

"But wait!" a Murphy apologist might say. "Murphy didn't include the delta V needed to rise above earth's atmosphere. That's 1.5 to 2 km/s! That makes the budget more like 18 which can be rounded up to 20."

An atmosphere does indeed add to delta V for departing a planet. On the other hand, an atmosphere is a big help for planet arrival. Park in a capture orbit with periapsis velocity just a hair under escape. Position the periapsis in the planet's upper atmosphere. Each orbit at periapsis, atmospheric friction slows the ship. This is known as aerobraking. Almost all of Mars' 5 km/s descent can dealt with via aerobraking. Here is Murphy's graph corrected for atmospheric influence:

Now it takes 13 or 14 km's to reach escape. But with descent taken care of with aerobraking it only takes another 1 km/s to reach Mars' surface. A more realistic delta V budget from earth surface to Mars surface is about 14 or 15 km/s.

And in fact numerous Mars landers and orbiters have used this method. I am stunned that Murphy, a self proclaimed space insider, has never heard of aerobraking.

Is a 6 km/s error a big deal? Since the exponent of the rocket equation scales with delta V, it's a very big deal. Murphy himself would tell you exponential growth can be dramatic.

Above graph assumes hydrogen/oxygen bipropellent. Each 3 km/s added to the delta V budget about doubles propellent needed. Each 5 km/s added nearly triples the amount. Murphy's 20 km/s delta V budget would need a little more than triple the propellent of the actual 15 km/s delta V budget.

In my opinion the limits to growth is the most important issue facing us. Can space resources raise the ceiling on our logistic growth? If so, it is worthwhile to invest in building space infra-structure. If not, expensive space infrastructure is a waste of money. We should look at the question seriously. That is why I get bent out of shape when a so-called authority makes common mistakes that would embarrass a freshman aerospace student.

Tom Murphy's arguments against space are often cited in discussions of limits to growth. When I find such a discussion, I will chime in that a bright high school student could tear apart Murphy's arguments. The most recent visit was at Mike Stasse's Damn The Matrix. As usual, the Do The Math crowd response was insults, appeal to authority, but no math.

Judging by Murphy's fans, his blog should be renamed "Don't do the Math. Take my word for it because I'm a Ph. D."

Saturday, February 1, 2014

Terraforming Mars vs Orbital Habs

Those who advocate Mars settlement like to say Mars can be terraformed. First I will take a look at what it would take to terraform Mars.

How much air do we need to add to Mars?

From NASA's Mars Fact Sheet, surface density of the Martian atmosphere is about .02 kg/m3. That is about 1.5% of Earth's surface air pressure of 1.27 kg/m3. Mars' atmosphere is virtually a vacuum.

Mars surface gravity is about 38% earth gravity. That means given an atmosphere of comparable temperature and composition, Mars atmosphere scale height is 264% earth atmosphere scale height. But Mars surface area is about about 28% that of earth's. 2.64 * .28 is about .75. To get comparable air density, we would need Mars' atmospheric mass to be about three quarters that of earth's atmosphere.

The total mass of the Martian atmosphere is about 2.5 x 1016 kg. Earth's atmosphere is about 5 x 1018 kg. So to make Mars surface air density earth like, we'd need 3.6 x 1018 kg of air added to Mars.

But do we need sea level air density? No, there are people who survive at higher elevations. This list of the world's highest cities show several places at around 5000 meter elevation. Granted the dwellers of the highest city La Rinconada, Peru don't live comfortably. But they demonstrate humans can endure air density half that of sea level. If half is sufficient, Mars only needs 1.8 x 1018 additional kilograms of air.

Would be Mars terraformers like to point at the frozen CO2 at the Martian poles. If Mars temperature is raised just a little, they hope the vaporized carbon dioxide would create a greenhouse effect that would cause more carbon dioxide to be vaporized. Their hope is that a runaway greenhouse effect could substantially boost Mars' atmosphere from frozen volatiles already in place.

According to Wikipedia, there is thought to be a 1 meter thick layer of CO2 at Mars north pole, a cap about 1,000,000 meters in diameter. At the south pole there is an 8 meter thick layer of CO2 over a cap having a 350,000 meter diameter. That's about 1.6 x 1012 cubic meters of CO2. Dry ice has a density of 1.6 thousand kg/m3. If all of that CO2 is vaporized (an optimistic assumption) that totals about 2.5 x 1015 kg of atmosphere. Short by almost 3 orders of magnitude, a miniscule contribution toward the needed 1.8 x 1018 needed kilograms.

Zubrin and McKay believe runaway greenhouse could boost Mars atmosphere to 300 to 600 millibars. Besides the polar dry ice, they also mention CO2 in Martian regolith. I believe most of Zubrin's optmistic estimates are influenced more by wishful thinking than hard data. But for the sake of argument I'll grant 300 millibars of CO2. 300 millibars of CO2 is not breathable. But let's say green plants combine Martian water and CO2 to make sugars and starches plus oxygen. Taking the carbon out of 300 millibars of CO2 leaves about 220 millibars of oxygen. Earth's 1000 millibar atmosphere is 1/5 oxygen, so perhaps a 220 millibar oxygen atmosphere would be breathable. But it would also be an extreme fire hazard. Apollo 1 taught us a pure oxygen atmosphere isn't a good idea.

Even with Zubrin's very optimistic scenario, it seems we'd still need to import 1.5 1x 1018 kilograms of nitrogen.

Can we add to Mars' air with comets?

Zubrin and McKay suggest  it'd take .3 km/s to nudge an ammonia asteroid in the outer solar system towards Saturn and then Saturn's gravity could throw the ammonia snowball Marsward.

"Consider an asteroid made of frozen ammonia with a mass of 10 billion tonnes orbiting the sun at a distance of 12 AU. Such an object, if spherical, would have a diameter of about 2.6 km, and changing its orbit to intersect Saturn's (where it could get a trans-Mars gravity assist) would require a DV of 0.3 km/s. If a quartet of 5000 MW nuclear thermal rocket engines powered by either fission or fusion were used to heat some of its ammonia up to 2200 K (5000 MW fission NTRs operating at 2500 K were tested in the 1960s), they would produce an exhaust velocity of 4 km/s, which would allow them to move the asteroid onto its required course using only 8% of its material as propellant. Ten years of steady thrusting would be required, followed by a about a 20 year coast to impact. When the object hit Mars, the energy released would be about 10 TW-years, enough to melt 1 trillion tonnes of water (a lake 140 km on a side and 50 meters deep). In addition, the ammonia released by a single such object would raise the planet's temperature by about 3 degrees centigrade and form a shield that would effectively mask the planet's surface from ultraviolet radiation. As further missions proceeded, the planet's temperature could be increased globally in accord with the data shown in Fig. 12. Forty such missions would double the nitrogen content of Mars' atmosphere by direct importation, and could produce much more if some of the asteroids were targeted to hit beds of nitrates, which they would volatilize into nitrogen and oxygen upon impact. If one such mission were launched per year, within half a century or so most of Mars would have a temperate climate, and enough water would have been melted to cover a quarter of the planet with a layer of water 1 m deep."

This scheme presupposes we could land a 20 gigawatt power source on a rock in the outer solar solar system. For comparison the Palo Verde Nuclear Power Plant, the largest nuclear power plant in the United States, produces about 3.3 gigawatts. So we're sending 6 Palo Verde Nuclear Power Plants out past Saturn. McCay's scheme stipulates using the comet's mass as reaction mass. So now we have a mining and transportation infra structure on the comet that digs up the ice and places this reaction mass in the nuclear rocket engine.

If we have the wherewithal to establish such infrastructure, we certainly have the ability to build habs on these rocks.

Asteroidal Real Estate

How much asteroidal real estate could 1.5 1x 1018 kilograms of air give us? An O'Neill cylinder 8 kilometers in diameter and 32 kilometers long would give us 804 square kilometers of real estate. Such a cylinder would have a volume of 1.6e12 cubic meters. On earth's surface, our air has a density of about 1.27 kg per cubic meter. So that volume at 1 bar density would be 2e12 kilograms of air.

1.5e18/2e12 = 750,000. Three quarters of a million O'Neill habitats. Recall each cylinder has 804 square kilometers of real estate. 750,000 * 804 km2 = 603 million km2. Mars' surface area is 145 million km2. So if we put the asteroidal resources to use where they're at, we get 4 times as much real estate.

Some would point out that O'Neill cylinders are very extravagant pieces of mega-engineering. I completely agree! It's my belief that humans don't need a full g to be healthy, I believe .4 g (a little more than Mars' gravity) would suffice. In which case the hab radius could be 1.6 km. Such a hab would have only  321 km2 of real estate but a volume only 2.6e11 cubic meters. 2.6e11 m3 * 1.27 kg/m3 = 3.3e11 kilograms. 1.5e18/3.3e11 = ~4.5 million. 4.5 million of the smaller O'Neill habitats. 4.5 million * 321 = 1460 million square kilometers. Or about as much real estate as 10 Mars planets.

If the goal is to provide more real estate and resources for humanity, terraforming Mars is an extravagant waste. We should ditch planetary chauvinism and go for the small bodies.

Robert Walker also takes a look at terraforming Mars.

Saturday, December 28, 2013

EML2 plane change

Most of my models use circular coplanar orbits. But Jon Goff points out many asteroids have a healthy inclination. Departing for an asteroid from the moon's orbital plane often involves a big plane change. And plane changes can be expensive (see comments in What about Mr. Oberth?)

But easy plane changes is a big reason I love EML2. This takes some explaining.

Let's look at a 60 degree plane change but no change in speed. I pick 60º because it's easy-- the original and new velocity vector as well as the delta V vector are all sides of an equilateral triangle.

If your speed is about 8 km/s (as in low earth orbit), a 60 degree plane change costs about 8 km/s.

With higher orbits, plane changes are cheaper. At GEO (about 36,000 km above earth's surface), orbit speed is about 3 km/s and a 60 degree plane change costs 3 km/s.

EML2 is about 63,000 kilometers above the moon's surface. It's moving about .17 km/s with regard to the moon. So a 60 degree plane change at EML2 costs .17 km/s:

But wait, it gets even better!

My favorite route from LEO to EML2 is the one found by Robert Farquhar:

The orbit is time reversible. A .15 km/s braking burn at EML2 cuts speed with regard to the moon to .02 km/s. The allows the ship to fall deep into the moon's gravity well. With the a little Oberth help at perilune, another .18 km/s suffices to send the ship to an 182 km perigee.

Here's a single burn at EML2 that cuts speed to .02 km/s as well as doing a 60 degree plane change:

The .16 km/s braking/plane change burn is only .01 km/s more than the .15 km/s coplanar burn Farquhar calls for.

Where else can you get a 60º plane change for only .01 km/s?

EML2 has a tiny C3 with regard to both the moon and earth. This confers a big Oberth advantage. And the easy plane change is icing on the cake.

Edit: Isaac Kuo has pointed out that earth has a 30 km/s vector coplanar to the ecliptic plane. A hyperbolic orbit departing earth might have a v infinity of 3 or 4 km/s. When you add a 3 or 4 km/s vector tilted 60 degrees to the 30 km/s vector, you are only inclined 4 or 5 degrees from the ecliptic plane. Still, this is a substantial plane change! This post is incomplete as I've only looked at departure from EML2. Hopefully, I'll soon have time to look at burns at perilune and perigee burns and the possible trans asteroid orbits.

Friday, December 27, 2013

Who needs humans?

This is in response to Quantum G's question "Why do humans need to return to the Moon to get resources to make "consumables and propellant", if robots can be sent to do that instead?"

Just let autonomous and/or teleoperated robots do all the work. Who needs humans?

Quantum G should try working in an actual mine. As an ASU student, I spent four summers working in the Phelps Dodge copper mine in Ajo, Arizona. At the top of every bulletin board was Murphy's Law: "What Can Go Wrong, Will."  And that was followed by many variations and corollaries of Murphy's Law.

Unlike a factory floor, mines are an uncontrolled, unpredictable environment. The unexpected can and does happen. When it does, human ingenuity is called for. You cannot write algorithms that anticipate every unforeseen problem.

Not that I'm against robots. See
Puppets, Telerobots & James Cameron,
Surgical Robots, and
Give NASA's SLS money to DARPA.
I believe improved robotics will be a major game changer when it comes to exploitation of space resources.

The moon is more amenable to tele robots than most locations in our solar system. At 384,400 kilometers from earth's surface, light lag latency is only 3 seconds. Since signal strength falls with inverse square of distance, lunar tele robots would enjoy much better bandwidth than machines on remote asteroids or Mars. Good bandwidth is important for immersive tele-presence as well as control of agile, dexterous robots.

And there are technologies that can mitigate a 3 second reaction time. For example Big Dog's balance or Google Car's collision avoidance.

Even so, a multitude of tasks are much easier with constant sensory feedback in real time. Things like finding a dropped hex nut. A 3 second light lag can make normally quick and easy chores time consuming and difficult. Robots controlled by humans in neighboring habs would be much more able than bots controlled from earth's surface.

And then there's the question of maintenance. Who maintains the robots?

Here is an article on mining giant Rio Tinto's "autonomous" robots. These driverless trucks move back and forth along well maintained and predictable routes. And they are closely monitored by nearby humans. Machines in less predictable environments such as the shovels are still human operated. And all the machines, whether "autonomous" or human operated, are maintained by humans.

Mines sans humans are still well beyond the state of art for earthly mines, much less mines in environments where we have zero operating experience.

Robots may reduce the need for human presence. But they won't completely eliminate the need for humans, not for a long while.

There is also important information to be gained from humans on the moon. What gravity do humans need to stay healthy? As I mention in What's the minimum spin hab?, this is still not known. If the moon's 1/6 gravity keeps humans healthy, that makes minimum spin habs for asteroid workers more than six times less massive. It would also indicate humans are okay living with Martian gravity.

Friday, December 20, 2013

What's the minimum spin hab?

This post was prompted by Robert Walker's comment: "I wonder what anyone here thinks about my idea for rotating carousels to provide gravity in the lunar colony? Not rotating entire hab, but just a thin shell of living quarters inside it, in a bigger hab if say a couple of hundred meters across, greenhouse domed, have like the living habs around the outside rotating continuously - perhaps on a track or something like that - at just the right speed for 1 g for the inhabitants. Smaller habs just rotate the entire room - and easier to construct than e.g. fairground rides on Earth because of the low gravity."

We know 0 g results in bone loss and other problems. We need gravity, but how much?

Is 2/5 g (Mars gravity) sufficient to keep us healthy? Or 1/6 g (moon gravity)? This is still not known. Our only data points are 0 g and 1 g. If a full g is needed, people on Luna or Mars bases would indeed need living quarters on rotating carousels.

On the other hand, if lunar gravity is sufficient, no carousel is needed on the moon or Mars. That would also drasticly cut the minimum sized hab needed to keep workers healthy in a microgravity environment like on an asteroid.

The amount of artificial gravity felt in a spin hab is ω2r where ω is angular velocity in radians and r is spin hab radius. Obviously if 1/6 g does the job, the hab radius can be cut by a factor of 6.

Another quantity to look at is ω, angular velocity. Earlier it was believed 1 revolution per minute was the top angular velocity humans could comfortably endure. This combined with assuming a full gravity resulted in proposals like the behemoth Stanford Torus. (2 * pi / 60 seconds) * 894 meters = ~9.8 meters/second^2 or about 1 gravity.

If spinning doughnuts nearly 2 kilometers across are a prerequisite for asteroid miners, I wouldn't expect asteroid mining habs in this century or the next.

But is 1 revolution per minute really the top ω workers can endure? Research by James Lackner and Paul DiZio suggests workers could become acclimated to higher angular velocities. If workers can get used to 2 rpm that would cut needed radius four fold. 3 rpms would cut radius 9 fold. Here is a table showing hab radius (in meters) that be needed for various angular velocities and gravities:

1 g 5/6 g 2/3 g 1/2 g 1/3 g 1/6 g
1 rpm
2 rpm
3 rpm
4 rpm

If lunar gravity is sufficient and workers can get used to 4 rpms, a 9 meter radius hab does the job!

Obviously a 9 meter radius spin hab is more doable than a 900 meter radius Stanford Torus.

While we're talking about effects of microgravity, let's take a look at cosmonaut Valeri Polyakov. From a SpaceDaily article: "Polyakov's space flight had lasted 438 days (bettering a year by more than two-and-a-half months). Yet upon return, his health was not much different than other cosmonauts' after a long flight. After those first steps, he completely readapted to gravity within two months. Moreover, his bone loss had been very low, only around 7 percent in some of his weight-bearing bones," 

 Granted, only a small fraction of us have the self discipline to adhere to Polyakov's exercise regimen. But he demonsrates that exercise can mitigate microgravity bone loss.

If you hope for humans on surface of other planets or in asteroidal habs, it would be good to know what gravity humans need and what angular velocity they could get used to.

Saturday, December 14, 2013

Arrgh! It's not the cost of the fuel

"What's the cost of propellant from earth vs getting it from the moon?" always comes up in discussions of lunar water. Or the cost of near earth asteroid propellant vs earth propellant.

Propellant is cheap, typically a small percentage of spacecraft expense. Spaceflight is expensive because vehicles are disposable. How much would a plane ticket cost if a 747 were thrown away each trip?

Well, how come we don't re-use our spaceships? It's due to constraints imposed by the rocket equation.

As delta-V budget  climbs, dry mass fraction shrinks. We can't eliminate engine or payload mass. We cut dry mass by making walls thinner and structure more tenuous.

Thinner walls mean fragility. Designing upper stages is like designing egg shells.

Upper stages are like cascarónes, confetti eggs. While cascarónes are fragile by design, upper stages are fragile due to the constraints imposed by the rocket equation and high delta-V budgets. An upper stage plunging into the atmosphere is like a cascarón plunging onto a friend or relative's head. But the conditions of re-entering earth's atmosphere at 8 kilometers/second are much more extreme than the back of her mom's head.

Given propellant depots at LEO, GEO, and EML1 or 2, ferries between orbits would have delta V budgets of 4 km/s or less. Moving between orbits, they don't have to endure re-entry. Much less difficult mass fractions and eliminating the extreme conditions of re-entry make re-usable ferries doable.

But how would these ferries by fueled? Tankers from earth would have a delta V budget of at least 9.5 km/s. If the tankers are throw-away, it is simpler and cheaper to just use the tanker to deliver the payload rather than fueling a ferry to deliver a payload.

However if the fuel source is the moon's surface or an asteroid at EML1 or 2, the tankers have lower delta V budgets and thus much less difficult mass fractions. Given reusable tankers to supply fuel, reusable ferries make sense.

Moreover, given propellant in LEO, an upper stage returning to the earth's surface doesn't have to re-enter at 8 km/s. Given propellant in LEO it can refuel and shed some of it's orbital velocity via reaction mass instead of aerobraking. Eliminating the 8 km/s re-entry makes re-use of upper stages much less difficult.

So it's completely missing the point to compare the price of earthly propellant delivered to the moon's surface vs propellant mined on the moon. The object isn't to get water on the moon's surface. The object is to get propellant at various locations in cislunar space so the delta V budgets can be busted into manageable chunks.

By breaking the tyranny of the rocket equation, reusable ships become possible. Given easily reusable space ships, the economies of spaceflight are completely changed. This is the potential of lunar (or NEO) water.