## Wednesday, September 19, 2012

### Beanstalks, Elevators, Clarke Towers

Planetary Beanstalks

Arthur C. Clarke well described a space elevator in his novel The Fountains of Paradise:
In the very decade that the first satellite was launched ... one daring Russian engineer conceived a system that would make the rocket obsolete. It was years before anyone took Yuri Artsutanov seriously. ...
Go out of doors any clear night and you will see that commonplace wonder of our age — the stars that never rise or set, but are fixed motionless in the sky. We ... have long taken for granted the synchronous satellites ... which move about the equator at the same speed as the turning earth, and so hang foerever above the same spot.
The question Artsutanov asked himself had the childlike brilliance of true genius. A merely clever man could never have thought of it — or would have dismissed it instantly as absurd.
If the laws of celestial mechanics make it possible for an object to stay fixed in the sky, might it not be possible to lower a cable down to the surface, and so to establish an elevator system linking earth to space?
When you build a bridge, you start from the two ends and meet in the middle. With the Orbital tower, it would be the exact opposite. You have to build upward and downward simultaneously from the synchronous satellite, according to a careful program. The trick is to keep the structure's center of gravity always balanced at the stationary point. If you don't, it will move into the wrong orbit, and start drifting slowly around the earth.

Besides popularizing the notion of Artsutanov elevators based on geostationary orbit, Clarke also invented geostationary communication satellites.

What is the altitude of a stationary orbit? Speed  of a circular orbit is (Gm/r)1/2. Speed is also ωr, where ω is angular velocity in radians. For a geostationary orbit, ω would be 2 pi radians/sidereal day. Using these two equations we can find radius of a body's stationary orbit:

ωr = (Gm/r)1/2
ω2r2 = Gm/r
r3 =  Gm/ω2
r = (Gm/ω2)1/3

Altitude of the stationary orbit is the orbit's radius minus the body's radius.

Stationary orbit altitudes of a few inner system bodies:

 Body Stationary Altitude Vesta 265 km Ceres 706 km Mars 17030 km Earth 35784 km

Vesta has a low stationary orbit because of it's low mass and high ω.

There are two accelerations at play: gravity and inertia in a rotating frame (the so-called centrifugal force).

Centrifugal acceleration is ω2r and gravity is Gm/r2. We can choose our units so that Gm as well as ω are 1. Then the acceleration gradient can be graphed like this:

The slope is steeper below the geostationary orbit. For the above and below portions to balance, the pink area must equal the blue area. Assuming uniform thickness, the blue lengths above need to be longer than the red lengths below geostationary orbit.

A friend pointed out "But why assume a uniform strand? The part near synch has the most tension, and so is thickest in most designs,"  Thickened portions can be modeled as several strands. Each strand would need to be asymmetrical to balance.

A 108,000 km strand above geostationary would counterbalance the 36,000 km strand from geostationary to earth's surface.

Ratio of tether thickness at stationary altitude to thickness at planet surface is called the taper ratio. Taper ratio varies depending on Gm, ω as well as tensile strength and density of tether material. This Wikipedia chart gives tensile strength and density of various materials. Equations from this The physics of the space elevator by P. K. Aravind can be used to find altitude of the elevator top as well as taper ratios.

 Body Stationary Altitude (km) Top Altitude (km) Taper Kevlar Taper Bucky Tubes Vesta 265 665 1.01 1 Ceres 706 1922 1.02 1 Mars 17030 65774 45 1.1 Earth 35784 143772 2.6e8 1.62

Tide-locked Moons

It is also possible to build bean stalks from tide-locked moons. For tide-locked moons, the stationary starting points would be L1 and L2. Here are some tide-locked moons sorted by altitudes to L1 and L2:

 Body L1 (km) L2 (km) Phobos 3.25 3.27 Deimos 16.63 16.64 Io 8655.52 8831.64 Europa 11987.17 12172.28 Ganymede 28737.39 29362.77 Callisto 49749.25 48241.32 Luna 56292.23 62789.73

These numbers come from equations on pages 133 to 138 of Szebehely's "Theory of Orbits - The Restricted 3 Body Problem".

With tide-locked moons, there are three accelerations: 1) gravity of central body, 2) inertia in a rotating frame (aka centrifugal force, 3) gravity of moon.

The angular velocity ω is 2 pi radians/moon's orbital period. We can set time unit as orbital period/(2*pi), length unit as moon's orbit radius, and mass unit as mass of central body. Then ω and Gm are 1 and the accelerations can be graphed like this:

The constant k is ratio of moon's mass to central body mass. On the left side of moon orbit, moon pulls away from earth so moon acceleration is shown as positive. On the other side, the moon pulls stuff towards the earth, so moon acceleration  is negative.

To counterbalance, the blue strands extending away from the moon must be longer than red strands dangling towards the moon. the asymmetry is even more pronounced on the EML2 beanstalk.

A length extending 234,000 kilometers from EML1 earthward would balance a 57,000 kilometer length from EML1 to the moon's surface. Liftport proposes a Lunar elevator somewhat like this. An 11 tonne Zylon tether would extend 264,000 km from the moon's surface earthward. That's a little shy of the length needed but their diagrams indicate a counterweight at the earthward tether end.

Here's a picture of the Liftport proposal:

Speed at apogees of red ellipses match ω * r. So virtually no delta V is needed for rendezvous with tether at apogee. The tether is within 3 km/s of Low Earth Orbit (LEO) and 1 km/s of Geosynchronous Earth Orbit (GEO).

Jerome Pearson et al have talked about lunar elevators. They point out a counterweight near EML1 has few newtons per kilograms, so a weight in that neighborhood would need to be quite massive. The chart on upper right of page 7 of this pdf indicates the counterweight mass at 60,000 km would be between 100 and 1000 times the tether mass. Here is a more detailed lunar elevator pdf by Pearson and friends.

Phobos Elevator

At 1.08e16 kilograms, Phobos is a large momentum bank. A Phobos tether could catch or fling many payloads with little effect on its orbit.

Mars fans like to point out that Mars' shallow gravity well allows a beanstalk made of conventional materials like Kevlar. They suggest a Mars elevator could be a gateway to the resource rich Main Asteroid Belt. To sling payloads to Ceres, a Mars elevator would need to be at least 46,350 kilometers tall. Taper ratio for Kevlar would be 45.

In contrast a Phobos tether less than 14,000 km can fling stuff to Ceres.

Kevlar taper for a Phobos tether is about 8, less than 1/5 of the Mars tether's taper.

Here is a graphic comparing taper and length of Phobos and Mars elevators capable of slinging payloads to Ceres:

A Phobos elevator accomplishes many of the same goals for a small fraction of the materials. It doesn't descend to Mars' surface, however. The Phobos tether foot is moving about .6 km/s wrt Mars' surface. So a small suborbital hop would be needed for a Mars ascent vehicle to rendezvous with the Phobos tether foot.  A Mars lander departing from the tether foot would need to shed .6 km/s, much less difficult than the typical 6 km/s.

A Mars elevator would need to avoid Phobos as well as Deimos. Not a problem with a Phobos elevator. The top of the Phobos elevator is below Deimos' orbit. And of course a Phobos elevator doesn't have to worry about collisions with Phobos.

Given a Phobos tether and a Deimos tether, it is possible to travel between the two moons with virtually no delta V. If a payload is released 937 kilometers above Phobos, it will follow an ellipse whose apo-aerion is 2942 kilometers below Deimos. At this apo-aerion, the payload is traveling the same speed as the Deimos tether at that altitude.

The eccentricity of this ellipse is (1 - (ωDeimos/ωPhobos)1/2) / (1 + (ωDeimos/ωPhobos)1/2).

Ellipse Peri-aerion is (1 + e)1/3 * Phobos orbital radius.

Ellipse Apo-aerion is (1 - e)1/3 * Deimos orbital radius.

Given two co-planar tide-locked moons orbiting a planet, there can be similar transfer ellipses between tethers. I like to imagine a system of tide-locked moons about a gas giant using such tethers. The tethers would need to lie outside of the gas giant's rings, though. Else the debris flux from the ring would likely cut the beanstalk.

A little bit of nay-saying (added 11-18-2012)

I'm not embracing elevators as the panacea that will open the cosmos. There are problems. Problems should be examined.

Throughput

A Spaceward article The Space Elevator Feasibility Condition looks at throughput. Elevator cars and their cargo add to elevator mass but not tensile strength. So unless the cars are a tiny fraction of elevator mass, they'll boost the taper ratio. How fast can the elevator cars move? If their horse power comes from solar arrays on the car, they may move fairly slowly. The distances are huge, it could easily take a car months to climb to its destination.

Initially the space elevator material must be delivered with rockets. If the mass delivered by rockets is hundreds of times the mass an elevator can deliver in a year,

The Space Elevator Feasibility Condition notes that throughput might not be enough to even maintain an elevator.

The longer the elevator, the more serious the throughput problem. It'd be much less of an issue in the shorter elevators like the Phobos or Ceres elevator.

Debris

Orbital debris could sever an elevator. This is a big problem for an earth surface to GEO elevator. This elevator passes through LEO which has a high debris density and this debris is moving about 8 km/s with regard to the elevator.

Tether Experiments is a page listing various tether missions. One of the missions was SEDS-2, a 20 kilometer tether deployed "to see how long it would remain intact in the face of collisions with space dust and other orbital debris. ... it was cut after only four days"

The other elevators I've looked at occupy volumes with a lower debris density. And the orbital velocities are more leisurely so the debris flux is more tolerable. But even if an impact is a long shot, it's a concern if a very large investment is at risk.

Balancing act during construction

This is mostly directed at the Lunar elevator. The Liftport elevator starts at EML1 and sends tether ends simultaneously moonward and earthward. A slight nudge from EML1 can send a mass along a chaotic orbit, sometimes wildly chaotic. Station keeping is important. During construction, this balancing act must be maintained while one end is traveling approximately 200,000 kilometers and the other end 60,000 kilometers. After the elevator is anchored to the lunar surface, this station keeping isn't necessary but it's unclear how long it will take for the anchor to reach the moon's surface. If the anchor impacts the lunar surface at near lunar escape velocity, it would likely vaporize. If the lunar anchor is lowered gently, the duration of the balancing act would be prolonged.

----

I'm not attacking these notions. Quite the contrary, I believe criticisms from a thoughtful Devil's Advocate can help a worthwhile idea more than cheer leading.

AlanSE said...

I mentioned this post here:

http://imgur.com/gallery/6cdch

In short, the Phobos throw seemed dicey to me. It's not instantly obvious how you could dangle something on that tether without it swinging around everywhere. But with certain assumptions, it's clear that it CAN be done, even if there will be significant deflection of the tether. Logistically it would still be difficult, or at least rather involved.

Hop David said...

Alan, I'm embarrassed to admit I don't know much about coriolis force. Googling it seems that it is indeed a concern for gravity gradient stabilized tethers. I need to read up on this. Thank you for your comments.

Hop David said...

There are ways ot deal with coriolis induced oscillations. An elevator car traveling up will push the tether in a retrograde direction, a car traveling down will push prograde.

By timing ascent or descent, oscillations can be dampened.

And a tether end swinging like a pendulum isn't necessarily bad. At mid swing the end is traveling fastest with regard to its anchor mass. Releasing a payload at midswing could be a way to enjoy some extra delta V. Also releasing a payload mid swing would take some of the momentum out an oscillation.

Joy said...

Thanks for the link on Centauri Dreams. Good work! I agree that Luna, NEAs, and Phobos are more accessible near term than Ceres. But I still think Ceres would be the sweetest and lowest hanging fruit for a Clarke elevator. The Phobos tether is pretty cool for tossing things around, but building a mass driver on Phobos to do the same thing might be easier.

Tensile strength not being a problem at Ceres would also allow bells and whistles such as a (mostly) nonstructural HV power cable from a solar array in Clarke orbit to the surface. (actually steel conductors wrapped in kevlar could be structural at Ceres) Climbers could draw power from a cable as well. Finally, a pipeline to the surface could expedite import/export of gases and fluids.

Cheers, Joy

Hop David said...

Thanks, Joy. A Phobos elevator could release a payload into mars atmosphere at a speed of .6 km/s wrt to mars surface. Someone from Mars surface could rendezvous with the tether foot with a small suborbital hop. A mass driver from Ceres couldn't do this. Power requirements for the mass driver would be huge and require a massive power source. An elevator on the other hand could provide delta V by borrowing from Phobos' orbital momentum.

You are correct a Ceres elevator has low enough stress you could have electric cables along its length. Thus the power source for the climbers becomes much less of an issue.

I believe gases and fluids would be a major export for Phobos but I'm wondering how a pipeline along the elevator would work. I don't think you could use a vacuum pump to lift the fluids and gases to orbit.

Quite a few asteroids have a healthy angular velocity. I believe elevators may be helpful for getting around the Main Belt

Joy said...

Hi David,
Re pumping fluids up a Ceres elevator:
Yes, you could not vacuum the export water from orbit, it would have to be pumped up the elevator from the surface. I have a column of water at Clarke orbit as having a base pressure of 76MPa ~ 10,000 psi. That is high pressure, but less than the Mariana Trench and well within off the shelf industrial design. You can buy water pumps online that can pump up to 36,000 psi. The pipe would have to be insulated and heated, but that technology was already done with the Alaska oil pipeline.
Cheers

Chris Wolfe said...

Hi Hop,

I'm having a lot of trouble following the math for tether taper ratios and masses for the Phobos tethers. Some references seem to be assuming that the force of gravity is uniform along the length of the tether and others simply say they arrived at their result through numerical integration without explanation.

Can you tell me where I'm going wrong, or provide your procedure for arriving at a taper ratio and (hopefully) a tether mass based on payload?

I have a spreadsheet demonstrating my method down to 10km steps using Zylon, but my results are dramatically different than most references. The taper ratio and mass ratio should be static regardless of payload and tether mass; they are after accounting for the station mass, but I still get a taper ratio of 276:1 and a mass ratio of 1,233:1 tether to payload. If I change the safety factor to 2 I get a taper ratio of 43:1, still several times larger than expected.

This is what I'm trying to do to get a numerical approximation that will fit in a spreadsheet. I'll walk through the process in words:

For the Phobos tether, assume the maximum tension is at the surface of Phobos and that the gravity of Phobos can be ignored. This is inaccurate but only by a very small amount. We also neglect any force due to the atmosphere of Mars. Minimum tension is at the foot, where force due to the cable mass is 0.
Step 1: Force due to any capture hardware at the foot is (gravity at foot m/s² - centrifugal acceleration at foot m/s²) * foot station mass kg = force N (Newtons or kg*m/s²); this force is static.
Step 2: Force due to the maximum payload mass at the foot is equal to (gravity at foot m/s² - centrifugal acceleration at foot m/s²) * payload mass kg = force N; this force is also assumed to be static.
Step 3: These forces must be resisted by the tether, whose required cross-section is: force N / material strength Pa (which is N/m² or kg/ms²) = cross sectional area m². Most sources want at least a factor of 2 for safety, but let's assume a factor of 3 to allow for strength loss in splices / joints, etc. Short form is now f / (strength / 3) or 3f/strength.
Step 4: Since this is a numerical approach, I need to find the mass of a short length of the cable based on the cross section found in the previous step. The mass of this length of cable is cross section m² * step length m * density kg/m³ = mass kg. Add this to the total tether mass for tracking purposes.
Step 5: The force produced by this section of tether is (gravity at section m/s² - centrifugal acceleration at section m/s²) * step mass kg = step force N. Add this to the total force for the next cycle.
I repeat steps 3, 4, 5 using a new radius r = old radius + step length until the new radius reaches the surface of Phobos.

Please feel free to copy the sheet and play with the numbers. Hopefully it is useful, but based on my results so far I also hope I've made a mistake somewhere and the real-world values will be better.

Hop David said...

Symeon, The expressions for tether width I get from Aravind's and Pearson's PDFs. If I remember right, those PDFs are linked to in the above blog post.

They use many steps and I have to admit I'm no wiz at integral calculus.

Just a couple of weeks ago I constructed a spread sheet where I take an approach similar to yours (Breaking the tether into small lengths and summing them). I'm pressed for time and energy (as always). I hope to look at your spreadsheet soon. It will probably be helpful when comparing to my own efforts.

In a future blog I hope to look at tether mass to payload mass ratios.

Chris Wolfe said...

Thanks for the reply. I appreciate all the effort you've put into collecting references and spreading information, both here and in the many space-related forums where you post. Between this site, Wikipedia and Project Rho an enormous depth of knowledge can be obtained.

I added a tab for the 'up' tether and also columns for V, Vesc and Vinf at each step.
It probably needs some additional thought, since making the tether longer seems to reduce the tension. Perhaps the max tension for that tether occurs when the payload is at the surface, rather than at the end of the tether.
One thing to consider: not all trips from Mars to Earth require the same Vinf, so it may be worth making the tether another 5,000km longer (~22,000km or almost to Deimos orbit) pushing the maximum Vinf over 4km/s in order to widen the launch and capture windows to and from Earth.
I will revisit my sheet when time allows, but for now consider the 'up' tab preliminary.

Chris Wolfe said...

Payload tension is highest when the payload is just leaving Phobos. I adjusted the sheet to test for max payload tension at each step, so each step of the tether is sized for the worst-case tension. The 'up' tether is still very achievable.

In early stages the tether would only be needed during arrival windows from Earth or launch windows from Phobos, so it could be retracted for the two-year idle period to extend its lifespan. A '4-year' tether could last over 40 years by doing this.

Using Zylon fiber and a safety factor of 3, the taper ratio is 11.34 and the mass ratio is 3.22 for the 7,980km tether and can launch payloads with a Vinf up to 3.27km/s. A 10.6t tether system could catch multiple 2.5t payloads from Earth as long as they arrive one orbit (~7h40m) apart and could return payloads at the same rate. A single 3.3t payload every three days could also be sent or captured. A tether of that mass could (in theory) be launched to Phobos on a single Falcon 9H with 2.5t of margin for deployment equipment. Costs for such a mission could be under \$300 million and could return tens to hundreds of tons of Phobos samples over the next few launch windows. Three payloads per day, 6t (excludes 500kg climber allowance), times the launch window (0-60 days depending on orbits); this system wouldn't be able to return samples until 2026 but that year has a launch window two months wide for this Vinf. That one window could return 360 tons of mass from Phobos to Earth orbit, roughly a 6.8:1 mass advantage over the 53t to LEO from the single launch.

Other windows where sample return to Earth isn't feasible would still allow the tether to catch a series of exploration vehicles and then relay them to the asteroid belt at the appropriate times. Four or five vehicles of 2.5t could be sent in a single Falcon 9H launch and then deployed throughout the belt; with the right instruments this fleet could form a VLBI array around 5.5AU in diameter for incredibly high resolution radio astronomy. This leg of the program could be done for perhaps \$700 million with immensely useful science return including asteroid and comet orbit determination, spectroscopy, parallax distance measurements of stars and many others.

Using the same parameters with a 10,000km tether gets you a taper ratio of 14 and mass ratio of 3.75, with a maximum Vinf of 4.0km/s. That would widen the 2026 return window to almost 100 days / 600 tons but would require an SLS mission or multiple commercial launches. It would also open up launches to main belt objects during bad alignments, overall increasing its usefulness. Let's call it about \$600 million, or \$1,000 per kg of Phobos samples. Even if half the 'sample' mass is heat shield and containment that's still \$2,000 per kg, cheaper than launching that mass from Earth.

That's all well and good, but what if we get really ambitious? A tether 31,330km long earns you a Vinf of 9.27km/s, which is enough to inject directly into a trans-Neptune Hohmann transfer orbit. This powerful tether would handle the same 2.5t / 3.3t payloads as the other two for less than 18t with a taper ratio of ~480 and mass ratio of ~5.4. This is only two Falcon 9H launches or one SLS launch; call it about \$700 million for the tether system and the same \$700 million for four spacecraft and you could send an orbiter to all four gas giants for around \$1.4 billion, with follow-on craft every 2 years for \$175 million each. We would have to be careful of Deimos. Launch windows vary from 1.9 years for Neptune to 2.2 years for Jupiter.

Hop David said...

Symeon, I've taken a look at your spreadsheet. I like that you've set tether thickness of a step by looking at total newtons from step lengths below. A more direct approach than the expression I use for thickness: e^(complicated exponent)*e^(horribly complicated exponent).
Going over it I agree with most quantities except cell B7. B7 is supposed to be Phobos' angular velocity in radians per second? You have 2.3192e-5 where I get 2.28e-4.
Except for cell B7, I like your spreadsheet much more than my awkward effort. I am hoping you'll change that cell and see if your numbers agree with mine.
My spreadsheet didn't have any safety factor. Pearson and Aravind use a quantity h they call characteristic height. To introduce a safety factor of 3, I've divided h by 3. With a safety factor of 3, 5.8e9 pascal tensile strength, 1560 kg/meter^3 density, foot radius - 3.697e6 meters, balance point radius 9.379e6 meters, I'm getting a taper ratio of about 61.

Chris Wolfe said...

Thank you so much; that was one number I didn't have much confidence in.
I updated with your value, or rather calculated the same value from the orbital period of Phobos. I get a taper ratio of 59.89, so it looks like our approaches are pretty close (within 2%). The 'up' tether values are less amazing now but still feasible.
It turns out that most rockets use a safety factor of 1.25 to 1.4; that might be reasonable to use for a prototype or experiment but I'd still like to see at least 2 for a structure that will catch or launch humans; I've set the starting safety factor to 2 for all tabs.

I corrected errors that resulted from the bad value I used and also added the gravity of Phobos. My idea of a tether launch to the gas giants turned out to be unreasonable, so I've removed that tab. I also broke up the three main forces (Mars gravity, Phobos gravity, centripetal force) into their own columns for clarity.

Would you mind if I deleted my comment on the 24th? It is based on bad data and could be misleading.

You and anyone else are welcome to use the sheet, formulas, ideas, etc. with or without attribution. I declare it to be in the public domain and absolve myself of any responsibility for damage or loss resulting from its use.
It could be adapted to other bodies with some effort. It's not perfectly accurate but I think it stands as a good validation of the formula used by your sources.

Hop David said...

Symeon, I don't mind if you delete your comment. In any case, it's your comment, you don't need my permission.

I would like to maintain contact with you, though. As I said I like the numeric methods in your spreadsheet more than what I use. Would it be possible to send me the spreadsheet as a Microsoft Excel document? My email is hopd@cunews.info

As I mentioned, I want to do some blog posts looking at tether mass to payload mass ratio. Would you mind if I mention you? With your permission I want to use your spreadsheet as my tool for looking at various scenarios. Although there are a few changes I'd like to make to the sheet.