(1) Kinetic energy:� KE = 1/2 mv²

(2) Acceleration due to motion at speed v in a curve with radius r:� a = v²/r

(3) Acceleration due to gravity from an object of mass M:� a = GM/r²

(4) Gravitational potential energy relative to the Earth:� PE = -GM_em/r
(Caveats:� zero arbitrarily assumed to be at r = infinity, assumes a perfect sphere, not accurate for points inside the Earth)

(5) Conservation of angular momentum:� mr x V = constant

(The "x" in this context is the vector cross product, whose magnitude is equal to the product of the magnitudes of the vectors times the sine of the angle between them; it is greatest when the two vectors are perpendicular.� At apogee and perigee, |r x V| = |r| * |v|)

From the above, we can derive some simple equations for orbits.� For a circular orbit, the gravitational acceleration is exactly equal to the acceleration required to curve the satellite's path into a circle.

(6) GM_e/r² = v_circ²/r
(7) v_circ² = GM_e/r
(8) v_circ = ( GM_e/r )^1/2

For a non-circular orbit, the satellite moves in a path described by the equation

(9) r = S/(1 - e cos theta)

where S is the semi-major axis and e is the eccentricity (if my memory is correct).� I found this is a pain in the posterior to analyze algebraically, and it doesn't make it very easy to see what happens if you abruptly change the speed of something (as you would with a high-thrust rocket burn).� I found it much easier to assume that every orbit is going to be changed at either the apogee or perigee (like a Hohman ellipse transfer orbit), that all burns will only change the tangential velocity (no up/down thrust; it's not anywhere near as efficient anyway) and that velocity changes happen instantaneously (a useful approximation).

If you grant those assumptions, it's fairly easy to use the conservation equations to calculate what your apogee conditions will be if you change velocity at perigee, and vice versa.� Here's my take on it (just for fun, because I like doing algebra, and it'll let everyone check my work or inspire someone else to go further and teach me something):

(10) Conservation of energy:� v_a²/2 - GM_e/r_a = v_p²/2 - GM_e/r_p

(11) Conservation of angular momentum: v_ar_a = v_pr_p

Assuming that we know the perigee conditions v_p and r_p, substitute and find the apogee conditions v_a and r_a.

(12) Express r_a in other terms:� r_a = v_pr_p/v_a

(13) Substitute into energy equation:� v_a²/2 - GM_ev_a/(v_pr_p) = v_p²/2 - GM_e/r_p

(14) Re-arrange as a quadratic equation in v_a:��v_a² - (2GM_e/(v_pr_p))v_a + (2GM_e/r_p - v_p²) = 0

From here we can plug this into the quadratic formula
x = [-B +- (B² - 4AC)^1/2 ] / 2A
with the coefficients:
A = 1
B = -2GM_e/(v_pr_p)
C = 2GM_e/r_p - v_p²

(15) v_a = [ 2GM_e/(v_pr_p) +- ( 4(GM_e)²/(v_p²r_p²) - 8GM_e/r_p + 4v_p²)^1/2 ] / 2

That ugly term 4(GM_e)²/(v_p²r_p²) - 8GM_e/r_p + 4v_p² turns out to be a perfect square, 4[GM_e/(v_pr_p) - v_p]².� This is fortuitous, and makes the solution of the quadratic very clean and simple:

(16) v_a = { 2GM_e/(v_pr_p) +- 2 [ (GM_e)/(v_pr_p) - v_p ] } / 2, which yields:

(17) v_a = v_p (degenerate case), or
(18) v_a = 2GM_e/(v_pr_p) - v_p (isn't it great when anything that simple comes out of such a mess?)
(19) r_a = r_pv_p / ( 2GM_e/(v_pr_p) - v_p )

I've cross-checked equation 18, and when you plug in the circular orbit velocity from equation 8 you get the same thing back out.� This is what you'd expect from a circular orbit (apogee == perigee).� Note that these equations are only valid if the total energy (equation 10) is less than zero, and as long as the system can be considered to be a two-body problem; for instance, get close to Luna and the extra gravity well starts throwing big errors into the assumptions.� I expect that solar tides will similarly confuse this simplistic analysis when you start getting out to trans-lunar distances, but it is probably still useful for checking the basics.

This gives everything we really need to calculate the simple cases, except numbers.� The one number we need is the product of the gravitational constant G and M_e.� G = 6.67*10^-11 N m² kg^-2, and M_e is 5.98*10²⁴ kg.� The product GM_e is 3.80*10¹⁴ N m² kg^-1.� This should make the rest pretty easy.

(I've got two more things for this thread under construction, mainly to address points and ideas from the Lunar Menu thread.� But it's probably going to be Wednesday before they're ready to post.)

Take (19) and multiply both sides by 2GM_e/(r_pv_p) - v_p:

(20) r_a [ 2GM_e/(r_pv_p) - v_p ] = r_pv_p

Multiply both sides by v_p:

(21) r_a [ 2GM_e/r_p - v_p² ] = r_pv_p²

Consolidate terms:

(22) (r_a + r_p) v_p² = 2r_aGM_e/r_p

Divide both sides by (r_a + r_p) and take the square root:

(23) v_p = { 2r_aGM_e/[r_p(r_a + r_p)] } ^1/2

We've been talking about salvaging Shuttle ET's for these concepts, but as things stand they are not taken to orbit.  Unless there is a method for getting an ET into orbit and keeping it there long enough for our purposes, there's not much point in speculating about the possibilities.  It would also be good to try to capture as much of the unused propellant as possible, particularly the hydrogen.

A figure that sticks in my mind is that the ET is taken up to 98% of orbital speed, and then discarded.  If we assume that the altitude where it's dropped is 200 km and "orbital speed" means what it would be at that altitude, this means that the orbital speed would be

(3.80*10¹⁴/(6371000 + 200000))^1/2 ~= 7604 m/sec.

This means there is a deficit of 152 m/sec to make up to get to a circular orbit at 200 km.  We will also want to raise the orbit while we're at it.  If we are going to put the tank up in the same orbit as the space station (to make it easy to salvage spent Progress freighters, perhaps) we need to boost to ~400 km altitude.  That takes another 57 m/sec kick at perigee (total 209 m/sec) per the calculation below.

That's not the only possible interpretation of the 98% number.  If we assume that the 98% refers to the velocity required to get to the 400 km apogee orbit from a 200 km perigee, we are talking about a perigee speed of

{2*6771000*3.80*10¹⁴/[6571000*(6571000+6771000)]}^1/2 = 7661.38 m/sec

This case requires about 154 m/sec boost to get to the transfer orbit.  We still need enough impulse at apogee to circularize at 400 km.  By equation 11,

v_a = v_pr_p/r_a = 7661.38 * 6571000 / 6771000 = 7435.08

To get into a circular orbit, we have to be moving faster.

v_circ = ( 3.80*10¹⁴ / 6771000 )^1/2 = 7491.44 m/sec

The delta-v required to circularize at 400 km is about 56 m/sec.

Suppose there's 15000 pounds of propellant left in the ET when the orbiter discards it, and our hypothetical boost motor burns this.  How well does it have to perform to do the job?

Required delta-v = 265 m/sec (case 1) or 210 m/sec (case 2).
Mass-ratio = 81000/66000 = 1.23
By the rocket equation, delta-v = 9.81 * I_sp * ln(MR).  This yields

I_sp = 265 / [9.81 * ln(1.23)] = 265 / (9.81 * 0.2070 ) = 131 (worst case)

This doesn't appear too difficult to do.  Let's assume there's 1000 pounds of H₂ and 14,000 pounds of O₂ remaining (very biased toward oxygen).  If we burn it at a 6:1 O:H ratio (typical for hydrogen engines) and get an impulse of 350 seconds (lousy for hydrogen engines, but maybe not for one built to the compromise of a retrofit to the ET) we have:

265 = 9.81 * 350 * ln(MR) => MR = e^{265/(9.81*350)} = 1.08024

This means we'd burn about 6016 pounds of propellant out of the total of 81000 pounds starting mass.  Of this, 1/7 or 859 pounds would be hydrogen - nearly all of the hydrogen.  This isn't what we'd like.

Suppose that we can run the engine very lean, 10:1 O:H ratio and get an impulse of 300 seconds.  Our rocket equation changes to:

265 = 9.81 * 300 * ln(MR) => MR = e^{265/(9.81*300)} = 1.09422

We'd burn 6974 pounds of propellant, but only 1/11 - 634 pounds - would be hydrogen.  We could salvage about 366 pounds of hydrogen.

If we assume a 12:1 O:H ratio and an impulse of 275 seconds, MR becomes 1.1032, we'd burn 7578 pounds of propellant and have about 417 pounds of hydrogen left (and about 7000 pounds of oxygen).  This hydrogen would normally vent off rather quickly, but with the right hardware it might be burned with some of the remaining oxygen and condensed for later use.  That would use 3336 pounds of oxygen to make 3750 pounds of water, leaving about 1.8 tons of O2.  That sounds like a reasonable haul:  almost 2 tons of water and nearly as much oxygen, on orbit.  Given that the quoted cost of launching stuff on the Shuttle is about $10,000 a pound, this salvage might be valued at over 70 million dollars.  I think you could do a lot worse.

isn't workable.

It could be done, but it would be a real pain.� One problem is that you couldn't send the tank to the moon using power transmitted from anywhere in the USA.� If you start from an inclined circular orbit and give the tank a shove near the northernmost part of its orbit, the orbit will remain tilted (not that big a problem) and the apogee will be at the southernmost part of the orbit (BIG problem).� It's kind of hard to get something to Luna when the part of your orbit that is at the same distance is off by 50,000 miles to the south.

You could address this problem by applying the power when the tank is crossing the equator, which keeps the apogee over the equator on the other side.� It would also tend to reduce the orbital inclination, because the force applied would be roughly eastward instead of in the direction of motion.� This runs into difficulties of geography.� There isn't all that much landmass on the equator compared to parts northward, and what is there isn't known for having lots of cheap and reliable electric power (or power, period).

I had envisioned this as something that could have been set up by a bunch of hacker-types with transmitting antenna arrays salvaged from old large-size satellite dishes and installed in an industrial park (perhaps renting roof and lawn space from other businesses).� That part flat out won't work.� If you have to put it someplace where there is no infrastructure, it gets many times as difficult and expensive.� I suspect that this puts it out of consideration.

As of now, I'm convinced that solar generators (PV arrays, solar-dynamic, or whatever) on the tank itself are the way to go.� Nevertheless, I think it's worth looking at the physics.

Suppose you've got an overhead orbital pass and you're trying to transmit power to the tank all the time when its ground track is less than 500 miles away.� If the tank is at an altitude of 250 miles, that gives a maximum slant distance of about 560 miles or ~900 km.� The tank is what, about 9 meters wide and 50 long?� The visual width is about 10^-5 radians wide by a maximum of about 5 times that high, assuming the tank is stabilized by gravity gradient in the up-down orientation; when the tank is overhead and presenting just one end, it's a circle about 2.2*10^-5 radians wide.

What about the ground station?� Suppose you've got something that's ultra-cheap:� several hundred old satellite dishes at about 1.5 meters wide, fed by something like microwave oven magnetrons (the actual thing won't work because they can't all be driven in phase, but that's another issue).� If you're using the same 2.45 GHz frequency (which is probably a good idea, because there are no other services operating there which aren't already enduring the interference of all the ovens on the planet) the wavelength of the output is about 12.5 cm.� To get a beam-width of 10^-5 radian requires an array width of about 1.3 * .125 m / 10^-5, or about 16 km wide.

Now that I get some numbers out of it, that doesn't look too practical either.� It appears that a scheme like this is going to need the ground station to be scaled down by a factor of 100 or so to be workable.� You could do this by cutting the wavelength by a factor of 10 (going to 24.5 GHz) and finding a way to make the receiver about 10 times bigger (90 meters across instead of 9).� This means that you couldn't put the receiver array beneath the foam insulation on an ET, but it would cut the dimensions of the transmitter array to about 160 meters.� This could probably be placed on the roof of a large warehouse; just about perfect.

Suppose you have the ground station, and the tank flies by.� You have the ability to put maybe 250 kw of power onto the receiver, of which perhaps 200 kw will be converted to electricity.� Suppose for a moment that this power can be converted to thrust against the tank with 100% efficiency by pushing against Earth's magnetic field.� What would that do to the orbit?

If the tank is at an altitude of 400 km (r = 6771 km), v_circ = ( 3.80*10¹⁴ / 6.771*10⁶ )^1/2 = 7491.44 m/sec.� 200 kw / 7491 m/sec = ~27 newtons.� Over a 1600-km long pass taking 214 seconds, this is ~5700 N-sec of impulse; pushing against a 30 metric ton tank, it increases its speed by .19 m/sec.

(1) v₀ = 7491.44 m/sec; r₀ = 6,771,000 m
(2) v_p = 7491.63 m/sec; r_p = r₀
(3) v_a = 2GM_e/v_pr_p - v_p = 7490.87 m/sec
(4) r_a = v_pr_p/v_a = 6,771,690 m

Looks like each kick would send the apogee up by upwards by almost 700 meters.� If you have enough ground stations favorably placed to give each tank 5 kicks a day, you'd be able to boost the apogee by a couple of miles a day.

This isn't quite as slow as it sounds, because the energy cost of climbing each mile goes down as r^-2; the altitude would increase a lot more for each kick as the apogee gets higher.� Countering this is the increased time for each orbit as it gets higher, so you have fewer passes.� It is still very slow, and subject to the ability to put ground stations in favorable locations.

Conclusion:� At this scale, beamed-power propulsion isn't a practical way of getting something to the moon.� It would take a lot more ground stations to get the transit time down to something reasonable; if you could depend on 5 kicks per orbit you'd be talking closer to 30 miles a day and a transit time measured in months.

Dean Ing coauthored a book (can't remember the other guy's name) The Future of Flight that used some alternative means of propulsion. Most useful to the discussion at hand was a reaction chamber heated by an external laser.

Took a glance at the book the other night, and the key portion is a clear window (guartz, I hope, unobtanium, I fear) that lets the laser light in to heat the propellant. Said propellant can be anything, within reason.

Ing felt that a free-electron laser would be best for atmospheric purposes (something about a possible need to change wavelengths on the fly). I don't believe that this would be so critical in space assuming both ends are in vacuum.

Keeping the beam on track is a nontrivial exercise, but in space, it's probably worth checking out... I suppose some of the work done in Livermore on SDI laser-kill would be applicable. At least the target isn't trying to dodge.

---

Funding was not even provided for the idea of boosting the expendable drop tank, an Air Force idea, into orbit, where they could be easily used for space station components.

---

Note that once you have a halfway decent number of stations, you can increase the 'system throughput' readily, each station kicking multiple successive tanks higher.

Tau, I think you really have something here.

------------------
Saintonge

I hear that ships like tramp freighters can often be had very inexpensively, but as I have no idea of how seaworthy they'd be for duty on the open seas nor how much maintenance they'd require.  I cannot hazard the slightest guess about the cost.

I'm trying to work up some numbers for propulsion using on-board power supplies.� More later.

quote:

---

Originally posted by Tau Zero:
I thought of ships, but the width of a typical ship is much less than the 160 meters array width of my second example.� This means that much of the transmitted power would spray past the receiver, and you are also back to the situation of "provide all your own infrastructure".� It might still have potential, but it's nowhere near as cheap.

I hear that ships like tramp freighters can often be had very inexpensively, but as I have no idea of how seaworthy they'd be for duty on the open seas nor how much maintenance they'd require.� I cannot hazard the slightest guess about the cost.

I'm trying to work up some numbers for propulsion using on-board power supplies.� More later.

---

Need a wider array? A couple of ships, a truss, and a common deck to make a big catamaran.

Infrastructure? Ships already have steam power plants, its just a matter of putting a generator in the cargo space.

Seaworthiness? Depends on the ship, I expect.

Sure, ground based stations may be preferable, but for the Southern Hemisphere, a few ships might just be the way to go.

I really think this could be done relatively cheaply.

------------------
Saintonge

Other thoughts. We already have H2 and O2 in cryro form, with a tendency to get pretty gaseous fairly quickly. (how quick, I don't know. What the overpressure allowed on the tank is, I don't know either.)

How'zabout using the H2 and O2 in a fuel cell to drive the infamous ion engine? This time, let's use a more suitable species for the ions. This would keep the mass of the engine to reasonable levels.

In addition, what about one of the booster packs the STS uses to kick satelites to polar orbit? Too tired to look up the specs, but I'm assuming that another 150 m/sec isn't all *that* tough. We'd need enough boost to get an orbit where the ion engine wouldn't have to cope with too much drag.

Oh yeah, how much attitude control will we need? Beyond that, how much attitude control is built in the ET? From what I saw, it didn't look like it had much of anything, beyond some "downward" firing separation rockets.

1. Would your catamaran deck be able to handle the twisting forces caused e.g. by the bow of one ship encountering a wave before the other?  What if one gets hit broadside by a wave and tries to roll?
2. Could you cycle the generators fast enough to bring up power for a pass lasting only 3 minutes?  The application might demand batteries.

quote:

---

Originally posted by Petethelate:
I like the idea of leaving the engine at home, but if you are talking about 24GHz, that sounds like a rather expen$ive $olution. Haven't mucked with the economics of RF devices, but while 2.4G is now considered pretty standard, 10X that isn't.

---

quote:

---

How'zabout using the H2 and O2 in a fuel cell to drive the infamous ion engine? This time, let's use a more suitable species for the ions. This would keep the mass of the engine to reasonable levels.

---

Would you be hurt if I asked you to go back and do the numbers?  That's what the equations are for, so you can check your ideas and see if they work or not.

quote:

---

In addition, what about one of the booster packs the STS uses to kick satelites to polar orbit?

---

quote:

---

Oh yeah, how much attitude control will we need? Beyond that, how much attitude control is built in the ET? From what I saw, it didn't look like it had much of anything, beyond some "downward" firing separation rockets.

---

Found this little link on ET conversion, allegedly from a 1994 Fidonet thread:
http://www.orbit6.com/et/ngfido94.htm
I haven't read it, at least some of the people in the discussion seem to be well-informed but they don't cite a whole lot of hard data.

Here's a ton of abstracts (with no links, unfortunately):
http://www.orbit6.com/et/et_abstr.htm
Technical papers: http://www.orbit6.com/et/papers.htm

It's going to take me a while to go through this.

(oops, deleted my other reply. Too bad about the unsuitability of the PAM.)

If I have time, I'll look up ion numbers and see what I can find out about the PAM.

I didn't mean to imply that the ion engine could get us from the "98% of orbital speed" to a usable orbit without any help.

Pete

Have you looked at the physics of electrodynamic tethers yet?� They push against Earth's magnetic field, which avoids the issue of reaction mass entirely.� If you have some source of electric power on board you could raise your orbit at will, but you could not change plane easily to perform a rendezvous.� I think the idea of boosting only to a low orbit, say 120 miles, and then using electric propulsion from there (assuming the system is powerful enough to overcome air drag) is a good one, and I should run some numbers for it.� You could run numbers too, there's a pretty good start up above, and Windoze has a calculator program that does all you really need to do.� (I doubt that using H₂ and O₂ in a fuel cell is going to deliver enough energy to give you much of an orbit change, and this appears to require sending up a set of fuel cells on every ET; the fuel cells would become dead weight as soon as one of the reactants was exhausted.� I'm pondering something like a solar-dynamic system using an inflatable "umbrella" mirror and a Stirling engine.� I'll post my scenario and numbers when I have them.)

As Tau says, the engine is good for 77e-3 newtons of thrust. Crunch that to get an accelleration of: (A = f/m) = 77e-3/36818kg = 2.09e-6meters/sec^2. Namely, gentle.

We need about 265 meter/sec delta v. Plug the numbers in time = 265/2.09e-6 = 4.01 years. (At this rate, the exhaust of the water vent from the fuel cell would become a major thrust factor.)

I do like the solar electric propulsion for the DS1--kind of nice to have to expend only reaction mass, but the ET is just too heavy.

1. An on-board power supply is not subject to problems with timing of orbital passes past a particular location on the ground.
   
2. An on-board power supply continues to work during the entire (illuminated part of the) orbit, whereas beamed power is only available when a ground station is in range.
   
3. An on-board power supply can be used to operate a drive all the way to Luna and even in circum-lunar operation.

quote:

---

Originally posted by Tau Zero:
Saintonge:� If you can find some numbers, they'd be very illuminating.� Things I'd worry about:

1. Would your catamaran deck be able to handle the twisting forces caused e.g. by the bow of one ship encountering a wave before the other?� What if one gets hit broadside by a wave and tries to roll?
2. Could you cycle the generators fast enough to bring up power for a pass lasting only 3 minutes?� The application might demand batteries.

As always, the devil is in the details.

---

I don't have any numbers on twisting moments, but thinking on it, what about letting the deck 'float' on top of the ship hulls. It just has to stay more or less level and pointed in the same direction. The ships could use cables to hold themselves together, and sideways thrusters to push them apart.

As to the generators, unequivocally yes. Gas turbines heat fast, and steam engines store their energy as pressure in the boiler. They can go from zero to full load in seconds.

quote:

---

Originally posted by Petethelate:
I like the idea of leaving the engine at home, but if you are talking about 24GHz, that sounds like a rather expen$ive $olution. Haven't mucked with the economics of RF devices, but while 2.4G is now considered pretty standard, 10X that isn't.

Tau Zero:
That doesn't worry me too much.� I seem to recall hearing of a device called the amplitron, which operates something like a travelling wave tube (uses conductor geometry to slow an RF wave to well under the speed of light, and uses this to transfer the energy of an electron beam to the wave); they are supposedly upwards of 80% efficient, and I doubt very much that it would be terribly hard to make a whole bunch of funny-shaped pieces of metal inexpensively given modern lathes.

Pete:
How'zabout using the H2 and O2 in a fuel cell to drive the infamous ion engine? This time, let's use a more suitable species for the ions. This would keep the mass of the engine to reasonable levels.

---

Not enough thrust, and not enough available delta-V for another.� (Work out the conservation of energy for an impulse of 5000 seconds, and how much delta-V you could get using the energy in that ton of hydrogen.)� But lack of thrust is the real killer.� You've got maybe half an hour to an hour to do your boost.� If you assume an hour to build up the first 180 m/sec, that's 0.05 m/sec²; against a mass of 38000 kg, that's 1900 newtons of thrust (around 400 pounds).� The xenon thruster on DS-1 yields 77 millinewtons flat out.� You'd be back in the atmosphere and burning up before you could do spit with an ion thruster.

Would you be hurt if I asked you to go back and do the numbers?� That's what the equations are for, so you can check your ideas and see if they work or not.[/B]

Still, if you did get the tank into more or less stable orbit first, Pete's scheme might work for the rest of the way. I don't have time to run the numbers, I'm late for class as it is.

------------------
Saintonge

quote:

---

Originally posted by Saintonge:
Still, if you did get the tank into more or less stable orbit first, Pete's scheme might work for the rest of the way.

---

quote:

---

I don't have time to run the numbers, I'm late for class as it is.

---

Doesn't look good. If you note above, I got 4 years to get 265 m/sec on the ET (assuming you could get 10kWyears worth of power. Even with a usable orbit (say, 200km, circular), you'd still need months go get to 400km.

BTW, power up concerns for shipboard generators shouldn't be an issue. It's not like the ET is just going to appear at some unknown azimuth, so even if it takes a few minutes to get up to full power, that's not a big deal.

Ptl

quote:

---

Originally posted by Petethelate:
Doesn't look good. If you note above, I got 4 years to get 265 m/sec on the ET (assuming you could get 10kWyears worth of power. Even with a usable orbit (say, 200km, circular), you'd still need months go get to 400km.

---

I just checked the DS1 site, and the quote I saw for the ion thruster is 5000 seconds impulse (*9.8 m/sec² = 49000 m/sec exhaust velocity), and 60% efficiency.� One joule of energy gets you 0.6 * 2/49000 = 2.4*10^-5 kg-m/sec of impulse.� The same joule of energy spent pushing against Earth's magnetic field at 80% efficiency gives you 0.8 / 7400 = 1.1*10^-4 kg-m/sec of impulse, or about 5 times as much.� Looks like a no-brainer.

Not sure I like the numbers on the power-in requirement. Yeah, 25GHz components are small, but you'd either need to have a hell of a system to get the heat out, or a very large bank of amplifiers to get 250kW. Keeping a bunch of amps in phase at this frequency is not a trivial job.

One way to keep this internal is to get 20kW for a continuous (more or less) boost. I'm assuming that nuke plants wouldn't be acceptable, but if you got, say, 20kW from solar panels over a 45 minute period, you'd get about .24m/sec. Thus, in a LEO, you'd get similar performance to the ground-based system. At higher orbits, you'd have a longer period in the sun.

This is a pretty large solar array (more back of the envelope calculations indicate that commercial solar panels yield about 10 watts/square foot), so you'd have to think about something that's already in orbit that would rendesvous with an ET and snag it. You'd still have to get the tank into a usable orbit.

Oh well, better get back to work.

quote:

---

How Long Do Satellites Live?

The lifetime of a satellite is strongly dependent on its altitude. At 300 km altitude it may last for 20-50 days (depending on the Sun's activity level) before it reenters and burns up. However, at 180 km, this lifetime reduces to mere hours.

---

quote:

---

Originally posted by Tau Zero:
Thanks for the data on the ship systems.� I'd wonder about thermal stresses and fuel consumption from having to keep steam up all the time, though.� Diesels would probably be better.

---

The thermal stress problem is a non-issue, I think. Steamships routinely keep at least some steam up, so they can manuver.

The problems with steam engines is that they are bulky and heavy. They are quite efficient in fuel usage. My guess is that optimum efficiency would be to use an afterburning gas turbine as the heater for the steam engine. The turbine provides peak power, and it's waste heat is captured for slower release by the boilers.

Remember, ships are big.

quote:

---

Saintonge:
Still, if you did get the tank into more or less stable orbit first, Pete's scheme might work for the rest of the way.

Tau Zero:
I doubt it.� By conservation of energy, total impulse for a given amount of energy is inversely proportional to the specific impulse.� Momentum P=mv, energy E=1/2mv², therefore P=2E/v.� When over half of the hydrogen is already used at I_sp=275, you're not going to get where you're going at I_sp=5000 and only 60% more energy; you're going to need something like 6 times as much energy.

---

I don't have time (again) to do numbers, but here's another idea: I remember something from a book "Power Supplies for Space vehicles." The author noted the advantages of generating power, then using the waste heat from the turbine to heat reaction mass.

quote:

---

Saintonge:
I don't have time to run the numbers, I'm late for class as it is.

Tau:
I have no class.�

---

"Thank you, I was hoping someone else would say that for me."

------------------
Saintonge

"She just left me. *sniff* She didn't even care enough to cut me head off or set me on fire. *sniff*"

quote:

---

Originally posted by Saintonge:
The problems with steam engines is that they are bulky and heavy. They are quite efficeint in fuel usage. My guess is that optimum efficiency would be to use an afterburning gas turbine as the heater for the steam engine. The turbine provides peak power, and it's waste heat is captured for slower release by the boilers.

---

quote:

---

I don't have time (again) to do numbers, but here's another idea: I remember something from a book "Power Supplies for Space vehicles." The author noted the advantages of generating power, then using the waste heat from the turbine to heat reaction mass.

---

quote:

---

Originally posted by Tau Zero:
...

Conclusion:� At this scale, beamed-power propulsion isn't a practical way of getting something to the moon.� It would take a lot more ground stations to get the transit time down to something reasonable; if you could depend on 5 kicks per orbit you'd be talking closer to 30 miles a day and a transit time measured in months.

---

Well, I'm back from my Space Resources Roundtable meeting. A guy in Houston more or less has a working system whereby a robot slowly crawls across regolith, turning it into polycrystalline, thin film solar cells (approximately 5% efficiency). I believe a single robot can produce about 10 MW of generating capacity per year (my notes are sketchy, and at home). Criswell has been talking about beaming power back to Earth from the moon for about 25 years at 250 W / m² using basically SAR technology. That would be another source of power to boost your tanks into orbit, but we are getting into a chicken and egg problem.

Anyway, according to Criswell, the idea of sending a bunch of robots to the 2 limbs of the moon to produce power to beam back to Earth pays off inside of 10 years, at which point we probably are generating enough power to shut down most power generation on Earth (except for storage during things like eclipses). We do need to put satellites in orbit to deflect power to places on Earth which are on the far side (as viewed from the moon).

quote:

---

Originally posted by Petethelate:
I vaguely remember the STS experiments on tethers, but not sure I heard much of the results (I do recall one jammed, but that's about it).

---

quote:

---

Originally posted by fortran:
A guy in Houston more or less has a working system whereby a robot slowly crawls across regolith, turning it into polycrystalline, thin film solar cells (approximately 5% efficiency). I believe a single robot can produce about 10 MW of generating capacity per year (my notes are sketchy, and at home).

---

quote:

---

Did any of you look up the weak stability boundary theory?

---

The idea of a tether got me thinking in a literal sense. People have been looking into space tethers for things like generating gravity by having a ship in 3 pieces, connected by a cable, one part having the crew, the other part holding most of the fuel, and in the middle some system meant to move the ship. From the center station the other pieces would be deployed.

The other idea was to use a tether to lower a satellite in to super high atmosphere to get samples, and then pull it back up, since we havent been able to get a ballon to stay up there and things. Problem to overcome would be the drag caused by the dangling line in an atmosphere. How about we have a relay system. Supplies go from earth to upper atmosphere, picked up by a "dangling hook" hauled out into space, attached to a vehicle optimized for space travel (use of tether length manipulating centrificul force can move an object) that would bring it to moon orbit. Setup a Soverjorn (sp?) type landing system (also one of red planet now...) for the supplies.

You drop it there, we pick it up.

Another idea. Large solar array around the moon (or earth, however it can work out) sending microwave radiation down to a powerstation on the moon, just closed steam powered electrical plant (just have to get the water) no atmosphere for the radiation to burn off on the moon.

Just some ideas.

R:I'm not even sure what that phrase is supposed to mean. I

I think he was referring to the Lagrangian stability points, but it's been busy for me, so haven't looked up anything either.

quote:

---

Originally posted by Mr. Zarquon:
People have been looking into space tethers for things like generating gravity by having a ship in 3 pieces...

---

quote:

---

How about we have a relay system. Supplies go from earth to upper atmosphere, picked up by a "dangling hook" hauled out into space...

---

I found some figures which probably ought to prompt revisions of at least part of the above.

According to http://seds.lpl.arizona.edu/ssa/docs/Space.Shuttle/et.shtml, the external tank is discarded at an altitude of 71 miles (114 km).� If we assume a target altitude of 160 nautical miles (184 statute miles, roughly 295 km), the required velocity at this perigee is 7707.34 m/sec.� 98% of this is 7553.19 m/sec, leaving a deficit of ~154 m/sec.� Circularization at 295 km requires an additional kick of 52 m/sec for a total delta-V of ~206 m/sec.

According to http://www.orbit6.com/et/carroll.htm, the amount of fuel remaining in the ET is less than we've been assuming.� This page quotes 3 tons of flight performance reserve, 1.5 tons of pressurized gas plus an unspecified quantity of trapped fuel.� Applying the required 206 m/sec boost to the tank with an initial mass of 80,000 pounds (66,000 lb dry weight, 9000 lb remnant fuel, 5000 pounds payload) with an impulse of 350 seconds requires 5240 pounds of fuel.� This would leave ~3760 pounds of remnant fuel, most of it gas rather than liquid.� This looks doable, more or less.

The aforementioned page on orbital decay mentions a rough lifespan of between 20 and 70 days for a satellite in orbit at 300 km, and "hours" for one at 180 km.� If we assume a lifespan of 30 days for the ET starting at 295 km without reboost and an exponential decay of the orbit with the decay rate at 180 km being 100 times the decay rate at 295 km, the initial rate of altitude loss appears to be around 2 km/day (I am having no luck constructing the equation in the back of my head and I don't have paper handy).� If the ET payload includes a 10 KW output engine (Stirling or otherwise), plus an inflatable mirror to feed heat to it (say, 50 KW_thermal into a 20% efficient engine to obtain 10 KW_electrical) and an electrodynamic tether deployed from the intertank section (perhaps blown out by a small mortar charge and reeling out its wire by inertia) and a few percent losses, a continuous 10 KW_e would provide about 1.25 Newtons of thrust.� Pushing against the remaining 75000 pounds (34100 kg) for 86400 sec/day (storage of heat to feed the engine during night passes is assumed), this would give 3.17 m/sec/day of boost.� If we assume that the difference in orbital velocity is equal to the applied delta-V (though with the opposite sign), I get this:

r₁ = 6371 km + 295 km = 6666 km
v₁ = (3.80*10¹⁴ / r₁)^0.5 = 7550.21 m/sec
v₂ = v₁ - 3.17 m/sec = 7547.04 m/sec
r₂ = 3.80*10¹⁴ / v₂² = 6671604 m

This is a gain of about 5600 meters, or almost 3 times the loss due to drag.� It looks like a 10 kilowatt generator and the appropriate tether system could haul an external tank out of a low parking orbit to a more permanent orbit; to do this, it would require perhaps 90 m² or less of mirror for its collector plus some area of radiator (which might be integrated with the intertank segment).

The advantages of the on-board power supply over beamed power are several.

1. There is no requirement for pass timing or orbital plane alignment.
2. The collector area is vastly smaller.
3. The solar power supply is not subject to decreasing performance as the altitude increases.

(Editted 27-11-2000 14:44 ET to correct errors.)

Speaking of which, umteen years ago, I belonged to L5 and they were mulling over different ideas for a cheap lift to space. The idea that I recall as the predessor to the spinning skyhook (as T0 has described) was the orbiting hoop. Perigee had to be at or below local ground level.

The main thing that comes to mind is that the hoop is a terrible idea. Really large, will have a very high relative velocity in the atmosphere, and is of dubious stability. Question for the experts: what merits if any did this idea have?