Space Elevators That Require A Great Deal Of Energy
Seems lame that our rocket scientists still haven't managed to effectively utilize those very same tidal energy forces to accelerate or even sustain an artificial satellite, as all of their satellites always seem to need a great deal of energy applied as a fair percentage of their mass in order to keep such from plunging back into Earth or of any other destination planet, thereby within a few years or possibly decades at best before becoming DOA. Of course, that may be because we still don't have squat situated at the efficient ME-L1 position, nor actually much of anthing other that's orbiting external to the nasty Van Allen zone except for what's located at ES-L1
With reguard to reapplied tether energy; Within the LSE-CM/ISS domain, the lifter(s) or transport pods may get the bulk of their energy (besides nuclear and of solar PV and of considerable solar/sterling energy conversion alternatives) from a rather extensive tether dipole, one element end of which is obviously attached directly to the moon itself (perhaps two or more of these) plus those dynamic tether elements associated with the CCM, while the other element is of what's been polarised by the Earth gravity-well. Situated between the dynamically regulated null are those massive counter-rotating flywheels for energy storage and redistribution, plus from whatever is obtained by the Earthly deployed dipole extension element, as that extended element or perhaps tentacle could reach to within 50,000 km of Earth (a bit closer if you dare).
3 Phase Tether Conduction Pathways are Possible
High frequency modulated over a lower frequency and/or over DC
This tether acquired energy may be transferred to the LSE lifters via tether conduction and/or via similar laser cannon and subsequent conversion to what the ESE could be using, though decades earlier at not 10% the investment as well as 20% more efficiently because there's no stinking atmosphere nor pesky clouds to deal with, as well as no commercial flights that could unknowingly travel directly through a rather substantial beam of energy. The alternative to transferring energy via laser cannon, of having a superimposed three pole tether conductor that's performing with a raw voltage potential of perhaps 100 kv per phase is certainly doable (within a bone dry vacuum the potential of accommodating even a million (1e6) volts per phase should easily be obtainable as well as sustainable), as to be starting off with the 1/6th gravity should permit such a conductive tether to coexist, especially of anything above 2r (0.0416 G).
If this 3 phase tether were to be conducting at the 1e6 volts, at a frequency of perhaps 50 khz, the overall energy transfer efficiency should be rather high, like better than 99.5%. A slight side benefit of accommodating this 3 phase energy transfer along the tether could become that of creating an artificial micro magnetosphere surrounding the tether itself, thus somewhat shielding it from the gauntlet of cosmic and solar flak, possibly even deflecting micro meteorites.
Since the basalt fiber tether itself offers such a terrific electrical insulator, especially if it were starting off at 3 meters in width, plus the obvious fact that it's existing within a near absolute vacuum, the notion of introducing three intentional conduction pathways, separated by as little as 10 cm for the 1e5 v and otherwise of 75 cm for the 1e6 v application should become worth investigating.
Do remember that the dynamic ME-L1 null rather uniquely offers an absolute zero G and thereby nearly absolute zero energy solution (even the ME-L1.1 CM/ISS sphere is perhaps offering not more than 0.000304 G), therefore minimal energy taken (other than acceleration and deceleration issues) for transferring whatever to/from the moon or even for exporting whatever towards Earth become merely a soft release function.
BTW; if this fully configured outfitted 300 meter CM/ISS sphere were to become a significant mass representing 50e6 t, at the Earthly gravity-well influence of .000304 G = 15.2e3 t worth of tether tension, as obviously this is where the two primary tethers along with these same tethers being associated with their interactive dynamic counter mass of the CCM comes into play, as to dynamically compensate all such things down to a dull roar, as either that or the deployment of the CM/ISS can obviously be something dynamically regulated by the amount of mass taken onboard and/or released, or from otherwise being situated closer to the ME-L1 (E/M null), whereas the tether dipole element deployed towards Earth plus whatever end-point mass still needs to come into this equation as much as for those primary LSE lunar gravity-well tethers. Absolutely nothing gets out of the loop, at least not for issues of mass nor of energy, whereas the energy taken in must equal energy out and/or stored, just as the tether tension per lunar gravity-well must balance out with the Earth gravity-well tension, else all is lost pending the capability of the tether itself and/or of whatever other artificial means of compensation (EMPD thrusters for navigation as well as dumping surplus energy on command).
This following is what I believe is just the very tip of this tether dipole iceberg.
Obviously there's been tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one.
Since I've learned from Marvin (Kirfox@pixie.co.za) that lunar recession is presumably taking a continuous 4 terawatts excluding friction, while I'm thoroughly confused about calculating something other upon lunar drag into similar terms of continuous energy requirement, thus I've decided to do a little more of my shock and awe reverse engineering, based upon yet another of my village idiot guestimates of lunar drag consuming another 1e12 W, a terawatt worth of required tidal forces that's having only to offset for drag in order to sustain the lunar orbit speed at zero recession.
Purely for coefficient argument sake; if we were to allot said 1e12 (terawatt) to represent the necessary energy in order to overcome a lunar drag coefficient.
1e12/38e12 = 26.3 mw/m2
In few other words, it seems as though mother Earth could be expending 5 continuous terawatts just upon the moon, plus whatever else in terms of tera BTU associated with the global shedding of energy, that's perhaps something less than what all of humanity as well as other life has been creating. For ease of doing just the rough human influence calculations, I like to apply an overall 10000 btu/human. At 6.5e9 folks, that's 65 tera BTUs. At 3.412 btu/w, 65e12 BTU = 19 terawatts and growing. Of course some of us folks use up far more than our fair share of Earthly resources, especially of certain warlords implementing actual shock and awe, and this is not to mention what mother Earth herself manages to let go of, more likely than not shrinking as a result.
If we were to further suggest that the space weather environment in which the moon travels about Earth is representing but 1/1000th the impact of a combined viscosity plus greater kinetic energy demands per m2 as that of the ISS orbit, while being roughly 11 fold faster and having to operate within greater viscosity than of the lunar trek.
If we were use an ISS surface reference area of 1000 m2, as well as a 1000 times factor for the greater viscosity and velocity multiplier:
26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp)
Thus by way reverse enginering (in place of my not identifying others with more correct information), it's taking 26.3 W/m2 worth of continuous energy applied in order to sustain the likes of ISS in a fixed (non-recession) orbit, in other words of overcoming drag as that based upon the consideration of the ISS environment demanding 1e3 greater effort per m2 than of lunar space, as that having to travel through essentially a thicker atmosphere and roughly 11 fold faster than the moon.
Obviously if lunar kinetics were merely 100 times lesser, then the reverse engineering calculations for the ISS compensation drops to a mere 2.6 W/m2. Obviously the ISS flight profile of leading and trailing surface area is not 1000 m2, nor is the viscosity of any constant.
Perhaps 26 W/m2 is simply too much continuous energy applied for ISS but, I'd have to bet that 26 W/m2 is not sufficient, thereby the atmosphere and/or space weather difference may actually become 400~500 times thinner for that of the moon traveling through it's environment at 1.025 km/s as opposed to the ISS having to travel at roughly 11 fold faster within a considerably greater viscosity.
Of course absolutely none of this is sufficiently correct, just my way of making due until others pitch in, but at least it gives myself something to go along with those 4 terawatts worth of recession energy, making the overall tally for continuous gravitational tidal forces per year applied onto the moon as representing perhaps 5 terawatts.
4e12 + 1e12 = 5e12 W (5 terawatts)
If there's 5e12 W to continuously draw upon, disregarding upon whatever solar PV and thermal/sterling conversions and/or induced solar plasma electrons and/or EMF factors taken from solar weather, as well as from tapping into our Van Allen zone of death, all of which should actually be worth quite a great deal, so much so that it seems exactly the sort of tether dipole extraction potential that I'm thinking can be safely directed and stored into those massive counter-rotating flywheels situated at the ME-L1 (mutual gravity-well null).
Just for a little further shock and awe argument sake, lets say that those additional energy conversion inputs (besides the overall recession and drag compensation energy) provides us with one additional terawatt resource, as now all toll were looking at 6 terawatts, whereas tapping into 50% of that perpetual energy cash yields 3 terawatts, more than enough energy to run 20 of those 100 GW laser cannons plus another terawatt to spare for the LSE-CM/ISS and then some. Obviously if one of the three terawatts is what's derived from being artificially converted/created, thereby only two of those three terawatts are those having to be extracted from the lunar recession energy.
5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted
We certainly can't take away everything, as that could reverse things by pulling the moon towards Earth, a seriously bad sort of thing to be doing. Although, if the bulk of energy taken is what's converted into laser cannon energy focused upon relatively small portions of Earth (not that Earth isn't already getting a little too hot under the collar), say quadrant target zones of as little as 1 km diameter along with a 10 km safety buffer zone, at least some of this extracted energy should return itself as a slight repulsion factor.
Screw Whatever Laws of Physics
I guess I'm not all that sure anymore, as it seems those laws of physics work just fine and dandy if you're supporting the status quo, only to discover upon their conditionals if you elect to oppose the ongoing ruse, the sting of the century. It seems that I'm not the one and only village idiot that thinks enough is enough.From: M. Luttgens (effects of orbit deceleration)
"To calculate the effects of the deceleration on the orbit of planets (or satellites)"
"According to LLR data, the Moon is receding from Earth at a rate of about 3.8 centimeters per year. Such increase of orbital distance, attributed to tidal effects on Earth, could mask the present small decrease of 1.87 cm/year."
Unless I've misunderstood (as usual), this above reference page seems to be suggesting that the coefficient of lunar orbital friction or drag is actually quite a high percentage upon the overall scheme of things, as 1.87 cm/year is nearly half as much as the 3.8 cm of reported recession, thereby the overall energy necessary in order to impose the actual 3.8 cm/year recession may in fact become a compensation worthy of 3.8 + 1.87 = 5.67 cm/year, which in turn may suggest 6 terawatts worth instead of my initial guestimate of 5 terawatts that was based upon adding one terawatt into what Marvin (Kirfox@pixie.co.za) had previously specified as to his calculations representing 4 terawatts worth of recession energy (excluding matters of friction).
Another interesting observation of the calculations resolved by others, it seems the mutual force existing between Earth and the moon amounts to 1.94e24 Newtones.
1.94e24 Newtones = 5.39e9 TW (as in 5.4 billion terawatts)
I'm not sufficiently knowledgable, nor do I have access to one of those do everything software calculators as to be suggesting that the previous notation of the lunar recession taking but an additional 5 terawatts is necessarily all that correct or not, as it only seems that it might require just a wee bit more energy than 5 terawatts in order to over come 5.4e9 TW of gravitational clutching.
OK, apparently according to some of those "spin" and "damage control" folks, even though all artificial satellites must utilize various propulsion schemes in order to sustain a given orbit, though for some unexplained reason that of a much heavier natural satellite, having a rough surface and lots of it, as in 38e12 m2 worth, requires absolutely no additional energy whatsoever to sustain a given orbit while having to travel itself through a viscosity of perhaps 6e9 atoms/m3, no less while making relative headway of 1.025 km/s.
As for understanding and predicting satellite orbits tend to go, it still seems rather odd to myself that we don't have a very specific number upon a lunar drag coefficient, basically we've got a whole lot of squat worth of supposedly honest lunar data. By the words of those specifying "zero energy" is required to sustain the lunar orbit seems only somewhat intentionally bogus to say the least. I don't believe there is any such thing as "zero energy", nor zero drag, at least not at the velocity of 1+km/s. Even a black-hole represents an equation of taking energy as well as there being considerations of drag. Unless we're talking about some sort of aligned magnetic repulsion (which can't possibly stay that way forever), as seems if it weren't for drag we'd all be either into infinity or headed right back at the Big-Bang beginning in no time whatsoever.
Of course, if our Earthly magnetic core should take another flip, and should the moon subsequently reverses its recession, then obviously the magnetic poles that may have been repulsing will in fact become attracting forces. This I'd have to understand, though oddly that's not being even suggested as a remote possibility by the sorts of "all knowing" wizards that otherwise placed our radiation-proof as well as meteorite proof astronauts on the moon, as in taking an environmentally friendly walk in the park.
Seems the moon has somewhat of a pathetic sodium like atmospheric surround, as well as a weak magnetosphere in addition to it's rough surface, and the solar wind that's a given push/pull as for representing a velocity neutral is not of the 6e6 atoms/m3 verity, more than likely a nasty soup of 6e9/m3 unless there's some serious flak to deal with, such as the November/October events of 2003 that could have temporarily pressed the viscosity index to an even nastier 6e12/m3 worth of solar flak, where at least some of those atoms are no longer merely hydrogen, but of iron. If we're to be talking of Pluto, then perhaps the 6e6/m3 of mostly hydrogen should apply, as otherwise the closer into the solar system nucleus the greater the viscosity, of which the closer to the sun somewhat benefits the likes of us and our moon by reducing those cosmic energy levels, which by itself seems like another fairly good indicator of the level of viscosity surrounding Earth and our moon (the lesser cosmic radiation the thicker the solar weather which in turn represents the bulk of whatever viscosity.
Exactly like a ship in the ocean, the friction coefficient of the forward facing area represents less energy demand than of what's trailing, same goes for aerodynamics. Towing a fully submerged sphere will easily demonstrate this, as opposed to a teardrop foil that will enable less energy applied to obtain greater momentum. Unfortunately, the moon is not representing any teardrop, though its weak atmosphere as well as minimal magnetosphere may help to create this desirable foil, though still dragging its butt will consume at least as much energy if not more than the frontal area.
Of course, absolutely none of this actually matters all that much if you're one of those damage control Borgs insisting upon worshiping a certain pagan God, especially if you're one of those that hates Cathars among others, unwilling to even consider the worth of anything lunar, let alone Venus being entirely out of the question.
BTW; here's another little something of worthwhile lunar tit for tats from Charles Ginenthal, regarding "THE ORIGIN OF THE MOON"
Obviously for those folks insisting that their God made Earth and our moon as the center of their perverse universe, thereby our moon was somehow of the very same composition as Earth, what can even Charles Ginenthal and of others like him possibly have to say, because regardless of the laws of physics, absolutely anything offered will be refuted during as well as long after he's dead and gone, regardless of whatever science truths. This being somewhat like asking the Pope about Cathars (what Cathars?), or Israel about those 6-Day war prisoners (what prisoners?), or that of certain WMD (what WMD?) and, don't even bother asking questions about flight-800 or the shuttle COLUMBIA.
As we manage to run ourselves out of moral fiber, as well as out of whatever natural petrol and of most other affordable energy resources, and remain too pathetically dumbfounded to safely utilize nuclear energy (like those smart-ass French have been doing for decades), and way too energy inefficient to rely upon wind and ocean tidal resources (too polluted and otherwise too greenhouse for efficient solar PV), thereby having insufficient energy for exterminating the remaining populations we don't happen to like; By properly using those ABL laser cannons of either near UV or near IR, or perhaps both spectrums so that at least humans aren't being blinded while attributing the least amount of thermal energy upon Earth (this is actually where I'd favor the near IR [750~800 nm] that shouldn't blind the nocturnal species), this energy transfer as a technological solution being where those multiple 100 GW cannons that can sort of light your fire from afar, just might do the trick.
If those before mentioned 20 or so 100 GW laser cannons are being efficiently transmitted and subsequently converted, I'd tend to think that an overall input/output conversion obtained on Earth might eventually reach 25%, thus 2 terawatts of input becomes 500 GW, which isn't all that bad for the hundreds of billions it may take to pull that one off, unless of course those were Chinese laser cannons and beam energy receiving stations, in which case we're probably talking about only a few billion at most.
Although, since I'm not entirely convinced that the one and only requirement will be for creating energy in the form of those utility grid electrons, as serious beams of this much potential WMD intensity targeting upon Earth could certainly have all sorts of other intentional uses and/or misuses. This is where I'd have to give the lock-box containing the final authorization and fail-safe key to the Dogon tribe, where at least that way there'd be no political agenda and/or ulterior religious nor government agency cult considerations that aren't well founded and deemed moral. Don't underestimate those Dogon, as they've been around far longer than most modern civilizations and, I'd just have to bet they're not out and about wasting valuable talents and resources upon locating WMD such as stealth donkey-carts, nor upon antagonizing other sufficiently nice folks into attacking them, hard to even imagine the Dogon imposing any of those trade tariffs and/or restrictions either.
The primary reason why I've elected the Dogon over those nicer and most likely smarter Cathars should be obvious; there are damn few if any Cathars remaining after the Pope finished them off.