From: gary@ke4zv.uucp (Gary Coffman)
Newsgroups: sci.space
Subject: Re: SPS feasibility (WAS: SPS fouling astronomy)
Date: 15 Aug 92 15:11:08 GMT
Reply-To: gary@ke4zv.UUCP (Gary Coffman)
Organization: Gannett Technologies Group

In article <Bsxo2q.MJy@fmsrl7.srl.ford.com> wreck@fmsrl7.srl.ford.com (R. Cage) writes:
>In <1992Aug13.075037.2707@ke4zv.uucp> gary@ke4zv.UUCP (Gary Coffman) writes:
[deleted]
>
>Also, you have the earth's energy budget wrong.  The disc area
>is about pi * (6.4e6 m)^2, or 1.28e14 square meters.  The solar
>constant is about 1360 W/m^2, so we get about 1.75e17 watts, or
>175,000 TW of incident sunlight.  Not 140 TW.  You're off by 3
>orders of magnitude!

Yeah I know. :-( I calculated in English units for Earth's area and
screwed up the conversion to square meters. Heat load isn't a significant
factor in any event, even with my wrong numbers it's still piddling.

>Adding another 4 TW of heat is literally a drop in the bucket.
>Spread it out and you'll have a very hard time detecting it.

Yep.

>>By contrast, for Earth based solar collection, there is no net increase 
>>in heat flux. The energy is striking the planet anyway whether we use it 
>>or not.
>
>3.)	You're assuming that the albedo of the earth's surface and
>	a solar collector are the same.
>	
>In a word, WRONG!  Solar collectors are a lot blacker than most things.
>
>If you counter that the collectors can be interspersed with light
>colored areas to reflect light and keep the balance the same, you
>can do that without solar collectors too.  It's not an argument
>for ground-based.

No, all energy that reaches Earth's surface is eventually re-radiated
to space. The time scale is different if it is immediately reflected
rather than being used as electricity first, but that doesn't matter
to the long term energy balance. It's still a piddling amount as you
noted above.

>>Now what's happening on the Moon? Gas lasers are very inefficient devices. 
>
>Free-electron lasers aren't.  Neither are amplitrons.
>
>>I don't have CO2 numbers at hand, but He-Ne efficiencies are around 0.1%. 
>
>I believe CO2 lasers have reached 20%; someone please correct me.
>
>>Now solar energy striking the top of the Earth's atmosphere, or 
>>the surface of the Moon, is 1 kW/m^2.
>
>Close enough.

Yeah it's 1.4 but round numbers are close enough for this calculation.

>>So our 1 TW delivered to the busbar on 
>>Earth requires a solar collector area of 1E14 square meters on the Moon.
>
>This assumes a system efficiency of 1%.  That is about 10x too low.  Solar
>cells are over 20%, conversion equipment to microwaves is well over 50%.

Actually that's an efficiency of 6% total from Lunar solar capture to
Earth busbar. The figure includes Lunar photovoltaics or thermal at
roughly 30% efficiency and CO2 lasers at 10%. It doesn't include any
transmission or conversion losses at either end. The number could be
2X either way and still be reasonably close. Since this is the Loony
Laser we're debunking, microwaves don't count.

>>requires a square 10,000 km on a side, or about 6,000 miles on a side. Sorry 
>>gentlemen, the Moon isn't that big.
>
>More like 3,000 km on a side.  Just about the area of Luna.

Numbers could be 2X either way. That's still a near *planetary* body
we're talking about plating with solar cells in any case. That's so
unreal I'm surprised *anyone* would seriously consider it.

>>Now if we discard the laser, the Moonbase 
>>with it's 6000 mile on a side collector, and simply use the same collector 
>>field we were going to use for the laser on Earth, we still gather in 1E10 
>>watts, or 10 GW, and we haven't spent an improbable fortune on the Moon.
>
>You can't use the same collector.  It's out of the sun a lot more than
>half the time, has to be built to deal with WATER and WIND, and a whole
>lot of other things that limit its life and output.  Vacuum and low G
>has the advantage that it's very friendly to most equipment.

No, it *is* the same collector. The one on Earth to catch the *laser*
beam from the Moon. It's already built to deal with the elements. It's
only 6 miles on a side.

>>A Lunar collector array would be in darkness 2 weeks out of 4, so storage
>>for two weeks would be required on Earth.
>
>Wrong.  You'd put collectors on both sides of the moon.  When power
>is only hitting the far side, it's the far side collectors which feed
>the transmitters.  Transmission lines are easy in 1/6 G and vacuum.

Plate the *whole* Moon? Really now. Transmission lines would have to
be superconductive in any case at the power levels we're talking about.
1/6 G doesn't really enter in, the damn things would have to be buried
below the solidly plated surface. Vacuum could be useful for the cryogenics.

>>Therefore, at least half of
>>Earth's capture arrays would be in daylight at any given moment. That's
>>at least 7 times better than the Loony Laser.
>
>Wrong.  Receivers not in direct view of Luna could be fed by
>reflectors (easy for lasers, a little harder for microwaves).
>If a receiver can take a feed from more than one reflector at
>a time, then interruptions from the reflector going behind
>Earth as seen from Luna will be avoidable.

Ok, now you're going to orbit huge reflectors as well as plate the
Moon with solar cells. Note that reflectors capable of handling
terawatts are tricky beasts at best. 

>>But solar *thermal* is workable *today* in certain cases.
>
>Yes, mostly peaking power or fossil displacement.  Not replacement,
>not base-load.  The whole point of Lunetta or SPS is base-load power.
>
>IMHO, if there is the capability to build that much collector
>area on the moon, a single catapult can put it into orbit and
>leave it in sunlight 24 hours per day, all month.  Having many
>feed angles and immunity from lunar eclipses and beam
>occultation by Earth is another big advantage.  This is why I
>think SPS has a better future than Lunetta.

A few fast breeders deal with the problem with *today's* technology 
for *trillions* less. I think that they're *all* loony.

Gary