There keep appearing arguments about the percentage of the Suns rays
that hit Earth which would be blocked by a sun shield positioned at the
L1 stable point.
I am presuming that there are some people on this forum skilled at
optics who could give a value for say a 100 km diameter disc which
reflected 100% of the rays away from Earth.
The actual disc would be slightly closer to the Sun so as to cancel out
the solar wind etc.
This is similar to an Eclipse with a partial and total area affected
although their probably will not be a total eclipse with an object of
this size at the L1 point.
I think it is only neccessary to calculate whether the shadow extends
past the Earth and allow for this in the calculations but otherwise
calculating the amount of sunlight hitting the shield is the amount
blocked.
Please correct these figures if they are wrong :-
Distance Earth to Sun = 150,000,000 km
Distance Earth to L1 = 1,500,000 km
Suns diameter = 1,400,000 km
Sun shield disc
===========
Sun Shield Disc diameter = 100 km
Possibly recalculate for a disc of 1000 km
and diameter disc 1,400,000 km
Roger Coppock - 30 Oct 2006 05:44 GMT
> There keep appearing arguments about the percentage of the Suns rays
> that hit Earth which would be blocked by a sun shield positioned at the
[quoted text clipped - 26 lines]
> Possibly recalculate for a disc of 1000 km
> and diameter disc 1,400,000 km
The solar constant is ~1367 Watts per Meter squared
http://en.wikipedia.org/wiki/Solar_Constant
Global Warming is now about 3 watts per meter squared,
which is 0.2% of the solar constant.
http://www.giss.nasa.gov/data/simodel/
(Global Warming will quadruple in a century or two.)
The cross sectional area of the Earth is 125,000,000 km^2
http://en.wikipedia.org/wiki/Solar_Constant
The area of sunshade in orbit needed to remove the current
global warming is, therefore roughly .002 * 125,000,000 km
squared or 300000 km squared.
A circle of 300000 km squared area has a radius of
sqrt(300000/p) = 309 km which is a diameter of
618 km. You had the right order of magnitude,
between 100 and 1000 km in diameter.
You would need about 38 of your 100 km diameter
reflectors to cover the 300000 km squared area
required to cover just today's greenhouse gas forcing.
300000/(p*((100/2)^2)) = 38.
It is far easier not to put the greenhouse gases into
the atmosphere in the first place.
Rick Jones - 31 Oct 2006 19:31 GMT
In sci.space.tech Roger Coppock <rcoppock@adnc.com> wrote:
> It is far easier not to put the greenhouse gases into the atmosphere
> in the first place.
I agree that it is better to treat the disease rather than the
symptom. However, I suspect that the "easier" aspect is only from an
engineering standpoint. A bunch of sunshades still permits John
Q. Sixpack to commute around in his 15 MPG rolling living room and AC
his 3000 square foot house.
rick jones

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Ed Kyle - 30 Oct 2006 06:15 GMT
> There keep appearing arguments about the percentage of the Suns rays
> that hit Earth which would be blocked by a sun shield positioned at the
[quoted text clipped - 26 lines]
> Possibly recalculate for a disc of 1000 km
> and diameter disc 1,400,000 km
It was kind of chilly today where I live. I had to wear gloves, a hat,
and a couple of fleeces to keep warm during my walk. I would
rather not have a colder climate forced onto me, and I would
certainly not be happy with dimmer sunlight. It would stunt my
tomato garden!
I'm not at all thrilled with the idea that someone, somewhere,
perhaps even outside the jurisdiction of my government, will
get to decide how much sunlight I receive. Please desist from
your plan to dim the sun or I will have my lawyer send you a
letter. ;-)
- Ed Kyle
Roger Coppock - 30 Oct 2006 06:15 GMT
> There keep appearing arguments about the percentage of the Suns rays
> that hit Earth which would be blocked by a sun shield positioned at the
[quoted text clipped - 26 lines]
> Possibly recalculate for a disc of 1000 km
> and diameter disc 1,400,000 km
I don't quite understand your optical references,
but I can check the size of your sun shield.
The solar constant is ~1367 Watts per Meter squared
http://en.wikipedia.org/wiki/Solar_Constant
Global Warming is now about 3 watts per meter squared,
which is 0.2% of the solar constant.
http://www.giss.nasa.gov/data/simodel/
(Global Warming will quadruple in a century or two.)
The cross sectional area of the Earth is 125,000,000 km^2
http://en.wikipedia.org/wiki/Solar_Constant
The area of sunshade in orbit needed to remove the current
global warming is, therefore roughly .002 * 125,000,000 km
squared or 300000 km squared.
300000 km squared is a shield with a radius of
sqrt(300000/p) = 309 km or a diameter of 618 km.
You would need 38 of your 100 km diameter shields
to replace it. 300000/(p*((100/2)^2)) = 38
It would be much easier not to put the greenhouse
gases into the atmosphere in the first place.
ianparker2@gmail.com - 30 Oct 2006 12:16 GMT
> There keep appearing arguments about the percentage of the Suns rays
> that hit Earth which would be blocked by a sun shield positioned at the
[quoted text clipped - 26 lines]
> Possibly recalculate for a disc of 1000 km
> and diameter disc 1,400,000 km
There is one important figure you have not mentioned and that is the
size of the Earth. If all the sunlight falling on the mirror is
destined to fall on Earth then the calculation is simple. The reduction
in energy is simply the amount falling on the mirror. Divide by the
amount falling on Earth and that is the proportional reduction. Size of
sun (in radians) is
1.4/150 (say 9 mills Rough and Ready)
9 mil projection at 1.4Gm (Lagrange) = 0.009*1500Mm or 1400km. This is
smaller than the Earth. Hence the Earth will intercept everything.
Earths size = Pi*r*r where r =20000/Pi km
or 400*10^6/Pi sqkm. Now area of mirror = 2500Pi sq km
It would be better I think to simply calculate sq km required for 1%
reduction. This is 1.3e6. Hence if our mirror weighs 55kg/kms this will
mean a mass of
70,000 metric tonnes for a 1% reduction.
There are 2 basic routes.
1) Try to get mass down 55kg/kms. and use a completely recoverable 2STO
with ion propulsion to L1 from LEO. This is doable.
2) A Von Neumann machine. There you are not so worried about getting
the weight down as there is plenty of material on the Moon/asteroids.
I would be extremely surprised if by 2050 BOTH these technologies were
not available. Hence the folly of politicians and people like Stern.
BTW - If we accept that poor governance is the main cause of poverty
carbon trading is absolute madness. North Korea produces very little
CO2 and its nuclear weapons program would get an enormous fillip from
any carbon trading scheme. Carbon trading will also encourage
fundamentalism and have a detrimental effect on the position of women
in Moslem countries. You do not need women if either :-
1) A sticky black substance comes out of the ground.
2) Mad economists like Stern donate you money.
Paul F. Dietz - 31 Oct 2006 14:26 GMT
> 1) Try to get mass down 55kg/kms. and use a completely recoverable 2STO
> with ion propulsion to L1 from LEO. This is doable.
One thing to point out is that you don't need mirrors,
you need devices that can diffract sunlight by a degree
(or so) so it misses the Earth. Small angle scattering
can be done with a sparse array of scattering elements,
so the mass could be much lower than a full mirror (even
a mirror with submicron holes). Something like a random
network of tangled carbon nanotubes, arranged between
think support elements, might do the trick.
You still have problems with CO2 acidifying the oceans,
and at high CO2 levels you get a change in the temperature
distribution even if global average temperature can
be controlled. Still, given the feckless response of
the world to calls to limit CO2 emissions I don't think
it would be responsible to retard a backup plan such
as this, if it had a real chance of working.
Paul