virus: energy crisis?

Eric Boyd (
Mon, 17 May 1999 17:51:50 -0400

Hi Virians,

I finally got around to doing something I'd been thinking about for a long while -- a comparison of the energy we could get from the sun to the energy we use.

I start with a decided extropian stance -- I assume that genetic engineering is going to produce "natural gas" trees, i.e. living organisms which can convert the sun's energy directly into some form of chemical energy humans use. (could also be "gasoline trees", if you want). In a sense, we already have plants of this type: certain types of corn are used to produce an alchol that can be burned as a fuel. What I'm thinking of, however, is more like a "maple syrup" kind of operation. Each tree will be hooked up to several "collection" tubes (attached to specially engineered "spouts" on the trees themselves), and as the tree produces natural gas, it is removed and collected and then distributed to human users. In this way, we can "grow" energy in the field, with a minimum amount of effort on our part.

So, with those genetically engineered organisms in mind, I began my analysis -- how much energy could we conceivably collect from the sun in that way?

Well, the frontal area that the Earth presents to the sun's radiation is a circle, with the radius equal to the Earth's radius, i.e.

Af = Pi*R^2 = 1.278 x 10^14 m^2
where R = 6,378,000 m

Now, only 29%[1] of the Earth's surface is land, and only about 20% of
*that* is suitable for argicualtral development like we're going to
need, so that leaves

Ap = A*0.29*0.2 = 7.412 x 10^12 m^2

To work with.

Now, direct sunlight striking a normal (which defines it's orientation: directly facing) plane carries about 1,350 watts / m^2, so assuming that the land above averages an incline to the plane of 20 degrees (this is due to the fact that the land masses are primairly
*not* on the equator)[2], and assuming further that only one over Pi
(=31.8%) of this land is "in the sun" at a time[3], we have that the amount of energy hitting that land is

E = Ap * cos(20) * (1/Pi) * 1350

Now, let's say that we can plant 10% of the usable land above with these new-fangled "natural gas" trees, and further that these trees are only 25% efficient.[4] It follows that we can convert the available solar energy above into

Etrees = E * 0.10 * 0.25

of useable chemical energy, continuously being "grown".

How does this value compare to world usage?

My data, sad to say, is a little dated, but I'll just multiply by large margins of safety, to be sure.[5]

My World Book encyclopedia says that in 1985, the USA used 74 quadrillion BTU's of energy (from *all* sources, including nuclear and hydroelectric).

Now 1 BTU = 1055 joules

Eus = 7.4 x 10^16 BTU/year * 1055 J/BTU / 365 days/year / 24 hours/day

                  / 60 min/hour / 60 sec/min
        = 2.476 x 10^12 joules/sec (watts) on average.

Now, if we assume that the US's energy consumption has doubled in the last 14 years, and that the US only represents 10% of current world energy usage, then the world now uses

Eworld = 2.476 x 10^12 *2 * 10

So, since 49.5 million MW is less than 74.8 million MW, we haven't reached the point of unsustainabilty *yet*. Indeed, I don't think we will ever reach it... this Earth based scheme forgets entirely about space-based sources such as solar-power orbitals, or potential break-throughs in fusion based nuclear power, not to mention such energy sources as hydro-electric and geothermal.

Conclusion: as long as we keep on the technology path, energy itself will not be a problem -- what will be the problem is the effect such large and wide scale energy production and use will have on the earth. Energy is violent -- and the Earth is fragile.


[1] Land area = 148.847 x 10^6 square kilometers

Ocean area = 361.254 x 10^6 square kilometers. 148.8 / (148.8+361.3) = 0.29, or 29%

[2] Note that this value is only an educated estimate. Is it a good one?

[3] You might think that 1/2 would be a better value -- but that would be forgetting that the curvature of the earth also inclines the plane in the longitudinal direction as well as the latitudinal direction. For those interested, the value of one over Pi comes from the fact that
(1) half the earth is in complete shade and (2) the integral of sin(x) over 0 <= x <= Pi is 2, while the area under the rectangle of height 1 from 0 to Pi is Pi (=3.1415). (the ratio of these areas relates the normal plane condition to what actually occurs on Earth)
This yeilds 1/2 * 2/Pi = 1/Pi as the final fraction of interest.

[4] Note that I think we can do considerably better on both of these counts if need be.

[5] Does anybody have more current values on hand?