"Robert Clark" wrote in message news ================================================== ================
================================================== ================
" wrote in message
...
================================================== ================
On Tuesday, June 20, 2017 at 10:31:48 AM UTC-7, Robert Clark wrote:

At a solar radius of 700,000 km away from the Sun, based on the light
intensity going inversely by the square of the distance, and with 1,360
watts per sq. meter (in space) at the Earth’s distance, or 1.36 gigawatts
per sq. km., I estimate this should give 60 terawatts per sq. km. at only
a
solar radius away from the Sun.

But in the post above, I had estimated that fully *on* the Sun’s surface
we
could collect 60 terawatts per sq. km. of power. Anyone have an
explanation
of this discrepancy?

I'm not going to go through the math for you, but think about how much of
the Sun your hypothetical power-collector can "see" from a radius away as
opposed to "on the surface" (which I take to mean just above the
photosphere).

...
Mark L. Fergerson

---
================================================== =================

The suggestion that close in to the surface at a solar radius away this will
limit the amount of the Sun's surface it can see is a good one. But if this
was the issue we would expect the solar radiance would be reduced even
further below what was available directly on the surface, not equal to it.

---
================================================== ===============
================================================== ===============

OK. I think I understand your argument now:

From the Earth's distance the lightrays are nearly parallel and you receive
the light from the entire solar disk. But when you make a comparison to a
much closer distance by just increasing the solar radiance according to the
closer distance you are *assuming* the illumination will be to the same
extent at that distance. In actuality though, a significant portion of the
solar disk will not be observable. See the image he

https://ibb.co/hionF5

That angle at center calculates to be 60° because its cosine is r/2r = 1/2.
So you see only 60°/90° = 2/3rds of the solar disk will be observable at
that distance.

I might estimate the radiance at that distance then as 40 terawatts instead
of 60 terawatts. However, because not all the rays are parallel at this
distance I'm not sure all portions of the Sun would make the same
contribution to the total so the answer might not be proportional.

Perhaps someone can calculate this.

Bob Clark



----------------------------------------------------------------------------------------------------------------------------------
Finally, nanotechnology can now fulfill its potential to revolutionize
21st-century technology, from the space elevator, to private, orbital
launchers, to 'flying cars'.
This crowdfunding campaign is to prove it:

Nanotech: from air to space.
https://www.indiegogo.com/projects/n...ce/x/13319568/
----------------------------------------------------------------------------------------------------------------------------------