On 23 Jan 2020 at 17:19, Greg Nokes wrote:

> Given a ring world, how wide would it have to be so that an observer
> in the, say, middle 5000 miles of it would not see the shield walls?
>
> IIRC from Niven´s work, the walls are like 1000 miles tall, and the
> floor is flat, so curvature would not play. I don´t think there is
> an easy mathy way to solve this...

Well, you can arbitrarily define an angle above the horizon that's
"too small" to be visible. And then you can work backwards from the
tangent of that angle to get the ratio between height (1000 mi) and
distance.

But that is only true in a vacuum. With air, atmospheric absorption
is gonna affect things long before that point.

Add in the fact that the line of sight goes thru all the layers of
the atmosphere and into "vacuum" and the calculation gets a bit
tricky.

I've got a similar problem I've set aside involving how the light
decreases on an alderson dic as the sun nears the horizon.

BTW, for anybody who cares, on a 1g Alderson disc, it's 48 hours from
sun-high to sunset and another 96 until sunrise (that's assuming that
at the 1 AU radius, the sun reaches a max of 45 degrees above the
horizon).

--
Leonard Erickson (aka shadow)
shadow at shadowgard dot com