This will give the actual mathy goodness needed to calculate this: https://en.wikipedia.org/wiki/Visibility

More generally (from the above linked article)

"At sea level, the Rayleigh atmosphere has an extinction coefficient of approximately 13.2 × 10−6 m−1 at a wavelength of 520 nm. This means that in the cleanest possible atmosphere, visibility is limited to about 296 km.”

So I’m thinking anything more than 600km wide would mean you cannot see them. As you approached the walls they would loom out of the haze, but all the way because there’s no curvature like on a sphere where you would see the peaks first; the walls would reveal themselves form the bottom up, I think since you would be closer to them than the tops. Also, the atmosphere doesn’t reach *nearly* as high as the walls. If you’re doing a rotation that provides 1G the atmosphere won’t reach more than a tenth of that height or so:

https://scied.ucar.edu/shortcontent/earths-atmosphere

"Air becomes so thin at altitudes between 100 and 120 km (62-75 miles) up that for many purposes that range of heights can be considered the boundary between the atmosphere and space. However, there are very thin but measurable traces of atmospheric gases hundreds of kilometers/miles above Earth's surface.”

1000km is MUCH higher than needed to contain atmosphere. you MIGHT actually be able to see the top of the wall with a telescope long before you could see the bottom.

This would be an interesting thing to model. Weather on a ringworld will be vastly different than on a planet, since most of our climate is due to convective equilibration of energy between the hot equator and the cold poles.

ALL of a ringworld is equator.

Maybe that 1000 km of wall contains gigatons of refrigeration equipment to make the rimwall areas ’the poles’ :-)

(it’s been *ages* since I’ve re-read Ringworld, should give it ago again sometime)

> On Jan 23, 2020, at 6:19 PM, Greg Nokes <xxxxxx@nokes.name> wrote:
>
> A question:
>
> 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...
>
>
>
> Sent from my iPhone
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Bruce Johnson
University of Arizona
College of Pharmacy
Information Technology Group

Institutions do not have opinions, merely customs