# Effects of the shadow of a city in the sky?

If there was a city using antigrav to maintain a position over Kolkata, and it had approximately the area of New York City (roughly 300 square miles, or 780 square kilometers), what altitude would it have to hold to avoid detrimentally shading the land below? For reference, the city must be high enough to function as a spaceport. It is roughly circular, with a radius of around 16 kilometers and a thickness of 5 kilometers at the edges, tapering at the planet-side and the space-side for an "axial" height of 10 kilometers.

Detrimental, in this context, means uncomfortably noticeable and/or effecting everyday life; think "partial solar eclipse."

• What does detrimental mean? It's not like anybody is farming in Kolkatta. This will be bad news for parks and potted plants, and electricity bill would go up. Does that count as detrimental? – Bald Bear Nov 30 '18 at 17:17
• Two big issues in calculating this firstly which definition of the area of New York city are you using there are a good half a dozen definitions for city area in common use honestly you would be better specifying the radius of the floating disk in actual units. Secondly I must second @BaldBear In asking what counts as "detrimental" ie what percentage of the sun's disk being eclipsed by the floating cities transit qualifies as acceptable. If it helps the dimming of ambient light in the area is proportional to the portion of the sun that is obscured so how much dimming is acceptable? – MttJocy Nov 30 '18 at 17:26
• I'm hesitant to link to Quora, but here's a similar post on Quora using a lunar eclipse that contains the calculation you'll need. To save you clicking a link, it's approximately 2*(m*S - s*M)/(S-M). S = sun's distance, s = sun's radius, M = moon's distance, m = moon radius – Punintended Nov 30 '18 at 17:27
• Very similar question: A 40km diameter alien saucer is floating 2km above the ocean for a long time. What are the effects on the sea ecosytem below? I retracted the close vote, this isn't a duplicate, but these are highly related. – kingledion Nov 30 '18 at 17:28
• @Punintended The problem here is this only calculates the distance needed to be outside the umbra eclipse (ie to be outside the area of totality) depending on the motion of the floating city over the surface a penumbral (ie partial) eclipse may well be "detrimental". – MttJocy Nov 30 '18 at 17:38

This is geometry question about subtending an angle.

First of all, let's figure out just how big our sunshade is. Looks like NYC is 784 square kilometers or so. Let's assume it's circular, which lets us use a bit of math to figure out that it's 16km in radius, or 32km in diameter.

If we're directly right under the middle of the umbrella, we can look straight up and see its center -- let's call this looking up at 0 degrees (if we look at the horizon, that'll be 90 degrees.) If we look over to the edge of the big black dot in the sky, we're looking at some angle between 0 degrees and 90 degrees. With a fixed radius, this angle is purely a function of altitude.

Let's start off with a nice equilateral triangle, so we look down 30 degrees from up (or up 60 degrees from horizontal) and see the edge. We can calculate an altitude of 27.65 kilometers.

Just how shady is that? Well, the earth rotates such that the sun takes about 12 hours to move across the 180 degrees of the sky, which means that it seems to move 15 degrees every hour. (This is also why a timezone is 15 degrees of longitude wide.)

Assuming it cuts across the middle of our sunshade, it takes 4 hours to traverse the 60 degrees from one side to the other. The other 8 hours of the day we still get the sun.

To solve for altitude given a desired X hours of shade, we can arrange our formulas to:

Altitude = cosine(hours-of-shade * 7.5 degrees) * 32 km

8 hours of shade is 16 km up.

A crippling 11.5 hours of shade is just 2 km up.

For reference, low Earth orbit (LEO) is usually considered to be between 160km and 2000km up. If our floating city is 160km up then it's not going to be very shady for very long -- I'll leave the calculation as an exercise for the reader.

For a sense of scale, this is like looking at a standard schoolbus that's a football-field away.

Hope that helps!

• This does not take into account penumbra vs umbra at a distance. – CaM Nov 30 '18 at 17:39
• @CaM Excellent point! – Roger Nov 30 '18 at 17:43
• So, for clarification, at the altitude needed to let the city function as a spaceport, the shade would be almost negligible for Kolkata itself? – G. B. Robinson Nov 30 '18 at 17:45
• @CaM Indeed this was the main limiting factor in me giving an answer here as the penumbra could matter a lot depending on your definition of "detrimental". I still remember the partial eclipse passing a few hundred km south of me in the UK when I was still too young to go and see it properly on my own but the weird lighting and noticeable drop in temperature was significant even to a human that can you know put on a coat. Imagining that happening daily would certainly have consequences for the ecosystem. – MttJocy Nov 30 '18 at 17:47
• 8 hours of shade?!? I hope never to live in @G.B.Robinson's novel! – user57423 Nov 30 '18 at 19:24

## I refer you to NASA...

NASA has a paper that does all of the math you need to do. In particular, they are determining the size of a shadow (both the penumbra and umbra) of an orbital body.

It is a 36-page paper with some complex algebra, but the results of their equations will give you the most accurate details of your floating city.

## The size of the shadow changes

Note, too, that the shadow size changes throughout the day and moves, so at noon, your shadow will be smaller than at 9 AM. The location of the shadow will also move. The time of year matters, too, since the sun's path shifts throughout the year.

So it can be possible, via the complex math involved, to determine the size and location of your shadow at any given moment, you'll need to plot that out for the entire year to see what parts of your surface city are in shadow at what times of the day for any given season of the year.

I was going to try to calculate this out, but the math went above my head and is involved enough that I think you're probably going to be better off hand-waving the results you want than trying to get scientifically accurate data for a given altitude.