I have a situation in my imaginary setting, and I am specifically interested in how this situation would affect the weather.

The situation is as follows:

Positioned above a large landmass of my world is an immense flat disc, at least two thousand miles in diameter. It is thin and has comparatively little weight or mass, but it is also opaque and does not let any light through.

This flat disc is positioned roughly 40-50 miles away from the surface of the world, hovering in place over the same landmass the entire time. This disc blocks any sunlight hitting it from reaching the surface.

In such a situation, what affects would this disc have to the weather of my world?

The disc in question is maintained and built through alien technology so advanced it is indistinguishable to magic. What I want is some input on how to describe the weather from the point of view of anyone stood underneath the disc - would the weather be calmer underneath this disc? Or would there be storms? Or would the weather be the same as usual? The disc is obviously not science based, but the I would like science based descriptions of how this disc would affect the weather. For all intents and purposes, imagine this world was identical to Earth, and the disc was hovering above America.

  • $\begingroup$ I am hoping that this question is better worded and easier to understand to my last question. This question IS NOT a duplicate, and answers to how such a situation would affect THE WEATHER are not available on this site. I would be very grateful to anyone who could provide me with some insight to how this disc would affect the weather (not sea ecosystems) $\endgroup$ – Jimmery Aug 20 '19 at 15:47
  • $\begingroup$ 2000 miles circumference is 630m across (e.g. about the same size as Kansas and Nebraska together), is that actually the size you were going for? $\endgroup$ – Morris The Cat Aug 20 '19 at 15:56
  • $\begingroup$ I meant diameter! thanks for spotting that one, I knew I was missing something $\endgroup$ – Jimmery Aug 20 '19 at 15:58
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    $\begingroup$ Instead of posting another question and flooding the old one with just complaints about who voted to close it, why don't you edit the old one and follow the normal reopening process? Attempting to circumvent the rules is never smart. $\endgroup$ – L.Dutch - Reinstate Monica Aug 20 '19 at 16:36

Ok 2000 miles are roughly 3200 km. Now with an earth sized planet of roughly 6300 km radius you have arctan(1600/6380)*2=28,15 degrees covered. This means that roughly 30/180*24 h= 4h are without sunlight. That’s roughly the time a solar eclipse can last. This of the time you won’t have sunlight. At that it will be at an lesser angle reducing the energy per area. Generally I’d assume that this will come with a lowered temperature in that specific area.

But as pointed out in the prior thread depending on the kind of object hovering it might also be that sunlight is captured in the morning if the material is reflective or the material emits absorbed light like a black body in all directions or is something in between. So there is also the possibility that the disc functions as cover because energy absorbed in the morning and evening by the earth is absorbed and emitted or reflected during the time the sun stand above your disc.

This possible thermal coupling between the earth and the disc makes it quite hard to guess the weather situation just based on the size. Let’s say the disk would be made of glass than it might function like a green house. But this might still be the case if the disk is black while absorbing and remitting radiation above the surface.

So I think you have 2 main choices here. If it’s a reflective material it’s possible that the temperature changes are more drastic and there is a time of cool down when the sun is blocked. If it’s more absorbent it might heat up the covered continent somewhat more even though there are 4 hours less of direct high sun light. In addition to that will the disk absorb light from the sun longer than the earth does ..so it might slowly release halve of the absorbed energy back to earth. Later is however more speculative on my side. After all the radiation is directed in all direction and not let through like in a green house. So cooling could also be be happening as with the mirror but the temperature rise in the morning evening might be not so noticeable due to the fact that the material re-emits some of the absorbed energy instead reflecting it.

All of this might result in changing weather situations. Like a greenhouse it might be more warm an wet or it might become quite hot in the morning and evening when the light is reflected, while during mid day the temperature drops. The result it rain might be coming down due to less water absorbed in the air when it gets cooler.

Than there is also a coupling with the surrounding due to convection ect. Is the surrounding less warm for example due to prolonged shadowing and lesser thermal coupling. In that case i suspect that is a chance for forming high and low pressure areas going for the outside to the inside. In case of a reflective material this might be different so. It’s simply very hard to say without knowing what the thermal coupling would be like.

Having a more noticeable effect would mean to increase the size even more.

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    $\begingroup$ And yes a solar eclipse will be accompanied by a noticeable temperature drop quora.com/… ...and we are talking about a similar time period here where the sun is completely blocked. However as pointed out above ..the vicinity of the disc might turn out to have a stronger thermal coupling with the surface. $\endgroup$ – World Peace Aug 20 '19 at 16:46
  • $\begingroup$ This is a good point. I hadn't considered the reflective (or lack thereof) of the underside of the object, but that would make a huge difference in how much solar radiation the area underneath it actually received, and how much the temperature would fluctuate throughout the day. $\endgroup$ – Morris The Cat Aug 20 '19 at 16:54
  • $\begingroup$ Yes it’s quite hard to tell without any extensive simulation of some sort. How big the impact of the thermal coupling might be. But we know that greenhouse gases have a considerable impact so it’s not unlikely for such disc to impact even if the spot underneath is not hit directly as the gas of atmosphere, the other free areas and the disc also are heated up besides the spot that has 4 less sun hours. $\endgroup$ – World Peace Aug 20 '19 at 17:12

A 2000 mile diameter disk is round-about 6.3% the disk-area of the earth. The usual back-of-the-envelope aspect of such a thing for intercepting sunlight is that it blocks 1/4 of the corresponding light. This is due to the thing spending most of its time at an angle. This means, this disk would intercept round about 1.57% of the total sunlight falling on Earth. CO2 is supposed to produce something less than 1%, depending on the stats.

Assuming it reflects the light it intercepts, that's all cooling.

It's going to get chilly.

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    $\begingroup$ The effects are going to be a lot more localized than that though. The entire earth might only lose ~2%, but the area directly underneath is going to lose something more like 60-70% at least. $\endgroup$ – Morris The Cat Aug 20 '19 at 17:05

It wouldn't affect the weather at all

The sun still hits the spot anyway. If the disk is locked with the rotation of the Earth, it only blocks the Sun when it's between the Sun and the point on Earth it's over. For the entirety of the rest of the day, that area will still get sunlight. Spot will still get sunrise, and sunset, it'll just have a solar eclipse every day at noon. Temperature might drop a little, considering the sun is hottest as noon, but we're talking 5-10 degrees here, not 30-40.

Fermi Estimate for length of the Eclipse: The Sun has a size to distance ratio of .009, the Moon has a distance ratio of .008, your disk has a ratio of 40, meaning it cover 4444x the amount of sky the sun does. Let's say the sun covers ~ .07 square inches, giving your object the area of 311 cubic inches, with a diameter of about 20 inches. That means it blocks twenty inches worth of Sun travel - which translates horribly when I try to force it to 'length of day'. I took the measurements about 3-4 inches from my eyes, which means the two feet is pretty significant. On the flip side, the sun moves fastest from out perspective at midday, so I'm extending my original estimate to about two hours on each side - four hours total.

Note: This would actually be effective at blocking out the sun above the Artic Circle or below the Anartic circle. But it wouldn't really affect the weather there anyway, given that it's already mostly subzero with terrifying blizzards.

  • $\begingroup$ I don't think this is right... A two thousand mile disk, forty miles above the surface of the earth is going to block a LOT more than 20 inches worth of the sun's travel. For reference: This is what a 1300 mile disk would cover: hurricanescience.org/images/hss/1979_tip_sizecomp_noaa.jpg $\endgroup$ – Morris The Cat Aug 20 '19 at 16:25
  • $\begingroup$ Possibly. I don't think my math is that off, but even a magnitude either way has pretty significant changes. And my 'twenty inches' is twenty inches about six inches in front of the human eye. $\endgroup$ – Halfthawed Aug 20 '19 at 16:32
  • $\begingroup$ Well, I'm not sure about your equation, but just drawing a quick picture of this object relative to the earth at that distance makes it look like it's going to block about 160 degrees or more of the sun's transit from horizon to horizon... imgur.com/a/23prs7e $\endgroup$ – Morris The Cat Aug 20 '19 at 16:41
  • $\begingroup$ 160 degrees is 10 and 3/4 hours. I know I'm being conservative, but I think that's too much. $\endgroup$ – Halfthawed Aug 20 '19 at 16:44
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    $\begingroup$ So call it 8 hours, that still completely changes your answer. No sun for 8 hours out of every day would have a pretty spectacular change in the weather. $\endgroup$ – Morris The Cat Aug 20 '19 at 16:51

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