How far away are clouds?

Factual Questions
1

I’m looking up at the sky. Directly above me, the sky is clear. But 30 degrees away is a cloud. How far away is the place that’s directly under that cloud? Obviously it will vary by the height of the cloud, but a rough idea?

What about the clouds on the horizon?
Standing right at the ocean’s edge, the (land) horizon is only about 3 miles away. How far away is the land underneath the clouds on the horizon?

I suppose I could “do the math” (eventually) but wondering if there’s a simple rule of thumb.

2

You’d need to know the type of cloud, at least, to estimate the height. Using an average would be pretty meaningless since the standard deviation is huge.

3

A cloud at 4,000 feet is 1.3 miles away on the ground at 30 degrees. That’s a really low, overcast day. A cirrus cloud at 20,000 ft is 6.5 miles away on the ground. Clouds generally only go up to 30,000 ft or so (I think). The nearer you are to the equator (well, wherever the sun is at that time of year), the higher the same type of cloud will be.

4

I’m looking at a little fluffy cloud near my garden and now have no idea how big or how far away it is. Thanks for ruining my evening. :wink:

Interesting. Why’s that?

5

Lucky you. Where I live it is very common to have a ceiling at 1000 or 1500 feet.

6

Clouds can be anywhere from 0 to 40-60,000 feet. 40k is usually as high as they get in temperate zones and 60k is in the tropics. When clouds are low enough to touch the ground, they’re called fog.

Noctilucent clouds can form over the poles up to 280,000 feet, but I assume the OP isn’t at a pole.

You could get familiar with different types of clouds and the altitude they form at, although they can range in thousands of feet of altitude so I don’t really see that helping. For easy math, you could assume you’re at the tip of a right triangle and make an estimate of the distance to the point under the cloud then solve for all sides.

7

Might I guess you’re a private pilot? The same is true around here (CA Bay Area), and a 1000 ft ceiling often means you can’t even do pattern work practice.

8

I recall my first long flight out to California. I was fascinated seeing clouds up close. I got even more excited seeing the shadow from them on the ground.

Funny thing. I always knew it was a bit darker on cloudy days. I never realized it was because of a huge shadow until that plane trip.

9

Can you let me know your working as I started trying to do the maths and got a bit stuck?

To firm up the problem…

Let’s say the earth is a uniform sphere of radius 6370 kilometres.
And completely covering the earth is a layer of cloud at 1830 metres (about 6000ft).

If I look at a cloud that is 30 degrees away relative to my POV, how far is the point on earth that is directly below that cloud?

As for the horizon, let’s say I’m 2 metres tall, standing at sea level, and that the horizon is viewed by looking straight ahead (is it?).
In this case we could say I’m in the centre of a chord which cuts the cloud layer. And using the equation length of chord = 2 sqrt( radius^2 - dist_to_centre^2), I get a distance to the cloud layer, on the horizon, of 152 kilometres.
Assuming this (surprising) result is correct, I still can’t figure how far along land those clouds are. :confused:

10

ETA: The OP’s post just above mine wasn’t there when I loaded the page. This is in response to the OP’s original post #1.

Back to the OP …

Your location is apparently London.

As said above clouds in your area can be at any altitude from zero (fog) to roughly 40,000 ft (high cirrus in summer).

For most purposes we can ignore the curvature of the earth.

Assuming a flat earth, simple trig says that the distance to the spot beneath a cloud will be cloud height / tan (look angle). Where “lookangle” means the angle above the horizon to the cloud from your location.

For a 45 degree look angle, the cloud is exactly as far away as it is high. So if it’s at 25,000 ft it’s about 5 miles up & therefore about 5 miles away.

for a 30 degree look angle it’s about 1.73 (or very very roughly twice) as far away as it is high.

For a 60 degree look angle, it’s about .57 (or very roughly half) as far away as it is high.

The relationship of distance and angle isn’t linear because the tangnet function isn’t linear.

For a 10 deree look angle the cloud is about 5.6 times as far away as it is high. At 5 degrees it’s about 11 times farther away as it is high. And at 15 degrees it’s about 3.7 times farther away as it is high.

So armed with these rough factors you can estimate cloud distances once you can estimate clould heights & look angles.

For very shallow look angles (“clouds on the horizon”), you’re not going to get a good answer. At low look angles a small error in angle estimation (looks like 2 degrees to me; or is it 3?) will double or halve the distance you’d compute. Likewise, it’s very hard to decide how high distant clouds are. The cloud type gives you some hints, but even then the uncertainty is a factor of 2 or 3 or even 10. e.g. Fair weather puffy cumulus could be anywhere from 1,500 ft to 15,000 ft. Cirrus is found anywhere from ~25,000 to ~40,000

If you have access to a local aviation weather report they will include cloud layer heights above the ground which will give you one half of the equation. See also here http://en.allexperts.com/q/Meteorology-Weather-668/2009/6/Cloud-base-heights.htm for a technique to estimate the height of the base of the lowest clouds if you know both the temperature & dew point. This technique works most places most times, but there are exceptions.
To bound the problem (based on the OP’s post just above mine) …

5 degrees is about as shallow as you can estimate accurately, and 8 miles up is about as high as clouds can get in your area. So ballpark the farthest away they can be is about 88 miles. And at that distance the curvature iof the earth isn’t zero, but it’s a lot smaller than the uncertainly in the other numbers & hence can (& should) be ignored.

11

A slight clarification … When I was saying a cloud was e.g. “half as far away as it is high”, I really meant, per the OP’s question, that “the spot on the ground directly beneath the cloud is half as far away as the cloud is high”. I tried to simplify the verbiage & some literal-minded folks might interpret that I was talking about the slant range from the observer to the cloud itself. Not so.