<p>Your cousin isn't as crazy as you think. The calculation in the linked question assumes no intervening trees, buildings, hills, or mountains, but also assumes no atmospheric distortion and no refraction. I'll ignore those last two factors, for now.</p>
<p>That calculation is how far away one can see the very top of a very tall cloud under those conditions. The rest of the cloud would be below the horizon. This is how even ancient people knew the Earth is spherical. A ship that sailed out from a harbor started to sink below the horizon until all that was left was the crow's nest at the tip of the mast, and then that too disappeared.</p>
<p>When a sailing ship came into harbor, all that would have been visible at first would have been the crow's nest. This would initially would have been indistinguishable from a piece of flotsam. A similar issue arises with your faraway cloud that you can barely see the top of. Except perhaps for the fuzziness caused by atmospheric distortion, it will be nearly indistinguishable from a much closer but much lower and much smaller cloud.</p>
<p>The one saving grace for your argument is atmospheric refraction. The Sun is still below the horizon when we see the sun rise, and it appears to remain above the horizon after it has actually set. Atmospheric refraction will typically enable you to see more of that faraway cloud than a simple straight line computation. On the other hand, atmospheric refraction might preclude you from seeing even the very top of that faraway cloud. The atmospheric refraction index is roughly proportional to pressure but inversely proportional to absolute temperature. A very high lapse rate might well cause light to bend upward rather than downward. (This is why predicting the exact time of sunrise and sunset is rather difficult, if not impossible.)</p>

CC BY-SA 4.0
2023-03-08T19:43:04.533Z
2023-03-08T19:43:04.533Z
David Hammen
<p><a href="https://earthscience.stackexchange.com/a/5257/239">The top of a cumulonimbus cloud is usually about 40,000 feet and can reach heights of over 60,000 feet, which would be visible for a distance of 245 or 300 miles, respectively.</a> from <a href="https://earthscience.stackexchange.com/questions/5255">How far away can you see a thunderstorm?</a></p>
<p>My crazy cousin said that at, say, 300 miles, said NOT could see this cloud formation - maybe just the top of it - because of the earth curvature ( distance: 300 miles - curve 11.3618 miles below sight (also 59990.14 feet). I am certain that he is wrong. But it is logical and the earth curvature calculation is legitimate.</p>

CC BY-SA 4.0
2023-03-05T19:36:03.280Z
I'm not clear on the confusion.  The very next sentence in the linked answer says "Of course, that would be the very top that would be visible at those distances...", which agrees with your question "maybe just the top of it".
Mark
<meteorology><atmosphere><geography>
My snarky cousin tossed a monkey into my ability to see large cloud at 300 miles HELP!
Jacques Taulard
# My snarky cousin tossed a monkey into my ability to see large cloud at 300 miles HELP!

(Asked by: username_0 on 2023-03-05)

[The top of a cumulonimbus cloud is usually about 40,000 feet and can reach heights of over 60,000 feet, which would be visible for a distance of 245 or 300 miles, respectively.](https://earthscience.stackexchange.com/a/5257/239) from [How far away can you see a thunderstorm?](https://earthscience.stackexchange.com/questions/5255)

My crazy cousin said that at, say, 300 miles, said NOT could see this cloud formation - maybe just the top of it - because of the earth curvature ( distance: 300 miles - curve 11.3618 miles below sight (also 59990.14 feet). I am certain that he is wrong. But it is logical and the earth curvature calculation is legitimate.

**Tags:** meteorology, atmosphere, geography

> **Comments:**
> **username_1**: I'm not clear on the confusion.  The very next sentence in the linked answer says "Of course, that would be the very top that would be visible at those distances...", which agrees with your question "maybe just the top of it".  

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## 1 answers

### Answer by username_2 on 2023-03-08. Score: 3

Your cousin isn't as crazy as you think. The calculation in the linked question assumes no intervening trees, buildings, hills, or mountains, but also assumes no atmospheric distortion and no refraction. I'll ignore those last two factors, for now.

That calculation is how far away one can see the very top of a very tall cloud under those conditions. The rest of the cloud would be below the horizon. This is how even ancient people knew the Earth is spherical. A ship that sailed out from a harbor started to sink below the horizon until all that was left was the crow's nest at the tip of the mast, and then that too disappeared.

When a sailing ship came into harbor, all that would have been visible at first would have been the crow's nest. This would initially would have been indistinguishable from a piece of flotsam. A similar issue arises with your faraway cloud that you can barely see the top of. Except perhaps for the fuzziness caused by atmospheric distortion, it will be nearly indistinguishable from a much closer but much lower and much smaller cloud.

The one saving grace for your argument is atmospheric refraction. The Sun is still below the horizon when we see the sun rise, and it appears to remain above the horizon after it has actually set. Atmospheric refraction will typically enable you to see more of that faraway cloud than a simple straight line computation. On the other hand, atmospheric refraction might preclude you from seeing even the very top of that faraway cloud. The atmospheric refraction index is roughly proportional to pressure but inversely proportional to absolute temperature. A very high lapse rate might well cause light to bend upward rather than downward. (This is why predicting the exact time of sunrise and sunset is rather difficult, if not impossible.)


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