# How loud would a 40 mile high speaker be? [closed]

I am doing a theoretical thought experiment for a presentation and I have a theoretical speaker that would be 40 miles diameter (circular), now this speaker would be above most of the earths atmosphere but that's by the by.

What I'm trying to determine or find is a formula I can use to determine output in DB for a speaker of diameter N (Of atypical output of 84 db at 20W for 1 metre) with input wattage W. Is there a way to do this, if I assume atypical values for other things like impedance?

## closed as unclear what you're asking by Rory Alsop♦May 3 at 17:22

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Why do you think SPL is related to size? 'output in DB for a speaker of size N' has no correlation. – Tetsujin Jun 25 '18 at 9:00
• I am not that knowledgeable about speakers; to my mind the size of the speaker would correlate to the maximum volume of air it could move and therefore DB? I am entirely prepared to be told this is wrong :) – Rob Jun 25 '18 at 9:03
• 40 mile high or 40 mile diameter ? -- you need to update your question with more parameters.... as this is theoretical, you also need to take atmospheric pressure into account, e.g. a speaker is designed to move air, and there is not a lot of air at 40 mile above earth surface... – Edwin van Mierlo Jun 25 '18 at 9:04
• Question tweaked, but again if there -is- a formula for size N wattage W, then.... – Rob Jun 25 '18 at 9:07
• I think, as a 'thought project', other than it bears some resemblance to Douglas Adams' rock band, Disaster Area, from THHGTTG it really needs a lot more work. Loudness, per se, is totally unrelated to size. Also, the mechanics of being able to even move something of that size without it collapsing would simply be insurmountable. – Tetsujin Jun 25 '18 at 9:21

## 2 Answers

Nothing, you hear nothing.

(is my best guess answer to this theoretical question, without going into significant scientific and mathematical calculations)

Get yourself a sheet of paper, and hold it vertically from the top, like this:

Now blow some air in the middle of the paper, and you will see that the paper is bending, more at the bottom, not so much in the middle, and very little at the top:

Now we have established that paper doesn't move when you hold it at one point, we can apply this knowledge to a speaker cone with a diameter of 40 miles (= 64km) with a voice coil at 20 miles high (= 32km)

First lets look at the athmospheric pressure by height:

So we can clearly see that at ground level, we have about 1000mb, at 10km we have about 200mb (one fifth of pressure), and going down rapidly from there up.

This air pressure is basically holding the paper cone of the speaker at the bottom, while there is no air pressure at the top, effecively doing this with the sheet of paper:

Now think about where the voice coil is. It is in the middle of the speaker, moving the cone.

However, as the cone has no air pressure at the top, for most of the cone, the paper cone will bend and only move in the upper regions of the atmosphere. There is no "resistance" to move, so it will move. The paper cone, like the sheet of paper, will bend. The atmospheric pressure at the very bottom will keep the cone from moving at ground level.

As you need movement (and air) to reproduce sound, the energy of the voice coil will be distributed to the upper and largest portion of the cone and not move the cone at ground level, nor in the part of the cone which actually has air to move (as we know now that the air is actually holding the cone still at the bottom).

Conclusion: you hear nothing.

Disclaimer: this answer is not scientific, and does not proof this with mathematical formula's. I think I have spend enough time on this already to write a theoretical answer to a theoretical question which is far from realistic.

• An interesting take, but I'm not convinced by the model of blowing on a sheet of paper, because the response to a steady stream of air can be very different from the response to a high-frequency oscillation. Speaker diaphragms are mass-spring-damper systems and generally have high damping factors (or you'd hear ringing), but most of it is due to electrical resistance instead of air friction. Barometric pressure affects the driver's movement only slightly. Then again, whether or not this holds for a hypothetical speaker with wildly unrealistic parameters is anyone's guess... – Marcks Thomas Jul 5 '18 at 15:54
• @MarcksThomas ... probably, maybe, possibly.... who knows... the question is very far from reality... :-) – Edwin van Mierlo Jul 10 '18 at 8:37

The size would dilute the energy pretty much: this is rather pointless. The one thing that larger size actually improves is low frequency emission but your question is not at all concerned with frequency.

It's also totally unclear what you mean with "speaker": a membrane? An opening? Horn speakers with large openings can have pretty good efficiencies with considerably smaller membranes.