• WANTED: Happy members who like to discuss audio and other topics related to our interest. Desire to learn and share knowledge of science required. There are many reviews of audio hardware and expert members to help answer your questions. Click here to have your audio equipment measured for free!

Spherical Full-Range Speaker

BigusDikus

Member
Joined
Mar 19, 2021
Messages
7
Likes
0
Hi,
looking to build a set of spherical speakers.
Let's say I use the Dayton Audio PS220. The Vas of that speaker is 90.2 liters. Using this calculator that leaves me with a sphere of lets say r=281.89mm. I'm using 24mm thick baltic birch plywood so let's keep the wall of that sphere the same. Which leaves me with a speaker with outer diameter of 611.78mm.
And that's it right? I simply slice the sphere correctly. Machine it out of that plywood, glue it all together, coat it with some nice laquer and bam I'm done, right?
(Sure I then need some fancy legs etc. but other than that I'm done, right?)
 

AnalogSteph

Major Contributor
Joined
Nov 6, 2018
Messages
3,334
Likes
3,278
Location
.de
Where did you get the idea that internal volume should = Vas? This calculator for CB speakers uses
Vbox = Vas / ((Qtc/Qts)² - 1)
With fs = 46.4 Hz, Vas = 90.2 l, Qts = 0.38 and desired Qtc = 0.707 I'm getting Vbox = 37 l undamped (empty) or 30 l damped (stuffed), giving f3 = 101 Hz, f8 = 65 Hz.
If you are actually shooting for 90 l undamped, you get Qtc = 0.537, with f3 = 103 Hz, f8 = 55 Hz... i.e. it drops off earlier but more slowly. This would generally be considered rather too large.
If you want minimum f3, that is 101 Hz between Qtc = 0.59 and 0.71... going for the middle of 0.65, we end up at 47 l (undamped) / 38.4 l (damped), with f8 = 61 Hz.
Dayton themselves suggest 0.5 ft³ = 14.1 l closed, not sure why as that would give a Qtc of 0.94 and substantial overshoot and definitely no f3 of 108 Hz. This would require Qtc = 0.84 (23 l undamped / 18.7 l damped).

BTW, something in this sphere calculator is broken... if I enter a = r (which should result in a half sphere of half the original volume), I get a remaining volume that's negative. Sanity check failed.
Lemme derive this real quick...
A segment of height h cut from a sphere of radius r has a volume of:
Vseg = π h² (r - h/3)
Problem is, we know segment radius a but not h.
However,
h = r - x, with x, a and r forming an orthogonal triangle, so
x² + a² = r², hence
x = √(r² - a²) and thus
h = r - √(r² - a²)

So if a = r, we get h = r; check. Vseg then becomes π r² (r - r/3) = 2/3 π r³ = 1/2 Vsphere; check. Looks good to go.

I used r = 24 cm, giving 57.9 l total.
Let's say a = 13 cm, then h ~= 3.826 cm, and Vseg ~= 3997.6 cm³ = 3.9976 l. 54 l remaining... nah, still a tad too big.
Trying r = 23 cm with 50.9 l next. Subtract Vseg = 4.6 l, 46.3 l remaining. More like it.
r = 22 cm gives 44.6 l, subtract Vseg = 5.45 l, so 39.15 l remaining.

Since r is the internal diameter, you can probably go with a smaller value for a... try 11 cm. Shouldn't make a huge difference though.
 
Top Bottom