Spherical Full-Range Speaker

[BigusDikus]

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]

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.