Whats the results if we build a cabin with exactly the same volume as the Vas of a driver? In a closed cabin. In an ported cabin?
i made a ported box for PYRAMID WC1220
it's Vas is 160 litres and i took 150 litres for enclosure volume and the result is good for this speaker with a Q=1,6
normalized graph my project pics
it's Vas is 160 litres and i took 150 litres for enclosure volume and the result is good for this speaker with a Q=1,6
normalized graph my project pics
Hi,
VAS represents the spring action of the spider + surround of a LS unit. VAS is an equivalent volume of air of a box that causes the same spring action as the Cms of the unit does.
So when placing a speaker in a closed box with the same volume as the VAS of the unit the total equivalent Cms halves and Q_box and F_box become sqrt(2) times that of the bare unit (Qts and Fs). That’s all. The related formulas are:
Q_box = Qts * sqrt (1 + ( VAS/V_box ) )
F_box = Fs * sqrt (1 + ( VAS/V_box ) )
or:
F_box = Fs * Q_box/Qts
VAS represents the spring action of the spider + surround of a LS unit. VAS is an equivalent volume of air of a box that causes the same spring action as the Cms of the unit does.
So when placing a speaker in a closed box with the same volume as the VAS of the unit the total equivalent Cms halves and Q_box and F_box become sqrt(2) times that of the bare unit (Qts and Fs). That’s all. The related formulas are:
Q_box = Qts * sqrt (1 + ( VAS/V_box ) )
F_box = Fs * sqrt (1 + ( VAS/V_box ) )
or:
F_box = Fs * Q_box/Qts
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