How many 6.5in driver equals one 8in driver

in series drivers will modulate each other due to difference between all drivers involved in Z(f) impedance vs frequency behavior, thus is not common! 

andy2,

How do you figure?

Four 6" cones have the same total area as a 12 inch cone.  If each cone can move 5% of its diameter, the four 6 inch cones move 0.3" EACH.  But the 12" cone moves 0.6" so it moves twice as far and this moves a greater volume of air.

in series drivers will modulate each other due to difference between all drivers involved in Z(f) impedance vs frequency behavior, thus is not common!

If you drive four 6.5in. all in parallel, the total impedance will be each/4.  If each 6.5in driver has an impedance of 4ohm, the total equivalent impedance will be 1ohm, which is too low.  That is why  you have to drive them in series.

I used to have a speaker with two 10in. woofers driven in series and they worked quite well.

Also there are series xover which actually relies on the impedance of each driver driven in series so it’s not that uncommon.

You could have a combination of series-parallel.  For example, two in series, then two of them in parallel to keep the total impedance manageble.

With my example above, if we consider the cones as "pistons" the increase in volume with double the distance traveled is 6 dB.

With my example above, if we consider the cones as "pistons" the increase in volume with double the distance traveled is 6 dB.

If the combination of smaller drivers equals the surface area of a larger driver, then the cone excursion is the same for both to move the same amount of air.

If ,a, is the surface area and ,d, is the cone excursion distance, then the amount of cubic inch of air moved is simply the product of those terms: a * d.

If you have two small drivers and each has a surface area of = SU, and the cone excursion is d, then the amount of air moved is (SU * d + SU * d).

And you have a single large driver with the same surface area as two smaller drivers, say BU.  The air moved is simply BU * d.

Put them together:

SU * d + SU * d = BU * d

(SU + SU) * d = BU * d

BU * d = BU * d.

See, d is the same for both cases.