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!

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.

This is where you error.  If the distance the cone can potentially move is based on a percentage of the individual cone area, the "potential d" is twice as long for the 12" driver and therefore it can potentially play louder.

And because you are only powering one driver, rather than four, it need not take any more amplifier power to attain that 6 dB increase.

andy, how do you define driver impedance of 4 Ohms?

below is an example of low frequency driver impedance, Z(f). 

pl keep in mind, as reactive/resonating behavior of every driver, even the same brand/model, is different. it is also dependent of acoustic loading of the speaker box, never the same for every driver.

to get 4 Ohms total four driver spkr system I used 16 Ohms drivers, in parallel.

parallel connected drivers have also direct connection to low output amplifier impedance, to dampen resonances.