How many 6.5in driver equals one 8in driver

All of these calculations omit the fact that an 8" driver will typically have more excursion than a 6.5" driver, and it is the volume of air moved, not the cone area, that determines how loud they can play.

If you have multiple 6.5in drivers in series, the cone excursion will be similar to a single larger drivers.

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.