That's exactly what he's saying. Things to consider: the sensitivity rating that you normally see is db @ 1m @ 2.83v. 2.83V equals 1 W into an 8 ohm load. It equals 2W into a 4 ohm load and 4W into 2 ohms. If 3 speakers all have the same sensitivity rating of 93dB @ 2.83V/1m and one is 8ohm, one is 4 ohm and one is 2 ohm, which is more sensitive? The 8ohm driver is only getting 1W and is creating the same amount of sound as the other two getting 2x and 4x the power respectively. It takes power to make power and as impedance decreases and current increases, resistive losses in the amp increase. Running an 8 ohm driver would allow the amp to run cooler and draw less power from the electrical system of the car compared to the same amp trying to get the same output from a 2 ohm driver with the same rated efficiency. Factor in that the amp will probably be able to produce more voltage into the higher load and the 8ohm driver will probably get louder overall as well.
Originally Posted by bdavies11
This is all true for mid/high freq drivers. If you want the driver to play lower, it gets a lot more complicated because cone excursion away from the rest position where Bl is strongest changes Bl and all the other driver parameters. The larger the excursion, the larger the change. Mid/high freq drivers barely move at all and their motor strength remains nearly constant across its power and freq range.
You are grossly over simplifying a very complex situation that you clearly don't understand. Let me know how that works out for you.
Ok then... Im going to go buy myself an amplifier that puts out .000000000000000000000015 watts @ 837,489 ohms for 14 cents (since it only puts out .000000000000000000000015 watts), hook it up to my 4" sony 837,489 ohm speaker and go shatter pioneers 180+ spl record with it.
Artificial Intelligence is no match for Natural Stupidity.
Never underestimate the power of stupid people in large numbers.
Life's tough...it's even tougher when you're stupid.