The other day I was thinking about the merits of small infinite baffle speaker boxes, and it occured to me that the cone restoring force provided by the air in the sealed box is different for an outward movement than an inward movement of the cone.
Consider this scenario: Assume atmospheric pressure to be 15 psi in round figures. The cone moves inward 1 inch and this reduces the box volume from 10 litres to 9 litres. The same outward movement would increase the volume from 10 to 11 litres.
The inward movement raises the box air pressure by a factor of 10/9 = x1.1111 = 16.6666 psi, an increase of 1.66666 psi. The same outward movement lowers the box air pressure by a factor of 10/11 = 0.9090909 = 13.636363 psi, a decrease of 1.363636 psi.
So there you have it, 1.666 psi increase, 1.3636 psi decrease. Sounds like a recipe for second harmonic distortion to me.
Consider this scenario: Assume atmospheric pressure to be 15 psi in round figures. The cone moves inward 1 inch and this reduces the box volume from 10 litres to 9 litres. The same outward movement would increase the volume from 10 to 11 litres.
The inward movement raises the box air pressure by a factor of 10/9 = x1.1111 = 16.6666 psi, an increase of 1.66666 psi. The same outward movement lowers the box air pressure by a factor of 10/11 = 0.9090909 = 13.636363 psi, a decrease of 1.363636 psi.
So there you have it, 1.666 psi increase, 1.3636 psi decrease. Sounds like a recipe for second harmonic distortion to me.
Circlotron:
I think that Richard Small wrote that these things should not matter as long as the volume displaced by the cone is less than 10% of the box volume. If you do the math, you will see that such a situation rarely happens.
For instance, a 4 inch speaker with a 1/4" excursion-very long for a 4 inch-will move about 1/690th of a cubic foot of air. So if you were planning an enclosure smaller than 1/69th of a cubic foot, you have a problem, LOL.
Incidentally, I believe the pressure on the cone increases with the square of the volume decrease. So if you had a situation where the cone movement decreased the volume to 90 percent of it's original value, the pressure on the cone would be increased by:
(1/.9) squared-or 123%.
For a situation where the volume is decreased down to 80%, the pressure increase would be:
(1/.8) squared=156%.
Or something like that.
Incidentally, a 12 inch speaker with a big 1/2 inch excursion only displaces about 1/40th of a cubic foot of air. Since the smallest volume you are likely to see a 12 inch in is about 1 cubic foot, you see that there is plenty of room.
Some have opined that these forces make a difference even when the speaker displaces less than 10%. Perhaps so. But for our 12 inch in the 1 cubic foot bx, the pressure on the cone is increased to 1.047 of it's normal value-up 5%. How much difference does that make? Maybe a matter of opinion. But it does not seem large.
I think that Richard Small wrote that these things should not matter as long as the volume displaced by the cone is less than 10% of the box volume. If you do the math, you will see that such a situation rarely happens.
For instance, a 4 inch speaker with a 1/4" excursion-very long for a 4 inch-will move about 1/690th of a cubic foot of air. So if you were planning an enclosure smaller than 1/69th of a cubic foot, you have a problem, LOL.
Incidentally, I believe the pressure on the cone increases with the square of the volume decrease. So if you had a situation where the cone movement decreased the volume to 90 percent of it's original value, the pressure on the cone would be increased by:
(1/.9) squared-or 123%.
For a situation where the volume is decreased down to 80%, the pressure increase would be:
(1/.8) squared=156%.
Or something like that.
Incidentally, a 12 inch speaker with a big 1/2 inch excursion only displaces about 1/40th of a cubic foot of air. Since the smallest volume you are likely to see a 12 inch in is about 1 cubic foot, you see that there is plenty of room.
Some have opined that these forces make a difference even when the speaker displaces less than 10%. Perhaps so. But for our 12 inch in the 1 cubic foot bx, the pressure on the cone is increased to 1.047 of it's normal value-up 5%. How much difference does that make? Maybe a matter of opinion. But it does not seem large.
Circloton:
Just as a lightheated addition to the discussion, I just looked up the tables. The difference difference in air presssure of that 12 incher in the 1 cubic foot box is only half as much as the difference in air pressure between sea level and 1000 feet above it. Somehow, I don't think we are dealing with big issues here, for most practical applications. 🙂
Just as a lightheated addition to the discussion, I just looked up the tables. The difference difference in air presssure of that 12 incher in the 1 cubic foot box is only half as much as the difference in air pressure between sea level and 1000 feet above it. Somehow, I don't think we are dealing with big issues here, for most practical applications. 🙂
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