- Thread Starter
- #16
Now getting to more finalized volume calculations. I'll go ahead and post this stuff
in case someone who's done a sonotube style sub sees anything funky about my calculations.
Here are the specs for what's required for this particular sub woofer for a sealed enclosure.
Net volume required after the volume of everything else in the box including the driver
are taken into consideration is .35 cu feet. JL recommended for my Jeep to bump that up by 10%
if possible. So, I will shoot for .385 cu feet or 665 cubic inches of free air space after it's all said and
done but realizing if I get into some trouble with space, I could go as low as .35 cu feet which is 605 cu inches
of free airspace, so we'll say between 605 to optimal 665 cubic inches.
TARGET FREE SPACE = 605-665 cu in
This drum shell I'm looking at is 13 7/8" in diameter.
Wall thickness for an 8-ply shell is .21" or about 7/16" for both walls
So the inner diameter will be 13 7/8" - 7/16" = 13 7/16" diameter
The radius of that is approx 6.7"
To get our cu in multiplier
pi x r x r or
3.14 x 6.7 x 6.7 = abt 142 cu inches per each inch length of tube.
So, starting with a tube length of 7", I get
7 x 142 = 987 cubic inches of air space before adding on the ends
or inserting the driver.
Subtract the displacement of the driver which is .047 cu feet or 81 cubic inches
987 - 81 = 905 cu inches of free space remaining
If the two ends are 3/4" deep, then the amount of cubic inches they will take up
will be 3/4+3/4=1.5" x our multiplier above which is 141 cu inches per each inch of tube length
1.5 x 141 = 212 cu inches will be used up by the bottom solid end and the top (sort of).
The top is a combination of about 2 5/8" of 3/4" thick wood and the top part of the driver
whose space is already subtracted. But I'll just consider the top to be a solid piece of wood
keeping in mind I have some fluff I can use if necessary.
So, the total remaining free space is:
905 - 212 = 693 cu inches of free air space. The minimum required space is 605 ci to optimum which is 665 ci
So I have between 88 ci and 28 ci remaining for extra support, wiring connectors, bolds for mounting, dynamat, etc.
Seems a 7" deep drum shell should suffice which is really good news considering my worse case scenario
photos above showing a 9" deep model which takes up more space than I care for. 2" less will be fantastic.
Any flaws in my logic ?
in case someone who's done a sonotube style sub sees anything funky about my calculations.
Here are the specs for what's required for this particular sub woofer for a sealed enclosure.
Net volume required after the volume of everything else in the box including the driver
are taken into consideration is .35 cu feet. JL recommended for my Jeep to bump that up by 10%
if possible. So, I will shoot for .385 cu feet or 665 cubic inches of free air space after it's all said and
done but realizing if I get into some trouble with space, I could go as low as .35 cu feet which is 605 cu inches
of free airspace, so we'll say between 605 to optimal 665 cubic inches.
TARGET FREE SPACE = 605-665 cu in
This drum shell I'm looking at is 13 7/8" in diameter.
Wall thickness for an 8-ply shell is .21" or about 7/16" for both walls
So the inner diameter will be 13 7/8" - 7/16" = 13 7/16" diameter
The radius of that is approx 6.7"
To get our cu in multiplier
pi x r x r or
3.14 x 6.7 x 6.7 = abt 142 cu inches per each inch length of tube.
So, starting with a tube length of 7", I get
7 x 142 = 987 cubic inches of air space before adding on the ends
or inserting the driver.
Subtract the displacement of the driver which is .047 cu feet or 81 cubic inches
987 - 81 = 905 cu inches of free space remaining
If the two ends are 3/4" deep, then the amount of cubic inches they will take up
will be 3/4+3/4=1.5" x our multiplier above which is 141 cu inches per each inch of tube length
1.5 x 141 = 212 cu inches will be used up by the bottom solid end and the top (sort of).
The top is a combination of about 2 5/8" of 3/4" thick wood and the top part of the driver
whose space is already subtracted. But I'll just consider the top to be a solid piece of wood
keeping in mind I have some fluff I can use if necessary.
So, the total remaining free space is:
905 - 212 = 693 cu inches of free air space. The minimum required space is 605 ci to optimum which is 665 ci
So I have between 88 ci and 28 ci remaining for extra support, wiring connectors, bolds for mounting, dynamat, etc.
Seems a 7" deep drum shell should suffice which is really good news considering my worse case scenario
photos above showing a 9" deep model which takes up more space than I care for. 2" less will be fantastic.
Any flaws in my logic ?