# simple box question

Discussion in 'Audio' started by Jonny-K5, Aug 24, 2004.

1. ### Jonny-K51/2 ton status

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im building a box and i have a simple math question, but i suck at math.
the size if this box will be 48"Lx18"Dx6"D.
id like to make this a wedge box. instead of 6" depth can i make it 3" at the top and 9" at the bottom? will this give me the same interior volume?

2. ### Jonny-K51/2 ton status

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3. ### Jonny-K51/2 ton status

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4. ### Jonny-K51/2 ton status

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5. ### spearchucker1/2 ton status

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Volume of cube= 6*18*48 = 5184 cubic inches
Volume of wedge shape= triangle + cube = (.5*6*18*48) + (3*18*48) = 5184 cubic inches

6. ### Jonny-K51/2 ton status

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[ QUOTE ]
Volume of cube= 6*18*48 = 5184 cubic inches
Volume of wedge shape= triangle + cube = (.5*6*18*48) + (3*18*48) = 5184 cubic inches

[/ QUOTE ]

so thats a yes right? thanks /forums/images/graemlins/thumb.gif

7. ### coreydRegistered Member

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take all your measurments, multiply them and then divide it by 1728, that will give you your cubic feet

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[ QUOTE ]
im building a box and i have a simple math question, but i suck at math.
the size if this box will be 48"Lx18"Dx6"D.
id like to make this a wedge box. instead of 6" depth can i make it 3" at the top and 9" at the bottom? will this give me the same interior volume?

[/ QUOTE ]

[ QUOTE ]
48"Lx18"Dx6"D

[/ QUOTE ]

Is 18" the height? How many cubic feet do you need? If the dimensions that you gave are for the outside of the box, the internal air space (which is what the speaker sees) will be less. Also you will need to compensate for speaker displacement, the amount of airspace occupied by the speaker when placed onto the box. If it is going to be a ported box you will also need to compensate for port displacement.

If the measurements that you gave are for the outside of the box, and if you use 3/4" think material to build it with - you will have 2.00 cubic feet inside of the box.

9. ### Jonny-K51/2 ton status

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[ QUOTE ]
Is 18" the height? How many cubic feet do you need? If the dimensions that you gave are for the outside of the box, the internal air space (which is what the speaker sees) will be less. Also you will need to compensate for speaker displacement, the amount of airspace occupied by the speaker when placed onto the box. If it is going to be a ported box you will also need to compensate for port displacement.

If the measurements that you gave are for the outside of the box, and

[/ QUOTE ]

yes 18" is the height. i know all the factors of building a box, just didnt know if going from 6" depth to a wedge of 3" at the top and 9" at the bottom would give me the same internal volume.

the dimensions i gave were for the outside, that was after i factored in the thickness of the wood. i am shooting for 2 cubic feet, sealed. the subs want 1 cu.ft. each including the displacement of the woofer.

thanks for the help. /forums/images/graemlins/thumb.gif