Speaker enclosure

Standby: I'm sure someone will just love to prove that wrong (which may be quite easy to do), and let us both know in no uncertain terms. ;)

Ok. Here's my answer:

Assuming that you gave the outside dimensions for a fully enclosed six sided box:

Subtract the volume of the 1.8cm wood:

8241.84 cubic cm in wood.

Are those inside dims?

Do you plan to make the box and then put in a woofer of unknown specs?

The woofer's specs will tell you how big the box needs to be. You can't work backwards from a box dimension unless you plan to vent it. There are all kinds of nomograms around to determine the volume of a box. You should be able to find a copy of Martin Colloms's work in your library. Anything by Alexis Badmaieff or Don Davis will help you along too.

I _ASSume_, that "a box with the dimensions..." would be outside measurements, particularly since he specifically gave the thickness (1.8cm) of the wood.

(interior W)(interior H)(interior D) = interior volume

(23 - 3.6)(40 - 3.6)(23 - 3.6) = (19.4)(36.4)(19.4) = 13,669.504 cubic centimeters

Any errors from there are conversion error by using something besides "cubic centimeters" to convert to the next "cubic whatever".

... it's the way I did it, but what the hell do I know?

OK ... not much, obviously! :)

... oh yeah!

Forgot ... we will now get into how the box is constructed; butt joints, miter joints, rabbet joints ... all of which could have some effect on the final answer.

... maybe. ;)

I get the same answer doing it that way. I'm still trying to figure out why the difference in my previous post. If I did it right, it should come out the same.

Spoke too soon... I only subtracted 1.8cm for thickness of the top and bottom in my first attempt. Should have been 3.6cm for both sides.

Well, s*it..isn't that obvious?? A lock-mitre bit of course...but which brand? Huh? Huh?

I may have jumped the gun a little on the question of inside/outside dims. From my perspective only the inside dims matter IOW. how much air is there in that box? Then the next question appears. The ratio/shapes of panels as to reduce standing waves at certain frequencies...but I could be talking s*it here..

What kind of joint?

If it's a miter, the inside dimensions would be:

volume= w*h*d