help me tune

[pj_chevy03 10+ year member]

so is trixter right.....

cuz there is a big *** difference between 3.76 and 7.62

and by the above comment db i'm geussin u agree with him

 

[DBfan187 5,000+ posts]

so is trixter right.....cuz there is a big *** difference between 3.76 and 7.62

and by the above comment db i'm geussin u agree with him

Well, Trixter is the mathematician, he's got the math skillz so.......
I just posted what WinISD said, but both Trix and I don't like WinISD - but most people say it's accurate, so... Pick one! It's your box.

 

[pj_chevy03 10+ year member]

well i have heard bad things about winisd so i'm gonna go with trixter......but why is the port so small.....it jus seems so small to everything else i have see.....is it because the 2" x 4" takes up more space than diamter wise than 4 " pvc

 

[DBfan187 5,000+ posts]

yes, it also could be a fact that your box is so small...

 

[pj_chevy03 10+ year member]

true dat....

 

[Trixter 10+ year member]

I use different formulas to find different aspects of building boxes. There are so many calculators and they all produce different numbers. I figure this way is the most accurate. It make take a few minutes longer, but it's worth the piece of mind.

Your 4" round port has more port area than a 2" x 4" slot port. The more port area you have, the longer you need to make the port to provide the same amonut of wind resistance. The amount of wind resistance is what tunes the box.

Take a small coffee straw about 1" and blow through it. No take a big McDonalds straw about 1" long and blow through it; it's a lot easier. Now how long do you have to make the larger straw to be as hard to blow through as the coffee straw? Same thing here.

 

[Trixter 10+ year member]

Dam double post!?!?!

 

[pj_chevy03 10+ year member]

so why would u not jus use like a 1" diameter port and thenu would have a real short port?????or are there disadvantages to them being too small....i am very new to learning about tuning boxes.....

 

[Trixter 10+ year member]

If the port is too small you will get port distortion. Almost like a whistle. You want a port big enough so that the air moves freely. Yet you want it small enough to still provide enough back pressure so the sub doesn't reach over excursion. I have heard people say that a good rule to fallow is to have a much port area as you have cone area. I don't believe that either. There's no way a single 10" sub should have a port of 78sq", unless it's in a really large box for comp. only, even then that's a lot. There is a formula to figure out min. port area but it kinda complex. If you really want it I could post it.

 

[pj_chevy03 10+ year member]

o.k. i think i'm ready to build my first ported box....but first i gotta get the mdf and that means about 50 miles (no hardware stores around here carry it so off to lowes). i'll post pics when its done. could be about 2 weeks though

 

[4DMAN00 10+ year member]

I get a box volume of 2.17cu'
For a 4" round port 12.3" long is 32Hz and for a 4" x 2" slot port 3.76" will produce 32Hz.

Use this very simple formula:

LV: vent length

R: radius of vent (Note: for a square vent R will be the sq.rt of A/Pi)

FB: tuning freq.

VB: box volume in cu"

4" round port:

LV = [(1.463 x 10^7 x R^2) / (FB^2 x VB)] - 1.463R

12.3 = [(58520000) / (3839754.24)] - 2.926

4" x 2" slot port:

LV = [(1.463 x 10^7 x 1.6) / (FB^2 x VB)] - (1.463 x 1.6)

3.76 = [(23408000) / (3839754.24)] - 2.3408

Have a question on this equation, for a square port why isn't the R squared? R = 1.6 and not R = 2.56. Because in my enclosure that make a big difference in port length about 20 inch difference.

 

[Trixter 10+ year member]

Have a question on this equation, for a square port why isn't the R squared? R = 1.6 and not R = 2.56. Because in my enclosure that make a big difference in port length about 20 inch difference.

You are correct. The equation should read:

7.5 = (37452800 / 3839754.24) - 2.3408

Good catch. That makes a litte more sence now. I'm on pain killers for a pinched nerve in my back so my thinkin' isn't 100% there. That's why I like to show all my math so it can be checked.

Also remember to subtract 1/2 in the slot port's width if you using the wall of the box as a common wall. Here the port length would really turn out to be 6.5"...if the box wall was used.

 

[pj_chevy03 10+ year member]

whoa so my port is supposed to be 6.5 in.... which length is it......and what do u mean by a common wall......i need a FINAL prt length....o.k wait is this correct??? common wall means u only use 3 other pieces of wood and use the back wall of the box as the forth wall of the port and in that case the port length would be 6.5......if i did not use the back wall as a wall of the port (and use 4 pieces of seperate wood for the port) then the length would be 7.5.....am i understanding this correct??????? so 6.5" port length with the box wall as a sie of the port???

 

[Trixter 10+ year member]

Yes. More often than not a slot port uses one of the outter walls of the box. In this case you need to subtract 1/2 of the ports width from it's length. If you were to simply put in a square port who's four side were its own, and not shared by the box, then you would not need to do this.

In your case, your port needs to be 7.5" long. If you use the side of the box as a side of the port then it would be 6.5" long.

I'll see if I can find something about it....

 

[pj_chevy03 10+ year member]

hey trixter could u help me on my thread "help me tune again" as well...thanx