displacement issue.. maybe some bad algebra too

[stocks 2]

box dimensions

38 * 13 * 15.98=

7894.12 in³ =

4.56835 ft³

tuned to 32hz

Bracing with 1 in dowel rod

up/down

r=.5 h=15.95

3.142 * .25 * 15.94 =

12.5 in³ =

0.007233 ft³

front/back

r=.5 h 12.97

3.142 * .25* 12.97 =

10.188 in³ =

0.005895 ft³

probably two of these.. perpendicular or parallel

so this leaves either 4.553884 for parallel or 4.555222 for perpendicular, up/down and front/back

plus sub displacement of .19

4.363884 parallel..........

and for port area, with 57.25 in² slotted, ~50% for aero is 28.625 in²

so a 6in aero port will be about 13.62 in long with the flare, with 28.26 in² without. round is also about 55% more efficient than slot so.. i think all my math is good, but someone may prove me wrong

so that's a displacement of 366.2 in³ = 0.211921 ft³

with displacements in consideration it is 4.151963 ft³

how would i bring this into my equation? average the non displaced and the displaced and take that and use it in the equation, or should i simply add the displacement of the port BACK into the box dimensions and forget about everything else?

yes, the winisd is modeled in a slotted enclosure but i did all the calculations separate, with the vent at 0

thanks in advance guys

[Adrian_D 520]

For modelling you have to use the volume with all the bracing, port and sub displacements subtracted from the total volume, which is 4.15 cf. Seems like you modelled it right in winISD

[KU40 173]

What do you mean by "round is 55% more efficient than slotted?"

[antonmiller 8]

What do you mean by "round is 55% more efficient than slotted?"

i am also cunfused about this.... where did you get your numbers on that?

[stocks 2]

and okay, i just wasn't sure how to get about the exact tuning i want. i guess i'll try it a little longer, maybe an inch or so and cut it if i need to

maybe being more efficient isn't what i should say.. less port area is required for the same airspeed

Edited October 16, 2009 by cranberryyumyum

[stocks 2]

This is a routinely asked question:

"How long does the vent tube need to be if I use an n" flared port?"

Granted, there are a few calculators out there that will calculate this for you, particularly if you're using some of the widely available manufactured ports. Here's a good example of a calculator available to you:

http://psp-inc.com/psp-inc.com/public_html..._calculator.cgi

But what if you have hand-formed your flared ends? Better yet, let's assume you are a dork like me and prefer to hand calculate a lot of things (believe it or not, the extra effort is worthwhile in the long run, but that's a rant for another day).

Without getting too into depth on the actual derivation of the formula, here's what you need to know.

Where:

Lv = length of the vent tube (in meters)

N = number of ports (unitless)

c = speed of sound (in meters/second)

Rm = mouth radius of the flare (in meters)

Rt = throat radius of the flare (in meters)

Rf = flare radius (in meters)

Fb = frequency of port resonance (in Hertz)

Vb = size of enclosure (in cubic meters)

V = Volume of air in the flare

Rm = Rf + Rt

and

V = pi*Rf*Rm*Rm - (pi*pi*Rm*Rf*Rf/2) + (2*pi*Rf*Rf*Rf/3)

and for a vent with two flared ends:

Lv = N*c*c*Rt*Rt / (4*pi*Fb*Fb*Vb) - 0.85*Rm - 0.613*Rm - 2*V / (pi*Rt*Rt)

and for a vent with one flared end:

Lv = N*c*c*Rt*Rt / (4*pi*Fb*Fb*Vb) - 0.85*Rm - 0.613*Rt - 2*V / (pi*Rt*Rt)

I KNOW you are happy to have a lot of equations thrown at you, but this is actually relatively simple to sort out, as far as equations go. First, let's fill in the blanks I know the answer to in our simulation. We'll assume I have already measured or calculated the radius of the flare, throat, and mouth.

Lv = length of the vent tube (in meters)

N = 1 port

c = 340 m/s at sea level

Rm = mouth radius of the flare (in meters)

Rt = 0.05m (approximately 2 inches)

Rf = 0.025m (approximately 1 inch)

Fb = 10 Hz

Vb = 0.284m^3 (approximately 1 cubic foot)

V = Volume of air in the flare

As you can see, we are down to only three things that need solving! First, let's solve for Rm. As mentioned previously:

Rm = Rt + Rf

Rm = 0.05m + 0.025m

Rm = 0.075m

Now we move on to solving for V. This one is a bit more complicated, but still easy when we know all of the variables.

V = pi*Rf*Rm*Rm - (pi*pi*Rm*Rf*Rf/2) + (2*pi*Rf*Rf*Rf/3)

V = 3.14*0.025*0.075*0.075 - (3.14*3.14*0.075*0.025*0.025/2) + (2*3.14*0.025*0.025*0.025/3)

V = 0.0004415625 - 0.000231084375 + 0.0000327083

V = 0.0002432

Ok, now we're getting somewhere! Let's get to solving for Lv where both ends are flared!

Lv = N*c*c*Rt*Rt / (4*pi*Fb*Fb*Vb) - 0.85*Rm - 0.613*Rm - 2*V / (pi*Rt*Rt)

Lv = 1*340*340*0.05*0.05 / (4*3.14*10*10*0.284) - 0.85*0.075 - 0.613*0.075 - 2*0.0002432 / (3.14*0.05*0.05)

Lv = 289 / 356.704 - 0.06375 - 0.045975 - 0.0004864 / 0.00785

Lv = 0.81020 - 0.06375 - 0.04597 - 0.06196

Lv = 0.63852m

Lv = 25.15"

At last, we know how long the vent tube must be. There is one last step which you might find interesting.

Let's say you model up an enclosure (who hasn't?). Using a straight port with a 2" radius, you notice that the vent speed is approximately 15m/s. What will be the new vent speed, now that we have flared ports? Using the values for Rt and Rm that we determined previously, we can easily calculate the change in exit velocity.

Vt = 15m/s

Vm = Rt*Rt*Vt / Rm*Rm

Vm = 0.05*0.05*15 / 0.075*0.075

Vm = 0.0375 / 0.005625

Vm = 6.67m/s

That's quite a change! From % perspective standpoint, that's a decrease of approximately 56%!

Hopefully you enjoyed the little math lesson. Whether this is practical for you to learn in depth or not, it is definitely worth reading and maybe picking up some theory. If someone asks you any questions about calculating flared ports, you can answer intelligibly.

Cheers!

okay so i have a few ideas for slot ports so i am gonna attempt to do one.. can anyone help me with the dimensions i need? i need help badly

well i just realized if my previous specs are internal i need to have 15.98 not 14.48.... so if it's this easy i'll be a little pissed

Edited October 17, 2009 by cranberryyumyum

[stocks 2]

nope, it isn't. i have to add port displacement i guess? or does winisd do that for you?

[KU40 173]

No winISD doesn't factor in displacement. You put in the net volume. Then when you go to actually build the box, you add in the port and sub displacement to get the gross volume needed.

[stocks 2]

alright thanks so much ku40. this site is always so much more helpful than ca