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View Full Version : Take 2: How much can I round the numbers...

sumone
12-10-2004, 01:38 PM
Box design part 2:

Sub
- Idmax 12" v3 d2 (http://www.imagedynamicsusa.com/website/tech/MAX12_V3_D2.jpg)

Gross Ported Volume Recommended
- 2.55cuFt
- Net Volume I'm gonna use arbitrarily: 2.50cuFt
- Calculated Net Volume I probably should use: 2.55 - 0.12(sub_displacement) - (pi * 2^2 * 15)/1728 [port displacement of a 4" diameter x 15" long round port]
- - 2.321cuFt

Tuning
- 28hz

2 Set Dimensions
- 20" long x 15" high (internal)
- Box depth will be determined after port displacement is calculated.

MDF Thickness
- 0.75in

Slotted Port Dimensions
- 2" wide x 15" high

Port length
Lv = Av*1.84*10^8/[Vb*1728*(Fb/0.159)^2] - 0.823*sqrt(Av)
- Av = 2" *15" = 30in^2
- Vb = 2.5cuFt
- Fb = 28hz
- - Lv = 36.6957"

End_Correction
- 2" (port_width) / 2 = 1"

Internal Port Length
- 36.6957 - 1" = 35.6957...let's say 35.7 inches?

Port Displacement
- 923.0625in^3

All Displacements
- 0.12*1728 + 923.0625 = 1130.4225in^3

Gross Volume
- 1130.4225 + 2.5 * 1728 = 5450.4225in^3

Calculated 3rd dimension: Depth
- 5450.4225 / 20 / 15 = 18.168075"

Port Wall 1 MDF length
- Depth - 2*MDF_Thickness - Port_Width
- - 18.168075 - 2*.75 - 2 = 14.668075

Port Wall 2 MDF length
- Internal_Port_Length - Depth
- - 35.7 - 18.168075 = 17.531925

Assuming this is correct, my question is, when dealing with numbers where the demical part is not divisible by 0.25, how much can I round up?

Damaso87
12-11-2004, 04:19 PM
Your end numbers are only as accurate as the ones you put in...ex:

4.65936 x 3 = 14

4.659 x 3.000 = 13.977

(Significant Figues)

sumone
12-11-2004, 10:06 PM
well yeah I know that, but who uses sig figs nowadays....

sumone
12-12-2004, 12:02 AM
cool. I guess that's what I needed to know...if slight rounding affects audition.

bcarpenterfhl
12-12-2004, 12:44 PM
dont mean to get off topic, but does slot displacement include the volume in the slot, or just the extra wood used to form the slot?

sumone
12-12-2004, 01:35 PM
volume in the slot + the volume of the port walls

sumone
12-12-2004, 02:14 PM
This is Jmac's approach (http://forums.caraudio.com/vb/showpost.php?p=546196&postcount=11):

1. Choose a number smaller than the Lv-end_correction (L) as the external depth (D)
2. first port wall = D - 2*mdf_thickness (call this P, represents internal depth)
3. second port wall = L - D (call this Q, represents length of wood for this second wall)

Q's area = Q * (slot_width + mdf_thickness)
P's area = P * (slot_width + mdf_thickness)

so the area of the port is (Q's area + P's area) or (Q + P)(slot_width + mdf_thickness)

the volume would therefore be that area times the internal box height, so

(Q + P)(slot_width + mdf_thickness)(internal_box_height)

Approach 2

Based off of Jmac's method, you can take it further and eliminate the guessing factor and work with numbers that are static:

Q = L-D, so then we could do:

(L-D + P)(slot_width + mdf_thickness)(internal_box_height)

P = D - 2*mdf_thickness, so we could do

(L-D + D-2*mdf_thickness)(slot_width + mdf_thickness)(internal_box_height)

you can cancel -D with +D within the first parenthesis to get

(L - 2*mdf_thickness)(slot_width + mdf_thickness)(internal_box_height)

which would be....

(Lv - end_correction - twice_the_mdf_thickness) times (slot_width + mdf_thickness) times internal_box_height

Approach 3 (http://forums.caraudio.com/vb/showpost.php?p=547246&postcount=31)

the stuff I did from that other post, which was still based off of Jmac :)

((internal_port_len - p_width) * mdf_thickness - mdf_thickness^2 + internal_port_len*port_width) * internal_height_of_box

^ in that case, internal_port_length is Lv - end_correction - port_width

jtomsic
12-12-2004, 07:34 PM
I hate you ... :up2somet:
roflmao :-D

sumone
12-13-2004, 12:22 AM
ok, well I decided to go with a net volume of 2.3cuFt. Problem is, after all the calculations, an L-slot port ain't gonna cut it. I need to make one where it's more like a maze instead.

so like before, "figuring out that third dimension" is a b*tch.

so anyone care to explain how to get displacement on a maze-like port?

...in the meantime....

here's what I've been thinking about literally all day...(I should be sleep right now so I can be on time for school but hey, this is important!)

the box I'm basing all this off of (http://www.geocities.com/xxinfamisxx/12recvented.jpg) (you gotta open a new window & copy/paste that url)

the full port length is represented by:

· length_through_opening: mdf_thickness

· full length sections: number_of_full_length_sections * (internal_depth - port_width + 2 * 1/2 * port_width + mdf_width_for_that_wall)

· leftover section: 2 * 1/2 * port_width + some length Y + end_correction

Simplifying into one equation, this yields:

Lv = mdf_thickness + (number_of_full_length_sections)(internal_depth + mdf_thickness) + port_width + some_length_Y + end correction

However, of this "leftover" section, it should probably be greater than the port_width. In other words, the leftover_section should be less than internal_depth - port_width, so that it doesn't "seal" up the port. And based on the definition of the leftover_section, which was (port_width + some_length_Y + end_correction), an inequality representing it could be:

internal_depth - leftover_section > port_width
internal_depth - (y + port_width + end_correction) > port_width
or
internal_depth - y - end_correction - 2*port_width > 0

If Lv=56.25, mdf_thickness=0.75, port_width=1.75, end_correction=port_width/2=0.875....

..if internal_depth was 14.5",

0.75 + (10 + .75)(num_sections) + 1.75 + y + 0.875 = 56.25

.75 + (15.25)(num_full_sections) + 1.75 + y + .875 = 56.25
15.25(num_full_sections) + y = 52.875

the iPart of 52.875/15.25 is 3; therefore, y would be 52.875 - 15.25*3, or
7.125

if I had chosen a depth of 10", num_sections would be
4, and y would be 9.875

However, since this selection of numbers does not satisfy the inequality of
internal_depth - y - end_correction - 2*port_width > 0; meaning that the leftover section is not longer than (internal_depth - port_width),

10 - 9.875 - 0.875 - 2*.75 > 0
-2.25 > 0

...cannot use this depth.

Another equation which must be taken into account is that the width for the port as a whole (as it spans horizontally in the box) must be big enough. In other words, the lengthwise dimension of the box would be:

some_space + length_for_sub + some_space + width_whole_port = lengthwise_dimension
some_space can be qualified as an arbitrary constant C, so we can say that the box is divided into non_port_as_a_whole_width & width_whole_port

width_whole_port can be represented by:

(num_full_sections+1_leftover_section) * (mdf_thickness + port_width)

what this means is that now, instead of the depth being dependent on the length & height, I'm saying length is dependent on depth & height. Cause the bigger the depth, the number of sections decrease, which in turn decreases the width_whole_port. The smaller the depth, the number of sections increase and in turns increases the width_whole_port.

...in order to figure out that third dimension, I could graph these different equations, filter out the ones that don't work, and select one which does.

on the TI... (window: xmin=5, xmax=20; ymin=0, ymax=20)

y1 = graph of number of sections based on depth
= iPart(52.875/(x+.75))

y2 = graph of the y_part of the left_over section
= 52.875 - y1(x)(x+.75)

y3 = graph of the inequality to see if the left_over_section is less than internal_depth - 2*port_width
= x - y2(x) - 0.875 - 2*1.75
- make the style of this line with "Above"...so that it shades everything above.

the displacements....
port_displacement would be:
(mdf_thickness+port_width)(depth)(num_full_section s) + (mdf_thickness+port_width)(y+port_width)

and all of that times internal_box_height

(previouly I had (y+port_width+end_correction) at the end, but end_correction is not included as part of the port displacement)

you could also graph that on the TI with:

y4 = port displacement
= (2.5)(x * y1(x) + 0.875 + y2(x) + 1.75)(13)
= 2.5(x * y1(x) + y2(x) + 1.75)(13)

probably wanna turn that graph off

so that'll give you your displacement, add that to sub displacement and Vb (1.5*1728) and that's total displacement. Divide by the two known dimensions (height & depth) and that'll give you your lengthwise dimension.

y5 = the third dimension
= (y4(x) + 1.5*1728)/13/x

However, you still have to make sure that this (lengthwise dimension - width_whole_port) is enough width for the sub.

y6 = whole_port_width
= (y1(x) + 1)(0.75 + 1.75)

y7 = the space for the sub: third dimension minus whole_port_width
= y5(x) - y7(x)

however after doing this, I figured out that that box has a sub_displacement of 0.0625cuFt, so you have to add that to the numerator in equation y5.
y5 = (y4(x) + 1.5*1728 + 0.0625*1728)/13/x

anybody care to comment or what I'm doing seems to be right?

sumone
12-14-2004, 04:21 PM
That post was needlessly complicated

yea I have a habit of overthinking stuff...

I see the way you think about it...and I'm working on some more overcomplexification of the simplicity :)

by the looks of it, it seems I was on the right path (at least the displacement was correct), but now since I got some simpler stuff to work with, that should make my process simpler :)