September 15, 2020 at 3:24 pm #11787
Would it be a problem to put the woofer and tweeter of the c-note kit into a enclosure with a volume of 11-12 liters?
But the specs of the woofer recommends a vented volume of 0.47 ft3 ~ 13,3 liters. https://www.soundimports.eu/en/dayton-audio-dsa135-8.html
What would happen to the quality of the bass and sound in general if I were to put the tweeter and woofer into another enclosure, with a volume of about 11-12 liters? I would keep the depth the same as the c-note bookshelf speaker, and the length of the tube the same also?
September 16, 2020 at 8:38 am #11791
You would want to re-tune it. Redesign the port for the larger cabinet. Keep in mind, when PE states a recommended box size, that really is just that…a recommendation. Sometimes, people change the box size for better power handling, tuning preference, etc. For example, there have been quite a few times, where WinISD will spout off a recommended size for the box that is enormous. Interestingly enough, it isn’t uncommon to be able to cut the box size in half with little loss in response.
September 16, 2020 at 3:21 pm #11792
September 17, 2020 at 2:12 am #11796
Are you ready for some math? This is from the QBASIC routines I wrote going on 30 years ago.
Ported enclosure – enclosure size given (existing box)
Minimum required info:
Internal volume of box in Cu.Ft. (Box vol)
Vas of Driver
Qts of Driver
Fs (or Fo) of Driver
Diameter of port in inches in decimal form (PD)
F3 = ((Vas/Box Vol)^0.5)*Fs
Fb = ((Vas/Box Vol)^0.32)*Fs
Port length = (1.463*(10^7)*((0.5*PD) ^ 2))/(Fb^2*Box vol*1728)-(1.463*0.5*PD)
This next section has to do with the peak response due to non-optimal box size. The peak may be positive or negative, indicating the level of the peak around the Fb in dB. There are 2 steps: the first is to calculate a temporary value (V) that will then be used to find the dB level. I will give you 2 ways to do the second part since QBASIC uses a different LOG function than standard calculators such as the TI35 and similar. (one is LOG and the other is Natural LOG, I forget which is which)
V = ((Vas/Box vol)^0.35)*Qts*2.6
Peak Response in dB = (LOG(V))*0.43429445*20 [this is the QBASIC version]
Peak Response in dB = (LOG(V))*20 [this is the Calculator version]
A helpful hint: most software calculators have conversion sections built into them. You just have to expand them and look under their menus. Often times they also include a Scientific layout that gives you access to the extra functions in the formulae.
Btw, in case you (the reader of this post) wonder what the caret symbol (^) means, it is to raise the preceding number to the power of the following number. So 2^3 is 2 to the 3rd power, or 2^3=8
Now, using all this info we start with the required info:
Box Vol 0.388 CuFt (11L)
Vas 0.28 CuFt. (7.93L)
Fs 51.81 Hz
PD 1.375″ (1-3/8″)
Plugging all that info into the formula gives you:
For a box size of 0.388 Cu.Ft.
Your target driver will have a
F3 of 44.01257
Fb of 46.67432
a 1.375″ diameter port at 3.728515″ long
A Peak response of -1.096577 dB above baseline
Obviously, since the Peak response is negative, it is below baseline.
Now running the box dimensions in the C-Note kit on PE, you end up with an internal volume of ~0.32 Cu.Ft. This calculates out to:
F3 of 48.46382
Fb of 49.64279
a 1.375″ diameter port at 4.068575″ long
A Peak response of -0.5108053 dB above baseline
In your case, a slightly larger box results in slightly extended bass response that is just a tad stronger than originally designed, even though it may seem opposite. Say the baseline is 87 dB. The C-Note would be 86.5 dB at 49-1/2 Hz and 83.5 dB at 48-1/2 Hz, while your box would be 86 dB at 46-1/2 Hz and 83 dB at 44 Hz. So very close, but at lower frequencies.
I hope I haven’t confused you too much. It can be a daunting task in the beginning. Once you dive in, it can be addicting. Take care, and ask away. I’ll share what I know.
September 17, 2020 at 5:36 pm #11799
September 18, 2020 at 7:32 am #11803
Does a tweeter need an enclosed volume as well? I mean, are the woofer and tweeter sharing the box volume, or am I calculating the final box volume of both drivers with the above formulas?
Thanks again, really appreciate it!
September 18, 2020 at 8:06 am #11806
That is a great question. Almost all tweeters are sealed units. So they do not need an enclosure. So if you were to open up a two way speaker, you will see the tweeter sharing the same space as the mid. That it perfectly acceptable. However, if you are building a three-way the mid and woofer would need their own enclosure. Typically the mid would be sealed and the woofer could be either sealed or ported.
September 18, 2020 at 8:15 am #11807
September 18, 2020 at 10:02 am #11808
Some follow-up questions, if you don’t mind 🙂
1) In your very educational calculations (thanks for that!) you conclude a port length of 3.73 inches. Is there any calculation / requirement for the distance from the “internal end” of the port tube to the front panel? See sketch
2) In the assembly manual for the C-note kit the recommended port length is 7″. I imagine that’s the reason the speakers have a relative long depth of 9.5″? Is it so that the longer port length you have, the more bottom you will get? Your calculation states a port length of 4.06″, which would allow me to make a more streamlined box, but what are the trade-offs?
Overall, I guess what I am trying to ask is, if the depth of the box has any importance (other than the required length of the port, of course) or if it is just the overall volume H x L x D that matters?
Thanks again 🙂
September 18, 2020 at 9:09 pm #11811
Re: tweeter – what Nick (123Toid) said. Can’t add anything more. 🙂
Re: port – you usually want at least the same distance (or more) as the diameter of the port from the wall to the port opening. So your ??” would at least be the same or more as the port diameter.
Not sure why they have a 7″ port other than tuning or phase inversion due to rear porting. Maybe Nick can chime in. (Also I haven’t had much chance to play with WinISD, so I don’t know how to mess with port lengths. I tend to go with the calculated “optimum” lengths rather than “forcing it”.)
September 18, 2020 at 9:49 pm #11812
September 18, 2020 at 10:39 pm #11814
When you look at how a speaker works, pushing air in first one direction and then the other in order to make the sound wave, and then knowing that the way the port works is to reinforce the target frequency/frequencies by being in phase with the sound coming from the front face of the speaker, and also realizing that no matter where the port is placed (front, rear, sides, top, bottom) the speaker is always acting the same on the port. That is, when the cone moves outward, it pulls air into the port from outside and when moving inward it will force air back into the port from inside. This is a constant fact. But where you place the port on the enclosure determines the phase shift the port needs to accomplish in order to have the port energy match the front face energy to actually reinforce each other. So, if the port is on the front, you want it to be in phase at the resonant frequency so they both push and pull in sync (technically the port is 180° since the back of the cone is pulling while the port is pushing). When you rear port, you want the rear port to be 180° (and again, technically this is 0° since the port is tracking the back of the cone) so that by the time it wraps around the speaker it becomes in phase with the front. The tuning of the port / length determines the output phase of the tuned frequency.
Short enough, yet long enough to give a basic idea, I hope. It’s a deep subject that actually can require quite a few chapters to really nail down.
September 19, 2020 at 8:13 pm #11820
Thank you for that, once again. I understand the first part fine, that makes total sense to me.
But you lost me on this second part:
“But where you place the port on the enclosure determines the phase shift the port needs to accomplish in order to have the port energy match the front face energy to actually reinforce each other. So, if the port is on the front, you want it to be in phase at the resonant frequency so they both push and pull in sync (technically the port is 180° since the back of the cone is pulling while the port is pushing). When you rear port, you want the rear port to be 180° (and again, technically this is 0° since the port is tracking the back of the cone) so that by the time it wraps around the speaker it becomes in phase with the front. The tuning of the port / length determines the output phase of the tuned frequency.”
I am sure you have actually answered my question, I just don’t understand the answer 🙂 For example, if I place a port on the rear, say 1,375″ diameter and 4″ in length as per your calculation, that would be optimal for that given setup. What would be different if I placed the port on the front in stead? A 180 degree bending tube, a different length? A different calculation?
I will be trying to install a port on the back of the speaker box. Are the calculations you have given me suitable for that? Do I need to take something additional into account, since you wrote that the reason PE had chosen a 7″ length for the port, even though the calculations states 4″, might be because of phase inversion due to rear porting?
Sorry I keep pushing on this, I just really want to understand. I might be talking about something totally different 🙂
September 19, 2020 at 9:45 pm #11821
To help with the two questions you have above.
Like @tvor-ceasar said, you typically will want about the port diameter away from the port wall. This simply means, if you are using a 1.5″ port you will want it to be at least 1.” away from the front baffle.
Your second question was what happens if you lengthen the port. This will give you a lower tuning frequency. However, that does not mean you will get more low end. You could be tuning the enclosure too low, which would actually lose bass performance. It also will affect cone excursion and rear port air velocity. That is one of the reasons using a program like WinISD is so important. It will allow you to see what would happen if you lengthen or shorten the port.
September 19, 2020 at 10:12 pm #11822
@123toid thanks! I think I will play around with the numbers a lot more before trying to understand exactly what is going on. Seems like the right approach in this case, but at least now I understand the addiction 😉
The videos you’ve linked me are great and very easy to follow step by step, but I have a habit of always trying to understand what happens behind the curtain. Not being able to interpret and validate computer readings and blindly following data output seems like a fast track to making bad designs 🙂
That’s one of the reasons this forum rocks!
October 1, 2020 at 9:07 am #11892
Hello again – Just wondering, can WinISD and the excel spreadsheet Unibox 4.08 (attached) do the same? I mean, do they work / calculate the same?
I have run into a problem where I cannot install WinISD on my laptop, because it is a company PC loaded with restrictions.
All I need it to figure out a good box size for the C-note kit drivers related to a good port length, I guess. The excel sheet seems to be able to do that with some help from @tvor-ceasar ‘s math above. But I have never tried WinISD so I don’t know what I am comparing to.
Sorry if I am asking an obvious question …
October 1, 2020 at 12:46 pm #11893
I don’t know what Unibox does for it’s calculations, since I’ve never used it.
WinISD – I am still learning how to use that as well. I’ve only done 2 drivers with it so far. However, in previous posts in which Nick ( @123Toid ) and I have given quick calcs to others or conferred back and forth, our end numbers have been pretty much identical.
For me, This Post explains pretty much where I got my info for calcs and also what I’ve done with it – I have written multiple programs in QuickBasic to answer my questions about the drivers I’m looking into. This is why, in an earlier post, I said that the minimum needs were Qts, Vas, and Fo/Fs. I put them in and pop out a scenario. Takes a few seconds. It actually takes longer to open up DosBox (needed to run QuickBasic) and set everything up than it does to enter information and check out the result. I am in the slow process of compiling them all into one for release to you all as a comparable quick check.
Now, even though those books were the big jumping off point, I have refined things over the years and am still doing so when I find reliable info on the web. I have many bookmarks that I will share when the Resources Thread gets up. (hint hint, Nick. 🙂 ) I don’t claim to be the last word, but a good jumping off point to get started.
I, like you, want to know how things work from the inside out. Once you know, then you can weed out the good information and opinions from the bad. Always the quest for me.
Also, I am trying to think of a way to answer your previous question to me. Sometimes it’s tough putting into words what I see in my head in such a way as to be easily understood. No offense meant. I have been told by others that I tend to speak “over people’s heads”, so I really do try to be thorough yet not too technical. I hope that comes across right.
October 2, 2020 at 8:36 am #11898
@tvor-ceasar thanks for that. yes, WinISD looks pretty useful from the tutorial videos that @123toid gave me. I just need to find another computer to try it out. But I figure, that maybe my task isn’t even so complicated that I need the program to figure it out. I was mostly just for playing around. The sheet seems to calculate some of the same things, but we’ll see.
Thank you for the consideration, my wife actually tells me the same if I become too technical about something in my field. But I welcome that “language” from others, if that is the best way for them to explain. I usually read the sentences many times anyway, simply because English is my second language. Also, I read them and think about them until I understand. If I hit a wall, I ask for further explanation. So, explain however feels natural to you, I don’t mind 🙂
October 4, 2020 at 7:47 pm #11904
I try not to use too technical language, unless I explain it. Otherwise, I feel too many people getting into this or people with language barriers, will get frustrated. Not to mention most of the people around me, don’t quite get the lingo. So it is nice to try to simplify it to something they can understand.
October 4, 2020 at 8:27 pm #11905
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