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View Full Version : Since we are on the subject of ports...



Hecta G
02-06-2008, 11:18 PM
In the rendering below, I have done a cutaway for easier viewing and explaining. The port is behind a 3/4" MDF. The outer tube is open on one end and has an OD of 4".

The inside tube has an OD of 2", both with a wall thickness of 1/16". Each tube is 6" long, and the arrows show the direction of air travel.

My question is how can you calculate effective port length when comparing to standard cylindrical ports?

http://sgcalligraphy.com/labrynth.jpg

Cravingbass123
02-06-2008, 11:19 PM
wtf???????????????????????????


why?

mrbomb
02-06-2008, 11:19 PM
interesting

matrxx dude
02-06-2008, 11:20 PM
diffferent size ends.

matrxx dude
02-06-2008, 11:21 PM
crazy ****

Hecta G
02-06-2008, 11:26 PM
Actually, it was something that I patented several years ago, and since I was young, had a crazy idea, built a prototype and actually worked.

My problem was why it worked. Until now I haven't wondered too much about it but with some of the guys here really REALLY know more about acoustics than I do. I approached from a mechanical side and was wondering if someone here could explain it to me.

I'm thinking about building some more to test and see if it makes any sense still.

BrianChia
02-07-2008, 12:30 AM
It's not really anything that new; take a look at bullhorns--they are a concentric folded horn, just like this but with exponentially increasing flare and loaded from the other end. But I have never seen this concept applied to ports.

twisztdauthorit
02-07-2008, 11:30 AM
interesting design, aslong as the airflow is smooth, i don't see why it wouldn't work. as for calculating it, well thats not my cup of tea.

Immacomputer
02-07-2008, 12:47 PM
So wait, you paid for a patent on something you don't understand and don't know how to design. Wow...

Hecta G
02-07-2008, 03:30 PM
So wait, you paid for a patent on something you don't understand and don't know how to design. Wow...

Your helpful...

as usual...

Immacomputer
02-07-2008, 05:31 PM
What's the patent number?

Hecta G
02-07-2008, 05:33 PM
What's the patent number?

5406637

loudaccent
02-07-2008, 08:51 PM
i took a look for the hell of it here it is
http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PALL&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.htm&r=1&f=G&l=50&s1=5406637.PN.&OS=PN/5406637&RS=PN/5406637

SilverSquared
02-07-2008, 08:54 PM
in the images there, WOW...I'll buy 4

Hecta G
02-09-2008, 05:00 PM
Does this mean that no one here can calculate it?

Immacomputer
02-09-2008, 08:15 PM
How about you do some measurements instead of us doing your work? I read through your description and patent and it sounds very basic. Who paid for that patent? What advantage does this have over a standard Helmholtz resonator?

Why don't you take some measurements to find the actual tuning frequency and then vary the height of the outside tube and see how it affects the measurements. Try to single out one variable at a time and work backwards to find a way to calculate and measure it. Making an electrical equivalent circuit would also help aid in pointing you in the right direction.

It's your patent; you do the work. I know that I wouldn't be doing jack **** as far as helping you along the way unless I share in the patent.

BrianChia
02-09-2008, 08:26 PM
Does this mean that no one here can calculate it?

No, I'm not trying to be a ****, but I just think that no one here really cares.

Hecta G
02-09-2008, 09:35 PM
Yeah, thanks..

Figured that much...

Hecta G
02-09-2008, 09:35 PM
Just asking to be lead in the right direction. Not asking anyone do to anything for me.

Hintzyboy
02-09-2008, 10:12 PM
It really shouldn't be too difficult. If my brain is working right, it's just basic trig. as long as this equation works out (A=area)

Ainner port = Aouter port-Ainner port


If that works out, then it's just a basic linear measurement. You just have to find the right size ports to make that work.

just as an example. For that design to work with a two inch inner port, you would have to use a 2.8" outer port. Then it would work like a standard cylindrical port, but it would take up less space outside the box.

If that's not what you're going for, then IDK. I know nothing about horns or any of that kind of stuff.

Hecta G
02-09-2008, 11:14 PM
Thanks Hintz,

Yeah I've done that and don't know much about horns myself. I have the basic difference in area and volume but trying to figure out the effect of each change in area.

I am assuming treating the first entry (small-large) as if it was a straight port, then taking the smaller port and it's length as if coming out of the enclosure. I may not even be describing this correctly.

I may just incorporate it into my next box and see what happens.

I think I just may post my build as i get to it. I'm sure there are a few here that could use a few giggles.

DarkFox
02-10-2008, 12:58 AM
yeah I was gonna say, this isn't anything new, just a diff way to do it. when people take a slot port and double it back on itself to get the length in a short enclosure, this is the SAME EXACT THING. just got to get it so you are left with the same area in the larger diameter, after you subtract the smaller diameter tube.... port length vs area calculations should work exactly the same.

Hecta G
02-10-2008, 01:29 AM
Thanks Fox,

It just hit with what you were saying. I was assuming on 2 different areas, but I think what you stated make them equal.

The difference between the larger and the smaller should equal the smaller. Let me know if that's what you are talking about.

Thanks!!!

James Bang
02-10-2008, 01:42 AM
I'd actually like to see this implemented.


I'd round out the outer/larger tube.

whizdumb
02-10-2008, 02:56 AM
Wow that is a really cool idea. I never would have thought of that...

Great for when you need a long port length but your box isn't deep enough. But how do you secure the outer tube?

Trying to figure out port length I think it would be a lot easier if each tube had the same effective volume. The one you described has the outer port being roughly 9.42 square inches while the inside port is only about 3.14 sq. in (not counting for tube thicknesses). Then you could just add up the effective lengths (adding some more for end correction) and you would have a pretty close approximation.

But yeah what you would probably have to do is build it and test to find out what the actual tuning is and go from there.

Immacomputer
02-10-2008, 09:28 AM
But it won't act completely like a ported enclosure as the mass of air in the outer tube will have its own point of resonance as will the inner port tube. I would bet that the distance away from the inner tube opening will have just as much of an effect as changing the length of either tube.

savagebee
02-10-2008, 09:31 AM
x2, I mean, wouldnt you also have to factor in the end correction that the inner and outer ports would cause where they sepearate?
I have no idea how that would even be possible

djman37
02-10-2008, 09:39 AM
what about when the green cylinder slides back far enough to become an enclosure itself? it looks like tuning wouldn't really change as the 'port length' doesn't really change unless you cut the green.