Hi,
I know that I can calculate the length of a port for a particular box volume, frequency and port area, using this formula:
- where L(ength of the port) and D(iameter of the port) are in metres, V(olume of the enclosure) is in litres and F(requency of port resonance) is in Hertz.
But I'm a bit lost in how to reverse the process (or the formula). You see, I'm looking to work out the port diameter to use when I already know the length of the port.
Why? you ask. I was just trying to see how easy it would be to use a hole cut in a box as a port - so the length might be 1 inch (25mm) or half an inch (12.5mm). Or I might have a circular 'cover' on the and of a pipe speaker and want to know what height to make the gap so that it resonates at the right frequency.
Just for interest at the moment...
I know that I can calculate the length of a port for a particular box volume, frequency and port area, using this formula:
- where L(ength of the port) and D(iameter of the port) are in metres, V(olume of the enclosure) is in litres and F(requency of port resonance) is in Hertz.
But I'm a bit lost in how to reverse the process (or the formula). You see, I'm looking to work out the port diameter to use when I already know the length of the port.
Why? you ask. I was just trying to see how easy it would be to use a hole cut in a box as a port - so the length might be 1 inch (25mm) or half an inch (12.5mm). Or I might have a circular 'cover' on the and of a pipe speaker and want to know what height to make the gap so that it resonates at the right frequency.
Just for interest at the moment...
Hi Jont,
I'm too stunned and lazy to work that out - I use Unibox for calculating such things.
Chances are a hole will not suffice as an effective port for most applications. In order to reduce length, diameter needs to be reduced as well. This can lead to "port noise" if the diameter is too small.
I'm too stunned and lazy to work that out - I use Unibox for calculating such things.
Chances are a hole will not suffice as an effective port for most applications. In order to reduce length, diameter needs to be reduced as well. This can lead to "port noise" if the diameter is too small.
It's been a while, but the last time I did this, I set up the normal formula in Lotus 123 and used the "Goal Searching" What-If feature to derive the permutations.
Syd
I'm currently fiddling around on Excel, trying various sizes until it fits. What I was looking at (that brought this up) was thinking of how to make surrounds WAF positive.
I've seen these bowls from ikea and thought that if I mounted a F/R driver in the 'base' and then fitted them to a flat base and hung that on the wall then it might work. After MJL's input, may with a little 'brim' on the edge, then they might work (as this would increase the length of the 'port'). Like this (primitive) sketch:
I've seen these bowls from ikea and thought that if I mounted a F/R driver in the 'base' and then fitted them to a flat base and hung that on the wall then it might work. After MJL's input, may with a little 'brim' on the edge, then they might work (as this would increase the length of the 'port'). Like this (primitive) sketch:
Attachments
Jon they're surrounds, make em sealed and loose the ulcerating addition and subtraction and all that math.
cool idea tho
cool idea tho
How would you solve
a*x^2+b*x+c=0
where:
x=D
c=-L
b=-0.8
a=2e6/(Vb*Fb^2)
doesn't that look like a quadratic root question?
Isn't there some sort of formula that we were supposed to memorize when we were ~14-15 years old, something like:
http://en.wikipedia.org/wiki/Quadratic_equation
note that there are two solutions, but one of them will be easy enough to throw out if you think on it a bit.
a*x^2+b*x+c=0
where:
x=D
c=-L
b=-0.8
a=2e6/(Vb*Fb^2)
doesn't that look like a quadratic root question?
Isn't there some sort of formula that we were supposed to memorize when we were ~14-15 years old, something like:
http://en.wikipedia.org/wiki/Quadratic_equation
note that there are two solutions, but one of them will be easy enough to throw out if you think on it a bit.
WinISD Beta is a good, fast program to figure that out 🙂
(*nix users - it runs flawless in WinE, at least in Ubuntu 😉 )
Cheers!
(*nix users - it runs flawless in WinE, at least in Ubuntu 😉 )
Cheers!
If you use the Helmholtz formula..
f = k(a/lv)^.5, k=c/(2pi)
for any given f and v the ratio a/l is constant
rcw.
f = k(a/lv)^.5, k=c/(2pi)
for any given f and v the ratio a/l is constant
rcw.
Moondog55 said:
It works out (after the math - or using an Excel spreadsheet) that the 'port' is roughly 84cm 'wide'. And so it would be less than a mil high. It probably wouldn't work as a port at all, more like an aperiodic enclosure.
I might have to try it out anyway - I can always just tighten the bolts a little more and it'll become sealed...
I would be concerned that with a gap that narrow you would get audible noise/ whistle.
I don't use small ports for that reason.
Syd
Cloth Ears said:
A restrictive port with very low Q is little or no better than a sealed box. Maybe you could fashion a little shelf on the back plate to extend the port a little. Is there a bowl which nests inside the one you are using? Note that your port is expanding so the area to use in your calcs would be (approx) the perimeter of a line in the middle of the length (the bowl brim in this case) and the width of the opening. This is a bit of an odd shape so the actual tuning frequency could be perhaps 20% lower or higher than predicted. This won't matter much if it is acting like an aperiodic, though.
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