Question on How to Measure Port length-2 diff. designs

[Texas Tech 10+ year member]

Here are two Simple Diagrams from Paint Program. I want to know how you would calculate the length of the port,given the measurements. Just what you would add up and the answer. Thanks

First, i have given the lengths measured down the middle of the port, and width of the port.

Second,I have given a length, and a width (3 inches). I also want to know what the length of the red line should be, or does it not matter.

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[slikrider20 10+ year member]

Doesn't matter..good question though.

And in your first design, end correction will throw off your tuning a yay bit, but i wouldn't even bother. Even moe's plans don't factor that in.

btw nice paint skills

 

[Texas Tech 10+ year member]

anybody know the length of the designs? there has to be someone who knows

 

[slikrider20 10+ year member]

anybody know the length of the designs? there has to be someone who knows

Last time i checked 12+7 was 19. i'm not sure though, better use the search function.

and i'm also not sure, but i think on the bottom one it's 15 not sure though

 

[fwb_1234 10+ year member]

what he really wants to know is how to calculate the "end correction"

 

[Texas Tech 10+ year member]

yea i know basic math, I mean is end correction not that big of a deal to calculate in. I figured for the first diagram it would be 12 +7 + 1/2(width) because for a "L" shaped port using the enclosures wall as a side to create a port you measure down the middle and add half ot he width for end correction. I was making sure that still applied if you Had TWO "L" shaped ports for two subs, one each having its own "L" Port (of course this is all illustrated in the first post on the firt diagram.

And all i know is you want to have the distance of the RED LINE on the second diagram ATLEAST the width of your port.

So, Is not calculating end correction on some designs okay to do? i know it does not effect your tuning drastically or anything.

Thanks for the help

 

[PowerNaudio 10+ year member]

here is an example of how you measure the port length for a T port.



another example.



i made these a while back so they are kind of crude. but get the point accross.

when calculating end correction. just add half the port width to the calculated port length. the diffrent wont be a critical one but the port will be the proper length for the calculated tuning.

on the second drawing that you posted the length of the port should be.

15" + 3" + .75" = 18.75"

 

[Immacomputer 10+ year member]

PowerNaudio, your pictures don't factor in end correction so they are not really correct. If part of the port is against a wall, it will use part of the air by the wall as part of the port.

This is what it should look like:

To the OP, if that gap at the end of the port line is 3 inches or smaller, the port will try to extend around the corner. If it is smaller, you will have issues with turbulence. Here is a picture of what I'm talking about:

For the second picture you posted, just make sure the red line is at least 3". You want to keep the back wall at least it's width away from the end of the port... more would actually be better.

 

[PowerNaudio 10+ year member]

port end correction does not take enclosure volume like you have in the drawing. so your drawing is wrong, because the blue area in that drawing clearly stated that it is port displacement. so the drawing i have is correct. if you look at the drawing it clearly explains what its is illustrating and what the drawing is for. wich is how to properly measure a bent port and port displacement. you can modify the drawing to illustrate port correction. but please use a diffrent color because blue is already for port displacement and the phisical port length, not for end correction

 

[Immacomputer 10+ year member]

But it is going to take up enclosure volume because the air there is being used by the port and is a part of the port. The sub is NOT seeing that airspace even though you may think it does.

 

[PowerNaudio 10+ year member]

But it is going to take up enclosure volume because the air there is being used by the port and is a part of the port. The sub is NOT seeing that airspace even though you may think it does.

sure, but that air is still air, not displacement being taken up. so sorry, but no. its not calculated as displacement.

 

[Immacomputer 10+ year member]

sure, but that air is still air, not displacement being taken up. so sorry, but no. its not calculated as displacement.

Can you tell me why the air in the port is not calculated as being net volume? Explain this to me.

 

[PowerNaudio 10+ year member]

Can you tell me why the air in the port is not calculated as being net volume? Explain this to me.

i never said the air in the port. i said the end correction is not taken into account as being displacement.

 

[Immacomputer 10+ year member]

i never said the air in the port. i said the end correction is not taken into account as being displacement.

So I'm guessing that means you can't answer my question then. I want to know why you don't use the air in the port for net volume. Please answer this.

 

[PowerNaudio 10+ year member]

So I'm guessing that means you can't answer my question then. I want to know why you don't use the air in the port for net volume. Please answer this.

because the air in the port ,is not part of the internal volume of the enclosure and seen as a mass by the sub, at all times by the sub.

the sub has two strokes a negative stroke and the positive. port end correction is only good for the negative stroke when air is being pulled in. when the sub its in its positive stroke the air is being pushed out. so end correction would have to be looked at from the out side of the enclosure, if there was a wall at that end. inside the enclosure that end correction is none existant on the positive stroke and thats why the end correction is not accounted for as port displacement. because its only good for one half of the subwoofers full stroke and not there for the other half. so it will make zero sence to account for it as a port displacement.