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

[PowerNaudio 10+ year member]

Hey you said on MY second picture the length would be calculated by adding 15+3+.75=18.75 Where did you get the 3" and .75" from. I know why you calculated 15 inches, because it is the actual length of the port, but i dont see why you added in 3 and .75 inches. Could you further explain that please

here is your image modified to help you understand it a bit better.

the values that i give you are assuming that the red line is 3", because you didnt provide any values for it. but you can go as small as 1.5" for the red line because the port is open on both ends which will still provide the proper port area for the port. any less and you will change the characteristics of the port.

so the math would be 15" + 1.5" + 1.5" + .75"=18.75"

And on this one

http://www.putfile.com/pic.php?img=3305668

You would basically just have to calculate one side of the "T" port" so on that diagram you would just do 10.25+5.75+1 (cuz that is half of the width [2"]) or would you do 10.25+5.75+2 because the width of the "T" port is really 4 inches?

for the end correction of my T port drawin. it would be 10.25+5.75+1, only if you taken into account the K factor before you made the drawing. you would calculate the port length and then add 1" to the physical length of the port to get the perceived length of the port.

now if you are designing the port then once you calculate the port length you would take the 1" "on a 2" wide port" from the length of the port, to properly account for the K factor "port end correction".

So to recall, so i feal assured on this, on MY first diagram, the T port with a wood board down teh middle to seperate the two ports, to calculate the length you would add 12+7+1.5(because that is half of 3,the width)

about your T port. my drawing is a true T port "no baffle separating each port". your drawing is not a T port, but two L shaped ports back to back, so you would calculate your T port length and width of the port at the end of the port, just like you would for a regular L shaped port. and 12+7+1.5 is incorrect. assuming that the port end is 3" from the nearest wall and 3" wide. it would be: 12" + 7" + 1.5" + 1.5" + .75 + 1.5"= 24.25" of perceived port length

when you draw your port you don't draw the port with port end correction, because port end correction is a perceived length. its not the physical length of the port. you draw the physical length of the port. the same way it would be built.

And also on your second diagram, the "L" port

http://www.putfile.com/pic.php?img=3305715

To calculate that length you would add 21.25+18.75+2 (half of the width)

yes to figured the perceived length of the port you would add 21.25+18.75+2 (half of the width) that will get you the perceived length, so you can find the tuning of the port. when you draw your port or build the enclosure. your port should look just like on the drawing.

well i hope that helped clear it up for you.

if not then ask away.

 

[PowerNaudio 10+ year member]

oh btw, i used scketchUP for both drawings then i modified them a bit with adobe photoshop.

 

[Texas Tech 10+ year member]

okay i am going to put all the diagrams into my paint program (so just a top view) and i will say the Effective Port Length, The Physical Port length, the port area, and how i calculated all of it. Thanks for the help and i will look later at the replies because i cant view the images because i am at school so when i get home i will look at the replies and ask some questions i might have

Thanks for your willingness to help me out

 

[Texas Tech 10+ year member]

oh, and the reason i am so interested in how to find the effective or percieved port length is because i am trying to put everything together to find out tuning frequencies of an enclosure.I will reply on the previous posts tonight

thanks for everything

 

[Texas Tech 10+ year member]

15" + 1.5" + 1.5" + .75"=18.75"

Why did you add .75 though. For the thickness of the wood?

I dont know if it is for the thickness of the front baffle, but on the diagram the 15 Inches of length starts at the very front of the port, so it does take into account the thickness of the baffle.

Assuming that the red lin is 3"....

And you do add the length of the red line (the red line) even though it is actually not any physical length? Is that like adding half of the ports width for the left chamber and half of the width for the right chamber so it adds up to be 3 inches.

 

[PowerNaudio 10+ year member]

15" + 1.5" + 1.5" + .75"=18.75"
Why did you add .75 though. For the thickness of the wood?

I dont know if it is for the thickness of the front baffle, but on the diagram the 15 Inches of length starts at the very front of the port, so it does take into account the thickness of the baffle.

Assuming that the red lin is 3"....

And you do add the length of the red line (the red line) even though it is actually not any physical length? Is that like adding half of the ports width for the left chamber and half of the width for the right chamber so it adds up to be 3 inches.

let me see if i can put this into words since i guess the drawing is a bit hard to figure out. lol

that straight port although straight is measure just like a T port but with out the port length on each side of the port.

ok the port starts at the out side opening of the port on the front baffle and ends on the inside end of the port inside the enclosure.

now since the port opening inside the enclosure is 3" away from the rear wall and facing the wall, the openings around the port turn into part of the port. and have to be accounted as port area.

so the port doesn't end on the physical end of the port but on the side of the port.

the red line in the drawing shows the distance from the wall to the port opening.

now you have to find the center point of that space. which is 1.5" from the wall and 2.25" from the outside edges of the port side wall. you have to account the port wall .75" as part of the the port length during a bend like that.

so you add the physical length of the port then from the end of the port to the 1.5" center of the space between the port and the rear wall. then a 2.25" from that center point to the side of of the port. including the ports mdf .75 wall. i had broken down that 2.5" in to two different values so you would get a better idea of where those numbers where coming from which where the port mdf wall and the 1.5" from the center of the port to the inside edge of the port.

if you look at the drawing that i modified of your drawing youll see the what the vales are measuring.

let me know if that explains it.

 

[Texas Tech 10+ year member]

physical length+length to center point from end of physical port and rear wall of enclosure+length of center point to one side of the port (including the woods thickness)

i got it now but i still have one question that i have to ask....

The measuring to the middle point makes sense for end correction and all,

but the measuring to one side is my question...why do you only measure to one side, why would you not measure to both sides, so instead of it being 2.25 it would be 4.5. But this will answer my question. You said accounted for port area, So that is only calculated in with port area?

So is 18.75 the effective length. That is the length i would put into my equation to find out what a port is tuned to.

And the Port Area would be 15(3)+1.5(2.25)

So it equals 48.375 square inches.

 

[PowerNaudio 10+ year member]

the reason you measure the port that way is not because of port end correction. is because the port will be in the middle of the enclosure with a sub on each side. the subs move the mass of air from each side, so you add the port length with the dimensions of one side only, if you where going to do it the way you suggested the port length would be too long and incorrect.

also on the T port drawing that i made youll notice there is a center deflector, these are necessary, to help the air flow properly instead of having it clash against the back of the wall, which will cause turbulence and make the port perceptively longer.

port correction is subtracted from the port before you draw it, its done while you calculate the port length.

so if the port is drawn or built up as 18.75 and is 3" wide to figure the tuning you would subtract 1.5" to the 18.75" of port length.

if you are calculating port length and your port calculations equal L:18.75" x W:3" , to account for end correction, you'll add 1.5" of port length.

 

[Texas Tech 10+ year member]

yea, i understand that.

So in a tuning equation i would put in 18.75 in as the length of the port for that diagram we have been discussing

 

[PowerNaudio 10+ year member]

if you have already accounted for port end correction yes, if you havent then add port end correction if you want to.

 

[PowerNaudio 10+ year member]

when you are doing a port like the one you have, where if it was longer has to split down to the sides, if its easier for you you can calculate the port as if it was two ports and half the width and you would split one port towards one side and the other port towards the other side. but you can leave the baffle in the middle out if you wanted to and replace it with the a deflector on the wall in the middle of the port.

 

[Texas Tech 10+ year member]

ooh okay so all that calculating, the 1.5 and 2.25 is not necessarily end correction, it is just necessary to calculate that for that type of port.However, that is added to the ports percieved length. So since the width is 3 inches on that diagram you would have to add 1.5 to the length for the total effective length. So it would be 20.25

Is end correction always half of the width.

 

[Texas Tech 10+ year member]

I got ya though, on calculating port length and the actual physical length of the port. I understand how the end correction is not in any way constructed into the box. The only time you would actually use the end correction would be to figure out port tuning. So when you build a box and the calculations say the percieved length(effective length i call it from JL Audio's website) is 15 inches and the width of the port is 4 inches, you would actually subtract 2 inches from that 15 inches. So when building your box you would make it to where your port length is 13 inches long.

 

[PowerNaudio 10+ year member]

I got ya though, on calculating port length and the actual physical length of the port. I understand how the end correction is not in any way constructed into the box. The only time you would actually use the end correction would be to figure out port tuning. So when you build a box and the calculations say the percieved length(effective length i call it from JL Audio's website) is 15 inches and the width of the port is 4 inches, you would actually subtract 2 inches from that 15 inches. So when building your box you would make it to where your port length is 13 inches long.

finally.

lol

 

[Texas Tech 10+ year member]

ooh okay so all that calculating, the 1.5 and 2.25 is not necessarily end correction, it is just necessary to calculate that for that type of port.However, that is added to the ports percieved length. So since the width is 3 inches on that diagram you would have to add 1.5 to the length for the total effective length. So it would be 20.25

is all that correct about the diagram though