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[urgent] Slot ported version of The Tango
[urgent] Slot ported version of The Tango
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Old 4th May 2017, 07:37 AM   #1
HumbleDeer is offline HumbleDeer  Belgium
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Default [urgent] Slot ported version of The Tango

Hey everyone,

I'm sorry if it comes across as a bit selfish or alike, but I'm in a rush for school. In two days I have to order my parts and give my teacher the plans, and I haven't gotten any luck with the time stretching designer I hired before.

I'm looking to replace the round port design by a slot port on bagby's design of The Tangos. I need them to perform equally well as the round ported ones but I don't know how to do it.

I'd appreciate it if someone could help me out, and save my GIP.

Thanks in advance,
Anna
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Old 4th May 2017, 08:12 AM   #2
Zvu is offline Zvu  Serbia
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Calculate surface of the circle using inner diameter of port tube. Use the same surface for slot port. Pipe depth remains the same regardless of port shape.
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Last edited by Zvu; 4th May 2017 at 08:21 AM.
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Old 4th May 2017, 01:51 PM   #3
HumbleDeer is offline HumbleDeer  Belgium
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I don't know and never understood how to do that. I don't have time to spend tedious hours learning it.
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Old 4th May 2017, 01:59 PM   #4
isaeagle4031 is offline isaeagle4031  United States
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What port diameter is specified?
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Old 4th May 2017, 02:13 PM   #5
HumbleDeer is offline HumbleDeer  Belgium
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I will elaborate at home, but the specifications say 2" port that's 7inches and 3/4 including the 3/4 inch MDF in the back.

I'm not able to figure out myself how to add a tuning port without messing up internal volume because that one needs to stay exactly the same.

Thanks for the help.
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Old 4th May 2017, 02:50 PM   #6
chris661 is offline chris661  United Kingdom
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Then you need a slot port with the same cross-sectional area as the 2" diameter round port, it needs to be 7.75" long, and once you've done that, you can play with the cabinet depth a little to get the volume the same as the original.

Hint: the volume of air inside the port is the same for the round or slot port, so the volume you've used is the difference between the volume of the wood for the slot port, and the volume of whatever material the round port was made of.

Maths lesson:
The cross-sectional area of a cylinder is just the area of the circle. Circles are unique in that a single number can be used to find a lot of different things. That number is the radius, the distance from the middle of the circle to the edge. The radius is also half of the diameter. If in doubt, draw it out - that'll make it pretty obvious.
Now, there are some simple equations we can use to find out various things about this circle once we know it's radius.
Lets say I've got one of those plastic ports that have a flare at one end, and I want to find out the length of a strip of foam I need to go around it.
Going all the way around is called the circumference, and it's simple to find. C = 2*(pi)*r
Where..
C = circumference
(pi) = 3.141592654 (a unique number that perfectly describes the ratio of a circle's circumference and diameter)
r = radius.
Since the diameter, d, is equal to 2*r, we can also write C = (pi)*d

So far, all of our units are a distance, and they're consistent. A circle with a radius of 0.5cm will have a circumference of 3.14159...cm. A circle with a radius of 0.5m will have a circumference of 3.14159...m.

The area of a circle is a little more tricky, since we're starting out with the radius, which is measured as a distance, and we're getting out an area, which is measured as distance^2 (squared)
A=(pi)*r^2, or A=(pi)*r*r. Those two equations mean the same thing,. Raising something to the power of 2 means you multiply it by itself. A power of 3 means you multiply it again. You'll learn more about that later in your academic career.

In your case, the port is 2" diameter. That means the radius is 1", or 2.54cm since you really ought to be working in Metric, and don't let the Americans tell you otherwise.
So, r=2.54cm. Put that into the area equation and see what you get.

Chris

Last edited by chris661; 4th May 2017 at 03:04 PM.
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Old 4th May 2017, 02:54 PM   #7
HumbleDeer is offline HumbleDeer  Belgium
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Hmm thanks that makes alot of Sense. So. If the round one is by corners 2*2 then the slot port can be anything multiplied that multiplies to the same as 2*2 like 1*4 or .5*8? Correct me if I'm wrong.
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Old 4th May 2017, 03:05 PM   #8
chris661 is offline chris661  United Kingdom
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Quote:
Originally Posted by HumbleDeer View Post
Hmm thanks that makes alot of Sense. So. If the round one is by corners 2*2 then the slot port can be anything multiplied that multiplies to the same as 2*2 like 1*4 or .5*8? Correct me if I'm wrong.
Grab some paper with squares on it and draw it out. A 2x2 square does not have the same area as a 2" diameter circle.

Chris
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Old 4th May 2017, 03:41 PM   #9
HumbleDeer is offline HumbleDeer  Belgium
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Quote:
Originally Posted by chris661 View Post
Grab some paper with squares on it and draw it out. A 2x2 square does not have the same area as a 2" diameter circle.

Chris
True, my bad. I was in a hurry haha. But it should be the same as 1x2, though, right? That's half, just like a triangle's half of its coefficient rectangle. Was my theory right though?

And afterwards would you mind checking what I tried?

Thanks.

Oh and hey, I'm making the box from the imperial plans and i'll do the port in the imperial plans too. Then I'll just rework it to metric.
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Old 4th May 2017, 03:42 PM   #10
jazbo8 is offline jazbo8
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[urgent] Slot ported version of The Tango
The lack of basic math skill is shocking, what are they teaching in school these days?!
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