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Old 13th August 2008, 08:39 PM   #1
steve71 is offline steve71  Australia
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Default From Hornsrep to the drafting table.. How do I calculate this angle?

I'm building a conical midbass horn loaded with an GPA 515G driver. It's based off of Erik's work at

http://www.volvotreter.de/pics/conic...n_setup_06.jpg

Going from the Hornsrep data to the build plan should be pretty straight forward except for this one angle "C". For the time being forget about getting the mouth and throat flush, I can calculate those angles.

For my construction (see first picture), the top and bottom walls (horizontal planes) of the horn are going to be somewhat simple wedges, but it's the sidewalls which will need to be cut at an angle to insure all walls of the horn are at 90 deg when taking a cross section.

In the second picture we're looking at a mock up of the horn from the listening position. The white note pad represents the horn throat and the lens cap is sitting in the horn mouth. The two cutting boards represent the walls of the horn. As you can see, if I cut the sidewalls of the horn at 90 deg, the horn cross section will not be square.

The black "square" is sitting there to represent the required square cross section. As you can see, if angle "C" is a simple 90 degs like the breadboard it will come out all wrong. It must be cut at unknown angle "C". Any ideas?
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Old 13th August 2008, 08:43 PM   #2
steve71 is offline steve71  Australia
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Pic # 2
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Old 13th August 2008, 08:53 PM   #3
Ron E is offline Ron E  United States
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Maybe a sketch would help, I am not sure how picture 2 relates to the project you reference.

You need three things for a square conical flare, right?

What is the mouth area?
What is the throat area?
What is the length?

From those I think we/you should be able to figure out the rest. You are doing it with butt joints, it appears, which is easier.
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Old 13th August 2008, 09:11 PM   #4
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Quote:
As you can see, if angle "C" is a simple 90 degs like the breadboard it will come out all wrong. It must be cut at unknown angle "C"
Maybe I'm not looking at this right, I can't quite get the picture. But if what you have laid out is what you want to duplicate, wouldn't it be:

tan^-1(opposite length/adjacent length)?

If angle "C" = angle "D" and "D" is represented by the red lines, the "adjacent length" looks to be right at 10 inches on the framing square.
Measure the distance between the corner of the framing square and the vertical cutting board for the "opposite length" and plug in to the above formula.

Or are my eyes deceived?
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Old 13th August 2008, 09:22 PM   #5
steve71 is offline steve71  Australia
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Quote:
Originally posted by Ron E
Maybe a sketch would help, I am not sure how picture 2 relates to the project you reference.

You need three things for a square conical flare, right?

What is the mouth area?
What is the throat area?
What is the length?

From those I think we/you should be able to figure out the rest. You are doing it with butt joints, it appears, which is easier.
I have those three things, and from that I can work out angles A & B. Angle C comes about because a 90 deg butt joint doesn't work.

I'll post up some more pics ASAP to makes things clearer. It's a hard thing to explain over the internet


Quote:
Originally posted by tsmith1315


Maybe I'm not looking at this right, I can't quite get the picture. But if what you have laid out is what you want to duplicate, wouldn't it be:

tan^-1(opposite length/adjacent length)?

If angle "C" = angle "D" and "D" is represented by the red lines, the "adjacent length" looks to be right at 10 inches on the framing square.
Measure the distance between the corner of the framing square and the vertical cutting board for the "opposite length" and plug in to the above formula.

Or are my eyes deceived?
The breadboards are just a mock up to demonstrate that a simple 90 deg butt joint won't work. They don't represent the correct angles, unfortunately.

If I can't find the math to work this out, then I'll have to resort to an accurate mockup and calculate the angle as per your maths above.

Thanks for taking the time to reply and BTW I like your signature!
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Old 13th August 2008, 09:54 PM   #6
steve71 is offline steve71  Australia
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Hopefully this will make things a little clearer.

I guess to simplify things, my question is really this...

At what angle do I need to cut the butt join in the first pic?

I know it looks like 90 deg, but in reality it's not.

There should be a mathematic equation relating the three angles A, B & C.
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Old 14th August 2008, 08:14 AM   #7
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Quote:
Originally posted by steve71
There should be a mathematic equation relating the three angles A, B & C.
Hi steve71,

There certainly is:-). Give me 24 hours to work through the trigonometry and hopefully come back to you with an answer. (I have to log off now).

Kind regards,

David
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Old 14th August 2008, 01:00 PM   #8
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Quote:
Originally posted by steve71
There should be a mathematic equation relating the three angles A, B & C.
Hi steve71,

If we assume for your conical horn that:

X1 = throat width
Y1 = throat height

X2 = mouth width
Y2 = mouth height

L12 = horn axial length

Then the “off-vertical” angle setting you need on your saw blade when cutting the tops and bottoms of the side panels is given by:

Angle = Arctan((X2 – X1) * (Y2 – Y1) / (4 * L12 * (L12 ^ 2 + 0.25 * (X2 – X1) ^ 2 ) ^ 0.5))

For example if:

X1 = 20
Y1 = 10

X2 = 40
Y2 = 30

L12 = 35

Then the required “off-vertical” angle = 4.49 degrees.

The attached sketch shows a not-to-scale cross-section view of the mouth-ends of the two side panels. For the purposes of indicating the required cutting angle, the two panels are shown parallel to each other in the drawing.

Hope this makes sense, and that it helps.

Kind regards,

David
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Old 14th August 2008, 06:16 PM   #9
steve71 is offline steve71  Australia
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Quote:
Originally posted by David McBean



Hope this makes sense, and that it helps.

Kind regards,

David

Wow David, I don't know what to say! Your post has certainly made my week. I've been chasing dead ends with this problem for a week now.

I don't know which is more impressive. The fact that you understood my question, or that fact that you knew how to solve it .

I just dusted of my old HP28S calculator and punched in the numbers from your example. Works like a charm!

Once again, thanks a million!
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Old 15th August 2008, 06:04 AM   #10
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Quote:
Originally posted by steve71
Once again, thanks a million!
Hi steve71,

No problems, glad to be able to help out :-).

Just to complete the picture, expressed in terms of angles (which was your original question), the formula becomes simply:

Angle = Arctan(Sin(AX) * Tan(AY))

Where AX is the half-angle between the left and right side panels of the horn, and AY is the half-angle between the top and bottom panels.

Kind regards,

David
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