This will be a slow long term project that I will be working on sporadically in between other projects and life's other necessities. The plan is to utilize 18 CSS LD25 tweeters and 18 Dayton DMA80-8 mids per side as well as two Dayton LS10 flat woofers back to back on the sides.
Comments
This will be the basic shape.
Bookmarked/Subscribed - this should be interesting . . .
Built one a few years ago. Lots of construction challenges! Have fun!
Is the front curve random or a certain radius as per a CBT standard?
Asking as I really have no clue if that's even a thing.
Yes, the front curve is always a portion of a perfect circle. The length is determined by the diameter of each driver and the space between each driver.
I'm sure Nick will nail this design. He might be an ugly_woofer but he has mad speaker skills!
I can't wait to see how he does the grill on these.
I wonder if a guy could devise a way that one person could move a speaker such as these?
Two thoughts, and hear me out here:
1. Shape it like an egg.
2. Give it a handle.
Bam! Problem solved!
That's gonna be bad ass.
DanP said:
"1. Shape it like an egg.
2. Give it a handle."
Egg shaped things can be rolled from place to place in a meandering/eccentric fashion while knocking over objects adjacent the periphery of its path of motion.
I think wheels with the above suggested handle might be a more suitable method for a single man to transport these items while a married man is likely to never have the need to begin consideration of such a project.
Ok, so does it matter what radius that circle is?
There's a ton of math equations in Don Keele's white papers, but the short answer is yes. The shading, height and angle, all can come into play.
I had some personal help from Jim Griffin when I built mine.
re:
The radius of the sphere is the question. The radius of the arc can be found by the following:
First you know the length of the arc section which is the drivers: 54 mm which is 2.126" x 24 = 51.024"
Add in the gap distance that you choose which is 23 x 0.0625" = 1.4375"
Add in the distance at the top and bottom of 1" each which totals 2"
So add 51.024" + 1.4375" + 2" = 54.461" for the total length of the arc which is the front baffle length.
Now the radius of the sphere is the length of the arc divided by (2 pi x 36/360) for this case. Thus we have: 54.461" / 0.6283 = 86.68" for your case.
You can also use a piece of graph paper, compass, protractor, and a pencil if you wish to envision what we just did.
Cool, thanks🤘🏼
I mocked these up in order to get the volume of the subwoofer section. I just use packing peanuts and boxes to fill it then put the contents into a box which is a much simpler calculation to figure.


Very creative way to measure net volume of a non-std box !
Curious, how did you go about cutting the curved indent track for the back (darker board) ?
Looking good! The weight of the sub will help with stability
I just used an auxiliary fence on a router with a 1/4" but an followed the front edge.
So I haven't made any progress except buying a couple drivers.
This happened. Apparently Freud has an issue with 9/32" bits fracturing. I ordered 3 Amana 9/32 bits, they should be here soon.
I've had some Yonico bits break, but never a Freud or Bosch
Problem solved,
I love Amana cutters. WTH happened? It looked like you lost your grip.
Nope, just making a light cut, and it broke off. Not even 1/8" deep. I've never had an Amana break, they cost a little more, but it's a quality product.
I have that same Freud bit. Did you buy a 3-pack of the Amana or just bought three so you'd have extra on hand?
I just bought three. I almost always buy supplies in multiples.
^ I share this view - you should see my reloading room and speaker driver closet - or maybe not . . .