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This gets back to reading . According to my post if you read it, I don't know the radius. I have rise/run of circular arch. No more. Just like with blueprints. They give you rise. They give you run. But to build it you need the radius. Which the plans don't ever give you. Answer.


(Rise)squared + (Span/2) squared
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2 x Rise
The architect gives you AB, and S. But you need the radius (R) to draw the arch.
Okay, but put it in the format shown above? Also, if you know the hypotenuse then the radius is easy. It seems you're being somewhat obtuse.
 

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Okay, but put it in the format shown above? Also, if you know the hypotenuse then the radius is easy. It seems you're being somewhat obtuse.
hypotenuse of what? You are given an arc segment from a circle. There is no hypotenuse to an arc. hypotenuse only applies to right triangles. Not circles. For example, notes on plans say top of arch 6'8". spring line, 6'0". opening 3'0". From that information you need the radius of the arch. Where is the hypotenuse of this image?
 

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Bunch of commercial framers. I can do the math and all but really could care less. I just call the builder and say this arch looks f#ck$n crooked!
 

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The radius is not given. That is the point.
Read. But I did give out the formula earlier.

(Rise)squared + (Span/2) squared
____________________________
2 x Rise
 

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Since you won't give me S I'll say it is 60". 60 x 60 is 3600 and 42.4 x 42.4 is 1800. Therefore, your radius is 42 13/32". It is a damn triangle if you go back and look at your post 99. Now, quite being smug and teach rather than agitate.
 

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:thumbsup:...:yes: That's right!! It ain't rocket surgery ..
I agree with you all day long that simple geometry is all that is required hanging most basic homes. Rocket Surgery?

However, If you're framing a big store front or a large roof a good handle on math is needed. Knowing how to calculate length of an arc helps when ordering material. My last 20 years I ordered all my own stuff. It started after I did a Target store and had to send a semi down to haul off the excess.

Length of Arc = Circumference x degrees / 360 or in the 84.8" diameter circle it would be 84.8 x 3.1416 x 90 / 360 = 66.6"

 

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I agree with you all day long that simple geometry is all that is required hanging most basic homes. Rocket Surgery?

However, If you're framing a big store front or a large roof a good handle on math is needed. Knowing how to calculate length of an arc helps when ordering material. My last 20 years I ordered all my own stuff. It started after I did a Target store and had to send a semi down to haul off the excess.

Length of Arc = Circumference x degrees / 360 or in the 84.8" diameter circle it would be 84.8 x 3.1416 x 90 / 360 = 66.6"

That is fine. But I guess you don't get it. We already have the chord length of the arc (width of opening). And also height above chord. But that doesn't give you the radius. Chord length (/2) and radius are not the same unless the arch is a perfect semi circle. Typically they are not. Sometimes they are elliptical.
 

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Yep that's right I don't get it but I still like you nerds at least you became drywaller and not government workers 💩 am I right Rick
 

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Second picture: A main corridor hallway after the cracks have been gone over possibly 2 more times since the wall was originally done, now primed and painted (still looks like crap)
Since we never use mesh tape, a bit of a stab in the dark, that might give a thought or 2 to try:

I'm looking at this with what I've seen happen using paper tape and FibaFuse tape on butt joints of my commercial jobs, and thinking that Maybe something similar could be happening to you, but is showing through as cracks for you because of the mesh not covering the butt joints, and any cracking happening behind the tapes, as much as paper and FibaFuse can cover joints.

The paper on the ends of some of the drywall board being made seems to 'lift' sometimes after coating, and sometimes after painting, leaving a 'peaked' seam that shows. More so with 1/2" board around here, but I've been seeing it happening more with 5/8" as well. Because the paper tape and Fibafuse tape could hide any cracking happening behind them from the board's paper letting loose better than mesh might, it just shows up as 'peaked' butt tapes for us. If we were using mesh, maybe it would show up as a crack(?) &/or peaking mesh tape and a crack(?)

To help keep the paper on the board ends down so it doesn't move and doesn't lift when hit with moisture, I've taken to skimming some setting compound (hot mud compound), like ProSet, over the butt joint ends and letting it set before taping. The odd butt joint still can lift, but not so bad as if I don't ProSet them. Then I'd likely have a lot more of them lifting at times.

If the paper on some ends already is loose enough, might have to cut it out before retaping, if don't want to keep having to do the same joints over and over, piling mud on top till things are held down(?)

I've also seen cracking happen on butt joints when FibaFuse tape is used because someone wiped down the tapes in ways that cut the fibers too much that were covering the seam. Has happened especially on the bottom 2 rows, where a bazooka was used to put on the FibaFuse. Tapes above that can often end up being put on by hand off scaffolds and lifts, often using paper tape there. Just food for thought, in case the taper is doing something differently when he's working above the 8' high flats, where you say doesn't show any cracking problems.

Also, is the taper is using a straight setting compound for putting on the mesh? If so, you'd think that might hold the board's paper ends down. But if he's using some kind of 'exotic' taping concoction, like part setting compound and part mud mixture....

On level 5ing such as that corridor, with that paint: Could help, I'd think. I'd think about asking for it.
 

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The architect gives you AB, and S. But you need the radius (R) to draw the arch.
That is fine. But I guess you don't get it. We already have the chord length of the arc (width of opening). And also height above chord. But that doesn't give you the radius. Chord length (/2) and radius are not the same unless the arch is a perfect semi circle. Typically they are not. Sometimes they are elliptical.
Complete BS sir. I asked you in several posts to provide a dimension for the length of the chord which you failed to provide. Then I created it and solved the equation. What on earth an ellipse has to do with the equation in your post 99 is just another of your attempt to baffle the subject matter. The State of California has certified me to teach math and a finisher in Colorado is challenging me?
http://scotthansen.net/Const_Tech/credential.pdf
If you post a proper equation then I can post a proper answer. Otherwise, go mix your mud and leave the teaching to those who have dedicated their lives to the craft.


 

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Complete BS sir. I asked you in several posts to provide a dimension for the length of the chord which you failed to provide. Then I created it and solved the equation. What on earth an ellipse has to do with the equation in your post 99 is just another of your attempt to baffle the subject matter. The State of California has certified me to teach math and a finisher in Colorado is challenging me?
http://scotthansen.net/Const_Tech/credential.pdf
If you post a proper equation then I can post a proper answer. Otherwise, go mix your mud and leave the teaching to those who have dedicated their lives to the craft.


I clearly did give arc chord length . As I said earlier it is the segment AB in this drawing. We are given Arc length (AB). And we are given arch height (s). But we are not given the radius (R). We need the radius to create the arch. And I gave you the formula for finding the radius provided you have height and width of the arch. And provided it is circular and not an ellipse.
 

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I clearly did give arc chord length . As I said earlier it is the segment AB in this drawing. We are given Arc length (AB). And we are given arch height (s). But we are not given the radius (R). We need the radius to create the arch. And I gave you the formula for finding the radius provided you have height and width of the arch. And provided it is circular and not an ellipse.
Clearly I'm not seeing something then so explain it? Lower the difficulty of learning? I don't see how you can solve an equation without a dimension? What number post did you mention a dimension?
 

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When do you ever get an arch that is made from a circle radius just don't work on most arches I would even say all but some must have a perfect radius none that I've done though
 
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