# Conical roof framing



## Paulsan (Jun 19, 2005)

Anyone have some insight into building a conical roof on an 8-sided gazebo?
Rafters, I assume, would have to attach to overhanging ceiling joists to allow for the circular base and soffit etc. but how would I go about fastening roughly 50 rafters at the apex? Rafter spacing will have to be no more than 6 inches so that I can maintain a smooth radius to the peak.
Any input would be greatly appreciated.


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## Paulsan (Jun 19, 2005)

*C'mon*

C'mon,
There must be someone viewing these threads that have some experience in this type of roof. I've been in the game long enough to to know that I can successfully frame, sheath and shingle this sucker. However, I am still looking for someone who has tackled this particular type of challenge to offer some time saving insight. I live and work in the Halifax area. From my experience with the local inspectors, if the job is done meticulously and there are no obvious structural issues, it will pass inspection because this sort of thing has no reference section in the code book. Any input is appreciated.
Paulsan


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## Teetorbilt (Feb 12, 2004)

50 rafters? How big is this thing? Rafters at no more than 6"s? What size rafters?
Can be done but is going to require some special equipment.
Let's start with some dimensions.


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## Tom R (Jun 1, 2004)

Not sure exactly what you're asking, - - and like Teetor says, - - 50 rafters sounds like an awful lot, - - but either way, - - I would do 8 rafters to the apex, - - then 'header' off in between them (near the top), - - then tie-in your 'tweeners' to the headers. Simple enough, huh??

Five rafters in between each would give you a total of 48, - - if your stuck on 50, - - start with 10 rafters, - - with 4 goin' to each header.

If you really wanted to be anal, - - you could figure and cut the slight (proper) arc on the headers, - - but the closer you install the headers to the apex, - - the less relevant it (the arc) becomes.


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## ConcreteGuy (Jun 10, 2005)

I can help you with "comical" roof framing.... :cheesygri


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## Tom R (Jun 1, 2004)

GCMan said:


> I can help you with "comical" roof framing.... :cheesygri


You mean that's not what we're talkin' about?? :cheesygri


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## Paulsan (Jun 19, 2005)

Thanks for the input.
You are right, I should have given more details. I figured that by limiting those details I might attract people who actually have experience framing a conical roof. It seems I have underestimated the resources here...
My customer is actually more 'anal' than myself, which I find curiously refreshing. He is on site daily, plastic hat and all, nose right in there looking at every cut. The dollar signs add up and the job satisfaction couldn't be better, so I don't mind at all that my work is being scrutinized.
As for details, each leg of the gazebo is 8"6". The architect matched the pitch at 12" with that of the main house roof. The roof will be clad with painted galv. sheeting, capped with a copper chicken or some such thing. My real question here is, being that I am under such close scrutiny, how can I calculate ( without the guesswork) the optimal location for header placement between what I will call my 'king rafters' , to allow for all the rafters required? And is there a formula for determing the circumference of a cone at any given point of it's rise? For example, say the radius at the top of the ceiling joists (outermost point where rafter attaches) is 18' .Then a 12'' pitch gives a rise of +/- 9'. Say I want to calculate the circumference 7' up towards the apex. How would I do this? I'm thinking this may be a good way to make a plywood template for the headers to give me a smooth transition to the peak. After all, he wants a cone shaped roof, not one that sort of resembles a cone.
Feel free to let me know if this seems to any of you like overkill.


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## Teetorbilt (Feb 12, 2004)

Paulsan, you have not 'underestimated the resources here', you have overestimated your own abilities.
The answer to most of your questions involve simple geometry. Much of your phraseolgy leaves much to be desired as most of us will not recognise what *you* choose to call building components.


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## Paulsan (Jun 19, 2005)

Teetorbilt said:


> Paulsan, you have not 'underestimated the resources here', you have overestimated your own abilities.
> The answer to most of your questions involve simple geometry. Much of your phraseolgy leaves much to be desired as most of us will not recognise what *you* choose to call building components.


Surely you understand what a king jack rafter is Teetorbilt. Is it such a stretch to call the common rafters (if they can be called 'common' on a conical roof) 'king rafters'?
It must have been my referring to the outside wall lengths as 'legs', right? I think you have a good idea what I'm TRYING to refer to.
All input is good input and I will check my simple geometry tables again for the appropriate formula.


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## Teetorbilt (Feb 12, 2004)

Paulsan, You are looking for the diameter of the top of a truncated cone. I can easily do this to 1 ten thousandth of an inch at any verticle place that you desire. I commonly work to these tolerences. The only difference is in the decimal point.


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## Tom R (Jun 1, 2004)

Paulsan, - - I don't know such a formula off the top of my head, - - but I'll give you a sure-fire 'barn-yard' method that I would use in this situation, - - (even under close scrutiny), - - first of all lets take for granted you were going to use what I'll call the '8-rafter' method (48 total), - - I would cut each header with a 22 1/2 degree bevel on each side, - - and whatever (compound) angle it takes so your rafters will be 'plumb-cut' (where they meet the header), - - just making sure the shortest measurement of the header is at least a 7 1/2" minimum (so it can recieve five 1 1/2" rafters) - - then install all 8 headers, - - then on any one header mark 5 evenly spaced lines (representing the rafter locations), - - now tap 5 nails down into the outside edge of your top plate, representing the bottom end of your rafter locations, - - now pull a string from the nails to the apex, - - and measure and record your height difference between your string and your header at each corresponding location, - - you only need to check location #'s 1, 2, and 3, - - because #1 will match #5, - - and #2 will match #4. Now from those recorded measurements you can determine your arc onto a piece of plywood, - - then just go ahead and use that for a template to make the other 7. Something tells me that if you were to draw an octagon, - - with each of the sides being the same length as your headers (plus 1 1/2", representing your rafter thickness), - - then found your middle and drew that circle connecting the points of your octagon, - - that would be your correct radius/diameter for your plywood 'arc-caps'.

OK, - - it's getting late, - - I've confused myself enough now.


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## Teetorbilt (Feb 12, 2004)

Confused me in the first few 'sentences'?


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## Tom R (Jun 1, 2004)

Sorry, - - I guess that brings us back to 'comical' :cheesygri 

Well, at least I know what I mean, - - err, - - I think.


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## Paulsan (Jun 19, 2005)

Tom R said:


> Paulsan, - - I don't know such a formula off the top of my head, - - but I'll give you a sure-fire 'barn-yard' method that I would use in this situation, -
> 
> Thanks for your advice, Tom. Seems perfectly understandable to me. Glad to have some positive feedback in the form of helpful advice.
> I found the formula that Teetorbilt advised me to use and I will try both methods and compare notes. If anyone cares to see the results, I would be happy to post pics.
> ...


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## Tom R (Jun 1, 2004)

Sure, Paulsan, let us see the results, sounds like a fun job, - - it's definitely worth 'figuring' the math, - - but the reason I prefer the 'barnyard' method, at least in this (rough) instance, - - is that math is a 'perfect' science, - - whereas wood framing members are much less than 'perfect', and often beg for compromise.

By the way, when I referred to really being anal, I didn't mean your structure should be anything less than 'conical', - - I meant that you don't really need to actually 'install' the 8 pieces of arced-plywood, - - you could actually just make one, and use 'it' all 8 times, simply as a 'location-reference', to let you know what height to nail-off your rafters.

Good luck with the project. :Thumbs:


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## Teetorbilt (Feb 12, 2004)

This is one instance where I might opt for dual sheathing. With an 8 1/2 ft base diameter, warping anything over 1/4" is going to be tough. At the top even this will prove impossible. Formboard would be the way to go and I would laminate it myself.


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## J2Jonner (May 24, 2005)

Actually via the formula below, an octagonal structure with 8.5 foot "legs" will have a diameter of 22' 2 1/2" between apexes. It wouldn't have a diameter of 8.5 untill about 6' 10" above his top plate. Pretty geeky aren't I? Sorry couldn't resist looking up that formula...  

2 * a * sin 67.5
----------------
sin 45


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## Joe Carola (Jun 15, 2004)

Paulsan,

Your sides are 102".

Diameter on the Gazebo will be 102/ cos (67.5) = 266.5388" or 22'2-9/16"

Your Circumference will be 266.5388 * (PI) 3.141593 = 837.3563" or 69'9-3/8"

Rafter Centers - 837.3563/50 = 16.74713" or 16-3/4" (Not including overhang)


You said, " Say I want to calculate the circumference 7' up towards the apex. How would I do this? "

You now know your Diameter is 22'2-9/16".

Since your Diameter is 22' 2-9/16" it's also your rafter span.

You want to go 7' up with a 12/12 pitch which means that you would have a run of 7' which you multiply * 2 to get 14' because you would come in 7' from each side.

Subtract the 14' from your 22' 2-9/16" which gives you 8' 2-9/16" or 98.5625" as the Diameter.

To get that Circumference you do the same as above.

98.5625 * (PI) 3.141593 = 309.6432" or 25' 9-5/8"

Joe Carola


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## Joe Carola (Jun 15, 2004)

Paulsan,

If you made a 48" Diameter platforn about 1' high or whatever size rafter your using, make sure it's as high as the plumbcut. 

48 Diameter would give you a 150.7964" circumference/ 50 = 3" centers. So the rafters would go right next to eachother.

Think of that 48" Diameter platform as the thickness of a big ridge. You would subtract that from the main diameter which is 22' 2-9/16 which would give you 18' 2-9/16 as the span.

Divide that by 2 and you get 9' 1-5/16" as the rafter run. You would cut all 50 rafters the same.

Now from that platform and up you do this.

Circumference 150.7964/9 = 16.75516" or 16-3/4" Centers from the platform and up to the top.

Joe Carola


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## ConcreteGuy (Jun 10, 2005)

> This is one instance where I might opt for dual sheathing. With an 8 1/2 ft base diameter, warping anything over 1/4" is going to be tough. At the top even this will prove impossible. Formboard would be the way to go and I would laminate it myself.


This commercial trained former rough-carpenter understands. We do that in formwork and falsework for concrete and masonry. Also, in lieu of the foam board, would luan work?


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