Replies

1. 
   MisterT | Jun 19, 2002 10:40pm | #1

   Joe Fusco He's our man if he can't do it no one can !!!

   Rah Rah.

   http://www.joefusce.com is the link I believe.

   Good Luck!

   Mr T

   Do not try this at home!

   I am a trained professional!

2. 
   newter | Jun 20, 2002 04:02am | #2

   I would run a hunk of 4 x 4 through the saw so that it is three sided and square cut the tops of the the rafters on a 12/12 pitch then the rest is just simple filling in of jack rafters. 

   Have fun with that last cap shingle.

   Todd - Framing Square

3. 
   TommyB12 | Jun 20, 2002 04:07am | #3

   Redstaines,

   Using the toothpick and raisin method, I determine your hip pitch to be 20.78/12.  That is by far the most accurate way to determine these things.

   You could start with 12/12 commons perpendicular from the center of the sides and just measure the hips too.

   Fusco may correct me.

   Tom

1. 
   Redfly | Jun 20, 2002 09:53am | #4

   At the grave risk of stepping into something, what, pray tell, is the 'toothpick & raisin' method?  i've looked in my Construction Principles, Materials and Methods as well as all the cookbooks we have, and can't seem to find it.

   I was thinking of using the native american technique of teepee style framing, i.e. binding the rafters together at the center with a bit of buffalo hide,  but I need a plausible explanation to satisfy my kid and his friends that his daddy really is a professional builder.

   thanks for all the suggestions!

2. 
   Joe_Fusco | Jun 20, 2002 05:24pm | #7

   Tommy,

   I'll comment on your findings.The 20.78 / 12 (1.73168) you give is based on the relationship between a 30, 60 and 90° right triangle and you are using the 60° tangent. It would be better to use the 30° tangent of .5773. Further, it relates to the ratio between "run" and "rise" and not run and hypotenuse or hip.The correct ratio is 1.154 if you use the run and 2 if you use the rise to determine the "run of the hip" not the hip length. A simple example would be if you had a three side roof and the sides were all equal length and their length were 24' you would know that a run of 12' would bring you to the center of the wall. To then find the center of the "roof" you would multiply 12' x (tan(30°)) and that would give you 6.93'. Again this is run to rise. if you take 6.93' and multiply it by 1.73168 you get. . . 12'.If you then take either the run or rise respectively you can find the hip run. 12' x 1.154 = 13.856' and 6.93' x 2 = 13.86'. Then taking whatever the rise is of the "roof" in this case a 12/12 or 6.93' and applying a2 + b2 = c2 you can find the hip length. 13.862 + 6.932 = sqrt(240.125) or 15.5'.You would also need to remember that the cheek cuts for the hip and jacks would be 30°.

   View Image

   Edited 6/20/2002 12:22:15 PM ET by J Fusco

   Edited 6/20/2002 12:23:22 PM ET by J Fusco

1. 
   Redfly | Jun 20, 2002 08:13pm | #8

   Now I remember why I always hired roofcutters to do my roof framing if it was anything other than a simple gable or dutch-gable roof.  My first thought was to do more or less what Dunc said, i.e., find the center point of the roof at the top plane of the walls by bisecting each side, measure the run from each corner of the triangle to that point, then use that number to calculate the length of each hip.  Is that more or less what we're talking about? 

   Thanks for all the input - these forums are a great resource that I've never really used before.  Its great to know there's such an active and friendly builder's community out there! 

1. 
   UncleDunc | Jun 20, 2002 08:39pm | #9

   >> ... by bisecting each side ...

   Not the sides. Bisect the angles. Bisecting the angles gives you the center of the inscribed circle, a circle that touches each side of the triangle. That means the shortest line from the center point to each wall is a radius of the circle and is perpendicular to the wall, so that all three rafters are the same length and the three roof planes have the same pitch.

   Bisecting the sides gives you the center of the circumscribed circle, one that touches each corner of the triangle. If you use that for the peak of the roof, you won't have equal pitch on all three planes. If your tree house were a 45/45/90 triangle, for instance, the roof plane rising from the long side would be vertical.

1. 
   Joe_Fusco | Jun 20, 2002 09:20pm | #11

   Uncle,

   I deleted my pervious post because why argue?

   View Image

2. 
   Redfly | Jun 20, 2002 10:27pm | #13

   I just got out a piece of graph paper to draw this out for myself (I'm much more visual that theoretical) and drew a perpendicular line from the centerpoint of each side of an isosceles triangle, and they met at a point equidistant from each corner.  Isn't that what I want to do - have all the hips the same length?

   What am I missing?

   Maybe it was all those spitballs I was shooting instead of listening to my geometry teacher.

3. 
   Joe_Fusco | Jun 20, 2002 10:53pm | #15

   Red,

   How about some numbers. What are the lengths of the sides of this thing? Don't worry about the angles between the wall, they will take care of themselves.

   View Image

4. 
   Redfly | Jun 20, 2002 11:35pm | #18

   Oops, I meant to say the base is 10'4", the sides are 9'9".

   Pretty close to equilateral, but far enough off to confuse me.

5. 
   UncleDunc | Jun 21, 2002 12:14am | #19

   >> How about some numbers.

   OK. Consider a triangle with one side 4' long and 2 sides each 10' long. Perpendiculars from the midpoint of each side intersect at a point 4.70' from the midpoint of the short side and 1.02' from the midpoints of the long side. The distance from the point of intersection to the corners is 5.10'. If a pitch of the two long sides is 12/12, the peak of the roof is 1.02 feet above the plane of the triangle. The length of the 3 equal hips is 5.20' The pitch of the third plane is 1.02/4.70, or 2.6/12.

   If the pitch of the short side is 12/12, the peak of the roof is 4.70' above the plane of the triangle, the length of the hips is 6.94', and the pitch of the two long sides is 4.70/1.02, or 55.3/12.

   So if the center of the circumscribed circle is used, the hips are the same length, but the pitches are not equal.

   Lines bisecting the angles intersect at a point 1.63' from the midpoint of the short side. Lines from that point perpendicular to the long sides intersect the sides 2' from the corners, and are also 1.63' long. If 12/12 pitch is selected, the peak of the roof is 1.63' above the plane of the triangle. The distance from the point of intersection to the two ends of the short side, and therefore the run of the two short hips, is 2.58'. The length of the short hips is 3.05' The distance from the point of intersection to the other corner is 8.17', and the length of the hip is 8.33'.

   If the center of the inscribed circle is used, the pitches are equal, but the hip lengths are different.

6. 
   UncleDunc | Jun 21, 2002 01:25am | #20

   Using your numbers, you have three choices,

   1. Equal hips of 6'3.3", 12/12 pitch on the long side, 9.9/12 on the short sides.

   2. Equal hips of 6'6.1", 12/12 pitch on the short sides, 14.5/12 on the long side.

   3. Two hips at 6'6.8", one at 6'1.4". 12/12 pitch on all three planes.

   Let me know if you want to see the calculations.

7. 
   Redfly | Jun 25, 2002 01:56am | #37

   Thanks Uncle.  I'm plan to use your calcs, augmented with some cool drawings by JFusco to build my roof.  I'll post a copy of the results

8. 
   Joe_Fusco | Jun 21, 2002 04:38am | #23

   Uncle,

   Abusive. . . Give me a break. I have no reason to be abusive to you.Your first post was difficult to decipher as was the last, but they prove to be correct. OK try these numbers. . .1 long side 10' 2-1/2"2 short side 7' 2-5/8".

   View Image

   Edited 6/21/2002 8:25:21 AM ET by J Fusco

9. 
   UncleDunc | Jun 21, 2002 06:03am | #25

   >> I have no reason to be abusive to you.

   I'm glad to hear it. So why do you keep saying my posts are hard to decipher but never specifying what needs to be clarified? That feels abusive to me. I like to think I could meet your standard for clarity if you'd just tell me where I fall short.

   >> OK try these numbers. . .

   >> 1 long side 10' 2-1/2"

   >> 2 short side 7' 2-5/8".

   This is the 45/45/90 case I mentioned earlier. Using the center of the circumscribed circle, the run of the 3 equal hips is 61.25", the peak of the roof is directly over the midpoint of the long side, so the plane of the roof rising from the long side is vertical, more like a gable end than a section of roof.

   Using the center of the inscribed circle, you can still make a normal looking roof. You'd have one short hip, a little under 30" run, and two long hips, somewhere around 70" run and the pitch of all three roof planes is equal, whatever pitch you choose.

   Edited 6/20/2002 11:28:04 PM ET by Uncle Dunc

10. 
    Joe_Fusco | Jun 21, 2002 07:30am | #27

    Uncle,

    I have no control over the way you "feel" about anything.

    As for my example, would you call it a triangular gable? Does it really have 3 hips or does it have 1 hip and two commons that are the same length?

    My first post was what I thought would be the simplest approach to framing a triangular roof a triangle with equal sides.

    As far as clarity, you might want to tell how you "get" the answers as opposed to just stating them.

11. 
    RogerMartini | Jun 21, 2002 12:22pm | #28

    Hey look, a discussion about framing hips!  Just like the robin is a sure sign of spring, this leads me to believe that this forum is returning back to normal.

    Ahhh, the good old days....

    :)Close enough for government work

12. 
    TommyB12 | Jun 21, 2002 04:29am | #22

    I got the answer right didn't I? 

    The toothpick and raisin method...

    I did the math and then assembled a model using toothpicks and raisins at the joints.  Then I eyeballed the hip, yep my eye tells me its 20.78/twelve, just like my figgerin.  Estimated time 43 seconds.

    Yes redstains stated it was an isoceles triangle,  which I assumed was an equilateral right triangle.  If it was something else, we would need alot more information.  And I would have to cut the toothpics.  Raisins on the other hand would still remain whole.  In a pinch you can use dried currants, but the shear strength is not as great.Tom

13. 
    Joe_Fusco | Jun 21, 2002 04:41am | #24

    Tommy,

    You da friggerin' man! LOL.You know what I always say. . . NAIL IT!!

    View Image

14. 
    UncleDunc | Jun 20, 2002 11:07pm | #16

    You only want all three hips the same length if all three walls are the same length. If all three sides of the triangle are the same length, both methods do in fact yield the same center. If the walls are different lengths, you can build a roof with the same pitch on all three planes, or a roof with the hip rafters the same length, but not both. For triangles where one angle is greater than 90 degrees, the center of the circumscribed circle isn't even in the triangle.

    When I was in school, an isosceles triangle was one that had two sides the same length and we called a triangle that had all three sides the same length an equilateral triangle. It's true that an equilateral triangle is a special case of isosceles triangle, but since there is a specific name for equilateral triangles, I assumed that your isosceles triangle was the more general case, with two sides the same length and the third side either longer or shorter.

    If you're talking about a 60/60/60 triangle, then of course you're right, along with Joe and everyone else who made that assumption, and I was insisting on a lot of unnecessary work.

2. 
   Joe_Fusco | Jun 20, 2002 08:43pm | #10

   Red,

   I didn't read Uncle Dunc's reply before I posted. Now that I have I wish I didn't. There would have been a time when I would have commented further, but that time has long since past.

   View Image

1. 
   UncleDunc | Jun 20, 2002 09:41pm | #12

   He didn't say it's an equilateral triangle, he said it's an isosceles triangle, which is one with two sides the same length.

   Edited 6/20/2002 2:43:25 PM ET by Uncle Dunc

2. 
   Joe_Fusco | Jun 20, 2002 10:50pm | #14

   Uncle,

   I'm aware what he said that's why I just could not believe the answer you gave. I can't even decipher what you said.

   View Image

3. 
   UncleDunc | Jun 20, 2002 11:24pm | #17

   >> I can't even decipher what you said.

   That happens sometimes. If you're not just saying this to be abusive, tell me the first thing you didn't understand and I'll try to clarify it.

4. 
   DrainSnake | Jun 21, 2002 06:41am | #26

   This is coming from a machinging background where compund angles aren't so very uncommon, I have a hard time seeing where anyone can speculate on framing grids without knowing the lengths of the sides of the triangles, and having a go at it from there. You have specified a roof pitch which will help, but it would be easier if you gave a target height for the peak of the roof. From what I can see you want a pyramid shaped roof, with its peak in dead centre. There is a name for that shape which escapes me right now (sorry). All I can say with the info you've give so far, is that I hope you have access to a really good double compound mitre saw, and a head for numbers.

   JAG

   View Image

   And When I must Leave the Great River, Oh Bury Me Close to its wave,And Let My Canoe and My Paddle, Be the only Mark over my Grave

   Zone 5b Brantford Ontario, Canada

4. 
   UncleDunc | Jun 20, 2002 10:44am | #5

   Bisect each angle. You have to measure or calculate half the angle. Drawing a line from the corner to the midpoint of the opposite wall will work for the unequal angle, but not for the two equal angles. The point where the angle bisectors meet is exactly under the peak of the roof. A line from the meeting point to each wall, perpendicular to the wall, represents the run of the single common rafter on each wall. (You don't have to frame it that way, of course, but it made thinking about the question easier.) The length of the rafter is 1.414 times the distance from the meeting point to the wall. The length of the hip rafters can then be found with the Pythagorean theorem.

   
   

   ---

   Edited 6/20/2002 3:47:24 AM ET by Uncle Dunc

1. 
   MisterT | Jun 20, 2002 01:09pm | #6

   You will also need to figure your birdsmouth, which consists of a plumb and level cut.

   These will determine you height at the out side of your plate.

   You Williamson need to pick an over hang distance. are you having a level or sloped soffit?

   You need to make sure you drop your hips (3) or you have to  find the hip bevel.

   if you will have a cathedral ceiling you need to determine the hip depth so you sheet rock lies in the correct plane.

   Your common and hip rafters will all need a double beveled plumb cut at the top.

   you will need to know the plumb cut angle for the hips.

   your hip jacks will all have a single bevel top plumb cut.some 1/2 right 1/2 left.

   (tip) when cutting hip or valley jacks get lumber that is long enough to give you the longest and the shortest from one board. when you make your top cut (single bevel plumb), the offcut will have the the correct cut for the opposite side of the hip/valley.

   for your sheathing you will need to know the angle at the hip(and the bevel if you are anal or if you won't be roofing over it)

   You will need to determine the length of you ridge board (0'-0" in your case, this is the one thing I could figure of the top of my head!)

   Help me out guys, am I forgetting any thing????

   Mr T

   maybe a pole and a buffalo hide would be easier!!!!???!!!

   remember to have fun! :)Do not try this at home!

   I am a trained professional!

5. 
   Hector45 | Jun 21, 2002 03:05am | #21

   I love math. This problem intrigued me so much that I followed the thread all day. I even made a couple little CAD models to verify everyone's calculations.  When Uncle Dunc pointed out that we were talking "isoscoles" not "equilateral" (I misunderstood the original problem. I'm not saying anyone else did) I began to wonder which was better - equal length hips or equal pitches. Then I realized.....

   #1 This is a treehouse

   #2 The triangle is close enough to equilateral that no one but Red will EVER know the difference, especially 20' up in a tree.

   #3 Unless the house is also triangular, the treehouse roof isn't going to be an exact match no matter what method you choose.

   #4 Given #1, #2 and #3, it makes sense to do the roof whatever way you feel is easiest. IMHO, use Dunc's number for equal length hips (round to the nearest 1/8") and let the slope fall where it may.

   #5 I'm sure this is hasn't enlightened anyone, but it does lead to my question.... which is: Why did you made the treehouse triangular in the first place?  I'm sure the reason was something like "circumstances dictated that this was the easiest/fastest/only way." I just can't imagine what those circumstances could be.

   Care to satisfy my curiousity, Red?

   PS Good luck with the roof. Hope the kids appreciate the thought and calculations that went into it.  (Yeah, right!)

1. 
   Redfly | Jun 21, 2002 08:45pm | #29

   I think you're right on all five points.  One reason I want to cut the rafters just right is that I may find myself relegated to sleeping in said treehouse, staring up at the roof, wondering what I did wrong, and I'd like at least to be able to admire my handiwork.

   My son, who badgered me into building it in the first place, could care less whether it even has rafters - a blue tarp would be fine with him.  Not a bad idea when I think about it.

   As to why the triangular shape, I kind of backed into into it, since we don't really have the ideal tree layout for a treehouse (huge cherry, elm or chestnut trees are perfect - we have 100' Douglas firs where the bottom branches are 40' off the ground), I decided to strap adjustable steel girdles around two adjacent firs about 16' off the ground and added a post about 10' away.  Then I build a triagular platform and laid decking over it.  Had I been thinking clearly about this, I would have made the triangle equilateral, but such is not the case.  But you're right, it's close enough to equilateral that nobody but me would ever notice.

   Thanks to everybody for all the great input!

1. 
   BillHartmann | Jun 21, 2002 10:15pm | #30

   " One reason I want to cut the rafters just right is that I may find myself relegated to sleeping in said treehouse, staring up at the roof, wondering what I did wrong, and I'd like at least to be able to admire my handiwork."

   In the Home Depot ad for the guy that does not even have a hammer, but promises his son a tree house. Look at the last second or two of the ad.

   The and the kid are in the tree house and the dad rolls he eyes when the kid thanks him.

   It looks to me that he is thinking "I hope that this stays up until moringing".

1. 
   JeffBuck | Jun 22, 2002 08:06am | #31

   OK.....as much as I love a good math problem and laying out the rafters to they fit 30 ft off the ground......It's a freaking tree house....and the roof is a freaking pyramid!

   Two words......STRING LINE! And another 4 words......GET ON WITH IT! So...using that great advice......make a guess...climb yer #### up there.......string it...level it....plumb it......mark it........and make it fit and get on with life!

   Really...some of the most impressive roofs in the world were built with a string, a level...and some dude on the ground with a picture in his head! Have fun too...Jeff   She's exotic ,but not foreign, like an old Cadillac......she's a knockout!

2. 
   Hector45 | Jun 22, 2002 06:23pm | #32

   Makes perfect sense now that you've explained it. 

   I hope the roof layout goes well.  Otherwise, you might be cursing under your breath "If I'd used TWO 10' support posts, this would have been so much simpler!"

   Cheers,

   Jon

1. 
   Redfly | Jun 22, 2002 07:43pm | #33

   Ff I'd have used 2 support posts, I might have ended up with a trapezium or somesuch and then I'd really be needing advice.

6. 
   Joe_Fusco | Jun 22, 2002 11:38pm | #34

   Red,

   I put this simple page together to show how to layout the roof you'd like to build. It might be helpful, good luck.

   View Image

1. 
   Redfly | Jun 23, 2002 09:16am | #35

   Joe,

   Wow!   After I build this roof and people ask me how I figured it so perfectly, I'll mumble something about hypotenusean cosigns and Pythagorean bisections and send them to your website for the truisms.

   I think I can do it using your "simple page".  Thanks bigtime!

   Roger Staines (aka redstaines)

1. 
   Joe_Fusco | Jun 23, 2002 04:10pm | #36

   Roger,

   Glad you can use it. Make sure you pass long a thank you to Uncle Dunc too, he was on it from the get go.

   View Image

7. 
   Joe_Fusco | Jun 25, 2002 03:36am | #38

   Roger,

   Just remember the "numbers" that Uncle gave are "point to point" numbers. They don't take into account any overhang, hip drop, H.A.P. or shortening. It's still important to mention that just about every rafter on this roof will be a "jack" rafter.

   You'll have to cut the cheek cuts for these jacks at whatever the plan angle of the hip is. In your case they could be 32° and either 26° or 29°. Since I need a "Tip of the Week" for my site I might as well use this one.

   You can calculate the unadjusted hip-jack for your roof by using the following equation.

   O.C./Cos(roof pitch angle) * Tan(plan angle)

   Using whatever your spacing is like 16" as "O.C." or on center and the roof pitch angle, if you use a 12/12 it would be 45° and the hip plan angle you get. . .

   16 / Cos (45) * Tan(32) = 14.14 or 14-1/8". This is the length of the first hip-jack coming from the 64° corner on either side. This is also the common difference meaning you just add this amount to the last to find the next.

   Another cool thing is you can find the length of any jack in any position by using (O.C./Cos(roof pitch angle) * Tan(plan angle)) * (Position) if you wanted to know the length of the third jack "position" would be 3. 14-1/8" * 3 or 42-3/8".

   Looking forward to the pictures.

   View Image

   
   

   ---

   Edited 6/24/2002 10:31:58 PM ET by J Fusco

1. 
   UncleDunc | Jun 25, 2002 04:10am | #39

   >> ... the "numbers" that Uncle gave are "point to point" numbers.

   Yes, that's true. Sorry for not making it more explicit.

2. 
   Redfly | Jun 25, 2002 10:16am | #40

   Joe,

   Your great math examples make me wish I'd paid attention during those geometry and calculus classes rather than just looking at Linda Di*****on (name changed to protect the innocent).

   Thanks again,

   Roger