Geometry Question

[jmundale]

Greetings all, new member here with a geometry question relating to a reciprocal roof design.

Im trying to find the length of the reciprocal roof beam between the supporting post and the intersection with the next recirpocal member. this diagram should help (trying to find distance in red)

Thanks!

[Greg Steckler]

Piece of cake once you know the roof pitch. Also is there a round ring (15') in the center? That is is it a closed panel?

[jmundale]

working on a 40:1 model... all my geometry issues stemed from me trying to reciprocate backwards... rise of the roof is appx 24" from the support to the intersection. I will post a picture when it is completed
(design inspired by Mike Ohlers '50 dollar and up underground house book' - thinking of a log/earthbag earth sheltered home using the post-shoring-polyethelyn method)

[Alberta Loggie]

Its been a while, but what you have drawn is an Isosceles Triangle (90/45/45) and both of the sides of the triangle (the outside you have marked as 15' and the one in question) should be the same length.

[iffy]

Its been a while, but what you have drawn is an Isosceles Triangle (90/45/45) and both of the sides of the triangle (the outside you have marked as 15' and the one in question) should be the same length.

That is true as long as you stay in a flat plane. If you take that series of right triangles and elevate the inner end of them by 2', something has to change. If you want to maintain the 15' center circle, then the legs must get longer, which will change at least 2 of the angles. If you want the legs to remain at 15', then the circle and at least 2 of the angles must change. That is assuming the 15' circle diameter is correct in the two dimensional drawing. I drew this out in autocad and the 15' diameter is correct.
Unfortunately, that is about as far as I can go, because I don't remember enough about trig to figure the angles and dimensions for a 3 dimensional drawing. I would probably fall back to the old tried and true method of doing a mock-up. You could cut 8 pieces of 1 x 1 or something like that and arrange them on a flat board so they make the outside perimeter. Then cut a 15" circle out of 2" thick stock (styrofoam comes to mind). Set that on the board so it is centered. Now, by trial and error you can use strings or sticks that represent the other two legs of each triangle. When everything fits right, just measure the length of the strings or sticks. However many inches that is will be the number of feet in your full size model.
Whew! I'm going to go get a caffeine injection.

[Greg Steckler]

iffy is right....the closer you get to the 15' circle the more the plane is skewed. But let's get back to the original question, "What is the length between the arrows?" ANS: if the elevation of the circle relative to the octagon is 24" then the rafter length between arrows is 18'-2 5/8". Now was that really what you wanted to know?

BTW: that is an extremely shallow pitch. Any snow load will put a huge thrust outward on the octagon. My suggestion is to at least double or 2.5x it.

(THIS ANS. IS REVISED BELOW)

[iffy]

Originally Posted by Greg Steckler

iffy is right....the closer you get to the 15' circle the more the plane is skewed. But let's get back to the original question, "What is the length between the arrows?" ANS: if the elevation of the circle relative to the octagon is 24" then the rafter length between arrows is 18'-8 5/8". Now was that really what you wanted to know?

BTW: that is an extremely shallow pitch. Any snow load will put a huge thrust outward on the octagon. My suggestion is to at least double or 2.5x it.

Greg, not sure how you got the 18' + measurement. His red arrow does not point to where the rafter intersects the circle, but does point to where it intersects the hypotenuse of the triangle. Since this is an isosceles triangle in the flat plane, the distance between arrows is 15'. When you elevate one end by 2' and look at the roof from the third plane you will now have a right triangle with one leg 15' and the other leg 2'. Using the formula for a right triangle A*A + B*B = C*C, you add 15*15 and 2*2 and then take the square root of the sum and that will be 15.13'. Of course elevating out of the flat plane starts playing havoc with angles or center circle diameter.

If it were me, since I am not smart enough to calculate the new angles and lengths, I would probably make a mockup cardboard or plywood to make the octagon, then cut a 15" circle out of 2" thick stock such a styrofoam and place it in the center, then start playing with sticks or string until everything fit, then measure the rafter lengths in inches and the full size will be that many feet long. The sticks representing the rafters could be left long and taped in place until they all fit and then take your measurements at the intersections. It could also be refined by making triangles out of cardboard using the rafter lengths in inches and when all the cardboard triangles are the same size and they all fit together, you have your final dimensions.

You are absolutely correct as to the insufficient pitch. The 2' elevation in 15' gives a 1.6-12 pitch, which is certainly not enough for shingles and is not recommended for metal. Not only will there be snow load problems, it will be very susceptible to leaks.

[Greg Steckler]

I stand corrected. 18' 2-5/8" is the length of the pitched rafter from the octagon to the tangent of the the circle (a more informative measurement I'm thinking). So all of the rafters are this length plus overhang at this shallow pitch.

(Thanks, iffy)

[iffy]

The red arrow was just in the wrong place. I agree that the measurement to the tangent point is the more useful. Still too flat