Anyone who can lend some insight into this problem please do...
Preface:
I am a building contractor and this problem relates directly to my trade and a real world mathematical puzzle...
I have roof (plane) that is clipped by a curved wall (cylinder)...the roof is actually comprised of 2 planes in a sense as it has a top and bottom plane and forms a 3D solid...consider this to be an inclined slab. The roof clips into the cylinder at a pitch of 5 and 12 or approximately 22.5 degrees (atan(5/12)).
Of course when a simple plane intersects a cylinder at any angle other than perpendicular to the axis of the cylinder, the shape can be described as an ellipse on a 2D plane coplanar with the intersecting plane.
What I need to figure out is how can I mathematically represent the intersection of the inclined slab and the 3D solid wall (cylinder), such that if I was to cut plywood or other flexible material and bend it around around my wall (cylinder)...I could in essence laminated it to the wall, and form a band or ledger around the wall whereby roof framing members would be hung from this band..
The shape required is an ellipse, but it is not the ellipse formed by the simple intersection of the single plane and cylinder...its an ellipse with a greater major radius and I am stumped on how to figure this out...
Any help or links or ideas on this?
Respectfully
Oneway
I'm not going to solve it directy. However, what I'd suggest, if you can get the engineer to buy off on it, is to block inside the radius, and use a Simpson hanger nailed into the face of the sheathing over the exterior radius wall. So, In essence, your ledger is inside the radius, instead of outside. I would think that with all the kerfing necessary to make a solid ledger conform to the shape of the ellipse, the ledger isn't going to have much left to supply the structural strength necessary to carry a roof load. I can't remember the exact designator of the Simpson connector we used, but we had a floorplan with a ski-jump type roof over the entry. The framers cut the rafters out of 2X12 in order to get a curved "2X6" rafter. Since the curve was built into the rafters, we needed a rafter hanger that would allow the angle of the bottom of the rafter. The engineer specified the hanger which basically face-nailed into the vertical wall, and had a tab that bent to accept the angle of the bottom of the rafter. It was bent, then nailed into the bottom of the rafter, and the face of the sides of the rafter. This way all you'll need to do is figure out where the rafters intersect the radius wall, place the hangers there, and then hang the rafters. It will help you avoid the nasty math you described.
I will recommend you do a search for Solid Edge 2D. Its a free CAD program distributed by Siemens that you can use to put drawings together, if you don't already have something.