Replies

1. 
   joeh | Jul 01, 2003 08:16am | #1

   Probably can't be done at all, but you might ask Joe Fusco. He's pretty good at oddball geometry & CAD.

   Joe H

   http://www.josephfusco.org/phpbb/portal.php

   
   

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   Edited 7/1/2003 1:25:13 AM ET by JoeH

2. 
   NathanW | Jul 01, 2003 08:37am | #2

   I'm no mathmetician, and I"ve never built something like this before. But, I did give this problem some thought about a year ago, as I did some bid work for a house that had a long radius wall, a tall one, intersected by a sloped flat roof. Lets see if I can remember the thought process and the possible framing solution I came up with...

   I think my plan was to...swing the 120' radius arc across ply to form the bottom plates. The wire rope used to swing the arc would have needed to be attached to a post that was firmly anchored at the center of the circle, and fairly tall, as you will see if I can spit this out with words. About six feet from the end of the wire, I planned to splice a second wire onto the main wire, so that a pole could be attached at the top and bottom, and when this pole was pulled tight, it would be plumb, and centered over the outside of the 120' arc. A pencil, pen or other writing tool would be attached to the bottom, then swing the arc. Then...here comes the elipse part...shore up your plywood that will be cut to form the top plates, at the same pitch as the roof, and level, across the length of the bottom plates. I hope that part makes sense. What you're doing is creating the roof plane, level and a few feet off the deck, and over the arc that was swung on the deck. Now, simply shorten the end of that pole, re-attach the pen or pencil, and swing the arc across the ply. I think I just caught something here. The string, wire or whatever needs to stay relatively level during the swing across the dummy roof plane, so the pole you use will need to telescope. Perhaps a telescoping grade stick would work to allow the wire or string to remain level, the pole to remain plumb, and the pencil to move up and down as required. Something else that may work well might be a tripod leg, but I'm thinking a 9' grade stick or larger might be the ticket. The roof I was going to do was only a 1/2 in 12 pitch, but it was over a wall that was about 100' long. I never calculated what the total rise from middle of wall to end of wall would be, and your 5/12 pitch may create so much rise that this method would impractical. This is really just an idea I had, about how to get those top plates stacking over the bottom. The problem after you get the top plates cut is...how do you frame this wall on the ground? And how do you transfer the layout? And calculate the stud differences? I thought that maybe if the bottom plate layout was done before the plate stock plywood was shored over the bottom plates, then the eliptical arc could be swung, cut out with a jigsaw, and then the layout transfered to the top plates with an accurate level, or plumb bob. The framing part is just a matter of supporting the weight of the wall along some points to prevent the top plates from collapsing under the weight of the studs.

   Hope this wasn't too confusing, and I'm not going to be at all offended if it is quickly shot down as impractical...

   Nathan

1. 
   MikeR | Jul 01, 2003 09:50am | #3

   Hi, I'm not a builder, so I can't give you details on the nuts and bolts of on-site aids for laying this out, but I am a math guy.  So what I can do is tell you the kind of ellipse you will need.  It sounds like once you know how to construct the ellipse, you can take it from there.

   As you noted, you're basically slicing a cylindar at some angle, so that gives you a cross section that is an elipse.  Well, one axis of the ellipse is easy, it's 26'.  That's the minor axis, because it's the shortest.

   So, what's the major axis (the long one)?  If the roof rises 5 units for a 12-unit run, then it will rise 10.8333... (10 5/6) feet for the 26' run.  I got that because 5 is to 12 as 10.8333... is to 26.  To check the math, 5 x 26 should be equal to 10.8333 x 12.

   So, the major axis of the ellipse projects downward onto a 26' line (as you noted), and projects outward (to the wall) a 10 5/6' line.  Well, that's a right triangle, with the major axis as the hypoteneuse.  So, we use Pythagorean's theorem to get that:

   The bottom squared plus the side squared is the major axis squared.

   (26 x 26)  + (10.83333 x 10.83333)  equals

   676 + 117.36111...  equals 793.36111...

   Now that's the major axis squared, so we need to take the square root to get the actual major axis:

   Square root of 793.36111... is 28.1666... , which is 28 1/6 feet, or 28' 2".

   So, you have an ellipse whose minor axis is 26' and whose major axis is 28' 2".

   Now, how do you construct that with a looped string and two nails?  OK.  All of this so far has been from my memory, and my mind is a little fuzzy at this late hour.  So for now, I will post this and see if you and others agree with those dimensions.

   Sorry for the theoretical treatment, but since I'm not a builder, I don't know the tricks of using graduated squares and all that to simply construct the answer.

   I'll do a little research to see if I can get you the looped string dimensions in a while.

   BTW, at first I thought knowing how tall the studs were along the ellipse would be a difficult thing to calculate, but now I suppose not.  The height of each stud is determined by its distance along the major axis line, just like on a nonelliptical roof.

   I hope to have more soon.  Also I'll double-check the math.

   Mike

3. 
   KRettger | Jul 01, 2003 10:04am | #4

   Hey Phranc,

   Cork in Chicago Here,

      Worked on a job called Oak Cliff Bible Fellowship on outskirts of Dallas Texas last fall and early winter. Project Forman was Mark Rodgers of Matrix Interior Const. He faced a similar dilemma and posted the question here on BT and got answer. Our project was heavy gauge steel studs, went up about 25ft. Had to frame in radius bottom track, project layout through two overhangs, and then end in an elipse as described in your post.

      Mark is the best layout man I have ever seen in my life. When it comes to figuring out layout, then justifying it to existing jobsite conditions, he is truly at the top of his game.

     I was one of the men actually doing the work and it's been so long I can't remember the actual formula's and layout procedures, just that  we used PLS5, and transit to create points to hit with framing and where to hit it at top plate.

      He's a friend of mine and I've got his email address and will send this thread as a link to his email in box, and then give him a call tomorrow during his morning break. I'm sure he'll get a kick out of someone coming up with the same quandary as he was in and looking for answer at BT just like he did. He was sweating bullets cause this was a multimillion $ contract he was running for us and this particular aspect of the job was an "architectural statement/specialty highlight" that really showcased the front entrance to a high dollar job, and no engineers, architects, etc. could figure out how to do. So he did the sensible thing and asked the men in the trades and doing the building here @ BT how to lay it out and they came through for him. It came out looking real great- I'm sure he and his notes on this will help out.

   If you want, email me your name and contact # and I will give it to Mark. Otherwise I'm sure you'll hear from him later on this thread ( I don't know how pressed you are schedule wise).

   Cork in Chicago.

1. 
   MikeR | Jul 01, 2003 11:11am | #5

   Phranc,

   Looks like you may have a good lead from Cork.

   If, instead, you just need the math, I have the second piece -- where to put the two nails, and how long the string is.  Instead of just figuring it out, I got it from a woodworking page done by a Rick Christerpherson at

   http://home.att.net/~waterfront-woods/Articles/Ellipse/stringellipse.htm

   He knew what kind of ellipse he needed (major and minor axes lengths) and from there figured out the string and nails. 

   It looks to me like your string will be 28' 2" (not counting the little bit on each end to wrap around the nail) and I believe your two nails will end up 10'10" apart (each one 5'5" from the center on the major axis).

   BTW, as you construct the plate, it will of course have some width to it.  Assuming that the 26' you quoted is the OD of the cylinder, you'll need, really, to construct two ellipses.  One at the OD and one at the ID.  You won't be able to just shorten the string and draw another ellipse.  It's probably best to just make the one ellipse, then use a string pinned in the center as a reference line.  Move the string around the ellipse and at regular intervals put marks x number of inches in from the original ellipse.  (Hope that makes sense.)

   Hope that helps.  Good luck!

   Mike

4. 
   markjrogers | Jul 02, 2003 12:50am | #6

   I think that the easy way to tackel the problem is to let the circle do the work for you....first lay down the base plate....the find the center of the radius at the outer most point....plumb up this point to the lowest elevation of the elipse....now square the point back to  the major axis line..remaining level as you square back.....with this done...snap a line from the high point to the low point along the major axis.....if you keep  the points of reference along the line you snapped, level, and square to the major axis, then as you move along points of the radius below. the elipse will form itself...use the angle line to set points of reference ,   if each point on the angled line is plumb with the point below, and is square off the major axis, level at that point, then the elipse shapes itself....please feel free to contact [email protected].... if you need futher assistance, or a diagram....Mark, Dallas TX