Replies

1. 
   Guest_ | May 16, 2000 01:58am | #1

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   Jack, are you seeking to learn the run of a 10" tread at 22.5 degrees off the perpendicular? If so, it's 10" divided by the cosine of 22.5 degrees, or 10.82".

   Hope this helps, Steve

2. 
   jack_keating | May 16, 2000 01:47pm | #2

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   Thanks Steve. I allowed for an inch plus for an overhang to allow for a 1x8 riser. By making the tread 10.82 it looks like my riser will stick out from the tread. It looks like I'm going to have to find some stair building books to teach me a thing or two!

3. 
   jack_keating | May 17, 2000 12:57am | #3

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   A client wants an angled set of steps cut into the outside corner of a deck. What I mean by cut into is, the bottom riser will be flush with the outside rim joist. There are three risers. The fourth riser is the rim joist of the deck itself. There is about 5 ft of deck from the house before you reach the stairs. Where the stringers attach there is a board about 24" perpendicular to the house there is a 22.5 degree angle then the next board is about 20" and a 22.5 degree angle and then the last board, which is parallel to the house is about 40". My question: Is there a formula or trick of the trade that one uses for the stringers at the 22.5 degree angles so that I can keep the 10" tread all the way around the stair? If this is too complicated to answer here are you aware of a book that deals with this situation? Thanks in advance.

4. 
   Guest_ | May 17, 2000 12:57am | #4

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   OK, Jack... I was figuring that your tread run on a common stringer was 10", and I forgot to mention that the 10.82" would be to the center of the 22.5 degree stringer, AND that its edges would have to be chamfered to install the risers.

   FWIW, the cosine of 22.5 degrees is 0.923879532, and if you can use all those decimal places, you're a better carpenter than I.

   Good luck, Steve