Replies

1. 
   UncleDunc | Mar 10, 2005 06:11pm | #1

   A Roman arch, if memory serves, is semicircular, so the arched portion of a 36" wide doorway would only be 18" high, plus the height of the brick. The portion of the doorway below the arch would be an ordinary rectangle, 36" wide by (60" - (18" plus the height of the brick)). Or perhaps less, if you want to run one or two courses of brick over the top of the arch. Here are some semicircular arches.

   http://www.a1specialistservices.co.uk/02.jpg
   http://www.bricksandbrass.co.uk/images/window/kilb010.jpg
   http://www.tulnoylumber.com/cbs-vfp/cbs25.jpg

   Your arch can be an arc of a circle but less than a semicircle, with a rise at the center anywhere between nothing and 18". Here's a flatter arch. It looks like about 48" wide and about 8" rise at the center.

   http://www.hollidaydevelopment.com/i/355-brick-arch.jpg

   Here's a much flatter one.

   http://www.montereymasonry.com/gallery/big/170_big.jpg

   And here's one that doesn't match any mathematical curve I know of.

   http://www.montereymasonry.com/gallery/big/171_big.jpg

   Here's an elliptical arch.

   http://homepage.mac.com/lynnejackson/.Pictures/Freihofer2/Freihofer%20Brick%20Arch%201.jpg

   Here's another elliptical arch with major axis vertical, a bit unusual.

   http://www.tandtmasonry.com/Brick%20Arch%20Fireplace.jpg

   Here's a gothic arch.

   http://www.creativespotlite.com/pen-ink/images/Brick_Arch_2.jpg

   I found all these at:

   http://images.google.com/images?q=%22brick+arch%22&hl=en&lr=&ie=UTF-8&start=0&sa=N

1. 
   MikeR | Mar 10, 2005 06:17pm | #2

   Uncle D,

   Hey, you da man!  Excellent... Thank you!

   Mike

2. 
   wilkesland | Mar 10, 2005 07:36pm | #3

   You may also find this link helpful. http://www.bia.org/html/frmset_thnt.htm

   It is the Brick Institute of America's Technical Notes regarding arch construction. I find this site to be very helpful when ever I have a question about masonry construction or design.

   Regards

   Barry