On Wed, 13 Aug 2008 23:17:41 -0400, George
wrote:

On Wed, 13 Aug 2008 21:31:47 -0400, mm wrote:

On Wed, 13 Aug 2008 16:03:38 -0400, George
wrote:

We have a roof problem that could most easily be solved by doing
somethign that would lower the pitch from 4:12 to 3:12. I keep hearing
that 4:12 is the "minimum" pitch for a shingled roof. But, that's also
what it was in the Stone Age.

Wish I could help. Come on, let us know why a shallow roof would
solve yoiur problem?

Hard to describe. It's an (old) L-shaped addition on the north side of
the house. The parallel section (ie, along the house) has a block
wall. The perpendicular section is wood frame, is wider than the
parallel section, and its edge is lower ... as it needs to be, to keep
the 4:12 pitch.

So, at the valley where the roofs meet, instead of neatly coming
together at the corner, the one roof ends some distance up from the edge
of the other. And, the second roof plane extends under that first one,
creating a very shallow cavity. That cavity seems be a source of decay,
AND an irresistable magnet for birds and squirrels.

If I raise the wall on the perpendicular section, the two roofs will
meet at the corner, and the cavity is gone. Only raising the wall
lowers the pitch.

This should be a question on a solid geometry test. Who says that
high school math was worthless?

HTH,
G