Posted - 10/08/2007 : 10:36:47
| Hi all,
We're currently having a debate here at home regarding the catchment of rainfall falling on the roof of a house.
I am trying to devise a formula for the calculation of how much rain different roofs are able to capture.
My problem is a flat-roof versus a pitched-roof.
Two houses have the exact same "floor" plan, ie- the foundation slabs are for example both 10m by 10m.
And considering the roof of both houses are vertically flush with each wall (no overhanging sections of roof).
House 'A' has a flat roof (therefore 10m x 10m), whereas house 'B' has a 'pitched-roof' where the roof originates at two opposing walls with an angle of 60degrees creating a triangular prism, meeting centrally over the center of the house.
The surface area of roof 'A' is a flat 10m x 10m = 100 sq.m
The surface area of roof 'B' is a 60deg pitched (equilateral triangle in side profile creating a triangular prism), resulting in a S.A of (10m x 10m)x2= 200 sq.m
Question: assuming 'lab' conditions, i.e- no wind,the amount of rain falling on both houses is exactly the same, and the 'water-capture efficiency' of both roofs are the same: will the house with the pitched roof and corresponding greater surface area capture more rain than the flat roof? and why?
One argument is that the rain falls in a vertical column over both houses, and because each house has the same 100 sq.m floor plan, the profile of each house pointing up into the oncoming rain is the same, therefore exposed to the same amount of rainfall each.
The opposing argument is that the rain captured is a function of the surface area of the roof , ie house 'B' has twice the S.A of house 'A' therefore it is able to capture twice the total volume of rainfall.
Can anyone clear this up for us and back it up with a simple formula?