posted 03-21-2012 08:39 AM ET (US)
It is often cited that gel coat resin should be applied to a thickness of 0.020-inch or 20-mils. If we have a gallon resin, that is a volume of 231-inch^3. Let us allow for some waste in the applicator and some over spray. Perhaps we can say that from a gallon we will get 200-inch^3 (or 0.11574-foot^3) of useful resin applied to a surface. If we apply at a uniform thickness of 0.020-inch, then the coverage area should be about200-inch^3 / 0.020-inch = 10,000-inch^2
To convert to foot^2 we divide by 144. The coverage area of a gallon at a thickness of 0.020-inch is
10,000-inch^2 x 1-foot^2/144-inch^2 = 69-foot^2
We can roughly approximate the surface of the interior deck of the OUTRAGE V-20 by its length and width. From the REFERENCE section information I find the dimensions to be
Length = 19-foot 10-inch = 19.83-foot
Width = 7-foot 4-inch = 7.33-foot
The surface area of the deck may be approximated as Length x Width:
19.83-foot x 7.33-foot = 145.4-foot^2
Perhaps we can reduce the deck area estimate somewhat to account for the taper in the bow. Let us say the deck will be 120-foot^2. What thickness of resin application will we get from our gallon of resin? We calculated earlier that a gallon of resin will have a useful volume of 0.11574-foot^3, and the area to be covered is 120-foot^2, producing a thickness of
0.11574-foot^3 / 120-foot^2 = 0.000964-foot or 0.0115-inch
We can cover the deck area to a thickness of 0.0115 or about 11-mills.