I'm impressed waterguy. You continually come up with the most impressive, in depth answers. You a boat builder/engineer/naval architect?
Where do you get this stuff? Bob S. On Sep 18, 2:46 pm, waterguy <[email protected]> wrote: > Your answer is found at 33 CFR § 183.53. Here's a > link:http://ecfr.gpoaccess.gov/cgi/t/text/text-idx?c=ecfr&sid=01a07e768bf6... > > First, you compute a factor. This factor is [boat length] times > [transom width]. If the transom is not the widest part of the boat, > substitute the width measured at the widest part of the stern quarter. > > If your factor is less than 52.5, you apply the horsepower ratings > shown in the table in 33 CFR § 183.53. That table provides > Factor :: Horsepower > 0-35 :: 3 > 36-39 :: 5 > 40-42 :: 7.5 > 43-45 :: 10 > 46-52 :: 15 > > If the factor is greater than 52.5, you apply a formula to the > factor. There are 3 possible factors depending on transom height and > hull shape. Your boat is not flat bottomed with hard chines, and it > probably doesn't have a 20-inch transom (my guess would be 15"). So > the formula you apply is: 0.8 times [factor] minus 25. > > If the capacity calculated is not a multiple of 5, you may raise to > the next higher multiple of 5. > > So, if your boat is 14 feet long and is 6 feet wide at the transom, > your factor is 14 x 6 = 84 > Applying the formula, 0.8 x 84 = 67.2 > 67.2 - 25 = 42.2 > > Raised to nearest multiple of 5 = 45 hp capacity. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "UnifliteWorld" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/unifliteworld?hl=en -~----------~----~----~----~------~----~------~--~---
