math fun.. what's the RPM of the motor? 30 ft square root x 1.56 is 8.544 mph as a basic potential hull speed, and energy demand starts going up sharply about 75% of that.. is 6.4 mph. 5280 ft per mile divide by 60 minutes is 88 feet per minute for 1 mph.. so 6.4 mph about 564 feet per minute, 8.54 about 751 feet per minute. (I'm really just "quick ballpark estimating" with this, no doubt others can do better!) 6.4 + 8.54, divide by two, about 7.47 mph, and about 7/8 of a 30 ft hull speed, if the waterline is about 30 ft.. is about where I'd expect to put a ballpark speed limit.. where aiming at max hull speed and motor under load being 7/8 of that, or 87.5%, motor under load might reach 80-85% of free spin rpm if geared-propped right. (is thinking not to stress it lots and get it heating up) 6" pitch is a foot every 2 revs.. 5280x2=10,560, divide by 60 (minutes) is 176 rpm.. per MPH if I'm thinkin straight.. so 176 x 7.47mph = about 1315 rpm? where a larger diameter 12" pitch at 658 rpm (more gear reduction) is probably better, more blade surface and less prop slip, but that gets real complicated too! how big a prop? what motor rpm? what gear reduction? maybe my ballparkin' is worth a darn as a starting point example to the math? motor RPM and what gear reduction is there to play with are pretty big factors. how much clearance from output shaft to hull for a prop radius-diameter too. theres lots I dont know, and lots of unknown variables too.. motor RPM and gear ratio is the first thing I'd be looking at. hope its somewhat a help there, again, I'm sure no expert either. --- On Mon, 2/27/12, Robert Jorgenson <falcolnxp@yahoo.com> wrote:
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