Martin Baxter wrote:
JimC wrote:

And you did take math, and geometry in high school? Consider a typical
crew of two people (assume total weight of 350 lbs) and a skipper of 230
lbs. (about my weight). Assuming that the skipper is sitting on the
captain's chair, about 1.5 ft forward of the center of the motor, and
about 16 feet aft of the center of mass. Assuming that the two crew
members are sitting two and three feet forward of the captain,
respectively, their total mass will be the equivalent of 350 pounds
positioned about 14.5 feet from the center of mass about which the boat
has a tendency to pitch. The motor, at about 220 pounds is about 17.5
feet from the center of mass. Squaring the distances, the relative
values of the rotational momentum of the skipper and crew are more than
twice that of the motor, despite the fact that they are closer to the
center of mass. Once again, Jeff, your theories are simply wrong. (If
you wish, I'll provide the calculations and "foot pounds" or whatever,
in greater detail.)

Jim:

I thought I'd run your numbers through my calculator, just for fun:

(14.5ft)^2*350lbs=7.36E4ft^2-lbs , that's 73,600 for the exponent
challenged.

(17.5ft)^2*220lbs=6.74E4ft^2-lbs , 67,400 ....

Maybe you think that 73,600 is more than TWICE 67,400, but I beg to
differ. To a physicist, they're essentially the same thing.

Point to Jeff.

Gawd, I don't believe other people are still reading this!

Thanks for doing the math, but the are a few problems here. Jim uses
his weight (230 lbs) when clearly this should be done with a
"standard" weight (160 lbs). Secondly, Jim invents the figure of 16
feet forward of the helm seat for the center of mass, which puts it
only 8.5 feet aft of the bow! Clearly, it has to be more like 11-12
feet forward of the stern. This has a huge affect on the r^2 part of
the formula, especially for the crew forward of the helm.