I've got a question, not a slam, about the GH37.  According to the spec
sheet on the manufacturer's website, the GH37 is about 36 feet long, 15
feet wide, displaces 47,000 pounds, and draws 2 feet 10 inches of water.
A Grand Banks 36 is a shade over 36 feet long, 12 feet wide, displaces
26,000 pounds, and draws 4 feet of water if you're lucky.  My minimal
knowledge of physics and stuff tells me that only one thing can account
for a boat with more or less the same length as a GB36 and almost twice
the displacement but having at least a foot less draft, and that one
thing is the three extra feet of beam.

I freely admit I was a terrible math student, but is that correct?
Approximately three extra feet of beam spread along about 36 feet of
hull can support the additional 21,000 pounds of displacement of the
GH37?  There's probably a simple formula for figuring this out, but my
unscientific mind wants to say 2 feet 10 inches seems to be a rather
optimistic claim for the draft of a 47,000 pound boat.  Or not.  I don't
know, which is why I'm asking.....

C. Marin Faure
GB36-403 "La Perouse"
Bellingham, Washington

The GH37 has a very large superstructure which can account for some of the
weight. She is designed for shoal draft and I believe the draft figures.

How about our GH37 owners chiming in?

Ron Rogers

I think what the original questioner was going after was Archimedes (? or
some old guy's) principal, that a floating object must displace its own
weight in water. Therefore a heavy object must displace a larger volume of
water than a lighter one.

Therefore how can a heavy GH37 draw only 2'10" compared to a lighter GB36 or
a Monk36 drawing a deeper 4'0" displace twice as much water? We understand
that a GH37 has a greater beam, 15' versus 13'. Is that enough to double the
volume of water displaced?

-- Jim

Jim & Rita Ague
M/V Derreen, Monk 36

The GH37 has a very large superstructure which can account for some of the
weight. She is designed for shoal draft and I believe the draft figures.