K Smith wrote:



This is a mea culpa.

I've made a mistake!! (yes, yes I know; again!!)

If there is anything I can say in mitigation, it's that I did realise
it on my own, get it checked this time:-) & have now tried to belatedly
correct it.


To measure surface area accurately you'll need to measure 1/2 the
bottom of the boat & that part of the sides (if a chine boat) below the
waterline, where ever possible reduce it to oblongs or squares, then
various right triangles when you run out of easy oblongs etc. Add them
all together, double it & that's the total surface area.

In boat design they use a planimeter to run over the lines.

To calculate the "displacement" (volume of the boat below the
waterline) you can actually get a pretty accurate measurement by using
"simpson's formula" even as your question seems to suggest, with the
boat still in the water by;

(i) Boats are usually designed on 10 "sections" i.e. notionally the
boat has 11 transverse stations/bulkheads across it equal distance apart
from the waterline bow. Say a 30 ft WL boat they'd be 3ft apart?? These
stations/bulkheads are not "real", although usually bulkheads are at a
station point, but whatever but you can easily measure with a tape what
the below waterline areas would be, even on a bigger boat.

(ii) You need to measure the "area" of each of those notional
stations/bulkheads, but just that area which is below the waterline.

This is WRONG it's not the "area" it's the perimeter length or
circumference!! Having made this error I carried it on through the rest
of the description.

(again designers with plans drawings etc run around the 1/2 shape X3 div
by 3 to average with a planimeter)

Damn!!! I even correctly described how designers measure the perimeter
length of irregular shapes, but once I had a mindset of "area", well
there ya go, I'm sorry again.

(iii) Once you know the below the waterline only area in sq ft

This should read "Once you know the below the waterline only perimeter
length in inches"

of
each of the boat's 10 notional stations, you multiple each by simpson's
multipliers 1,4,2,4,2,4,2,4,2,4,1 (11 notional below waterline
stations/bulkheads gives 10 equal length sections of the boat)

(iv) Add all the answers together so you now have the sum of functions.

(v) Use simpson's formula to work out the boat or ship's current
displacement per;

2 X 1/3 X sum of functions of 1/2 areas X (inverted scale)sq X the
common interval X 64 = displacement in ponds of salt water, or for fresh
water use 62.2 as the last figure.

This is the formula as used in boat design, so it is a bit more yuk
than you need, all you need is;

(a) 2 is to account for only using 1/2 the below waterline
station/bulkhead area, you can leave it out if you measured the full sq
ft of each area before using his multipliers.

(b) 1/3 is just part of the formula.

(c) Sum of the functions is explained in (iv) above. (but
designers tend to just use 1/2 then multiply by 2 see (a))

(d) Inverted scale squared doesn't bother you because you
can use feet as a direct measure, whereas a designer might be using say
1/2"
to the foot in their drawings. So make sure your notional below the
waterline areas or 1/2 areas if you choose, are in sq ft.

This whole paragraph needs correction because clearly if you have the
sum of functions in inches then you don't need to adjust for scale. So
you just use you answer from (iv)

(e) Common interval is the length in feet of each section,
again say it's waterline length of 30 ft the "common interval" is 3.

(f) At this point the formula should have delivered you the
boat's below the waterline volume in cubic feet, the 64 is just the
weight in pounds of a cubic ft of salt water, or 62.2 for fresh water;
to give you the displacement in lbs (weight of the boat).

The rest looks pretty much OK, so I hope not too many of you have been
working in vain on this:-) Of course I know I know I know nobody even
read it:-) but so what?? it was wrong:-) I can't help but correct it for
the record.

Sincere apologies again.

K


K




Gary