Gary Warner wrote:
I've had a few random questions rattling around the

brain for a while - figured I'd pose them there.



When a sub dives to a certain depth, is that depth

measured from the bottom of the sub, the middle,

or the top. In other words, if it dives to 200 feet,

does that mean there is 200 feet of water above

it or that the bottom of it is 200 feet below the

surface?

Depth below the surface, but pass on which part of the boat they
measure from. It's really just a measure of pressure on the hull not
"depth" as such, but the result is the same; unless it's in warm fresh
water:-)


On modern gas car engines and when it's cold

outside, is it still better to let them warm up a

bit and how warm (how long) is necessary?

Modern?? with modern multigrade oils??? A little warmup while you fix
the phone & belt up etc is OK but don't leave it just idling. The best
thing is to get it up to thermostat temp as soon as possible & to do
that it's best making some power (it takes forever if just left idling).
On the other side don't jump in a cold engine & take it to max power,
revs etc, that's not sensible either.




Anyone have a good way to find the total surface

area of my boat hull while it's in the water? It's

a 22' boat, not very deep V, "square" transom. I

did some estimating, but wonder if there are any

creative ways to get more accurate.

From your thread title I think you want the "volume" of the hull below
the water line??? not the area??? Also your question seems to suggest
"while it's in the water"?? you want to work this out without pulling
the boat out??

For the surface area, most of the anitfoul paints, either on the tin or
a pamphlet, have a simple method to work out how much paint you'll need.
But they're not too accurate.

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

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

(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).

K




Gary

"K Smith"

From your thread title I think you want the "volume" of the hull below
the water line??? not the area??? Also your question seems to suggest
"while it's in the water"?? you want to work this out without pulling
the boat out??


Sorry, not worded clearly on my part.

I'm looking for the area. Reason is just for curiosity. Reason I want the
area
is to calculate the pressure per square inch. Example: Boat is 4000 lbs,
area
where it touches the water is X. So approximate pounds/square inch is Y.

No, I don't need to do this IN the water. What I was trying to say is that
I only want the area that TOUCHES the water. The remaining area of the
hull would not be "supporting" the boat while in the water.

It's just a winter month curiosity exercise.

Thanks.

"Gary Warner" wrote in message
...
Sorry, not worded clearly on my part.

I'm looking for the area. Reason is just for curiosity. Reason I want
the
area
is to calculate the pressure per square inch. Example: Boat is 4000 lbs,
area
where it touches the water is X. So approximate pounds/square inch is Y.

No, I don't need to do this IN the water. What I was trying to say is
that
I only want the area that TOUCHES the water. The remaining area of the
hull would not be "supporting" the boat while in the water.

It's just a winter month curiosity exercise.

Thanks.



Do not need the wetted area of the boat for those calculations. Just how
far under water is the square inch. And take the average depth of the
location and multiply by the pressure at depth. And the pressure is a
little under 1/2 psi per inch of depth.
Bill

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