How can I determine, in advance, the freeboard in a 14' jon boat with
the following information?

1-Flat bottom, square front.
2-14' long, 70" beam, 21" side depth, 16" transom height.
3-Total capacity of persons and gear = 820 pounds.
4-Usual load of persons and gear = 475 pounds.
5-Small outboard motor = 50 to 70 pounds.

So, based on the above data, am I gonna see that water lapping at the
edges or not?

-KCB

Let's run the numbers. Archimedes' priciple states that the volume
immersed in the water will be equal to the total weight of boat, motor,
gear, etc. Assume the boat is perfectly rectangular 14 ft by 6 ft,
that's a total bottom surface area of 84 sqft. Water weighs approx 64
lbs per cubic ft. You don't say how much the boat weighs but lets
assume 300 lbs or so. So, total weight equals about 300 (boat) + 475
(people and gear) + 70 (small OB and fuel) = 845 lbs. Divide 845 by 64
to get cubic feet of water, about 13.2 by my calculator. So,
immersion in feet will be 13.2 divided by 84, about .157 feet or 1.89
inches. Round it up to 2 inches for good luck and then subtract from
your lowest point, 16 inches at the transom for an answer of 14 inches,
or 19 inches of freeboard at the side (21 minus 2).

Hope that helps.

As a very quick sanity check

14 x 6 = 84 square feet

if submerged 3 inches (1/4 foot) then you have

84 /4 cubic feet = 21

21 x 64 = 1344 pounds flotation.

Since in a flat square cross section each inch of submersion will
require the same weight two inches submerged = 896 lbs approx.

There is still the issue of weight distribution on the boat that could
mean a significant depth at the stern since the outboard is actually
outside the boat perimeter.

If the front edge of the boats bottom is at the water surface because
of weight towards the stern the depth at the stern will be about 4
inches.

wrote:
Let's run the numbers. Archimedes' priciple states that the volume
immersed in the water will be equal to the total weight of boat, motor,
gear, etc. Assume the boat is perfectly rectangular 14 ft by 6 ft,
that's a total bottom surface area of 84 sqft. Water weighs approx 64
lbs per cubic ft. You don't say how much the boat weighs but lets
assume 300 lbs or so. So, total weight equals about 300 (boat) + 475
(people and gear) + 70 (small OB and fuel) = 845 lbs. Divide 845 by 64
to get cubic feet of water, about 13.2 by my calculator. So,
immersion in feet will be 13.2 divided by 84, about .157 feet or 1.89
inches. Round it up to 2 inches for good luck and then subtract from
your lowest point, 16 inches at the transom for an answer of 14 inches,
or 19 inches of freeboard at the side (21 minus 2).

Hope that helps.

+++++++++++++++++++++++++++++++++++

It certainly did. Thanks. Since the boat only weighs 233 pounds, with
your calculations the transom freeboard and side freeboard should be
about another 1/2 inch more---14.5 at the transom and 19.5 at the side.
Guess that will relieve a little anxiety about having to use the
coffee can to bail. (Still would rather have a jon boat with a 48"
side depth and transom but would probably need a ladder to board. Now
THAT would be some freeboard!)

-KCB