Quote:
Originally Posted by bencoder  another uninformed, but possibly slightly more scientific opinion:
it would add 3% to the volume of water you displace, which means that the added depth would be closer to the cube root of 3%.. about 1.5%, although it'll be less than 1.5% since the size(at the waterline) increases as you submerge the boat.
Assuming the above figure of 3% more dense is correct  |
If I may borrow Gord's symmetrical cubic boat say 1 metre in length, 1 metre in breadth, perfectly square waterplane that remains constant down to the keel. It also weighs exactly 1030 Kg. 1 cubic metre of saltwater weighs 1030 Kg, so it's easy to determine the draught of the boat - 1 m.
As Gord alluded to, sea water is not uniform; variations in salinity and temperature give it a range of density (1020 - 1030 kg/cubic metre) - I chose 3% as the worst case scenario.
If I took my cubic boat into fresh water (which weighs 1000 Kg/cubic metre) it would still need to displace 1030 Kg of water. 1 Kg of fresh water = 1 litre = 1000 cubic centimetres. 1030 Kg = 1030000 CC of water. Since length and breadth remain the same at 1 m or 100 cm, the draught calculates as 103 cm or 1.03 m; a 3% increase in draught for a 3% difference in density.
As has been pointed out, boats tend to have a more complex form than my cubic boat, but contrary to what Gord stated, a boat needn't be a perfect cube to have a consistent waterplane area. TPI or PPI is tied directly to the waterplane area at the actual waterline. The OP's boat with 4'6" draught at the max change of 3% will only sink by about an inch and a half. There are many boats that would not show a significant change in waterplane area from their waterlines to 1.5 inches above. Assuming the boat doesn't have a reverse bow or significant tumblehome (which would cause PPI/TPI to increase), one could assume that the full 3% might represent the worst case scenario.
So for a rough estimate, I stand by my 3% figure.