Originally Posted by Brent Swain
Mild steel is 60,000 PSI tensile and compression strength . That is 11250 per linear inch for 3/16th plate. Multiply that by the 96 inches in the side of one of my twin keels. That is 1.08 million pounds per side, times four keels sides.
How are you going to break that with a boat under 20,000 lbs?
Just saw two navy 100 footers in Heroit bay. 3/16th hull plate on 100 ft navy ships . And you say the same plate thickness is too light for a 36 foot pleasure boat? You say that a boat which could survive 16 days pounding in 8 to 12 ft surf on the west coast of the Baja, or pounding across 300 yards of Fijian coral reef in big surf, or colliding with a freighter, or hitting the sharp corner of a sunken barge at hull speed, all with minimal damage, is not strong enough? Now that's a stretch!
My hulls are all single thickness, 1/8th for the decks cabin, etc, 3/16th for the hull ,1/4 for the keel sides, and half inch for the keel bottom ( with 4500 lbs of lead ballast poured on top)
Layering steel is a big mistake, guaranteeing corrosion between the layers unless totally sealed.
A good whack with a sledge hammer and a centre punch on lower parts of keels, etc, where corrosion is most likely, is a good starting point on buying a steel boat. If it doesn't give, you have enough thickness there.
Structural failures of steel boats under 40 feet are extremely rare. Your "Invisible " fractures have zero chance of ever causing any problems in steel boats under 40 feet in their lifetimes.
How does such "invisibly fractured" steel compare in strength to a copper fastening in red cedar every six inches, or six inches of plastic?
Okay, let's address the math here....
First we need know that to figure the force of the impact, we are going to have to calculate the Impulse of Force, which is the force of the impact in PSI in this case over the period (length of time) of the impact. In order to know the force we have to know the weight, the speed and the size of the impact point, and to get the Impulse on that we need to know the duration.
From there it is a simple little calculus problem, and we can get it all done in just a few minutes. So, we figure that the length of one of Brent's keels is about 60 inches long by 6 inches wide at the point of contact. ( I tried to look up some of Brent's designs on sailboatdata.com like I do with Bob's or Roger Long's but I found this photo, and I am just kind of guessing on the actual length, at the end of the story it won't matter.
So we have a weight (per Brent) of 20,000 lbs
A hull impact area of 360 square inches maximum.
We will designate a forward hull speed of 10 knots which is 26.465 per sec
A downward force of 954417 N on the keel, and that is not a full calculation, but there are too few here who would understand the math to make it worth the effort.
So that translates to 214561.47 PSI
Score Fukushima Debris Field 1 BS hull Zero
In other words a full on 10 knot collision would poke a hole in your boat, and that force is the force that would occur along the entire 360 square inches, angle it slightly and it goes way up. I know my math is dirty on this because the formula I used was not the full formula, which gets to be fairly complex because I would have to estimate too many of the numbers. I used
Impact Force(F): 2 m vt
Which is really too simple. The actual formula for calculating keel impact force is longer and more complicated and I would have to calculate too much stuff after a very long day at work.
However for those interested the whole formula is here....
Guide Specifications and Commentary for Vessel Collision Design of Highway ... - Aashto - Google Books