1981 FORMULA F233 RESTORATION

Questions about boat repairs with our resins and fiberglass: hull patches, transoms and stringers, foam, rot etc.
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Re: 1981 FORMULA F233 RESTORATION

Post by NOTHING ELSE MATTERS »

Joel, she is a 24* dead rise and yes, there is room, if you remember i was going to have a sub floor under the tank with a drain pipe by the keel and one stringer on each side 4-5" high. Also i was going to put 3/4" x 4" plywood stiffeners about 8" away from the length of the keel.

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Re: 1981 FORMULA F233 RESTORATION

Post by tech_support »

Also i was going to put 3/4" x 4" plywood stiffeners about 8" away from the length of the keel.
no need for this if you glass that "sub floor" to the hull. If you glass that "subfloor" to the bottom, you have basically made a stringer. Also with 24 deadrise, the keel itself is a stringer. No problem moving the main stringers out to 32" apart, they will then only be 10" for the keel stringer structure (glassed in subfloor).

The span for your sole will now be wider, that only means you might need a stiffener of framing under the sole in this wide spot.

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Re: 1981 FORMULA F233 RESTORATION

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Joel, got the shipment today, thank you very much. I'll be in OBX all next week, when i come back will start working on this old girl again, can't wait.

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Re: 1981 FORMULA F233 RESTORATION

Post by NOTHING ELSE MATTERS »

Question for the pros.
How much foam do i need on a 5000 lbs boat? I'm trying to figure out some storage compartments on my boat.

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Re: 1981 FORMULA F233 RESTORATION

Post by Cracker Larry »

How much foam do i need on a 5000 lbs boat?
A lot 8O Foam supports 60 pounds per cubic foot, so that would be 83 cubic feet. A 2 gallon kit of foam makes 8 cubic feet, so you would need about 21 gallons of liquid.
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Re: 1981 FORMULA F233 RESTORATION

Post by NOTHING ELSE MATTERS »

That's what i figure, 11 kits=5280 lbs, but that seems a lot of foam or maybe not, i never used this product before, so :help:
With another words, in order for the boat to float if swammed, i have to have the amount of the boats weight same with foam buoyancy or more? :doh:
What i mean is that the boat weighs 5000 lbs, if filled with water weighs more, how much buoyancy do i need now?

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Re: 1981 FORMULA F233 RESTORATION

Post by wadestep »

Don't forget that the boat itself displaces water. It would be hard to calculate, but you can subtract the bouyancy of the boat from the extra flotation required. I'd bet this would cut down the amount of foam required significantly.
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Re: 1981 FORMULA F233 RESTORATION

Post by wadestep »

NOTHING ELSE MATTERS wrote:That's what i figure, 11 kits=5280 lbs, but that seems a lot of foam or maybe not, i never used this product before, so :help:
With another words, in order for the boat to float if swammed, i have to have the amount of the boats weight same with foam buoyancy or more? :doh:
What i mean is that the boat weighs 5000 lbs, if filled with water weighs more, how much buoyancy do i need now?
If your boat was infintessimly small, yet still weighed 5000 lbs (like a mini black hole) (impossible), then you would need 5000 of bouyancy. However, your requirement should be significantly less than that. I don't know how to figure this out without sinking the boat and then weighing it underwater. there must be another way... :lol:
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Re: 1981 FORMULA F233 RESTORATION

Post by NOTHING ELSE MATTERS »

wadestep wrote:
NOTHING ELSE MATTERS wrote:That's what i figure, 11 kits=5280 lbs, but that seems a lot of foam or maybe not, i never used this product before, so :help:
With another words, in order for the boat to float if swammed, i have to have the amount of the boats weight same with foam buoyancy or more? :doh:
What i mean is that the boat weighs 5000 lbs, if filled with water weighs more, how much buoyancy do i need now?
If your boat was infintessimly small, yet still weighed 5000 lbs (like a mini black hole) (impossible), then you would need 5000 of bouyancy. However, your requirement should be significantly less than that. I don't know how to figure this out without sinking the boat and then weighing it underwater. there must be another way... :lol:
Now you got me more confused. I know about the physics that what ever you put in the water looses as much weight as the weight of the water displaced, that is what i'm trying to understand.

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Re: 1981 FORMULA F233 RESTORATION

Post by tech_support »

You do not need enough to displace 5000 pounds, but you do need enough (assuming you want positive flotation) to displace what the rest of the boat does not. The materials that make up the boat have a volume (wood/glass/gas/tanks/etc) that volume displaces water. So what you actaully need is enough buoyancy foam to float the difference between the boats weight and its submerged displacement. Now the problem :) what is that volume of all the parts in the boat? :doh: You cant answer that unless you have a very detailed spreadsheet or a 3D computer model. The easy way: assume the original designers did this calculation correctly when they deigned your boat, and they knew how much foam they needed to give positive flotation. So you can fill that same spaces they filled and have the same amount of buoyancy.

In your case, you are moving things around, but you can still measure the volume of the areas that used to have foam. I would replace that same volume, plus a little to be conservative.

All of this is just to say that you have to use 5,000 pounds of buoyancy :) Of course it doesn't really hurt anything to use more foam.

In the case of your boat, they were using the foam to support the sole and stiffen up the hull - so you may find that they used more than they needed for just buoyancy :!:

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