Michael J Edelman wrote: SNIP > Given a hypothetical kayak of 10 ft^3 volume, we could get 64 lbs of ballast > from 0.133 ft^3 of water or .012 ft^3 of lead. Assuming the rest of the space > is air filled, the boat has the same displacement in either case, but a greater > resistance to heeling with the lead. Your points are well taken. One little thing, though. You'd need 1.0 ft^3 of water to get 62.4 lbs. of ballast. (0.133 is one divided by 7.48, the number of gallons per cubic foot) The same ballast using lead would require about 0.13 ft^3. - i.e. One ft^3 of water weighs 62.4 lbs. and 1 ft^3 of lead weighs about 490 lbs. Lead probably weighs a good bit more than this, since the 490 figure is for steel. (My book didn't list the weight of lead.) Let's say we wanted 62.4 lbs. of ballast. Then we'd need 0.13 ft^3 of lead, which could fit in a 6" diameter by 7.8" long tube strapped to the inside centerline of the hull. Or - we'd need 7.5 gallons of water strapped down somewhere. A cubic foot of water (or anything for that matter) takes up a lot of space! Having said all that, I agree that for most applications, the time needed to assemble and install the ballast system might be better spent working on paddling skills - braces & rolls - although that extra 50 FT-LBS. or so of righting-heel torque could be very helpful to a roll in rough water. Craig Olson Bellingham, Washington *************************************************************************** PaddleWise Paddling Mailing List Submissions: paddlewise_at_lists.intelenet.net Subscriptions: paddlewise-request_at_lists.intelenet.net ***************************************************************************
Craig Olson wrote: > Michael J Edelman wrote: > > SNIP > > Given a hypothetical kayak of 10 ft^3 volume, we could get 64 lbs of ballast > > from 0.133 ft^3 of water or .012 ft^3 of lead. Assuming the rest of the space > > is air filled, the boat has the same displacement in either case, but a greater > > resistance to heeling with the lead. > > Your points are well taken. One little thing, though. You'd need 1.0 ft^3 of water > to get 62.4 lbs. of ballast. (0.133 is one divided by 7.48, the number of gallons > per cubic foot) The same ballast using lead would require about 0.13 ft^3. - i.e. > One ft^3 of water weighs 62.4 lbs. and 1 ft^3 of lead weighs about 490 lbs. Lead > probably weighs a good bit more than this, since the 490 figure is for steel. (My > book didn't list the weight of lead.) Whoops. Right. I had a beer with lunch. But the relationships hold. The specific gravity of lead is 11.43, I think. > Let's say we wanted 62.4 lbs. of ballast. Then we'd need 0.13 ft^3 of lead, which > could fit in a 6" diameter by 7.8" long tube strapped to the inside centerline of > the hull. Or - we'd need 7.5 gallons of water strapped down somewhere. A cubic > foot of water (or anything for that matter) takes up a lot of space! > > Having said all that, I agree that for most applications, the time needed to > assemble and install the ballast system might be better spent working on paddling > skills - braces & rolls - although that extra 50 FT-LBS. or so of righting-heel > torque could be very helpful to a roll in rough water. Of course, those of us who use folding kayaks don't really know how to roll- since we never have to. ;-) But I do recall Derek Hutchinson writing how he used a lead filled tube to add ballast for photography from the boat. Even the experts use it! -- ---------------------------------------- Michael J Edelman mje_at_mich.com http://www.mich.com/~mje http://www.mich.com/~mje/kayak.html http://www.mich.com/~mje/scope.html *************************************************************************** PaddleWise Paddling Mailing List Submissions: paddlewise_at_lists.intelenet.net Subscriptions: paddlewise-request_at_lists.intelenet.net ***************************************************************************
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