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ContinuousWave: Post-Classic Whalers
Dauntless 15 Lifting Bridle
|Author||Topic: Dauntless 15 Lifting Bridle|
posted 05-13-2004 10:03 AM ET (US)
Has anyone fabricated a lifting bridle for use with a hoist? I will be putting the boat in the water at many clubs via a hoist and am keen to know if anyone has gone through the brain damage of fabricating a bridle and determining what the correct lengths are so the center console is not harmed. Thanks.
posted 05-14-2004 12:49 AM ET (US)
It's possible to figure out where the center of gravity of
boat+motor is from the following information:
Total weight of boat+motor+trailer. (Wbmt)
You want the bridle eye over the boat+motor CG, or maybe
I think. Somebody check my math.
If d comes up negative (possible, but not likely) the CG
And thanks to Kawika, who posts here sometimes, for pointing
Once you decide where you want the eye, you can figure the
When sizing the cable and hardware, remember that the sum
posted 05-14-2004 10:22 AM ET (US)
On thinking about it some more, I supect it is likely that
the load on the bow cable would exceed the weight of the boat.
Do the math on cable loads, or find someone who can. It's
relatively simple vectors (well, relatively simple if you
are an engineer ;-).
posted 05-14-2004 12:26 PM ET (US)
Okay, since I'm not an engineer, here's a simple sample of a bridle on a boat close in dimensions to a Dauntless 15. In this sample, it's assumed the center of gravity is 1/3 of the way forward from the stern eyes to the bow eye, and that the hoist line is over the CG.
What are the loads on the lines?
posted 05-14-2004 12:45 PM ET (US)
Oops... found an error and just updated drawing.
posted 05-14-2004 09:34 PM ET (US)
Based upon your drawing with the dimensions shown and the center of gravity being 1/3 the overall distance from the stern, with a 10' vertical height from the horizontal plane of the lifting eyes, I concur that your dimensions are correct. I believe, however, that the load would be split equally between the front and the rear. therefore, the bow line would need to support 900 lbs and the two stern lines would need to support 450 lbs/ea (900 lbs). This would mean that the tension in the bow line would be 1272# and the tension in each of the stern lines would be 520#.
If you kept the harness dimensions the same, then the tension in each of the lines would be:
Bow Line = (Total Weight/2) x 1.414
Maybe someone with rigging experience could advise what safety factor to apply to the static tension since the motion of the crane will unquestionably add additional load during the lift. There may also be additional load to break the surface tension of the water. Knots and wear to the lines will also significantly affect the working load of the lines.
If it were my rig, I would be VERY conseervative. The cost of rope and fittings is a lot less than a new whaler. By the way, although I believe these numbers to be correct, I TAKE NO RESPONSIBILITY for the accuracy of these numbers.
posted 05-17-2004 07:26 AM ET (US)
Thanks, do I understand that the rears should be 5.8' and the bow, 10'?
posted 05-17-2004 08:33 AM ET (US)
No! That was just an example that assumed the CG was 1/3 of the way forward from the stern eyes. In that example, the 10' distance to the bow eye and the two 5.8' distances to the stern eyes, are the distances at the top of the gunwales. They are only theoretical and go right through the console. The lift lines in that example are 11.6' stern and 14.14' bow.
Go weigh the boat as Chuck described and find where the CG on YOUR boat is measured from the stern forward. Measure from stern eye to stern eye, and from a line between the stern eyes to the bow eye. Measure from a line across the gunwales to the top of the center console windshield frame.
Then come back and post those numbers. At that point, we'll have something to work with.
posted 05-17-2004 11:24 AM ET (US)
The load would not be split equally front to year.
The vertical load on the bow line is
1800*(5.0/(10+5.0)) = 600
The vertical load on each of the stern lines will be
(1800*(10/(10+5.0)))/2 = 600
The total load on the bow line will be
Again, somebody check my math AND my engineering. Statics
Also note that this example harness is much "taller" than
Also note that extremely conservative design would be an
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