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Author Topic:   Dauntless 15 Lifting Bridle
David Mc posted 05-13-2004 10:03 AM ET (US)   Profile for David Mc   Send Email to David Mc  
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.
Chuck Tribolet posted 05-14-2004 12:49 AM ET (US)     Profile for Chuck Tribolet  Send Email to Chuck Tribolet     
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)
Tongue weight of boat+motor+trailer. (Tbmt)
Total weight of trailer only. (Wt)
Tongue weight of trailer only. (Tt)
Distance from the axle center line to where tongue weight
was measured. (D).

You want the bridle eye over the boat+motor CG, or maybe
just a touch forward of it.

The CG of b+m is d inches in front of the axle, where

d = D * (Tbmt-Tt) / (Wbmt-Wt)

I think. Somebody check my math.

If d comes up negative (possible, but not likely) the CG
is BEHIND the axle).

And thanks to Kawika, who posts here sometimes, for pointing
out that this was possible (his idea, my math, I take all the
blame).

Once you decide where you want the eye, you can figure the
cable lengths with a tape measure.

When sizing the cable and hardware, remember that the sum
of the loads on the cables will be greater than the
weight of boat+motor. If the eye is close to the deck,
the load in one cable could far exceed the weight of the
boat. Do the math.


Chuck

Chuck Tribolet posted 05-14-2004 10:22 AM ET (US)     Profile for Chuck Tribolet  Send Email to Chuck Tribolet     
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 ;-).


Chuck

Moe posted 05-14-2004 12:26 PM ET (US)     Profile for Moe  Send Email to Moe     
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.

http://www.engr.udayton.edu/staff/lriggins/Whaler/bridle.bmp

What are the loads on the lines?

--
Moe

Moe posted 05-14-2004 12:45 PM ET (US)     Profile for Moe  Send Email to Moe     
Oops... found an error and just updated drawing.

--
Moe

davej14 posted 05-14-2004 09:34 PM ET (US)     Profile for davej14  Send Email to davej14     
Moe,

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
Each Stern Line = (Total Weight/4) x 1.16

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.

David Mc posted 05-17-2004 07:26 AM ET (US)     Profile for David Mc  Send Email to David Mc     
Thanks, do I understand that the rears should be 5.8' and the bow, 10'?

David Mc

Moe posted 05-17-2004 08:33 AM ET (US)     Profile for Moe  Send Email to Moe     
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.

--
Moe

Chuck Tribolet posted 05-17-2004 11:24 AM ET (US)     Profile for Chuck Tribolet  Send Email to Chuck Tribolet     
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

600*14.14/10 = 848

The total load on each of stern lines will be

600*11.6/10 = 696

Again, somebody check my math AND my engineering. Statics
class was 35 years ago.

Also note that this example harness is much "taller" than
my dealer-built Montauk harness, and the total load on the
line goes up rapidly as the harness becomes shorter.

Also note that extremely conservative design would be an
excellent idea. Boats weren't meant to fly. ;-)


Chuck


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