Using a snatch block for a Bahamian Moor - SailNet Community
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post #1 of 13 Old 04-07-2009 Thread Starter
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Using a snatch block for a Bahamian Moor

I'd be interested in your thoughts on this... An "improved" method of Bahamian Moor is detailed in this post - The Better Bahamian Moor - which uses a snatch block down low to minimize the threat of wrapping your secondary rode around your keel. I have a spare Ronstan RF6721 block (https://store.catsailor.com/pc/viewP...dproduct=10799) which "deforms" at 4000lbs. Our secondary anchor rode is 1/2" line with 20' of 5/16" chain. I have NO CLUE as to the loads that would be placed on the snatch block. Any thoughts on whether this particular block might work for this function? Passport 40, probably weighing about 28,000lbs or so. Our primary anchor is a Rocna 25kg with an all chain 5/16" rode. The secondary is a Fortress FX23. Thoughts?

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Chris

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post #2 of 13 Old 04-07-2009
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It sounds like a pretty good approach.

I'm obviously reluctant to gainsay Don Casey, but if I were going to follow his suggestion, I would use our stern anchor to set the downwind/current anchor, paying it out over the stern roller. Once the desired scope was achieved, I would set it hard under power (forward) to make certain it set it well. I would not want to rely on merely winching it in as he suggests.

Then I would pay the remainder of the stern anchor rode out while we continued to advance forward to set the upwind/current anchor. After backing down and setting the upwind/current anchor (minding the stern anchor rode all the while), I would then lead the stern anchor rode up forward to the bow (outboard, of course) and follow his instructions about affixing the snatchblock.

Let us know if you try it!


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post #3 of 13 Old 04-07-2009 Thread Starter
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I agree with you - I would also set the stern anchor under power. We don't have a roller so I'd have to cleat it on a winch. I wonder how much strain on the cleat I would have by running it up to 75% for 15-30 seconds though. I'm primarily intrigued by the whole snatch block idea though. I'm thinking 4,000lb until deformity is reasonable in all but the heaviest of weather, but I have no idea about how to calculate the load.

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post #4 of 13 Old 04-07-2009
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Chris,

When setting from the stern, don't forget that 75% throttle ahead has a lot more force than 75% throttle astern (assuming fixed blade prop).

As for the force on the snatch block, I don't think it's an issue. The way I understood it, the lazy rode will not be putting any load on the block, it's just acting as a fair lead to keep the angle on the slack road low enough to avoid fouling the underbody of the boat.

When the wind/current shifts, the lazy road becomes the working road, but the pull on the road is straight (i.e. it does not deflect around the turn of the block) so there is no load on the snatch block. Does this make sense?

Here I am assuming a combination chain/rope road.


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post #5 of 13 Old 04-07-2009
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Okay, I just reread Casey's description. He does seem to advocate snugging both roads so that they're always under load.

If you are concerned about the loads on the snatch block, you could try as I suggested above, i.e. never load more than one road at a time and let all the load come to the mooring cleats. This will result in a somewhat larger swing circle.


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post #6 of 13 Old 04-08-2009
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You should bear in mind that, if you're going to set the stern bower under power that you'll now have a snatch block holding it underwater at...just about the depth of your prop? God forbid you have to bring either rode in under a strain. As you can tell, I'm not much of a fan of rodes married beyond reach and certainly not underwater.

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post #7 of 13 Old 04-09-2009
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I'll be interested in your reports on that technique, Chris. One thing that would concern me would be if the boat decided to do a few 360's over a few days you might wind up with a tangled mess above the snatch block. Don Casey mentions using the dinghy to spin the big boat and untwist the rodes, but that can be impossible with a strong tide flowing. My approach to that part of the problem has been to retrieve the down current anchor using the dinghy and then do laps around the upstream anchor's rode until the twists are gone.
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post #8 of 13 Old 04-09-2009
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Chris

Why are you putting the stern out first? Does it matter?

I was thinking that if you put the stern out first, then the bow, you'll have to back down on the bow anchor and the stern rode will be very close to the prop & rudder. If you put down the bow first and then the stern, when you power forward to set the stern, you have the length of the boat before you might fowl the bow rode. Certainly you will have someone to tend the rode either way, but having the distance from the bow to the prop seems like a good measure of safety.
Am I missing something (well, besides THAT)?
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post #9 of 13 Old 04-09-2009
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Chris,

I would be VERY interested in knowing how that works!!! Please relay back. I like the concept. Regarding the load on the block, even if it fails the scope has not changed and you are still relying on the anchor rode. Seems a win-win to me. I certainly would want to keep an eye on the block for wear/barnacles that might chafe the anchor line over time. I also believe that you will have to deal with a lot of turns/wraps/fouling of the rodes. But that cannot be helped if you use that type of anchoring.

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post #10 of 13 Old 04-09-2009
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Quote:
Originally Posted by JohnRPollard View Post
Chris,
As for the force on the snatch block, I don't think it's an issue. The way I understood it, the lazy rode will not be putting any load on the block, it's just acting as a fair lead to keep the angle on the slack road low enough to avoid fouling the underbody of the boat.

When the wind/current shifts, the lazy road becomes the working road, but the pull on the road is straight (i.e. it does not deflect around the turn of the block) so there is no load on the snatch block.
I think the problem is going to come when you are pulling 90 degrees to both anchors. Then the snatch block is going to have a significant load on it. I have no idea either how to calculate this, but I bet SD does.

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