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For the engineers. Working load SK cable

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albert
Captain

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USA
262 Posts

Initially Posted - 03/31/2004 : 00:38:28

1/4" 7x19 304 stainless has a working load of 1300lbs. 316 stainless has a working load of 1160lbs.

The keel is 1500lbs.

What is the working load on the cable?

Albert Iturrey
al@comhertz.com
Abacus 1984 C25 #4679

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Leon Sisson
Master Marine Consultant

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1893 Posts

Response Posted - 03/31/2004 : 01:15:50

Albert,<blockquote id="quote">quote:<hr height="1" noshade id="quote">The keel is 1500lbs. What is the working load on the cable?<hr height="1" noshade id="quote"></blockquote id="quote">I'm no mechanical engineer, but I'd guess the working load on the C-25 swing keel cable when just below fully retracted would be around 750lbs to just over 1,000lbs. That guess includes "normal" rough wave action, but does not include shock loads from bumping hard bottom or riding up over submerged obstructions.

In the ICW at Boca Raton, I ran up over some sort of submerged rubble which resulted in the swing keel free falling probably over a foot before the cable caught it again. The resulting "WHAM!" made the whole boat and rigging ring like a bell, but nothing broke.

-- Leon Sisson

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CGood
Deckhand

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Response Posted - 03/31/2004 : 14:33:32

Let 'd' equal the horizontal distance between where the cable exits the hull and where it connects to the fin.
Let 'f' equal the angle between the fin and horizontal.

If d=0 when f=0, then Fcab=Fpin= 1500#/2 = 750#
If f=90, then Fcab= 0# and Fpin= 1500#

These are the max/min forces the cable should see in a static sense. Any angle in between will have forces that fit within this range.

If the assumption that d=0 when f=0 is not accurate, significantly higher cable forces are possible: Fcab = 750#/SIN(q).

Let 'q' equal the angle the cable would form with horizontal with the fin was fully horizontal.

Since SIN(q) approaches 0 as q approaches 0, Fcab approaches infinity (because the denomiator of the above equation approaches 0) - meaning this is an angle you’d never be able to reach in real life.
For example, if q=60: Fcab = 866#; q=30: Fcab = 1500#

As mentioned in the article above, this is all statics (not to be confused with statistics). When dynamics are taken into account, these forces can generally be multiplied by the g-force to which the fin is subjected. If you feel twice as heavy in your seat when you crash over a wave, the fin also feels twice as heavy. Slack occuring in the cable during such an event can cause an even greater multiplier.

IMHO, you would want a safety factor of at least 4 (2x diameter based on the effect of area) to 10 depending on your comfort level. However, this would be based on the failure point. I would expect that the working load would already have some significant safety factor built in for workmanship, wear and situations such as these. Before going to a 1/2" cable which sounds rediculous, I would learn more about what factors go into rating the cable's 'working load' in the first place.

CG

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Champipple
Master Marine Consultant

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6855 Posts

Response Posted - 03/31/2004 : 15:46:26

Does the buoancy (sp?) of the water have anything to do with this? Afterall it is submerged.

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CGood
Deckhand

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Response Posted - 03/31/2004 : 22:44:17

I don't know the dimensions of the fin, but let's say it's about 4' by 1' by 4". That's about 1 1/3 cubic feet. To my best recolection, water is about 7#/cu.ft. This would mean a displacement of less than 10 pounds. In other words, 1500# becomes 1490# effective. A nominal impact and can therefore be assumed out of the analysis.

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ClamBeach
Master Marine Consultant

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3072 Posts

Response Posted - 03/31/2004 : 23:32:59

Since the cable is perpendicular to the keel at full retraction, I think about 900-1000 lbs is all you will see in static load. (by my eye,the assumption d=0 when f=0 is pretty close)

However, note that the distribution of mass in the keel is not uniform, it is considerably heavier on the aft or bottom end. This is countered somewhat by the fact that some of the keel mass extends forward beyond the pivot pin when in the retracted mode.

"I would expect that the working load would already have some significant safety factor built in"
From the charts I looked at there appears to be about a 5:1 yield/swl ratio.

There's about 7-1/2 gallons of water in a cubic foot.. about 65 lbs. Still not enough displacement to make a huge difference.

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