What is your Bigness Factor?

[Andrew Troup]

To those who say displacement must be important because of inertia, I think you are looking only at the demand side: the demands inertia places on the anchor.

And certainly, displacement increases that demand.

However displacement (in most cases) reduces the supply, by the same factor.

An external force, from wind or water, which can get a 20 tonne boat up to 1/2 knot in ten seconds, will (in the same time) speed a 10 tonne boat up to double that , ie one knot. (from v=u+at)

Then when the boat comes up against the resisting force of the anchor, the same situation applies in reverse: the light boat moving faster will require the same decelerating force acting for the same time as the heavy slow boat.

It's not that inertia is not important, but, at least in theory, it cancels out.

It's a bit like a skinny person being able to do chinups despite having puny arms: their strength disadvantage is cancelled out by their weight advantage.

- - - - - -

In the anchoring cases under discussion:

The 'cancellation' proposition fails when the boat is lifted by a swell.
Gravity then becomes a factor in supplying inertia, because gravity is caused by mass: the more mass, the more force. The other external forces we've considered are not affected by mass.

So, as I mentioned a while back (possibly in another thread): if the waves are big enough (and of long enough period) to impart gravitational potential energy, my feeling is that the heavy boat will make bigger demands on the anchor than a light boat of the same dimensions.

[Delfin]

Originally Posted by conachair

Err, displacement is in there. A very brief look at the dyn_mix sheet came nowhere close to revealing how he calcs it but throwing a few figures in suggested that it is non linear.
Quite a complex sheet, did you get anywhere figuring out his equations?

In very simple terms dismissing displacement seems flawed. F=ma, linear, twice the force = twice the acceleration, always. The effects of underwater profile through the water on acceleration is not linear, twice the force will not = twice the acceleration over time, depends on the velocity.
Adding linear and non linear together doesn't result in a linear function.
Anyone care to comment? Always lots to learn but dismissing displacement seems rash even in simplistic models.

I'm ready to be wrong

Incidentally, Conachair, if you look at the dynmix calculations, you'll see that increasing the displacement DECREASES what he refers to as "Dynamic Overtension". I presume this is because the heavier boat will be less lively in the waves and as a result put less strain on the rode in a swell, but since F really does equal ma, it makes sense.

Unless it all just looks like "blah-blah" to you, then it doesn't.

[Delfin]

Originally Posted by Andrew Troup

In the anchoring cases under discussion:

The 'cancellation' proposition fails when the boat is lifted by a swell.

For that to be true, the more massive vessel would need to be lifted the same as the less massive vessel of the same dimensions, which is not the case. The lifting force remains equal to the mass times the acceleration. If the force is the same but the masses differ, the more massive object will accelerate less. The net effect on the anchor would the same however because the force will have to be countered by the rode length, its composition, and ultimately the anchor itself.

Originally Posted by Andrew Troup

Gravity then becomes a factor in supplying inertia, because gravity is caused by mass: the more mass, the more force.

Yes, gravitational attraction is affected by mass. But it is not gravity that is moving the vessel, it is a wave. Gravity really doesn't have anything to do with it.

Originally Posted by Andrew Troup

So, as I mentioned a while back (possibly in another thread): if the waves are big enough (and of long enough period) to impart gravitational potential energy, my feeling is that the heavy boat will make bigger demands on the anchor than a light boat of the same dimensions.

Seems that way, but it isn't true because F really is equal to ma.

[Snowpetrel]

Originally Posted by Andrew Troup

Hey, awesome, SnowP. I beat Jedi ! (again....)

But not Hawk (21ish) - so nicely in the middle of the Big Bruce Boys...

Is K temporarily off your case BTW, or just asleep? (Just kiddin')

On a slightly more serious note: I guess you're expecting to include a few more gradations for anchor factor, eg say 1.0 for Bruce (and CQR, maybe?)

I have a feeling fishermen should be a bit more than 1.2, and dreadnought a fair bit more (which might leave QM2 looking a bit under-endowed, but I guess the massive chain is partial compensation...), but this is just a crude proportioning hunch.

Do you think there should be any factor for hull and especially rig windage, especially given that there's quite a lot more breeze at the top of a tall rig?
In comparison with a motor sailer with a short stick - although arguably that would tend to have more hull and cabin windage....
( I guess a gaff rigged ketch, say, has a lot more windage than mast height alone would suggest, too)

Hmmmm ....

When Mirabella V dragged in France (246', lying to a 600kg x 2.2= 1320 lbs shiny Bruce knockoff:

246^2/1320= 46 !!!) WHY DIDN'T THEY CONSULT SNOWPETREL???

Afterwards, the classification society rules were criticised for taking heed only of her hull and superstructure windage, taking no account of her rig.

Reworking the calcs to account for the frontal area of the 90m mast and the three furling headsails, (but without factoring in the fact that in the F6-7 blowing when she dragged, it would have been 45 knots at the top of the rig)

it was reckoned her main anchor should have been 1080kg (over one TONNE! for a sloop...)

1080kg x 2.2= 2376 lbs:
246^2/2376= 25.5 .... Bingo!

I think you're onto something here, mate -
"very conservative" takes some heed of the altitude/windspeed problem, and what's more, it seems appropriate for a boat which cannot row out a kedge at the drop of a hat ....)


It's hard to know where to stop, isn't it, once you start...

Heres another formula to try out. This one is for wave loads on a boat and includes displacement/ length ratio as a proxy for lighter boats

(Disp (t)/LWL(m))x100
Anchor wt (kg)

The result should probably be under about 5? with lower being better. the 100 factor is just to bring the result up to a normal range without a load of decimal points ahead of it, mixing units does the same and is more convenient.

I have a speadsheet of results and some thoughts to follow, but K (bless her) has suggested I do other things today...

I have played with the exponent in my other formula, reducing it to ^1.66 seems to skew the results badly in favour of smaller boats, Prehaps the real value should be 1.9? maybe the slightly higher factor by using 2 accounts for some of the effects of increased displacement? As Alain makes mention that in some wind loading scenarios (trapezoidal gust) displacement can increase loads by up to 20%.

Anyway thanks for all the inputs, and debate, I have certainly learn't alot and been forced to think hard about what really goes on at anchor.

Cheers

Ben

Now.. to those dishes, then the shed

[Delfin]

Originally Posted by Snowpetrel

Heres another formula to try out. This one is for wave loads on a boat and includes displacement/ length ratio as a proxy for lighter boats

(Disp (t)/LWL(m))x100
Anchor wt (kg)

The result should probably be under about 5? with lower being better. the 100 factor is just to bring the result up to a normal range without a load of decimal points ahead of it, mixing units does the same and is more convenient.

I have a speadsheet of results and some thoughts to follow, but K (bless her) has suggested I do other things today...

I have played with the exponent in my other formula, reducing it to ^1.66 seems to skew the results badly in favour of smaller boats, Prehaps the real value should be 1.9? maybe the slightly higher factor by using 2 accounts for some of the effects of increased displacement? As Alain makes mention that in some wind loading scenarios (trapezoidal gust) displacement can increase loads by up to 20%.

Anyway thanks for all the inputs, and debate, I have certainly learn't alot and been forced to think hard about what really goes on at anchor.

Cheers

Ben

Now.. to those dishes, then the shed

Ben, I'm not so sure about the new formula. Unless I have made a mistake, Delfin, who most think has a more than adequate anchor comes in at 4.78 which means that if I carried the size recommended for her, I would be over 5.

Regarding "overtension", Fraysee indicates that these loads have a maximum value of 20% on loading only with all chain at full extension and then only with trapezoidal gusts...if you read his comments on nylon rodes or play with his dyn_mix spread sheet you can see these overtensions exist, but would have no material effect on anchor choice. In fact, calculated overtensions go down with a mixed rode when displacement goes up, so I'm not sure what is going on there.

[Andrew Troup]

Originally Posted by Delfin

For that to be true, the more massive vessel would need to be lifted the same as the less massive vessel of the same dimensions, which is not the case. The lifting force remains equal to the mass times the acceleration.

They are lifted the same distance, I think, whether it's a long swell, or a V.E.R.Y long 'swelling', like a tidal wave, or for that matter the tide coming in: this is because the lifting force is not constant, as you imply. It, too, changes with mass.

When a swell lifts the centre of mass of a 1 tonne boat to a maximum of say 1 foot above mean sea level, the upwards force is provided by the buoyancy of the boat.#

I contend that the same swell would lift a 2 tonne boat the same 1 foot. The force available has doubled.

# Buoyancy is equal to displacement, provided the boat has had time to reach equilibrium.
- - - - -

Your other point about the resulting 'demand' on the anchor:

I mention gravity because, having been lifted a certain distance by a swell, the vessel is then free to travel down the sloping face, under the influence of gravity. Because the acceleration due to gravity is constant, a heavy boat will accelerate at the same rate as a light boat*

So the inertial 'demand' placed on the anchor by the heavy boat (in this unusual situation) will be double, as the speed is the same and the mass is double

I don't think this is very applicable to short period waves, the sort you would find in any other than an open roadstead anchorage

On short waves, even if they're high, the entire vessel is not on the wave all at once: the bow may be lifting but the stern is dropping. Furthermore, the response is not instant, so it would surprise me if centre of mass of the vessel would have the time or the inclination to vary much in altitude.


*(provided they have the same underwater resistance - but at low speeds this is a small fraction of the available force, probably not significant)

[Snowpetrel]

You both have good points. I am afraid I dont have the time to explain properly. But this is only to judge how the anchor and rode might be loaded in large waves. Equal wavelength to the lwl or so. Causing the vessel to be caught in rhe orbital currents. In this case a lighter longer boat woild have a clear advantage over a heavy short boat, but becasue the load is cyclic it can be taken care of (in most cases) by dynamic rode strech...damn phone keypad...

[Delfin]

Originally Posted by Andrew Troup

They are lifted the same distance, I think, whether it's a long swell, or a V.E.R.Y long 'swelling', like a tidal wave, or for that matter the tide coming in: this is because the lifting force is not constant, as you imply. It, too, changes with mass.

When a swell lifts the centre of mass of a 1 tonne boat to a maximum of say 1 foot above mean sea level, the upwards force is provided by the buoyancy of the boat.#

I contend that the same swell would lift a 2 tonne boat the same 1 foot. The force available has doubled.

Andrew, the force of the wave is the force of the wave. The fact that the wave hits a 60 ton boat doesn't change the force relative to a 30 ton boat. And from practical experience, I can tell you that Delfin presents more boat to the wake of some lunatic running the through the anchorage in a Skidoo than the Olson 30 next to me. But it is the Olson 30 that bucks over the wave straining at the rode, while we barely move. The energy transferred to two vessels of the same dimensions is equivalent regardless of their displacement, and no, the same wave would not lift a 2 tonne boat the same as a 1 tonne, as you can observe yourself by looking at different boats at anchor. If this were not the case, then the heavy boats presenting more boat to the wave would be bucking around equivalent to the anchored canoe, but that clearly isn't the case.

As Snow Petrel notes, cyclic loads can be slightly higher on a heavy boat, (I think that is what he is saying, but since he is typing with his toes on his cell phone its hard to say) and that is no doubt true. However, it has such a minor effect relative to other factors that it doesn't affect the anchor choice enough to worry about. That is why no one pays much attention to displacement when they are making recommendations on ground tackle.

[Cotemar]

I am thinking you guys are thinking too much.
If the zinkoyd cravistat gets any more torque than you will have too much cravistat in your zinkoyds thereby reducing you’re anchoring to theory instead of the real world we live in.

Maybe just go out for a nice enjoyable sail and grab a mooring and relax.

[Bash]

Originally Posted by Cotemar

If the zinkoyd cravistat gets any more torque than you will have too much cravistat in your zinkoyds thereby reducing you’re anchoring to theory instead of the real world we live in.

That's only true until the cravistat frangulates.

[Delfin]

Originally Posted by Cotemar

I am thinking you guys are thinking too much.
If the zinkoyd cravistat gets any more torque than you will have too much cravistat in your zinkoyds thereby reducing you’re anchoring to theory instead of the real world we live in.

Maybe just go out for a nice enjoyable sail and grab a mooring and relax.

Amen, brother.

p.s. I had a zinkoyd once, and camomile lotion seemed to help.

[Cotemar]

Originally Posted by Bash

That's only true until the cravistat frangulates.

You do have a point their

[Cotemar]

Originally Posted by Delfin

Amen, brother.

p.s. I had a zinkoyd once, and camomile lotion seemed to help.

I am with you. Camomile lotion is the best cure for Zinkoyd I have ever seen.

[Andrew Troup]

Delfin

It's evident you didn't read or reflect carefully on what I wrote in my last post, but I guess it would be churlish not to try one last time.

eg your "wave hitting...". I quite agree, but this has nothing to do with lifting on a long swell, which is the ONLY thing my post considered

and "looking at different boats at anchor..."

If you were to do this in an open roadstead exposed to a long swell train, boats of different mass would go up and down harmoniously, all floating on their marks. So would a ping-pong ball, for that matter.

Where would you draw a demarcation between rising to a tide and not rising to a chop?
Because you're talking about situations within that demarcation, and I'm talking outside it.

[Andrew Troup]

Snowpetrel:
I'm not considering orbital currents. That's trickier, and in any case I think it falls into the same category as the majority of forces, which are largely independent of displacement (because of the "supply matching demand" balance referred to earlier)

I'm considering the boat travelling, under the influence of gravity, down the flank of a long swell.

It would do this regardless of the orbital motion of the water particles: they will add to the initial acceleration, through underwater drag (rather than the gravitational force I'm talking about), but in inverse proportion to mass (which I think is why a surboard initially accelerates quicker without anyone on it).

The situation I'm singling out is more like the situation of a skier. Heavy skiers do not travel faster than light skiers on a given slope, but they punch a bigger hole in the wall when they 'hit the daylodge'.

I think 'travelling downhill due to gravity' is the probably the one mechanism where a 2 tonne boat applies roughly twice as much force as a 1 tonne boat when it gets pulled up short by the chain. And I think it applies only in roadsteads, ie virtually unlimited fetch.