Is there a mathematical formula to determine how much weight your bow can handle?

[Sam Woodbridge]

Is there a mathematical formula to determine how much weight you can hold in the bow before it’s too heavy? Is there something based on the length and weight of sailboat to determine it?

This question arises after much thought and input from other people concerning weight of anchor and carrying as much chain as possible in the bow of a sailboat.

The sailboat in particular is a 2018 Beneteau 38.1.

Thank you for any and all responses. I’m also at work so my returning replies might be slow.

Daily reader Sam.

[Alan Mighty]

how about:

Pitching moment of inertia = mass x (distance from Longitudinal Centre of Gravity)^2

[chrisr]

define 'too heavy' ?

if you thinking when the boat starts to trim by the head, then you never allow this (except maybe when you are doing a rating measure but that's another story)

however short of this imho it's a matter of experience. large vessels have calculations for stress and trim, but with small craft it's not necessary

as a rule you don't want weight in the ends...as close to central as possible. with stuff like anchors & chain it may not be possible to keep it out of the bow, so the next step is to keep it to a minimum

no need to carry 200m of chain when 100m will do
use 8mm high tensile chain instead of 10mm ordinary
etc

no need for any calculation...just use your experience and common sense

cheers,

[Sam Woodbridge]

Thank you both for the replies. (Mighty Alan - Chrisr)

I feel I understand you both. And maybe what I’m asking is begging for it. But I’m not one to learn the hard way (when I can help it) I just want the right amount of weight on the bow of the boat without having to experiment to much. In my mind and from others I want a lot of chain, but not always in life is more better.

[GordMay]

Pounds per inch immersion = (WPA x 64) ÷ 12

MOMENT TO TRIM 1 INCH (MTI):
The MTI is the moment, expressed in foot-pounds, that will change the fore and aft trim of the yacht one inch.
For a displacement hull, the MTI is, roughly (but close enough for all practical purposes), 0.35 times the square of the waterline area divided by the WL beam.

➥ Sail Calculator Pro v3.54 - 3200+ boats

➥ https://goodoldboat.com/brewer-by-the-numbers/

[Sam Woodbridge]

Originally Posted by GordMay

Pounds per inch immersion = (WPA x 64) ÷ 12

MOMENT TO TRIM 1 INCH (MTI):
The MTI is the moment, expressed in foot-pounds, that will change the fore and aft trim of the yacht one inch.
For a displacement hull, the MTI is, roughly (but close enough for all practical purposes), 0.35 times the square of the waterline area divided by the WL beam.

➥ Sail Calculator Pro v3.54 - 3200+ boats

➥ https://goodoldboat.com/brewer-by-the-numbers/

Whoa! GordMay thank you. When I get time I’ll check out the links and see if I can get this figured out.

[Cheechako]

Really gut feel is enough. And how much does your waterline change with a sample weight up there?
Whenever I had a worry about too much chain and weight in the bow of a boat, I went with 150 ft of chain with a 5/8 nylon rod attached instead of 200 ft of chain. The rope rode very seldom saw the light of day. But it does depend on your typical cruising grounds.
Maybe weight in the bow will give you a longer waterline length and you'll be faster! :>)

[chrisr]

Originally Posted by Sam Woodbridge

Thank you both for the replies. (Mighty Alan - Chrisr)

I feel I understand you both. And maybe what I’m asking is begging for it. But I’m not one to learn the hard way (when I can help it) I just want the right amount of weight on the bow of the boat without having to experiment to much. In my mind and from others I want a lot of chain, but not always in life is more better.

an empirical way would be to mark your waterline and then place say 3 x 20L jerry cans of water on the bow (60L = 60kg) and then see by how much the bow has sunk.

you know how much the chain weighs, so you can calculate how much the bow will sink for a given length of chain

nb : this method is not perfect owing to the change in shape of the bow as it sinks, but it'll be close enough

cheers,

[Sam Woodbridge]

Thank you Chrisr and Cheechako for your replies.

[psk125]

Gord's formula is a good tool, but the actual amount X pounds of chain will make a boat's bow dip will vary according to how thin the boat is in the bow sections and how much flare there is forward. As ChrisR suggests, a long, narrow bow with no flare will tend to dip more than a short, wide bow that has lots of flare.

[zstine]

the above equations will tell you how much the bow will sink / pitch forward on her lines. if your boat squats a little, then perhaps you can get away with a bit more weight. my boat was perfectly even on her lines when I added a windlass, battery, anchor and 150ft of chain... probably 250-300lbs. she sank 2.5 to 3 inches in the bow and it was noticeable. the change in sailing was noticeable. the rig also pitches forward effecting ballance... however, I added solar on an arch and a 25 gallon aux fuel tank aft. with that weight, I was pretty near even again, just a bit lower.

[TrentePieds]

Sam:

You say: "This question arises after much thought and input from other people concerning weight of anchor and carrying as much chain as possible in the bow of a sailboat."

Given that you concern is whether the weight of your ground tackle would be "too much", I think you are coming at your "problem" from the wrong angle. Your HAVE to have adequate ground tackle, of course, whatever the weight of ADEQUATE ground tackle may turn out to be.

A modern 40 footer like yours, sailed on the coast of Alaska or BC, might well carry the following for a bower anchor: 1) A "new style" anchor like a Rocna weighing 44lbs. 2) About 50 feet of 3/8" chain having a breaking strength of about 7K lbs. for a weight of, say, 75 lbs and 3) 300 feet of 9/16" 3-strand nylon rope with a breaking strength (when new) of about 10K lbs for a weight of, say, 22 lbs. So the total weight of ADEQUATE principal ground tackle for your boat, which is what you carry "up forward", will be about 140 lbs. That's the weight of a woman. So why are you worried?

We can argue about the precise numbers. What's important here is just that you see the logic.

So look into the question of whether the gear I've proposed really IS adequate. Getting a pat answer from anyone of us, will not serve you adequately in the long run. As skipper you have to understand the reasoning behind the decisions.

A boat like yours, lying head to wind, in 40 knots of wind would exert a pull on the ground tackle of something like 1,200lbs. But a boat like yours is not likely to lie still. She will yaw about. As she ranges out to the extremity of the yaw and begins to come back again, the pull might rise to, say, 1,600 lbs.
So you need to dimension your gear to be able to tolerate that PLUS a safety factor. The displacement ("size") of the boat has little to do with it. "Windage" (wind resistance) has a LOT to do with it.

So go to the literature and figure out what YOU need in the water and weather in which you propose to sail.

What about the length of the rode then? Obviously, that will depend on the depth of water in which you intend to anchor. Your draft determines the minimum depth of water in which you can anchor. The particular waters in which you will need to anchor will have only certain places where it is at all feasible to anchor. In Alaska and BC that is certainly true! I would suggest that you consult the charts and pick, and remember, places that at LOW tide have a depth of between 10 and 30 feet. Then consult the tide tables for those places so you will be in a position to determine, taking the rise and fall of the tide into account, whether 350 feet of rode will be ADEQUATE for those water depths.

So don't worry about the weight of the tackle. In terms of operations, as others have pointed out in different words, that parameter will turn out to be trivial!

TrentePieds

[Sam Woodbridge]

Thank you for your time and replies TrentePieds, Zstine, and Psk125.

All your input does help and I sincerely appreciate the time put into your replies.

Thanks again - Sam

[Cadence]

There is no simple formula. To many variables for it to be simple.

[Bargain Bill]

As a nuclear physicist, I'd have to get approval from NASA to release the formula.

In lieu of that i'd follow chrisr's approach.
1. Determine (and mark) the current water level at the bow.
2. Figure out how much chain you want and what it weighs.
3. Put that much water in jugs over the chain location.
4. note the change in water level and adjust chain estimate for your needs.