Calculating the boat speed thought the water

[Sandy]

A long time ago I saw a formulae for calculating the speed through the water by timing an object passing from the bow to the stern something like:

The time in seconds divided by the square root of the length
​

Does anybody know the actual formula?

(I don't use a plotter so must be mad)

 

[tillergirl]

One foot per second = 0.592483801295896 knots. So its simples init?

 

[LadyInBed]

I reckon
Bow to Stern length ÷ time from B to S = ft per sec
ft per sec × 3600 = ft per hr
ft per hr ÷ 6080 = knots
I will leave the rest to you

 

[FWB]

You double metres/second to get knots.
So if your boat is 8m and it takes 3 secs for the thing you chucked in the water at the bow to pass the stern, you're doing about 5kts.
9m boat then 6kts etc.

 

[AliM]

The square root bit that you half remembered is to calculate the hull speed of your boat (in knots), which is something like 1.3 Times Square root of waterline length in feet. It is only a rule of thumb and the factor of 1.3 varies a bit when you ask anyone else.

All that is completely irrelevant to your original question - the posts above answer that!

 

[mjcoon]

A long time ago I saw a formulae for calculating the speed through the water by timing an object passing from the bow to the stern something like:

The time in seconds divided by the square root of the length
​

Does anybody know the actual formula?

Are you taking the mick? The square root comes in when calculating maximum hull speed for a displacement (non-planing) hull. (Something like 1.6 times square root of the wetted length expressed in feet gives result in knots.)

Measuring actual speed will be linear with length...

Mike.

 

[LadyInBed]

From my notebook (copied from several sources)

Maximum Speed

1.4*sq root of waterline length (ft)

Ave speed / day

20* sq root of waterline length (ft)

Wave Length

5.12*(time in secs between crests)2

Wave Speed

3.03*time in secs between crests

Volume below waterline (in cu ft)

Wt in Tons * 35

[35 (34.973) cu ft of Seawater weighs 1 Ton]

[1 cu ft of Seawater weighs 64 lbs]

USYRC's 'Capsize Risk Factor'

Beam / ( Disp / (.9*64)) ^.333

(Values less than 2 are good).

DISPLACEMENT/LENGTH RATIO:

Displacement in Tons / (0.01*Waterline Length)3

The D/L ratio is a non-dimensional figure derived from the displacement in tons (of 2240 lbs)

divided by 0.01 LWL cubed, or Dt/(.01 LWL)3.

It allows us to compare the displacement of boats of widely different LWLs.

Some examples of various D/L ratios follow, but are generalities only,

as there is often a wide range within each type.

BOAT TYPE D/L RATIO:

Light racing multihull 40-50

Ultra light ocean racer 60-100

Very light ocean racer 100-150

Light cruiser/racer 150-200

Light cruising auxiliary 200-250

Average cruising auxiliary 250-300

Heavy cruising auxiliary 300-350

Very heavy cruising auxiliary 350-400

>300 best for GRP >400 best for Steel

POUNDS PER INCH IMMERSION (PPI):

The weight required to sink the yacht one inch.

It is calculated by multiplying the LWL area by 5.333 for sea water or 5.2 for fresh.

The PPI usually increases as the hull sinks into the water as the LWL area is also increasing

due to the shape of the hull above water.

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, or:

For example; a boat has a LWL Area of 165 sq ft and a Beam WL of 8 feet.

The MTI is 0.35 x 165 x 165 /8 = 1191 ft-lbs., say 1200 for rough figuring.

Now you hang a 100 pound dink 18 feet abaft the CB. You've added 1800 ft-lbs of aft moment

so the boat will trim 1800/1200 = 1.5 inches down by the stern.

However, the boat does trim about its CF and, as that is usually abaft amidships, the stern will move less than the bow. You might find that she trims 5/8" down by the stern, and 7/8" up by the bow, making a total trim change of 1.5 inches.

Obviously on a cat things are slightly different.

once the new waterline area is calculated you can apply the figures to SA/D

(taking into account the new displacement with the gear onboard)..

this gives a reasonable view of the potential change.. mainly a monohull formula though


SAIL AREA/DISPLACEMENT RATIO:

The SA/D ratio is the sail area in sq ft. divided by the displacement in cubic ft to the 2/3 power,

or SA/D.667 .

Ratios below 14 are suited for motor sailers,

from 14-17 for ocean cruisers and

from 16-18 for typical coastal cruisers.

Ratios over 18-20 are seen on racing dinghies, inshore racers and ocean racing yachts.

The more extreme screamers can have very high SA/D ratios indeed; A 60 foot design boat, had a SA/D ratio, based on 100% fore triangle, of well over 30, depending on her displacement at the moment.

Her displacement varied widely as she could carry 8,000 pounds of water ballast in tanks on the windward side

 

[Sandy]

Are you taking the mick?

No... It was a half remembered formulae that I saw a long time ago hence the request for clarification.

 

[Sandy]

Thanks all

All sorts of useful information there to help this non plotter, compass wielding, paper chart user old salt to consider over a glass of Highland Park.

 

[onesea]

Does anybody know the actual formula?

(I don't use a plotter so must be mad)

Sailing without plotter you must resort to old fashioned techniques..

Drink beer throw bottle overside, when out of sight drink another beer, throw bottle over side.

In this manner when the wind is light and your going slow you enjoy a slow lesirely beer. When the wind is up and your going quick your drink faster and do not notice how much the wind has picked up...

This also has the advantage of leaving a trail so as you can calculate your leeway and as on the plotter you have a track you can follow back to your departure point.

It is important that you carry sufficient beer to complete the passage, if not you have to resort to whisky gin and the whisky. This is not advisable as arrivals can become a thing of the past, waking up and thinking where am I and how did I get here is not something to be done regularly.

The moral of the tail is don't run out of beer!

 

[LadyInBed]

Slight drift from speed, but in 't old day I used to deploy a blob buoy on a 20m line to get a wake course.
Ah! They were the days

 

[DJE]

You double metres/second to get knots.
So if your boat is 8m and it takes 3 secs for the thing you chucked in the water at the bow to pass the stern, you're doing about 5kts.
9m boat then 6kts etc.

Keep chucking stuff overboard and she will gradually get faster!

 

[stephenh]

I have my own book of acquired 'useful info.' and digging deeply into it I found :

Dutchmans Log -

Distance in feet x 0.6 divided by time in seconds = knots

Any help ?

 

[nigelmercier]

You double metres/second to get knots...

No, you are way off: 1 metre per second = 1.94384449 knots

 

[Giblets]

Perhaps something like the Knotstick would be handy.

http://www.knotstick.com/index.htm

(No connection - just thought a clever idea [although a tennis ball log would be cheaper])

 

[nigelmercier]

Just had a thought. If one had an old mechanical stopwatch, then the dial could be re-calibrated to a fixed distance on the boat.

 

[Robert Wilson]

Just had a thought. If one had an old mechanical stopwatch, then the dial could be re-calibrated to a fixed distance on the boat.

Clever! Might try that.:encouragement:

 

[DJE]

Just had a thought. If one had an old mechanical stopwatch, then the dial could be re-calibrated to a fixed distance on the boat.

You might as well go the whole hog:

 

[awol]

A long time ago I saw a formulae for calculating the speed through the water by timing an object passing from the bow to the stern something like:

The time in seconds divided by the square root of the length
​

Does anybody know the actual formula?

(I don't use a plotter so must be mad)

Exactly what kind of engineer are you?

 

[mawm]

1 Kn = 1 nautical mile/h
= 1852/3600 m/s (metres/sec)
= 0.5 (approx) m/s

Therefore if you travel 5m in 2 sec your speed is 5 Kn......I think.