Wind speed, sail size and tension

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yvanboniec
yvanboniec
16 posts
16 posts
8 Mar 2012 9:07pm
Good morning,

I would like to know wether a chart with the following axis exist

on one axis, there is the wind speed,

on another axis, there is the sail size/surface,

and, in between, there will be the corresponding weight or tension on the sail (in lbs or kg) due to the wind speed

Has anyone come accross that type of table before? If so, where can I find it?
jsnfok
jsnfok
WA
899 posts
WA, 899 posts
8 Mar 2012 9:42pm
really, like really, you need a chart to tell you what to rig????
d1
d1
WA
304 posts
d1 d1
WA, 304 posts
8 Mar 2012 9:58pm
Mu! There are a many more factors than just wind speed. You will need a multidimensional graph, and even then it will be impossible to correctly account for sail design variances and individual preference. Trial & Error is your friend.
decrepit
decrepit
WA
12872 posts
WA, 12872 posts
8 Mar 2012 9:58pm
I've seen a formula somewhere that has force on one side and wind speed on the other.
But a sail is more complicated than that, it's foil produces lift, that needs to be taken into account as well I think.

Just googled this, if it's any use?



The Generic Formula

For using the actual sustained wind speed expected (were we to actually determine it) :

Force, F = A x P x Cd

A = The projected area of the item

P , Wind pressure (Psf), = .00256 x V^2 (V= wind speed in Mph)

Cd , Drag coefficient, = 2.0 for flat plates. For a long cylinder (like most antenna tubes), Cd = 1.2.
Note the relationship between them is 1.2/2 = .6, not quite 2/3.


yvanboniec
yvanboniec
16 posts
16 posts
9 Mar 2012 1:53am
In the meantime, I found this

F = 1/2 * air density * (wind speed) ² * Aerodynamic coefficient * Surface.
If wind speed is 20 m/s and the sail surface is 6.4, then I get
= .5 x 1.293 x (20) ² x 1.1 x 6.4
= 0.5 x 1.293 x 400 x 1.1 x 6.4
= 1820 newton = 185 kg or 407 lbs


not sure about the drag coeficient, you chose 2, I chose 1.1. found conflicting information about what to select.
yvanboniec
yvanboniec
16 posts
16 posts
9 Mar 2012 3:33am


I am doing this to know what rope / cord strenght I should use (what tensile in lbs or kg) ? and if I need to do several loops/rounds when I attach the sail to the clew with the rope. I am trying to shorten the time I spend riging my sail.
KEVKEV
KEVKEV
74 posts
74 posts
9 Mar 2012 4:08am
maaaaaate...just loop it round once and yank it out...any length of outhaul rope will be fine..go to your local shop and get a length....spend more time sailing than worrying about formulas and stuff.....
redsurfbus
redsurfbus
304 posts
304 posts
9 Mar 2012 6:22am
You are not going to break any decent outhaul/downhaul rope at the clew (unless its old and worn.) If you are so worried then get some marlow formuline which lasts 10x that of normal braided dynema in my experience. I used to replace my downhaul rope about 5x per year until I got some formuline, never had it snap and rig big race sails for 100+ sessions per year and it has lasted 2 years (I should really replace it but it shows no sign of wear)
firiebob
firiebob
WA
3182 posts
WA, 3182 posts
9 Mar 2012 9:43am
Yvan, talk to the other guys where you're learning, listen, watch and learn. Also Redsurfbus is on the money re line

d1
d1
WA
304 posts
d1 d1
WA, 304 posts
9 Mar 2012 11:02am
yvanboniec said...

In the meantime, I found this

F = 1/2 * air density * (wind speed) ² * Aerodynamic coefficient * Surface.
If wind speed is 20 m/s and the sail surface is 6.4, then I get
= .5 x 1.293 x (20) ² x 1.1 x 6.4
= 0.5 x 1.293 x 400 x 1.1 x 6.4
= 1820 newton = 185 kg or 407 lbs


not sure about the drag coeficient, you chose 2, I chose 1.1. found conflicting information about what to select.


I like the scientific approach, but it is wrong to assume that this force will be applied at the clew end. The clew of a properly down-hauled sail shouldn't take much stress. Rope strength shouldn't be an issue - the issues I've seen are usually due to abrasion over many years.

As a rule of thumb, and this is related to your other related posting, it is typical to specify the amount of outhaul in units of length (typically centimetres), rather than force. Some sails specify absolute amounts, and sometimes a deviation from neutral. In the latter case, pull clew with two fingers until you can feel significant increase in tension. This is neutral. Any more, and you have positive outhaul. Any less, and you have negative outhaul. There may be variations on this theme.

redsurfbus
redsurfbus
304 posts
304 posts
9 Mar 2012 6:33pm
cheers firiebob!

Another thing I thought of was dont focus on these forces. The outhaul and tension of a sail is a 'feel' gained from experience.

As a quick guide - if you are placing your boom clamp mid cut out then the sail printed stting will often be correct (not always), placed higher to the top of the cut out, add a couple of cm to the extension on the boom.

If the sail feels twitchy then you have too much outhaul on, but will go upwind very well. If it feels slow to plane and heavy you may have too little outhaul on, this can be shown as the sail will touch the other side of the boom when powered up (this is called negative outhaul and used by speed sailors when bearing off the wind).

The best thing for you to do is set your sail and spend a session adjusting the outhaul from max to mid to min in a constant wind, this will give you an idea of the different feel that the sail gives at different settings.

As for rigging, I race with people who use only 1 line of rope through the clew eyelet and nothing more, as part of adjustable outhaul setup. I prefer a double line through the eyelet. Some booms have a loop and go set up (http://www.k-bay.co.uk/_product_43824/Alloy_Tail_Piece_62cm_-_Standard), rather than pulleys there is a hole that the line comes from, a cleat and a 'rimmed catch' (cant think of what to call it) on the other side of the mid point. All you have to do is put the loop through the clew and over the catch, then tighten by pulling more line through the cleat........takes seconds to rig the outhaul this way.
decrepit
decrepit
WA
12872 posts
WA, 12872 posts
9 Mar 2012 8:41pm
yvanboniec said...



I am doing this to know what rope / cord strenght I should use (what tensile in lbs or kg) ? and if I need to do several loops/rounds when I attach the sail to the clew with the rope. I am trying to shorten the time I spend riging my sail.


funnily enough the maximum force on the outhaul rope has very little to do with wind strength!
It's more about your physiology, height, weight, and arm length mainly.
Think about it, that's what's hanging on the boom, so controls the force on the rig.

Attach a spring balance to your harness and swing from it at the same angle you would normally sail at, (mast upright, with arms extended from the boom.
Bounce around a bit to mimic going over chop. That will give you force on the boom at harness lines, normally 1/3 along the boom arm. I guess that means that 1/3 of the load is on the clew rope?
This load could be exceeded in catapult, sudden gust situations I guess.

There's more force on the clew than you'd think, as a couple of my mates have found out by modifying the clew end of the boom. The breaking strain of 3-4mm rope is quite high.
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