J. Dillon

08-23-2007, 02:24 PM

Does any one know the lbs pressure on sail per Sq ft, say in a 5,10, 15, 20 25kt etc. wind ?

JD

JD

View Full Version : Wind pressure on sails ?

J. Dillon

08-23-2007, 02:24 PM

Does any one know the lbs pressure on sail per Sq ft, say in a 5,10, 15, 20 25kt etc. wind ?

JD

JD

Thorne

08-23-2007, 03:51 PM

I'm no expert on this, but I *think* that you have to be a lot more specific to get any hard numbers. Wind pressure on sails is really complex, and involves the speed of the sail through the air, the size of the sail, direction of the wind in relation to the sail, etc.

And of course for small boats, unless they are on the hard / rocks, they will tilt and spill the wind in many conditions. Sail tension and twist also should factor in.

From http://www.otherpower.com/cgi-bin/webbbs/webbbs_config.pl?noframes;read=16416 -

"

Re: wind psi ?

Posted By: Brian

Date: Monday, 19 May 2003, at 5:25 p.m.

In Response To: wind psi ? (http://www.otherpower.com/cgi-bin/webbbs/webbbs_config.pl?noframes;read=16407) (Chuck)

I believe Ed used this equation to come up with his numbers, so I'll just give you the equation so you can figure it out at different elevations.

The equation to find the air pressure at a certain velocity is 1/2(rho)(v^2) where rho=the density of the air at your site and v=the velocity of the wind. To get an accurate density number, just hit Google or any other search engine to find the density of air at your elevation and that will get you pretty close to the actual pressure at some wind speed v.

Then like Ed said, just multiply the pressure by the area the wind is acting on to find the force acting on the object. Just be sure to watch your units and be sure they cancel out. The equation won't work if you mix US with SI units unless you do the appropriate conversions."

And even more fun stuff here - http://web.mit.edu/2.972/www/reports/sail_boat/sail_boat.html

http://web.mit.edu/2.972/www/reports/sail_boat/flow_over_airfoil.gif

And of course for small boats, unless they are on the hard / rocks, they will tilt and spill the wind in many conditions. Sail tension and twist also should factor in.

From http://www.otherpower.com/cgi-bin/webbbs/webbbs_config.pl?noframes;read=16416 -

"

Re: wind psi ?

Posted By: Brian

Date: Monday, 19 May 2003, at 5:25 p.m.

In Response To: wind psi ? (http://www.otherpower.com/cgi-bin/webbbs/webbbs_config.pl?noframes;read=16407) (Chuck)

I believe Ed used this equation to come up with his numbers, so I'll just give you the equation so you can figure it out at different elevations.

The equation to find the air pressure at a certain velocity is 1/2(rho)(v^2) where rho=the density of the air at your site and v=the velocity of the wind. To get an accurate density number, just hit Google or any other search engine to find the density of air at your elevation and that will get you pretty close to the actual pressure at some wind speed v.

Then like Ed said, just multiply the pressure by the area the wind is acting on to find the force acting on the object. Just be sure to watch your units and be sure they cancel out. The equation won't work if you mix US with SI units unless you do the appropriate conversions."

And even more fun stuff here - http://web.mit.edu/2.972/www/reports/sail_boat/sail_boat.html

http://web.mit.edu/2.972/www/reports/sail_boat/flow_over_airfoil.gif

Bruce Taylor

08-23-2007, 10:12 PM

http://i197.photobucket.com/albums/aa291/BDSTaylor/Windpressure.jpg

Figment

08-24-2007, 10:23 AM

That one's a keeper, Bruce.

Very nice.

Very nice.

Woxbox

08-24-2007, 05:32 PM

I used to believe that the force increased as a square of wind speed, but recently read that there's a cubed function at work. Which is it? that chart looks a lot more like it's squared.

donald branscom

08-24-2007, 05:49 PM

Wind pressure? I don't think it is like wind blowing on the side of a barn, It is more like an airplane wing that is vertical.

One side is creating lift and the other side is the low pressure side.

How much lift is it creating. Not like Bernoulli's effect. It is Hump effect.

From the net:

Upper surface flow is indeed faster than the lower, so much so that transit time at the upper surface in typical normal flight is always LESS than at the lower. Although Bernoulli's law is sound and well proven, the premise of equal transit time is invalid and without foundation in known physics. Thus the most popular explanation, world-wide, of wing operation is false, and easily shown to be so.

The falsehood is not due to Bernoulli's law, which is well proven, but rather due to falseness of the principle of equal transit times.

Hump theory does not allow for balsa toy gliders with flat wings, which have equal upper and lower surface path lengths. Lift is, without question, due to pressure being greater at the lower surface than at the upper surface. Thus according to Bernoulli's law, air transit time at the upper surface must be less than at the lower. This refutes the hump theory argument that upper and lower transit times must be equal. Hump theory also does not allow for inverted flight of aerobatic airplanes, which have equal upper and lower wing surface curvatures.

For mathematical test of hump theory, consider one the most popular trainers, the Cessna 150 or 152, which has wings of 160 square feet total area, upper surface path about 1.6 percent longer than the lower, and can fly, flaps-up, at 55 miles per hour, or 81 feet per second. We can calculate lift according to hump theory. Let's assume near sea level air density of about .0024 slug per cubic foot, and assume no change in below-wing pressure or flow velocity, as hump theory explanations usually do. If upper surface velocity is increased by 1.6 percent in order to attain equal transit time, the change is .016 times 81, or 1.296 feet per second, making upper surface velocity 82.296 feet per second.

Now let's plug that information into the Bernoulli expression which says pressure difference is equal to one half density, times difference between initial velocity (airplane airspeed) squared and upper surface velocity squared. Pressure difference between upper and lower surfaces then is (1/2)x.0024 x [(81 squared)-(82.296 squared)]. The upper surface pressure reduction then becomes .2544 pounds per square foot. Multiplying this by wing area of 160 square feet gives total lift of 40.7 pounds, a small fraction of the 1600 pounds rated gross weight of the airplane. (I will end it here to limit the discussion to your specific question.)

This should help awnswer your question and you could just extrapolate the rest.

And remember this was at 55mph.

One side is creating lift and the other side is the low pressure side.

How much lift is it creating. Not like Bernoulli's effect. It is Hump effect.

From the net:

Upper surface flow is indeed faster than the lower, so much so that transit time at the upper surface in typical normal flight is always LESS than at the lower. Although Bernoulli's law is sound and well proven, the premise of equal transit time is invalid and without foundation in known physics. Thus the most popular explanation, world-wide, of wing operation is false, and easily shown to be so.

The falsehood is not due to Bernoulli's law, which is well proven, but rather due to falseness of the principle of equal transit times.

Hump theory does not allow for balsa toy gliders with flat wings, which have equal upper and lower surface path lengths. Lift is, without question, due to pressure being greater at the lower surface than at the upper surface. Thus according to Bernoulli's law, air transit time at the upper surface must be less than at the lower. This refutes the hump theory argument that upper and lower transit times must be equal. Hump theory also does not allow for inverted flight of aerobatic airplanes, which have equal upper and lower wing surface curvatures.

For mathematical test of hump theory, consider one the most popular trainers, the Cessna 150 or 152, which has wings of 160 square feet total area, upper surface path about 1.6 percent longer than the lower, and can fly, flaps-up, at 55 miles per hour, or 81 feet per second. We can calculate lift according to hump theory. Let's assume near sea level air density of about .0024 slug per cubic foot, and assume no change in below-wing pressure or flow velocity, as hump theory explanations usually do. If upper surface velocity is increased by 1.6 percent in order to attain equal transit time, the change is .016 times 81, or 1.296 feet per second, making upper surface velocity 82.296 feet per second.

Now let's plug that information into the Bernoulli expression which says pressure difference is equal to one half density, times difference between initial velocity (airplane airspeed) squared and upper surface velocity squared. Pressure difference between upper and lower surfaces then is (1/2)x.0024 x [(81 squared)-(82.296 squared)]. The upper surface pressure reduction then becomes .2544 pounds per square foot. Multiplying this by wing area of 160 square feet gives total lift of 40.7 pounds, a small fraction of the 1600 pounds rated gross weight of the airplane. (I will end it here to limit the discussion to your specific question.)

This should help awnswer your question and you could just extrapolate the rest.

And remember this was at 55mph.

Woxbox

08-25-2007, 09:05 PM

Well, this thread could go on forever. But anyhow -- as far as what's going on to create lift in sails and wings, what the textbooks never seem to mention is that regardless of which theory you favor, the air/wind changes direction after being intercepted by the sail or wing.

This is very obvious with a helicopter, the wash being hard to ignore. But a plane in flight does the same thing -- throws the air downward. Once you consider that, its easy to see Newton's law coming into play -- a lifting force equal to the downward force of the displaced air. Thus the plane goes up and the boat moves forward.

I'm not trying to start a techhical argument here, just adding agreement to the fact that the textbooks do a lousy job explaining how wings work. Those flat-winged paper and balsa planes do prove that fancy curved sections are not needed to fly. The fancy curves just add a great deal of efficiency.

This is very obvious with a helicopter, the wash being hard to ignore. But a plane in flight does the same thing -- throws the air downward. Once you consider that, its easy to see Newton's law coming into play -- a lifting force equal to the downward force of the displaced air. Thus the plane goes up and the boat moves forward.

I'm not trying to start a techhical argument here, just adding agreement to the fact that the textbooks do a lousy job explaining how wings work. Those flat-winged paper and balsa planes do prove that fancy curved sections are not needed to fly. The fancy curves just add a great deal of efficiency.

MiddleAgesMan

08-26-2007, 01:35 AM

I pondered the same question when I was trying to decide how to rig my Gazelle. I wanted free-standing spars but Colvin's original design called for aluminum tubes supported by stays. I took a lesson from Hurricane Hugo and decided flagpoles would be fine without stays.

I drove through the Charleston area right after Hugo hit and saw trees down, power poles snapped, even a church steeple on the ground. I also saw many aluminum flagpoles standing, unharmed.

I probably oversimplified but I calculated the stress on a bare 8 inch diameter flag pole in a Cat 4 hurricane was greater than what that same pole experiences in 20 knots of wind carrying 400 square feet of sail. I based my numbers on what I had read about wind forces increasing as a square of the increase in wind speed. I disregarded how heeling reduced the exposure and came away from the exercise confident in the ability of the poles to do the job without stays. They did.

I drove through the Charleston area right after Hugo hit and saw trees down, power poles snapped, even a church steeple on the ground. I also saw many aluminum flagpoles standing, unharmed.

I probably oversimplified but I calculated the stress on a bare 8 inch diameter flag pole in a Cat 4 hurricane was greater than what that same pole experiences in 20 knots of wind carrying 400 square feet of sail. I based my numbers on what I had read about wind forces increasing as a square of the increase in wind speed. I disregarded how heeling reduced the exposure and came away from the exercise confident in the ability of the poles to do the job without stays. They did.

Bruce Taylor

08-26-2007, 04:52 PM

Jack hasn't clarified his original question, but I assume he just needs to calculate loads on his sails, for the purpose of determining press of canvas for a particular vessel.

Martin's formula (as used in that chart from Skene's, posted above) is as follows:

P = V ^2 x 0.0043

where P = pressure in lbs / ft ^2

V = wind speed in knots.

Multiply P by the sail area and you can roughly approximate the forces loading the CE, and use that info. to do certain tasks (like estimating the trimming moment on the hull, in a given wind, as described by mmd in another thread on the forum: http://www.woodenboatvb.com/vbulletin/upload/showthread.php?t=12184&page=21 ).

Martin's formula (as used in that chart from Skene's, posted above) is as follows:

P = V ^2 x 0.0043

where P = pressure in lbs / ft ^2

V = wind speed in knots.

Multiply P by the sail area and you can roughly approximate the forces loading the CE, and use that info. to do certain tasks (like estimating the trimming moment on the hull, in a given wind, as described by mmd in another thread on the forum: http://www.woodenboatvb.com/vbulletin/upload/showthread.php?t=12184&page=21 ).

donald branscom

08-28-2007, 09:03 AM

Jack hasn't clarified his original question, but I assume he just needs to calculate loads on his sails, for the purpose of determining press of canvas for a particular vessel.

Martin's formula (as used in that chart from Skene's, posted above) is as follows:

P = V ^2 x 0.0043

where P = pressure in lbs / ft ^2

V = wind speed in knots.

Multiply P by the sail area and you can roughly approximate the forces loading the CE, and use that info. to do certain tasks (like estimating the trimming moment on the hull, in a given wind, as described by mmd in another thread on the forum: http://www.woodenboatvb.com/vbulletin/upload/showthread.php?t=12184&page=21 ).

What does this( ^ ) symbol represent in your formula?

Thanks.

Martin's formula (as used in that chart from Skene's, posted above) is as follows:

P = V ^2 x 0.0043

where P = pressure in lbs / ft ^2

V = wind speed in knots.

Multiply P by the sail area and you can roughly approximate the forces loading the CE, and use that info. to do certain tasks (like estimating the trimming moment on the hull, in a given wind, as described by mmd in another thread on the forum: http://www.woodenboatvb.com/vbulletin/upload/showthread.php?t=12184&page=21 ).

What does this( ^ ) symbol represent in your formula?

Thanks.

Bruce Taylor

08-28-2007, 09:58 AM

What does this( ^ ) symbol represent in your formula?

Thanks.

It's an old ASCII-based way of saying "Raised to the power of". I haven't figured out how to do superscripts in the forum's editor!

Thanks.

It's an old ASCII-based way of saying "Raised to the power of". I haven't figured out how to do superscripts in the forum's editor!

ssor

08-30-2007, 08:41 PM

Power equals mass times velocity squared. The mass of air is approximately constant therefore velocity is the variable. Generally air pressure relative to velocity is measured normal to the flow. Somewhere around 15 pounds per square foot at 60 MPH sticks in my mind. Building spec's or something.

ittiandro

04-01-2017, 11:50 AM

http://i197.photobucket.com/albums/aa291/BDSTaylor/Windpressure.jpg

If I use this chart, the total windforce at12knts/hr on a 8. 0 m2sail(=85sqft)would be about 85 lbs, but if I use the P=v^2x.0043 formula ,I obtain about 52 lbs! Am I missing something?(I am calculating for a windsurfing sail, but the principle should be the same.The discrepancy is an internal one, for a boatsail.

Thanks

Ittiandro

If I use this chart, the total windforce at12knts/hr on a 8. 0 m2sail(=85sqft)would be about 85 lbs, but if I use the P=v^2x.0043 formula ,I obtain about 52 lbs! Am I missing something?(I am calculating for a windsurfing sail, but the principle should be the same.The discrepancy is an internal one, for a boatsail.

Thanks

Ittiandro

Woxbox

04-01-2017, 12:00 PM

I believe that chart assumes the wind hits the surface measured head-on. The wind typically strikes a sail at an angle, therefore the force is less. The situation is highly dynamic. Trim the sail in, and the force increases, ease it out and it decreases. This is obvious when under way, but not so easy to put hard numbers to.

Sailing upwind, a 100 sq. ft. sail might only present a profile of 20 sq. ft. or something like that to the wind, which is to consider how much air is actually intercepted by the sail. Straight downwind, boom all the way out, the sail will then catch 100 square feet of air.

Sailing upwind, a 100 sq. ft. sail might only present a profile of 20 sq. ft. or something like that to the wind, which is to consider how much air is actually intercepted by the sail. Straight downwind, boom all the way out, the sail will then catch 100 square feet of air.

ahp

04-01-2017, 12:43 PM

From a mechanical engineer's hand book I found that a 100 mph wind against a flat surface exerts 26 pounds per square foot. Wind pressure varies as the square of the velocity. At 10 mph the pressure would be 1/100 th, or 0.26 pounds per square foot. Does the wind blow flat against the sail?

I wanted to know if my garage door would cave in during a hurricane, worst case.

I wanted to know if my garage door would cave in during a hurricane, worst case.

skaraborgcraft

04-01-2017, 01:02 PM

Jim Michalak did an essay on it. There are lots of variables, angle of heel, angle of attack, boat speed, its a constantly moving figure, but good to get a figure for a broadside wind to determine such things as a mast size.

J.Madison

04-01-2017, 01:17 PM

If you are looking for rig loads, use the stability curve of the hull. The maximum righting moment of the hull equals the max loading on the rig. Rig load can never exceed stability righting moment or the boat would roll over and spill the force until equalizing.

The standard formula is: Vertical shroud load= Righting moment @30 degree * 1.5/ Half beam of hull

You should be able to find the righting moment @ 30 for various hulls. It does not change much among monohulls of a set waterline length. 1.5 is a factor to get max righting moment from that at 30 degrees.

The standard formula is: Vertical shroud load= Righting moment @30 degree * 1.5/ Half beam of hull

You should be able to find the righting moment @ 30 for various hulls. It does not change much among monohulls of a set waterline length. 1.5 is a factor to get max righting moment from that at 30 degrees.

cracked lid

04-01-2017, 04:19 PM

Power equals mass times velocity squared. The mass of air is approximately constant therefore velocity is the variable. Generally air pressure relative to velocity is measured normal to the flow. Somewhere around 15 pounds per square foot at 60 MPH sticks in my mind. Building spec's or something.

Power isn't mass * v^2. That's double kinetic energy, since kinetic energy is ½ m*v^2. Power is the rate at which work is being done, which is equivalent to the rate of the change in energy of the system. In the case of constant Force, it's Force times velocity.

However, that's really wrong way to approach this, as you need to look at the impulse exerted on the sail by air. Impulse is Force (on the sail) *time = mass (or air) * change in velocity (of air). The mass of air interacting with the sail is time dependent and is equivalent to the cross sectional area of the sail in the air stream multiplies by the length of the air column that passes through this are in some time frame (that's the volume)multiplied by the density of the air.

Force *time=density*Area*length*change in velocity

Thus Force/Area=Pressure=density*(length/time)*change in velocity = density*apparent windvelocity*change in wind velocity, and since the change in windvelocity due to the sail shape is also dependent on apparent windvelocity, the Pressure increases with the square of the apparent windspeed as indicated by the graph based on Martin's formula.

What Martin's formula gets right is that it depends on the square of the wind speed and a multiplicative factor. What it get's wrong is the argument that it comes from a Bernoulli effect. The multiplicative factor hides an error of scale in the derivation that shows up in Donald's calculation. It's not a difference in air speed over the opposite surfaces of the airfoil that matters, but rather Newton's 3rd law with momentum conservation. When the airfoil exerts a force to push the air one way the air exerts an equal force in the opposite direction on the airfoil, which is termed lift. Angle of attack allows flat surfaces to provide lift, and an airfoil shape increases efficiency by reducing drag.

Power isn't mass * v^2. That's double kinetic energy, since kinetic energy is ½ m*v^2. Power is the rate at which work is being done, which is equivalent to the rate of the change in energy of the system. In the case of constant Force, it's Force times velocity.

However, that's really wrong way to approach this, as you need to look at the impulse exerted on the sail by air. Impulse is Force (on the sail) *time = mass (or air) * change in velocity (of air). The mass of air interacting with the sail is time dependent and is equivalent to the cross sectional area of the sail in the air stream multiplies by the length of the air column that passes through this are in some time frame (that's the volume)multiplied by the density of the air.

Force *time=density*Area*length*change in velocity

Thus Force/Area=Pressure=density*(length/time)*change in velocity = density*apparent windvelocity*change in wind velocity, and since the change in windvelocity due to the sail shape is also dependent on apparent windvelocity, the Pressure increases with the square of the apparent windspeed as indicated by the graph based on Martin's formula.

What Martin's formula gets right is that it depends on the square of the wind speed and a multiplicative factor. What it get's wrong is the argument that it comes from a Bernoulli effect. The multiplicative factor hides an error of scale in the derivation that shows up in Donald's calculation. It's not a difference in air speed over the opposite surfaces of the airfoil that matters, but rather Newton's 3rd law with momentum conservation. When the airfoil exerts a force to push the air one way the air exerts an equal force in the opposite direction on the airfoil, which is termed lift. Angle of attack allows flat surfaces to provide lift, and an airfoil shape increases efficiency by reducing drag.

cracked lid

04-01-2017, 04:32 PM

All that being said, since what you care about is the pressure your sail cloth has to stand up to, you can hit the high limit of this by treating your sails as a flat surface with the wind blowing into them as if you are running directly downwind, but using the speed of wind over water vs. relative to the boat. Then just use the calculator at: http://www.engineeringtoolbox.com/wind-load-d_1775.html and be sure to check your units.

If your sail area is in ft^2, multiply it by 0.092903 before entering it into the area box.

If your wind speed is in knots, multiply by 0.514444 and if in miles per hour multiply by .447

Air density changes with temperature, decreasing .0042 kg/m^3 per degree Celsius increase of temperature with a value of 1.225kg/m^3 at 15C.

If your sail area is in ft^2, multiply it by 0.092903 before entering it into the area box.

If your wind speed is in knots, multiply by 0.514444 and if in miles per hour multiply by .447

Air density changes with temperature, decreasing .0042 kg/m^3 per degree Celsius increase of temperature with a value of 1.225kg/m^3 at 15C.

john welsford

04-02-2017, 03:07 AM

I've used the chart in Skenes with the stability that I've calculated, to work out the sail area when designing. It seems to work fine, take the figure for the designed "full sail windspeed", work out the vertical centre of area, that gives you the lever arm, given the two one can compare the heeling moment generated with the stability. I use Kg/M, and if I get the two to coincide at the chosen windspeed I'm pleased, if not, I'll look to reduce or increase the sail area to get that figure to match.

John Welsford

John Welsford

Pitsligo

04-02-2017, 10:19 AM

I really like Mr. Taylor's graph, but isn't there also the factor of temperature? A cold, dense wind, during autumn sailing, kicks us around noticeably more than does a wind of the same measured velocity on a warm summer day. What temperature is that graph predicated on, and how do temperature changes modify it?

Taken to extremes, this is why the opening "storm" in "The Martian" wouldn't actually have caused so much damage as was necessary to incite the story: the atmosphere on Mars is so thin that even with very high velocity winds it doesn't pack any punch. (Sheesh! Now I need to design a larger rig for Bucephalus, for when we emigrate to Mars...)

Alex

Taken to extremes, this is why the opening "storm" in "The Martian" wouldn't actually have caused so much damage as was necessary to incite the story: the atmosphere on Mars is so thin that even with very high velocity winds it doesn't pack any punch. (Sheesh! Now I need to design a larger rig for Bucephalus, for when we emigrate to Mars...)

Alex

IHWillys

04-02-2017, 10:47 AM

If I use this chart, the total windforce at12knts/hr on a 8. 0 m2sail(=85sqft)would be about 85 lbs, but if I use the P=v^2x.0043 formula ,I obtain about 52 lbs! ...

The formula using mph has a factor .004(as on the chart). The formula using knots has a factor of .0053, not .0043.

The chart uses mph.

Using 1.15 conversion factor knots to mph results in 13.8 mph for 12 knots.

13.8 looks to cross at .75 lb per sqft on the chart(post 13).

P=v^2x.004 results in .76

Using 12 knots in the formula for knots:

P=v^2x.0053 results in .76

So either way, 85 sqft in 12 knots or 85 sqft in 13.8 mph, 64.6 lbs of pressure.

Ken

The formula using mph has a factor .004(as on the chart). The formula using knots has a factor of .0053, not .0043.

The chart uses mph.

Using 1.15 conversion factor knots to mph results in 13.8 mph for 12 knots.

13.8 looks to cross at .75 lb per sqft on the chart(post 13).

P=v^2x.004 results in .76

Using 12 knots in the formula for knots:

P=v^2x.0053 results in .76

So either way, 85 sqft in 12 knots or 85 sqft in 13.8 mph, 64.6 lbs of pressure.

Ken

cracked lid

04-02-2017, 05:24 PM

I really like Mr. Taylor's graph, but isn't there also the factor of temperature? A cold, dense wind, during autumn sailing, kicks us around noticeably more than does a wind of the same measured velocity on a warm summer day. What temperature is that graph predicated on, and how do temperature changes modify it?

Taken to extremes, this is why the opening "storm" in "The Martian" wouldn't actually have caused so much damage as was necessary to incite the story: the atmosphere on Mars is so thin that even with very high velocity winds it doesn't pack any punch. (Sheesh! Now I need to design a larger rig for Bucephalus, for when we emigrate to Mars...)

Alex

Air density changes with temperature, decreasing .0042 kg/m^3 per degree Celsius increase of temperature with a value of 1.225kg/m^3 at 15C, which seems to be the common baseline temp for air pressure calculations. Since the pressure scales linearly with density, you simply take your value for the pressure from the chart and multiply it by (1-(.0042*difference in temp from 15C/1.225))= (1-0.0034*(temp in Celcius-15)) or if you prefer Fahrenheit, (1-0.0019*(temp in Fahrenheit-59))

https://lh3.googleusercontent.com/fy9TKXxy__WM0uw-Ygc8NjLVS159o6LY_2RT6BFffQwV0pm3OFCln23ooOnRDurjC8 gMn_Px0ivPV0EJeGC3ALXr9tRecE-BAc3JXPlF3FeGQ1XWzciUN4TbFBWwtlQbUJBlE6dKIfzu0Wqz-udus18F-4G4CzvlXZqIl_csDR9cbuYwa8ngoQYu7YpyJlTAOkX1WchzfGA aL5l-KVPhn_KYnzQT-TWv__JD2t6teXMXNNBNkq5JAPeyGEyeOb-jf5bQmkL0JBATkUhtJTyXzZjuurI8MSOul_4qtJxGqsLiCibfW Ty_KQUarZOt2acyjzklNePKDhVJw3Aya7RixQ7e_97C6_nYmzg 8u4RiI7ImZIQdPUi3hXn5pXDgHw1zlmk_IwMyxvAR888qna8zX 8IA3aBA1IZWiHAF5Am_39AArBCwmuZBgr7GPETMtl0AU4TVwew aalrGWwIWFEcbMIjbFM0QhyxCfXR476zlAC3qcjtPXbW2szfK8 S9DblHpPc5AKYMrtU-UZq8CAKuhboxV2vJVcbsv3pNUgmG4SdMomWNJXjUHeN-jB24PWKtxwR9lfEd_gh1h0Gcr8Ux9wfXYNR7vB3-U6RcXJc7QWNe_x1D1X7nI=w523-h195-no

Taken to extremes, this is why the opening "storm" in "The Martian" wouldn't actually have caused so much damage as was necessary to incite the story: the atmosphere on Mars is so thin that even with very high velocity winds it doesn't pack any punch. (Sheesh! Now I need to design a larger rig for Bucephalus, for when we emigrate to Mars...)

Alex

Air density changes with temperature, decreasing .0042 kg/m^3 per degree Celsius increase of temperature with a value of 1.225kg/m^3 at 15C, which seems to be the common baseline temp for air pressure calculations. Since the pressure scales linearly with density, you simply take your value for the pressure from the chart and multiply it by (1-(.0042*difference in temp from 15C/1.225))= (1-0.0034*(temp in Celcius-15)) or if you prefer Fahrenheit, (1-0.0019*(temp in Fahrenheit-59))

https://lh3.googleusercontent.com/fy9TKXxy__WM0uw-Ygc8NjLVS159o6LY_2RT6BFffQwV0pm3OFCln23ooOnRDurjC8 gMn_Px0ivPV0EJeGC3ALXr9tRecE-BAc3JXPlF3FeGQ1XWzciUN4TbFBWwtlQbUJBlE6dKIfzu0Wqz-udus18F-4G4CzvlXZqIl_csDR9cbuYwa8ngoQYu7YpyJlTAOkX1WchzfGA aL5l-KVPhn_KYnzQT-TWv__JD2t6teXMXNNBNkq5JAPeyGEyeOb-jf5bQmkL0JBATkUhtJTyXzZjuurI8MSOul_4qtJxGqsLiCibfW Ty_KQUarZOt2acyjzklNePKDhVJw3Aya7RixQ7e_97C6_nYmzg 8u4RiI7ImZIQdPUi3hXn5pXDgHw1zlmk_IwMyxvAR888qna8zX 8IA3aBA1IZWiHAF5Am_39AArBCwmuZBgr7GPETMtl0AU4TVwew aalrGWwIWFEcbMIjbFM0QhyxCfXR476zlAC3qcjtPXbW2szfK8 S9DblHpPc5AKYMrtU-UZq8CAKuhboxV2vJVcbsv3pNUgmG4SdMomWNJXjUHeN-jB24PWKtxwR9lfEd_gh1h0Gcr8Ux9wfXYNR7vB3-U6RcXJc7QWNe_x1D1X7nI=w523-h195-no

whiskeyfox

05-10-2017, 05:43 AM

In typical upwind trim there will be a very noticeable range of pressure distribution across the sail with the maximum pressure differential just behind the leading edge.

This website shows simulation results that illustrate it nicely:

http://www.elvstromsails.com/tech-layline/fsi-analysis

What I find very interesting is that the stress in the fabric does not appear to show the same distribution. It is instead completely dominated by the stress concentrations at the sail corners.

This website shows simulation results that illustrate it nicely:

http://www.elvstromsails.com/tech-layline/fsi-analysis

What I find very interesting is that the stress in the fabric does not appear to show the same distribution. It is instead completely dominated by the stress concentrations at the sail corners.

Peerie Maa

05-10-2017, 06:21 AM

I've used the chart in Skenes with the stability that I've calculated, to work out the sail area when designing. It seems to work fine, take the figure for the designed "full sail windspeed", work out the vertical centre of area, that gives you the lever arm, given the two one can compare the heeling moment generated with the stability. I use Kg/M, and if I get the two to coincide at the chosen windspeed I'm pleased, if not, I'll look to reduce or increase the sail area to get that figure to match.

John Welsford

That is entirely appropriate. Wind on the broadside and square to the sail plan is the worst case for stability.

If one is looking for force to drive the boat forward, you need to look at lift vs drag data.

Here is a starter for ten (plus 100) https://en.wikipedia.org/wiki/Forces_on_sails

John Welsford

That is entirely appropriate. Wind on the broadside and square to the sail plan is the worst case for stability.

If one is looking for force to drive the boat forward, you need to look at lift vs drag data.

Here is a starter for ten (plus 100) https://en.wikipedia.org/wiki/Forces_on_sails

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