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As we all know, with no motor, and no energy from the water or crew a sailboat gets all its moving energy from the wind.

Whilst exploring the aerodynamics of just how the sails do this amazing feat, I thought it might be instructive to analyse this statement using basic physics on a real life situation, so here goes.

As a model, I'm using a Farr40 which will do 7 kts in 12 kts TWS.

From independent research, I believe that it takes a force of around 2,000 Newtons (~200Kg weight) to drag this boat through the water at that speed. 

Wind is a mass of air moving relative to the water, and any moving mass contains kinetic energy. This energy can be deduced by looking at the Dynamic Pressure of the wind against the area of the sails presented to the wind.

Let's start with the sail area:

With a mast of 16.5 metres, a jib foot of around 5 and a main food of around 6 metres, the total area is 90 M2  

But the boat is going upwind, with a True Wind Angle (TWA) of 40°, so the area presented to the wind is Area x cos(TWA) or 68M2 .

The Dynamic pressure of a fluid is given by P=1/2(rho)V2Where rho is the density of the fluid. 

Air density is around 1.2 Kg/M2  (dry air at at NTP)

12 Kts wind speed is 6 M/sec, so plugging in rho of 1.2 and V of 6  we get a Dynamic Pressure of 23 Pascals or 23 Newtons/square metre.

This pressure on the sails produces  23 x 68 = 1540 Newtons.

The spreadsheet is here

Whilst this is in the right ballpark as we're looking for 2000 Newtons, it's still far short of the answer as it would require 100% conversion of wind energy to force, and that would require all the energy of the wind : effectively reducing the wind speed to zero, which obviously doesn't happen.

So where have I gone wrong?


 

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Did you ‘go wrong’ when you calculated that the wind causes a pressure on the sails rather than creates lift? Foils create lift forces many times greater than the drag. 20...40 times? Also the volume

Did you just prove that you scientifically can’t sail to your rating?

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Did you ‘go wrong’ when you calculated that the wind causes a pressure on the sails rather than creates lift? Foils create lift forces many times greater than the drag. 20...40 times? Also the volume of air affected by the foil extends far beyond the edges of the foil.

In my sailing experience the winningest racers use zero physics knowledge. They trim the sails to where the boat wins the race. Same with pilots: trim the airplane to not crash.

And your mistresses pics, please, newbie...

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37 minutes ago, El Borracho said:

Did you ‘go wrong’ when you calculated that the wind causes a pressure on the sails rather than creates lift?

I don't think so. The only source of energy is the kinetic energy of the wind. Sure the sails and foils generate lift, but that all has to be powered by the wind force. Remember conservation of energy: it can't be created, only converted. The wind energy came from the sun, the sails convert the wind energy into thrust (and heel and leeway).

(sorry, no mistresses)

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I did not dig into your spreadsheet. But my guess is your “dynamic pressure” is typically called induced and parasitic drag. The induced drag on properly trimmed sails produced prodigious amounts of lift ... a portion of it in the direction of travel.

Perhaps work on the DDW case first as it is largely force from drag. 

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Pointless theorizing - there's a lot of that in this place.

Work on your boat or, if it's warm enough, go sailing.

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15 minutes ago, SloopJonB said:

Pointless theorizing - there's a lot of that in this place.

Theorising is under-rated, remember Copernicus, Newton, Einstein, Rutherford... They would have been great sailors.

It's mid summer here in Oz, but I can only race three days a week, meanwhile, I ponder.

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20 minutes ago, El Borracho said:

But my guess is your “dynamic pressure” is typically called induced and parasitic drag.

No, that's a whole different kettle of fish. Drag, whether induced or parasitic is a resistive force generated by an aerofoil. 

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Perhaps your calcs should be done in power, not force. Estimate how much power is in the wind near the boat. 

Why is coefficient of lift not in your scribblings? Would seem important for calculating a force going to weather.

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21 minutes ago, Fleetwood said:

Where did your 2000N drag number come from?

Good question. From two sources

1. A mate in naval engineering is working on an 80 ft racing boat and they deduced 6000 at 10 kts

2. Another reputable source towed a 32 footer at 6 knots and measured 1400 on a strain gauge.

So my 2,000 is a guesstimate. I'm searching for confirmation and may well get our boat towed with a string gauge.

 

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2 hours ago, Sailor Al. said:

As we all know, with no motor, and no energy from the water or crew a sailboat gets all its moving energy from the wind.

Whilst exploring the aerodynamics of just how the sails do this amazing feat, I thought it might be instructive to analyse this statement using basic physics on a real life situation, so here goes.

As a model, I'm using a Farr40 which will do 7 kts in 12 kts TWS.

From independent research, I believe that it takes a force of around 2,000 Newtons (~200Kg weight) to drag this boat through the water at that speed. 

Wind is a mass of air moving relative to the water, and any moving mass contains kinetic energy. This energy can be deduced by looking at the Dynamic Pressure of the wind against the area of the sails presented to the wind.

Let's start with the sail area:

With a mast of 16.5 metres, a jib foot of around 5 and a main food of around 6 metres, the total area is 90 M2  

But the boat is going upwind, with a True Wind Angle (TWA) of 40°, so the area presented to the wind is Area x cos(TWA) or 68M2 .

The Dynamic pressure of a fluid is given by P=1/2(rho)V2Where rho is the density of the fluid. 

Air density is around 1.2 Kg/M2  (dry air at at NTP)

12 Kts wind speed is 6 M/sec, so plugging in rho of 1.2 and V of 6  we get a Dynamic Pressure of 23 Pascals or 23 Newtons/square metre.

This pressure on the sails produces  23 x 68 = 1540 Newtons.

The spreadsheet is here

Whilst this is in the right ballpark as we're looking for 2000 Newtons, it's still far short of the answer as it would require 100% conversion of wind energy to force, and that would require all the energy of the wind : effectively reducing the wind speed to zero, which obviously doesn't happen.

So where have I gone wrong?


 

In determining the usefulness of such data? It reminds of the aerodynamic calculations that conclusively prove bumblebees can't fly, while ignoring the fact that they do. It was an exercise designed to demonstrate how easy it is to mess up in mathematical modeling but was used to demonstrate that scientists are damned fools.  

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8 minutes ago, Mark K said:

In determining the usefulness of such data? It reminds of the aerodynamic calculations that conclusively prove bumblebees can't fly, while ignoring the fact that they do. It was an exercise designed to demonstrate how easy it is to mess up in mathematical modeling but was used to demonstrate that scientists are damned fools.  

When I hear that notion, I always watch this...

 

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a) You shouldn't consider the projected area of the sail plan. Consider the whole area.

b) Your equation for lift should be = 1/2 rho V^2 x Cl where Cl = lift coefficient. Cl will be >1

c) your 2000 N is a WAG.

d) But lift =/= forward propulsion force. Only some component of the lift vector is forward. Lots is sideways.

The ORC VPP physics is all in the public domain - including a lot of the VPP formulations including sail lift coefficients

https://www.orc.org/rules/ORC VPP Documentation 2016.pdf

The drag formulations are also in the file. You could write your own VPP if you were so inclined.

image.png.5105292ab27914db399b3bf6e1eb928b.png

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1 hour ago, Sailor Al. said:

Good question. From two sources

1. A mate in naval engineering is working on an 80 ft racing boat and they deduced 6000 at 10 kts

2. Another reputable source towed a 32 footer at 6 knots and measured 1400 on a strain gauge.

So my 2,000 is a guesstimate. I'm searching for confirmation and may well get our boat towed with a string gauge.

 

So your back of the envelope calc gets you to within ~30% of your WAG - that's not too bad.....

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or C.A. Marchaj or Tom Whidden, or .... get on a Farr 40 and tune up side by side.   BTW, that 7 knots in K nots of breeze sounds a little low unless you were calculating being shoved dead down wind like a lump.   

This is for a Cookson 12m.  Theoretically slower than a Farr 40. ^_^

455923913_ScreenShot2021-01-18at8_25_18PM.png.108ff01915dd0acee0a24ed45c580480.png

 

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91tG3RaSTCL._AC_UL320_SR320,320_.jpg

41wLX56xyzL._SX357_BO1,204,203,200_.jpg

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Two biggies:

1. The drag is much less, probably around 1100 N at 7 knots

2. Calculate the rig force using apparent windspeed...

Gross numbers are around 4000 N for sailforce and 4:1 for sideforce:forward force.

(Sail area is around 100 m2 upwind, design wind calcs use 4 kgf/m2 for initial calcs)

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1 hour ago, Zonker said:

a) You shouldn't consider the projected area of the sail plan. Consider the whole area.

 

Thanks for checking the numbers.

I think I do have to use the projected area because the Dynamic pressure is a vector (with direction), so the force on the boat is affected by the angle. Consider the extreme case, with the sail plan parallel to the wind: force would be zero. 

  "b) Your equation for lift should be = 1/2 rho V^2 x Cl where Cl = lift coefficient. Cl will be >1"

Disagree: the equation is for total force which can't be greater than the kinetic energy go the wind. CI is about the aerodynamics of how the sail extracts the force.

"c) your 2000 N is a WAG." WAG == Wild Assed Guess? Well, not quite.

Your diagrams come from the aerodynamics of how the sail shape generates lift. I'm just looking at the raw physics where total energy is constant. 

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12 minutes ago, Left Shift said:

This is for a Cookson 12m.  Theoretically slower than a Farr 40. ^_^

From your12 kt  polars, VMG of 5.55 at 39° is 7.14 BS (5.55/cos(39)). My 7 kts was just a rough starting position. 

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I would like to throw another book into the mix.  "Symmetry of Sailing" by Murray Ross.  Murray was a Kiwi "Mr. Science" for the Cup effort in 1986-87.

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10 minutes ago, Frogman56 said:

Two biggies:

1. The drag is much less, probably around 1100 N at 7 knots

2. Calculate the rig force using apparent windspeed...

Gross numbers are around 4000 N for sailforce and 4:1 for sideforce:forward force.

(Sail area is around 100 m2 upwind, design wind calcs use 4 kgf/m2 for initial calcs)

1100N may be right, I don't have any experimental data. What is your source for 1100?

Yes, correct, you calculate rig forces on AWS, but for the physics of using wind energy to overcome water drag, then we have to stick to TWS 

Not sure how you get 100 m2 sail area, spreadsheet supports 68 m2

And what is your source for 400 N for sail force please?

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I agree with the recommendation for Tony Marchaj's books.

"Sailing Theory and Practice"' and "Aero-Hydrodynamics of Sailing" should give you all you need.

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47 minutes ago, Sailor Al. said:

From your12 kt  polars, VMG of 5.55 at 39° is 7.14 BS (5.55/cos(39)). My 7 kts was just a rough starting position. 

So you are using wind pressure in a sailing mode that is entirely driven by airfoil lift?

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3 minutes ago, Left Shift said:

So you are using wind pressure in a sailing mode that is entirely driven by airfoil lift?

For the moment, I'm trying to stay away from the intricacies of airfoil lift, and just look at the raw mathematical physics of a system that has a single source of energy (the Dynamic pressure of the wind )  which results in the boat overcoming the 2,000 N of hydrodynamic drag of the hull.  

My current algorithm yields 1500 N, but relies on a totally unrealistic 100% conversion of the wind energy. 

 

 

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Nope. The VPP document I pointed to gives you all the equations you need to derive sail lift, drag, side force, hull resistance etc.

It's typical in the hydro/aero community to separate your lift and drag components of a foil or whatever, and them sum all the drag forces, all the lift forces etc. 

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Sailor Al,

1. 1100 N educated guess from detailed vpp for parametrically similar boat with 38 ft lwl. Actual for F40 may be a bit less, between 1000 and 1100.

2. TWS for the aero is not useful. Use awa and aws, because that is what the airfoil 'sees'.

3. F40 sail area report as 98 to 102 m2 depending on calculation method. For example, class data has mainsail area 63 m2 and jib area 34.6m2; Farr 40 oz website has upwind sail at 102.8 m2. Lets call it 100!

4. 4 kg/m2 is standard NA practise, more or less.

 

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I fear you are trying to get realistic numbers out of a mathematical model that is far too simplistic to be capable of doing anything of the kind. You are not going to be able to do anything useful until you are successfully estimating lift and drag for both the components In the air and the components in the water. 

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Jim,

What Sailor Al might be looking at is the base level of the physics.

If we can understand how a well understood yacht sails upwind near 'max rm' aka ' design wind', (i.e. understand at a component level) this can help us with our understanding . ..say 'why does the speed go down at least 0.15 when one crew leaves the rail?'

Or, more importantly, ' how do i improve the aero l/d once max rm is utilised?

Or, indeed, 'how is that 400 kg sideforce generated, on the hydro side?

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11 hours ago, Sailor Al. said:

Theorising is under-rated, remember Copernicus, Newton, Einstein, Rutherford... They would have been great sailors.

It's mid summer here in Oz, but I can only race three days a week, meanwhile, I ponder.

Einstein was a sailor. It certainly wasn't his strength. 

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13 hours ago, Sailor Al. said:

As we all know, with no motor, and no energy from the water or crew a sailboat gets all its moving energy from the wind.

Whilst exploring the aerodynamics of just how the sails do this amazing feat, I thought it might be instructive to analyse this statement using basic physics on a real life situation, so here goes.

As a model, I'm using a Farr40 which will do 7 kts in 12 kts TWS.

From independent research, I believe that it takes a force of around 2,000 Newtons (~200Kg weight) to drag this boat through the water at that speed. 

Wind is a mass of air moving relative to the water, and any moving mass contains kinetic energy. This energy can be deduced by looking at the Dynamic Pressure of the wind against the area of the sails presented to the wind.

Let's start with the sail area:

With a mast of 16.5 metres, a jib foot of around 5 and a main food of around 6 metres, the total area is 90 M2  

But the boat is going upwind, with a True Wind Angle (TWA) of 40°, so the area presented to the wind is Area x cos(TWA) or 68M2 .

The Dynamic pressure of a fluid is given by P=1/2(rho)V2Where rho is the density of the fluid. 

Air density is around 1.2 Kg/M2  (dry air at at NTP)

12 Kts wind speed is 6 M/sec, so plugging in rho of 1.2 and V of 6  we get a Dynamic Pressure of 23 Pascals or 23 Newtons/square metre.

This pressure on the sails produces  23 x 68 = 1540 Newtons.

The spreadsheet is here

Whilst this is in the right ballpark as we're looking for 2000 Newtons, it's still far short of the answer as it would require 100% conversion of wind energy to force, and that would require all the energy of the wind : effectively reducing the wind speed to zero, which obviously doesn't happen.

So where have I gone wrong?


 

Hack!

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One half rho v squared A Cl.  A is the total area of the sail.  Produces lift roughly perpendicular to the ambient flow.  If you are in steady state going speed v, the flow the sails see is not the true wind speed, it's apparent speed.  In 12 true, the apparent wind is gonna be about 17 or 18 kts, not 12.

If steady state, the driving force = drag. End of story.  Unless you think Newton was wrong.

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9 hours ago, Sailor Al. said:

Yes, correct, you calculate rig forces on AWS, but for the physics of using wind energy to overcome water drag, then we have to stick to TWS 

 

If you are using basic physics (conservation of energy  or momentum) you should be able to redo the calculations in any frame, and still keep it all balanced. If you have set up the problem correctly, then you should be able to work in either true wind (frame pinned to average water position) or apparent wind (frame pinned to boat.) and still get to the answer.

This may help you find where the error sits.

Have fun.

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9 hours ago, Sailor Al. said:

1100N may be right, I don't have any experimental data. What is your source for 1100?

Yes, correct, you calculate rig forces on AWS, but for the physics of using wind energy to overcome water drag, then we have to stick to TWS 

Not sure how you get 100 m2 sail area, spreadsheet supports 68 m2

And what is your source for 400 N for sail force please?

No.  The sail feels apparent wind, not true wind.  Period.

How do you think AC foilers go 40 kts in 12 kts of true wind?  It ain't by calculating forces using 12 kts!

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12 hours ago, Sailor Al. said:

Thanks for checking the numbers.

I think I do have to use the projected area because the Dynamic pressure is a vector (with direction), so the force on the boat is affected by the angle. Consider the extreme case, with the sail plan parallel to the wind: force would be zero. 

 

The dynamic pressure is not a vector quantity,and has no direction. It's a pressure, or a amount of energy per unit volume.... if you don't find its invariant under rotation you are calculating it incorrectly.

 

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16 hours ago, SloopJonB said:

Pointless theorizing - there's a lot of that in this place.

Work on your boat or, if it's warm enough, go sailing.

yea don't try to understand how it works but just do it

 

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13 hours ago, Sailor Al. said:

Yes, correct, you calculate rig forces on AWS, but for the physics of using wind energy to overcome water drag, then we have to stick to TWS 

 

That doesn't make sense. If you're an airplane wing moving through air at 400kts you don't use the TWS to determine your lift and drag force vectors. Even though it has thrust it is just another lift and drag vector.

You're also not looking at the lift component of the hull forces on the water, mast as part of the mainsail, hull acting as its own foil, etc.

 

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5 hours ago, ALL@SEA said:

Einstein was a sailor. It certainly wasn't his strength. 

But apparent wind had to influence his thinking. ?

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4 minutes ago, Amati said:

But apparent wind had to influence his thinking. ?

I seem to recall reading that he was more interested in watching the wake... there's another thread in the making here...

 

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11 minutes ago, ALL@SEA said:

I seem to recall reading that he was more interested in watching the wake... there's another thread in the making here...

 

Who was the physicist that wanted to ask God why turbulence even existed? Feynman?

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To quote a naval architect of my acquaintance:  Multiplying one approximation by another approximation rarely results in greater accuracy.

(She happened to be referring to PHRF ToT conversions, but her thought certainly applies even more concisely here.)

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9 minutes ago, Left Shift said:

To quote a naval architect of my acquaintance:  Multiplying one approximation by another approximation rarely results in greater accuracy.

(She happened to be referring to PHRF ToT conversions, but her thought certainly applies even more concisely here.)

AKA cumulative error.

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10 minutes ago, Left Shift said:

Maybe because "god" wasn't actually involved?

IIRR, that was Feynman’s point....

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12 minutes ago, Left Shift said:

To quote a naval architect of my acquaintance:  Multiplying one approximation by another approximation rarely results in greater accuracy.

(She happened to be referring to PHRF ToT conversions, but her thought certainly applies even more concisely here.)

Engineers at a certain large airplane manufacturer used to joke that airplane design was “guessing to 4 decimal places”. :rolleyes:

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6 minutes ago, Ultraman said:

The OP needs this book.

A Degree in Physics might help him understand what's in it too...

IMG_20210119_1112492.jpg

IMG_20210119_1113067.jpg

Did Ross approve the foto on the dust jacket?    He does use up a bit of print on twist... although there seems to be more camber at the head of the main...<_<.... now I’ll have to find the book and look up twisted flow strategies.  Oh well, I’m retired!:lol:

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1 hour ago, Amati said:

Engineers at a certain large airplane manufacturer used to joke that airplane design was “guessing to 4 decimal places”. :rolleyes:

I did a Transatlantic race with a retired Boeing engineer.  He does not fly in airplanes.

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1 hour ago, Amati said:

Did Ross approve the foto on the dust jacket?    He does use up a bit of print on twist... although there seems to be more camber at the head of the main...<_<.... now I’ll have to find the book and look up twisted flow strategies.  Oh well, I’m retired!:lol:

I watched the video posted above https://www.youtube.com/watch?v=bCwtcDNB15E&amp;feature=youtu.be about the location of the downwash and the upwash on an airfoil and reminded me of a trick we used to do in solid breeze when we didn't want to reef.  It only worked if the main had two full battens.  

The trimmers would sheet on fairly hard, quickly drop the traveler and get the two upper battens to pop to weather.  Then we would trim in as usual.  The lower 2/3rds of the sail would be as "normal and I'd use the lower leach as my weather helm control.  And we would fly up hill.  Many pickle dishes to prove it. 

I assumed it was because we were generating "anti-heeling" lifting forces from the inverted section.  Now I'm wondering if we weren't demonstrating exactly what the guy in the video was preaching?

Harder to do with the fathead mains we have now, but for those sails, we have basically taken the shape out of the top section with super stiff battens.  

Hmmm.  Will ponder some more.  

 

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Back after a good night's sleep to see lots of replies 

10 hours ago, Frogman56 said:

1. 1100 N educated guess from detailed vpp for parametrically similar boat with 38 ft lwl. Actual for F40 may be a bit less, between 1000 and 1100.

2. TWS for the aero is not useful. Use awa and aws, because that is what the airfoil 'sees'.

3. F40 sail area report as 98 to 102 m2 depending on calculation method. For example, class data has mainsail area 63 m2 and jib area 34.6m2; Farr 40 oz website has upwind sail at 102.8 m2. Lets call it 100!

4. 4 kg/m2 is standard NA practise, more or less.

1. I will have to go and test on the water. 1000 seems light on
2. No TWS is the correct parameter. See later comments on frame of reference.
3. I have found our ORC Cert and agree, F40 sail area is 98 m2  and have updated the spreadsheet accordingly 
4. 4 kg/m2 seems high: My 1/2ρV^2 formula gives  23 Pascals which is around 2 Kg/m2

6 hours ago, JohnMB said:

If you are using basic physics (conservation of energy  or momentum) you should be able to redo the calculations in any frame, and still keep it all balanced. If you have set up the problem correctly, then you should be able to work in either true wind (frame pinned to average water position) or apparent wind (frame pinned to boat.) and still get to the answer.

Yes, and I am choosing to use the Earth (or water as I'm assuming no current) as my frame of reference in which the wind is at 12 Kts (TWS).

 

3 hours ago, JohnMB said:

The dynamic pressure is not a vector quantity,and has no direction. It's a pressure, or a amount of energy per unit volume.... if you don't find its invariant under rotation you are calculating it incorrectly.

On further consideration, I agree it's not a vector, but orientation IS significant. If you present a  sail parallel to the wind (TWA 0, "in irons") , it will not me subject to any dynamic pressure, at 90 ° it will be subject to 100% of the pressure, so at 40°TWA it will be subject to sin(TWA) of the pressure (not cos as I initially used, spreadsheet updated).

image.thumb.png.8e5292a4043dda755e10ac162e4dc407.png

1 hour ago, Ultraman said:

The OP needs this book.

A Degree in Physics might help him understand what's in it too...

Yes it does. Mine is a BSc. Mathematical Physics, Birmingham UK.

I don't have Garretts book in my library unfortunately, so can't check his offering.

 

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27 minutes ago, Sailor Al. said:

Mine is a BSc. Mathematical Physics, Birmingham UK.

I don't have Garretts book in my library unfortunately, so can't check his offering.

 

Mine is B.Sc. (Physics), University of British Columbia.

You can pick the book up used on Amazon for C$15 or new even for C$40!

It was published in 1987, but I think most of the theory still applies, even if materials and design has moved on a lot since then.

Worth the effort to find and read this book, although it is a hefty tome sure to induce some Zzzs.

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32 minutes ago, Sailor Al. said:

Yes, and I am choosing to use the Earth (or water as I'm assuming no current) as my frame of reference in which the wind is at 12 Kts (TWS).

The point about changing your frame of reference is that it sometimes helps you figure out what you have missed. If you don't get the same answer in the other frame of reference then this might point you in the right direction for your error.

As several people have noted its is common practice to calculate sail forces using apparent not true wind, there is probably a good reason for this, and exploring it might give you more insight into the problem.

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13 minutes ago, Ultraman said:

Worth the effort to find and read this book, although it is a hefty tome sure to induce some Zzzs.

I have accessed an on-line copy, and see he's using the conservation of momentum argument to explain the upwind performance. As a fellow physicist, you will quickly spot the fallacy of the argument. I will save my $C18.

image.thumb.png.6accaa610c957454dce3171fa9d0bb46.png

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1 hour ago, Left Shift said:

I watched the video posted above https://www.youtube.com/watch?v=bCwtcDNB15E&amp;feature=youtu.be about the location of the downwash and the upwash on an airfoil and reminded me of a trick we used to do in solid breeze when we didn't want to reef.  It only worked if the main had two full battens.  

The trimmers would sheet on fairly hard, quickly drop the traveler and get the two upper battens to pop to weather.  Then we would trim in as usual.  The lower 2/3rds of the sail would be as "normal and I'd use the lower leach as my weather helm control.  And we would fly up hill.  Many pickle dishes to prove it. 

I assumed it was because we were generating "anti-heeling" lifting forces from the inverted section.  Now I'm wondering if we weren't demonstrating exactly what the guy in the video was preaching?

Harder to do with the fathead mains we have now, but for those sails, we have basically taken the shape out of the top section with super stiff battens.  

Hmmm.  Will ponder some more.  

 

Masthead? 3/4?  What kind of camber distribution on the previous sail? (Flatter or more curvy towards the tip?)

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33 minutes ago, Sailor Al. said:

I have accessed an on-line copy, and see he's using the conservation of momentum argument to explain the upwind performance. As a fellow physicist, you will quickly spot the fallacy of the argument. I will save my $C18.

image.thumb.png.6accaa610c957454dce3171fa9d0bb46.png

Do you know the theories of lift? There is more than one. And none of them by themselves are perfect. They are also interrelated.

This book doesn't look worth owning. I agree.

Your initial attempt made bad assumptions and faulty guesstimates. The physics isn't any more complicated than lift. The overall system is simply aerohydrodynamic rather than aerodynamic. The couple between the two fluids results in what is termed a "drag angle" where the lwer the angle the more efficient.

Yes, I am trying to confuse you.

You cannot take the true wind. That is simply not what is happening. Read Einstein. He has the answers.

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21 hours ago, El Borracho said:

Did you ‘go wrong’ when you calculated that the wind causes a pressure on the sails rather than creates lift? Foils create lift forces many times greater than the drag. 20...40 times? Also the volume of air affected by the foil extends far beyond the edges of the foil.

In my sailing experience the winningest racers use zero physics knowledge. They trim the sails to where the boat wins the race. Same with pilots: trim the airplane to not crash.

And your mistresses pics, please, newbie...

if we coul get an L/D on a sail of 40 we would no longer need auxiliaries. try 3 to 5 for lots of sailboats as a whole

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20 hours ago, Sailor Al. said:

Theorising is under-rated, remember Copernicus, Newton, Einstein, Rutherford... They would have been great sailors.

It's mid summer here in Oz, but I can only race three days a week, meanwhile, I ponder.

At least one of those (Einstein) was a sailor. I'm not sure how great though.

Albert Einstein, Sailor - National Maritime Historical Society (seahistory.org)

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24 minutes ago, fastyacht said:

if we coul get an L/D on a sail of 40 we would no longer need auxiliaries. try 3 to 5 for lots of sailboats as a whole

Decode please: "L/D", "sail of 40", "auxiliaries", "3 to 5"

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6 minutes ago, Doug Halsey said:

At least one of those (Einstein) was a sailor. I'm not sure how great though.

It's all there in  your reference: "His exploits as a sailor are not as well known and for good reason—Albert Einstein was a terrible sailor!" 

Thanks

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19 minutes ago, Sailor Al. said:

It's all there in  your reference: "His exploits as a sailor are not as well known and for good reason—Albert Einstein was a terrible sailor!" 

Thanks

He invented Relativity. Applies to sailing.

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50 minutes ago, fastyacht said:

Do you know the theories of lift? There is more than one. And none of them by themselves are perfect. They are also interrelated.

This book doesn't look worth owning. I agree.

Your initial attempt made bad assumptions and faulty guesstimates. The physics isn't any more complicated than lift. The overall system is simply aerohydrodynamic rather than aerodynamic. The couple between the two fluids results in what is termed a "drag angle" where the lwer the angle the more efficient.

Yes, I am trying to confuse you.

Yes, theories of lift are many and various, and the subject of my major interest, but for this thread, I'm not focusing on how the sails generate the thrust, but on the bigger picture of how much energy is available from the only source available (the TWS) compared with the energy required to physically push the boat through the water (the hull drag).

I loved your throw away line: "The physics isn't any more complicated than lift"! 

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29 minutes ago, Sailor Al. said:

I loved your throw away line: "The physics isn't any more complicated than lift"! 

Well, it ain't.

30 minutes ago, Sailor Al. said:

Yes, theories of lift are many and various, and the subject of my major interest,

 

And you don't know what L/D means?

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Just now, Sailor Al. said:

Hey, don't be nasty. The more you know, the more you know you don't know.

Not true. The less you know, the more you know you know.

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Just now, fastyacht said:

Not true. The less you know, the more you know you know.

That's just the corollary. 

OK L/D is lift to drag ratio, and is of great interest to aeronautical engineers building aircraft wings, where drag is the force opposing the engine's thrust (perp to the lift).

When looking at the performance of a sail, the only source of thrust is that generated from the forward component of total lift, and so drag, perpendicular to total lift provides the  heeling and leeway force. 

 

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5 minutes ago, Sailor Al. said:

That's just the corollary. 

OK L/D is lift to drag ratio, and is of great interest to aeronautical engineers building aircraft wings, where drag is the force opposing the engine's thrust (perp to the lift).

When looking at the performance of a sail, the only source of thrust is that generated from the forward component of total lift, and so drag, perpendicular to total lift provides the  heeling and leeway force. 

 

Nope on both counts. Well close. But lift and drag are measured wit respect to the incident far field flow direction, neither of which are alinged with the thrust direction. But good try.

Look it is really simple. I don't feel like pulling books down (right now) but just go find 12 meter stuff. IT is publishd all over the place including resistance. Tons of tank tests were run for decades. You know the drag. You know the speed (also publshd) you know the sail area, etc. Power is force integrated over a period of time.

Do some deconstruction analysis and you might learn something. (Yes, some of us myself included are sometimes cynical old codgers who did this very thing 35 years ago already. IT is more fun when you have to struggle. Trust me)

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You began the thread with units of force. Now, above, you wrote energy. I would probably work in power, though.

I know why TWS is used in polars and the like. But AWS seems better suited for this calculation.

Your idea of dragging a hull thru the water with a spring scale to establish a driving force is deeply flawed.

An aside: The blades of a wind turbine are very slender compared to the volume swept yet they extract energy from the entire swept disc. Even if transformed to the helical path vs. axial. Somehow this notion applies to your effort.

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3 minutes ago, El Borracho said:

Your idea of dragging a hull thru the water with a spring scale to establish a driving force is deeply flawed.

Why is it deeply flawed? I know it won't be 100% because the hull will perform a bit differently when heeled (in 12 kts with good crew work, it will heel around 15°), but I can't see why  towing it at the nominated speed wouldn't generate pretty much the same drag as experienced when sailing. I'm only interested in getting to about 10% accuracy.

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14 minutes ago, El Borracho said:

....

An aside: The blades of a wind turbine are very slender compared to the volume swept yet they extract energy from the entire swept disc. Even if transformed to the helical path vs. axial. Somehow this notion applies to your effort.

https://en.wikipedia.org/wiki/Betz's_law#:~:text=According to Betz's law%2C no,is known as Betz's coefficient.

^ this derivation might be a good starting point if you pay attention to your reference frames.

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12 minutes ago, Sailor Al. said:

Why is it deeply flawed? I know it won't be 100% because the hull will perform a bit differently when heeled (in 12 kts with good crew work, it will heel around 15°), but I can't see why  towing it at the nominated speed wouldn't generate pretty much the same drag as experienced when sailing. I'm only interested in getting to about 10% accuracy.

Socyou going to pretnd you can estimate heeled yawed resistance?

Jist findcthe damned old metre boat curves. They are out there...heeled resistance included.

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36 minutes ago, Sailor Al. said:

Why is it deeply flawed? I know it won't be 100% because the hull will perform a bit differently when heeled (in 12 kts with good crew work, it will heel around 15°), but I can't see why  towing it at the nominated speed wouldn't generate pretty much the same drag as experienced when sailing. I'm only interested in getting to about 10% accuracy.

My guess is that you are dividing the boat into two parts: The sails, and everything else. To measure the force accurately everything else must be in the state it would be in on a beat in 12 kt wind. A Farr 40 has exceptionally large topside exposure to the wind when heeled, for example, so (somehow) arrange for that heel angle while towing. Plus the hull is making leeway so consider (somehow!) attaching a bridle to tow it at that angle. Set the rudder too... And of course wait for a 12 kt breeze...

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3 hours ago, Amati said:

Masthead? 3/4?  What kind of camber distribution on the previous sail? (Flatter or more curvy towards the tip?)

First one we did it on was a 36' masthead with a slightly short hoist #3 up.  The second was a 40' frac with a full hoist #3 or short #4 with diamond stays stiffening the top. 

Both rigs had a ton of bend laid on.  About 18" on the mast head and 27" on the frac, so the mains were pretty bladed out, obviously.  But the top still had camber to pop out.  The frac boat we never reefed below 32 TWS with that set up.  Fast and lovely to drive.  

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52 minutes ago, Sailor Al. said:

Why is it deeply flawed? I know it won't be 100% because the hull will perform a bit differently when heeled (in 12 kts with good crew work, it will heel around 15°), but I can't see why  towing it at the nominated speed wouldn't generate pretty much the same drag as experienced when sailing. I'm only interested in getting to about 10% accuracy.

For starters, springs are deeply flawed.  Ever go to a real weighing station?  where weights matter?  The big letters on the read out always say XYZ Manufacturer - NO SPRINGS

The again tank tests are also deeply flawed.  Look up "The Persistence of Vorticity".  After the first run, the water is too turbulent to take a good reading for hours.

What on earth good is 10% accuracy?  I can calculate the speed of light using my microwave to about 10% accuracy.  What use is that other than a parlor trick?

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