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Go Back   SailNet Community > Skills and Seamanship > Learning to Sail > Why is beam reach (or near to it) the fastest point of sail?
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Thread: Why is beam reach (or near to it) the fastest point of sail? Reply to Thread
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Topic Review (Newest First)
04-07-2012 05:26 PM
eyytee
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by tarmes View Post
Does anyone here know of a resource that explains exactly why the fastest point of sail for most boats is the beam reach (or close to it)?

If you search for boat polars charts:

Click to View Search Results for polar charts boat - Google Search polar charts boat - Google Search


you will find that with sufficient windspeed the fastest course is broad reach. Especially for fast efficient boats or ice boats. It all depends on how your hydrodynamic efficiency compares to your aerodynamic efficiency. You can define a L/D ratio for both parts of the boat (below and above water). Combined they give you the polar charts and the fastest course.
04-06-2012 03:17 AM
tarmes
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by brehm62 View Post
I think this is what you were saying and I agree. However, even though I tried to say it as simply as possible I still ended up with three fairly complex paragraphs. And without benefit of diagrams this can still be misinterpreted. I think this thread has included a lot of saying the same thing in different ways.
Actually I found that very clear. Although maybe that's because I now understand the principal
04-06-2012 03:12 AM
tarmes
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by RichH View Post
Here's a graphical representation of the aerodynamic output of a sail in either a closehauled or beamreach orientation.
Thanks, with that it become obvious....
04-05-2012 11:21 PM
brehm62
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by Minnewaska View Post
To start, if you are sailing directly downwind, the faster you go, the lower the apparant wind on your sails. You are out running the wind. So far, so good? The faster you go, the less you have to work with heading downwind.
This is correct from a boat point of view. In reality of course, energy from the wind is being transferred to the boat and then bled off in hull drag. But, yes, from the sailor's point of view, the wind dies off as you accelerate downwind. There is less push on the sails and rigging. Naturally though, the rudder will still have good authority because you are moving. It's interesting that this is why a jibe is such a trap. You are heading downwind and the apparent wind seems very light, not any cause for worry. But if you veer off then the apparent wind speed picks up and hits your sails with unexpected force. You suddenly have your hands full.

Quote:
If you sail upwind, you actually create more wind to work with (increased apparent wind), because you compound the wind as you begin to move into it.
Well, again, from a boat centered point of view this is correct. The increased wind will push harder on the sails and rigging. There will appear to be more energy available to drive the boat forward.

Quote:
Therefore, it stands to reason that you will get more boat speed sailing into the wind, then away from it.
Yes, indeed, IF (and this is a big if) the apparent wind angle stayed the same.

Quote:
So why a difference in close hauled and a beam reach? A beam reach remains the lowest point of sail, where you are still being pulled by the lift of the sail, rather than being pushed from behind. Therefore, you may still be creating some increase in apparent wind.

Now the vector part. If you thnk about where the boom is on a close haul vs. a beam reach, this may begin to make sense. Since you are creating a wing with your sail, the boom is much closer to the center of the boat on a close haul, as the wind is coming more directly at our bow. On a beam reach, your boom is eased out so the front of the wing (sail) is facing the wind coming from the side.
I think what you are saying is something like the following. If the boat were stationary and facing perpendicular to the wind then the wind would be coming directly over the side of the boat. And, in this position, 100% of lift from the sail would be used to push the boat forward. Here we have maximum potential (but of course the boat isn't moving yet).

As the boat begins to accelerate forward, the apparent wind will increase and this would theoretically give additional lift to the sail. And, it would if the wind angle stayed the same. However, it doesn't. As the boat accelerates the apparent wind will come more and more from the bow. It will shift forward. And, the apparent wind will shift more forward the faster the boat goes. This is bad because, as the wind shifts forward, it becomes less useful. Why? Because lift is perpendicular to the wind and therefore as it shifts forward an increasing amount of the apparent wind would only generate side force rather than forward propulsion. Also, the increasing side force will have to be countered by the keel and rudder so this will increase drag.

So, why is close hauled worse? As you pull towards the wind the apparent wind vector shifts even more towards the bow, so even though theoretically the apparent wind would increase, there is less and less lift to move the boat forward. More and more of the wind's energy simply goes to pushing the boat sideways which would result in more heel and more leeway but not more forward motion.

I think this is what you were saying and I agree. However, even though I tried to say it as simply as possible I still ended up with three fairly complex paragraphs. And without benefit of diagrams this can still be misinterpreted. I think this thread has included a lot of saying the same thing in different ways.
04-05-2012 12:22 PM
DRFerron
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by Minnewaska View Post
Donna,

...

I will only address the apparent wind moving forward comment. I meant the angle of apparent wind on your wind indicator will move toward your bow (forward) as the boat begins to move.
Thank you.
04-05-2012 12:02 PM
RichH
Re: Why is beam reach (or near to it) the fastest point of sail?

Here's a graphical representation of the aerodynamic output of a sail in either a closehauled or beamreach orientation.

The "F" is the approx. resultant FORCE output due to the aerodynamics, "X" is the resultant in the direction of the boat and is the SOLE component vector that is responsible for boat speed. The distributed 'arrows' are the vectors of pressure gradient, the F vector is an approximate 'resultant' of all the 'arrows'.

You can see in the 'chart' between the two dwgs. that X1 is smaller than X2, beam reach is faster than closehauled.

The 'pumkin seed effect' AND the slip of the boat's lateral resistance toward leeward is ignored in this illustration, as 'relativistic math' would be quite complex for such a discussion; but however, since the Y vector is greater when closehauled - the pumpkin seed effect is actually GREATER when closehauled because the output in the Y direction (across the beam of the boat) is greater than when at a beam reach.

Simple Speak: just compare the 'length' of the resultant X Vectors in the drawing --- the beam reach produces a LARGER vector (all due to 'trigonometry').

FWIW - lower than beam reach (a high broad reach -- ~125°-135°) is actually faster on most boats because the sail is still or 'can be' in an aerodynamic flow regime and that resultant X vector is even larger than at when on a beam reach, and with less 'slip' , etc.
04-05-2012 04:53 AM
Minnewaska
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by DRFerron View Post
I have to think about this, thank you. I'm still not seeing how the apparent wind is blowing in the same direction the boat is moving (how I interpreted "apparent wind [moving] forward"). When I see the diagrams of true vs. apparent wind, the apparent wind is always at an angle slightly offset from true, but still not moving forward. Same with when I stick my hand out of the window of a moving car. I understand what apparent wind is, I just don't know the deeper theory behind it. I perhaps am not understanding your definition of "forward." But that's OK. I'll figure it out at some point.
Donna,

This thread has become something else. I'm not going to bother addressing a reply above that said I was wrong and then described my exact point in different words.

I will only address the apparent wind moving forward comment. I meant the angle of apparent wind on your wind indicator will move toward your bow (forward) as the boat begins to move.
04-05-2012 02:49 AM
Rockter
Re: Why is beam reach (or near to it) the fastest point of sail?

Brehm62 :

Let me see if I can explain this.
Wind Power = 1/2 * Rho * Velocity^3 * Area

Rho is the density of air which is about 0.0807 lbs / cubic ft.
If we are using cubic feet then we need Velocity in feet also.
Let's say the wind is blowing 6 knots.
That is about 6.87 mph or 10.08 feet / sec.
Let's just round that down to 10 ft/sec.
Let's say we have 200 square feet of sail area.

Okay: 1/2 * 0.0807 lbs / ft^3 * (10 ft/sec)^3 * 200 ft^2
0.04035 lbs / ft^3 * 1000 ft^3/sec^3 * 200 ft^2
8,070 lbs (ft^3 / ft^3) * ft^2/sec^3
8,070 lbs ft^2/sec^3
Now we have to divide by gravity which is 32 ft/sec^2
8,070 lbs ft^2/sec^3 / 32 ft/sec^2
252.1875 lbs (ft^2/ft) /(sec^3 *sec^2)
252.1875 lbs ft /sec
1 Horsepower is 550 ft lbs /sec so this is a little less than half a horsepower
0.46 HP


Well explained that man!
A word of caution.
My High School maths teacher woulde be jumping up and down like a hairdresser at THREE horrible sins within your explanation....

1 : Never, but never put units in the middle of equations. Write them above and below the equations if you wish but never within the equation.

2 : The unit you use for force should be the lbf ... do not forget the "f" after it.

3 : Worst sin of all, NEVER pluralise a unit. Units have, by definition, no plural. If you put an "s" after a unit, it denotes "second", so let's have none of this "lbs" stuff ever again.


Apart from three sins, it was a good explanation.

The thread has become rather confrontational. For me, I just put the sails up, choose a course, and trim the sails until they just stop flapping. My ship is fastest when the relative wind is ahead of beam.

One of the beauties of sailing, methinks. I am fastest when the wind is slightly against me.

Now there is a pleasant thought.

On my beloved Loch Ness, the wind is either right on my nose, or right on the back of my neck, as the land masses either side channel the wind right at me, or behind me. Unless I choose to sail across the Loch, that is. It's only about a mile wide.
04-05-2012 02:23 AM
brehm62
Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by reed1v View Post
Downwind with spinnaker can allow you to exceed wind speed
Yes, if you are using an asymmetric spinnaker on a broad run. If you are using a symmetric spinnaker running dead downwind then, no.

Of course, if the wind is not constant then whenever the wind drops you would still briefly be moving faster than the wind.
04-05-2012 02:19 AM
brehm62
Re: Why is beam reach (or near to it) the fastest point of sail?

Let me see if I can explain this.
Wind Power = 1/2 * Rho * Velocity^3 * Area

Rho is the density of air which is about 0.0807 lbs / cubic ft.
If we are using cubic feet then we need Velocity in feet also.
Let's say the wind is blowing 6 knots.
That is about 6.87 mph or 10.08 feet / sec.
Let's just round that down to 10 ft/sec.
Let's say we have 200 square feet of sail area.

Okay: 1/2 * 0.0807 lbs / ft^3 * (10 ft/sec)^3 * 200 ft^2
0.04035 lbs / ft^3 * 1000 ft^3/sec^3 * 200 ft^2
8,070 lbs (ft^3 / ft^3) * ft^2/sec^3
8,070 lbs ft^2/sec^3
Now we have to divide by gravity which is 32 ft/sec^2
8,070 lbs ft^2/sec^3 / 32 ft/sec^2
252.1875 lbs (ft^2/ft) /(sec^3 *sec^2)
252.1875 lbs ft /sec
1 Horsepower is 550 ft lbs /sec so this is a little less than half a horsepower
0.46 HP
This is all we have to power the boat; there isn't any other source. Will we get more than this if the boat is moving? No. This is all there is.

But ... (someone might protest) HP doesn't move the boat, force does. Well, this is true. So, would it be possible to get unlimited force from a fixed amount of HP? No. The amount of force available is dependent on the speed of the boat. Let's say we sail 10 ft per second perpendicular to the wind.

252.1875 lbs ft/sec / 10 ft/sec
= 25.21875 lbs of force
If we sail faster we get less force, not more.
This is all boilerplate physics and engineering. There is no way to get around this.

Could we go faster than 10 ft per second? Yes, if the total hull drag is less than 25 lbs of force at that speed then we'll have thrust left over to go faster. Could we go faster than 10 ft per second downwind? Clearly not because at 10 ft per second there would be no relative wind so force would be zero. Okay, so why can't we use relative wind to sail faster upwind than when reaching?

When we are reaching, 100% of the lift force of the sail is in the direction we want to go. So, under normal circumstances this would give us maximum speed.

When we angle upwind then only part of the lift force is in the direction we are traveling. But what about relative wind? We keep hearing how the sails generate force in regard to relative wind, not true wind. Now, being quite familiar with physics I know that it makes no difference if we view wind relative to the boat, a fixed point, or to the water. So, let's try relative wind.

The simplest way to illustrate this is to just make assumptions. Let's assume that the wind is blowing out of the north at 10 ft per sec. Let's assume that we can travel in any direction at the same 10 ft per sec. Now, I'm sure at this point someone is screaming that we can't just make that assumption. Actually we can. The way you normally do this process is you make an initial assumption and see what happens. Then, if you wanted a good model you would take the result and use it as the second starting point and repeat this over and over until the results changed very little. This is iteration. However, in this case, I don't think we'll have to iterate this over and over to see the truth. So, let's start with our assumptions and see what happens.

I'm going to take a constant vector for the wind of 10 ft per second blowing south and this is -90 degrees. East is 0 degrees, and North is 90 degrees. We aren't going use west but if we did it would work the same as east. Now, if the boat could travel at a constant velocity of the same 10 ft per second in any direction, what would the relative wind be at each point?

Angle of travel : Rel Wind velocity
90 : 20 (north, directly into the wind)
80 : 19.92
70 : 19.7
60 : 19.32
50 : 18.79
40 : 18.13
30 : 17.32
20 : 16.38
10 : 15.32
0 : 14.14 (east, perpendicular to the wind)
-10 : 12.86
-20 : 11.47
-30 : 10
-40 : 8.45
-50 : 6.84
-60 : 5.18
-70 : 3.47
-80 : 1.74
-90 : 0 (south, directly downwind)

We see exactly what we would expect: the velocity is highest when pointing directly into the wind and lowest when running dead down wind. If you are still having problems with the notion that the boat can travel at this speed in any direction then just imagine that we are starting from a tow or by using the motor.

Now with the boat speed and course fixed what could we actually get out of the sails? What we care about here is that lift (which gives us thrust) is perpendicular to the relative wind. And, if the relative wind is not perpendicular to our direction of travel then what we care about is that part of the wind vector that is perpendicular. So, what do we see?

Angle of travel : Relative size of lift component
90 : 0 (north, into the wind)
80 : 0.09
70 : 0.17
60 : 0.26
50 : 0.34
40 : 0.42
30 : 0.5
20 : 0.57
10 : 0.64
0 : 0.71 (east, perpendicular to the wind)
-10 : 0.77
-20 : 0.82
-30 : 0.87
-40 : 0.91
-50 : 0.94
-60 : 0.97
-70 : 0.98
-80 : 0.996
-90 : 1 (south, directly downwind)

Here we can see that as we head into the wind that the actual part of the wind that produces lift gets smaller and smaller. If you prefer percents just multiply by 100. However, to see what the total is we need to multiply this by the relative wind speed.

Angle of travel : Forward thrust from sails
90 : 0 (north, into the wind)
80 : 1.74
70 : 3.42
60 : 5
50 : 6.43
40 : 7.66
30 : 8.66
20 : 9.4
10 : 9.85
0 : 10 (east, perpendicular to the wind)
-10 : 9.85
-20 : 9.4
-30 : 8.66
-40 : 7.66
-50 : 6.43
-60 : 5
-70 : 3.42
-80 : 1.74
-90 : 0 (south, directly downwind)

Here we can clearly see that thrust is at a maximum near 0 degrees which is a track perpendicular to the wind (or a beam reach). For the sticklers here, this is a linear comparison. The actual force due to wind velocity is squared so the curve would be even sharper. However, squaring the numbers won't change the fact that it is largest near the middle or perpendicular to the wind.

Does this explain why we can sail fastest on a beam reach?

Note: I can show more intermediate steps if necessary. You get the combined wind velocity by using the Law of Cosines as the angle sweeps from 0 to 180 degrees. You get the vector angle by using the fact that for all triangles the interior angles always add up to 180 degrees. And, for an equilateral triangle the remaining two angles are the same so each one is half of 180 minus the sweep angle. Then you subtract the wind angle from the boat's track angle to get the relative wind angle. The lift is 90 degrees to this angle. Then you use cosine to get the portion that is in the same direction as the direction of travel. Then you multiply this by the relative wind velocity. Again, if you wanted to be more precise you would square the velocity first. However, this is negated in practice since drag is also squared.
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