USCG Exam Question: Apparent Wind

[OldFrog75]

Having had 24 hours to contemplate this question, I remain in the camp that although we trim our sails for apparent wind, point of sail definitions are based upon true wind.

When in irons, the bow is facing directly head to true wind. When sailing DDW, the stern is facing directly to true wind. When on a beam reach, true wind is 90 degrees off the beam.

It stands to reason that close reach would be defined as true wind ahead of the beam while broad reach and running would be defined as true wind abaft of the beam.

I couldn't find a single source that defines points of sail as based upon a boat relative to apparent wind. I think it would complicate things since any POS definition would also require a further explanation of the definition of apparent wind.

[tstano]

Originally Posted by goboatingnow

All sailing directions refer to the wind as experienced on the boat, Since only when stopped can you experience " True " wind. It follows that all wind you experience on the boat is apparent wind. Thats what every boat sails in. it never sails in the " true wind"

I thought I read there were some cats that could match wind speed traveling DDW.

[jheldatksuedu]

Originally Posted by C.Karo

True wind or direction can only be experienced if you are stationary. If you have movement, it becomes relative or apparent wind. Any Questions? No, forget it. I am finished. I am heading out to get my fill of apparent wind....

I you sail your boat on a beam reach (measured 90 degrees to apparent wind) you are sailing downwind, and a beam reach is when you are not sailing up or down wind, you are not losing or gaining anything to the wind. That's what makes it a precise point of sailing, you are precisely not gaining or losing. It has to be measured against true wind, even if you can't do it easily while on a sailboat, it can still be defined that way and it's the only way it makes any sense to define it. Go sail across a bay on your beam reach and then sail back again on a beam reach and you will find yourself significantly downwind when you return. The better your boat the more downwind you will be. Do the same thing measured against true wind (even if it is hard to figure out your exact direction to head) and you will return to "precisely" where you started.

From here on out when for some reason I don't want to gain or lose to the wind, I'll choose the more difficult to judge beam reach having some wind in my face (basically equal to the boat speed) knowing that I'm real close to a true precise beam reach.

[cal40john]

Originally Posted by tstano

I thought I read there were some cats that could match wind speed traveling DDW.

With regular sails, no. If you're travelling DDW at TWS the apparent wind is zero, therefore no force on the sails.

Cats can make better than TWS VMG downwind by reaching with regular sails.

Theoretically you can sail DDW faster than the wind using a propeller in the air driving a propeller in the water (Propeller spinning is making apparent wind). There have been examples made that work on land (propeller driving wheels), but I have yet to have heard of this on a boat working to exceed TWS travelling DDW.

[thinwater]

Originally Posted by OldFrog75

Having had 24 hours to contemplate this question, I remain in the camp that although we trim our sails for apparent wind, point of sail definitions are based upon true wind.

When in irons, the bow is facing directly head to true wind. When sailing DDW, the stern is facing directly to true wind. When on a beam reach, true wind is 90 degrees off the beam.

It stands to reason that close reach would be defined as true wind ahead of the beam while broad reach and running would be defined as true wind abaft of the beam.

I couldn't find a single source that defines points of sail as based upon a boat relative to apparent wind. I think it would complicate things since any POS definition would also require a further explanation of the definition of apparent wind.

This is the first comment that gets to the heart of the problem. The vector business is triva an obvious ONLY if we have one definition of beam reach. I have seen respected references for each definition:

1. Beam reach = apparent wind on beam.
2. Beam reach = true wind on beam.

As a former performance cat sailor (my current cat cannot exceed true wind speed) I can tell you that definition 1 is basically useless on a fast cat; you are close hauled most of the time and NEVER keep the apparent wind less than 30 degrees forward of the beam (even while going quite deep). In fact, with the reacher up it is common for good crews to tack down wind, never jibing. We've all seen this. In other words, AC boats broad reach down wind, very, very fast.


Thus, while using definition 2 may require a bit more thought on the part of the sailor, it is a far more useful definition. It also suggests the wave direction, which def 1 does not.



Without the CG defining the term it is simply a poor question. I would have guessed "4", assuming the more common and conservative definition, knowing all the while it was a poorly written question.


The correct question would have defined the apparent wind angle as 90 degrees rather than relying on archaic and ambiguous terminology.

[OldFrog75]

Originally Posted by thinwater

This is the first comment that gets to the heart of the problem. The vector business is triva an obvious ONLY if we have one definition of beam reach. I have seen respected references for each definition:

1. Beam reach = apparent wind on beam.
2. Beam reach = true wind on beam.

Thus, while using definition 2 may require a bit more thought on the part of the sailor, it is a far more useful definition. It also suggests the wave direction, which def 1 does not.

Without the CG defining the term it is simply a poor question. I would have guessed "4", assuming the more common and conservative definition, knowing all the while it was a poorly written question.

The correct question would have defined the apparent wind angle as 90 degrees rather than relying on archaic and ambiguous terminology.

It might be reverse engineering but I think the fact answer #1 is false implies the CG defines beam reach as relative to true wind.

[jheldatksuedu]

Originally Posted by cal40john

Theoretically you can sail DDW faster than the wind using a propeller in the air driving a propeller in the water (Propeller spinning is making apparent wind). There have been examples made that work on land (propeller driving wheels), but I have yet to have heard of this on a boat working to exceed TWS travelling DDW.

I've always wanted to build a "sail" boat with a savonius windmill and a transmission driving a propeller, you should be able to go straight into the wind, or any other direction, downwind would still be lousy. The "blades" would be cloth sails on a framework so you could reduce sail, otherwise it sure would be a problem in a storm.

I had a friend in Kansas, that lived off the grid with all kinds of windmills making power to run his house.

[fryewe]

You guys are all trying to confuse me, aren't you?

Here's the way I thought through this...

I assumed I was in a car, sitting still, rather than a boat, with the true wind directly on my left, perpendicular to the car. The true wind and apparent wind are the same speed and direction, directly from my left.

I accelerate to a constant speed in the direction I am pointing, creating a wind component due to my speed directly on the nose. The true wind hasn't changed, but the vector sum of the true wind and the wind created due to my motion, which are perpendicular to each other, is the hypotenuse of a right triangle whose magnitude has to be larger than either of the component winds. This is the apparent wind.

The direction of that vector (the apparent wind direction) has to be forward of the beam, because it is the sum of a wind on the beam and a wind on the nose. It has to be between the two.

Now, if I transport myself from my imaginary car to my boat, and am sailing...I would not be on a beam reach, but a close reach, with the apparent wind forward of the beam. If I want to sail on a beam reach, I would turn away from the apparent wind, placing the apparent wind on the beam, which places the true wind abaft the beam.

The true wind now becomes the hypotenuse of the triangle, and has to be greater than the apparent wind.

So, from the USCG question...
...you are sailing on a beam reach which means:
-the apparent wind is on your beam
-the true wind is abaft your beam, and
-if the measured (apparent) wind speed is 25 knots, the true wind speed is greater than 25 knots.

The turn to make isn't a one step computation, even if true wind is constant in both speed and direction, because as you turn the self generated speed vector changes in both speed and direction.

[med]

Originally Posted by fryewe

You guys are all trying to confuse me, aren't you?

Here's the way I thought through this...

I assumed I was in a car, sitting still, rather than a boat, with the true wind directly on my left, perpendicular to the car. The true wind and apparent wind are the same speed and direction, directly from my left.

I accelerate to a constant speed in the direction I am pointing, creating a wind component due to my speed directly on the nose. The true wind hasn't changed, but the vector sum of the true wind and the wind created due to my motion, which are perpendicular to each other, is the hypotenuse of a right triangle whose magnitude has to be larger than either of the component winds. This is the apparent wind.

The direction of that vector (the apparent wind direction) has to be forward of the beam, because it is the sum of a wind on the beam and a wind on the nose. It has to be between the two.

Now, if I transport myself from my imaginary car to my boat, and am sailing...I would not be on a beam reach, but a close reach, with the apparent wind forward of the beam. If I want to sail on a beam reach, I would turn away from the apparent wind, placing the apparent wind on the beam, which places the true wind abaft the beam.

The true wind now becomes the hypotenuse of the triangle, and has to be greater than the apparent wind.

So, from the USCG question...
...you are sailing on a beam reach which means:
-the apparent wind is on your beam
-the true wind is abaft your beam, and
-if the measured (apparent) wind speed is 25 knots, the true wind speed is greater than 25 knots.

The turn to make isn't a one step computation, even if true wind is constant in both speed and direction, because as you turn the self generated speed vector changes in both speed and direction.

All very good - but you are using a definition of beam reach which does not correspond to that which almost everyone else is using.

What most folk understand by a beam reach is that you are heading at 90 degrees to the direction of the true wind (as opposed to any other kind of wind). Even ASA sailing schools all have that diagram on their websites which show that.

The direction of the true wind of course does not change just because you are going faster or slower.

[Cadence]

Originally Posted by med

All very good - but you are using a definition of beam reach which does not correspond to that which almost everyone else is using.

What most folk understand by a beam reach is that you are heading at 90 degrees to the direction of the true wind (as opposed to any other kind of wind). Even ASA sailing schools all have that diagram on their websites which show that.

The direction of the true wind of course does not change just because you are going faster or slower.

It is all academic, 26.25. Have no idea it's practical value.

[El Pinguino]

Has anybody mentioned 'ground wind' yet? Just wundring ...

[goboatingnow]

Originally Posted by med

All very good - but you are using a definition of beam reach which does not correspond to that which almost everyone else is using.

What most folk understand by a beam reach is that you are heading at 90 degrees to the direction of the true wind (as opposed to any other kind of wind). Even ASA sailing schools all have that diagram on their websites which show that.

The direction of the true wind of course does not change just because you are going faster or slower.

Sorry , a beam reach is where the sails are trimmed to roughly the wind coming at right angles to the boat as you sail along

That's apparent wind.

Points of sail refer to that experienced by the boat -again that is apparent wind.

On a boat moving , true wind is a mathematical construct. It doesn't exist in reality , so the boat doesn't sail in it.
Dave


Sent from my iPad using Cruisers Sailing Forum

[beebopbogo]

As you can see by these two vector diagrams (One with the boat 90 degrees to true wind, the other 90 to apparent) the coast guard defines a beam reach as relative to the apparent wind.

If you choose to define a beam reach relative to true, you will get the answer wrong on the test.

Here's a reminder of the original question as spit out by the USCG test. They clearly state number 4 is correct:

Question 8. You are sailing at 8 knots on a beam reach in an apparent wind of 25 knots. Which statement is TRUE?:
1. The true wind is a little abaft your beam, at just under 25 knots.
2. The apparent wind at the top of your mast will be slightly stronger than 25 knots and slightly farther forward than the wind at deck level.
3. If you turn to a close reach, the apparent wind will reduce in strength.
4. None of the above are true.
You chose 1. The CORRECT answer was number 4 , None of the above are true..

[monte]

but theyre different boats. The top one looks like a motorsailor and the bottom one is clearly an asymetrical dinghy..

[beebopbogo]

Hey don't knock the asym dinghy! The pointing ability on those one leg races is AMAZING!