where is the relative wind velocity on this boat???

[albertlee]

http://www.phys.unsw.edu.au/~jw/sailing.html

 

under the "How can boats sail faster than the wind? " section, it speaks about the relative wind speed.

 

What is that??

[swansont]

http://www.phys.unsw.edu.au/~jw/sailing.html

 

under the "How can boats sail faster than the wind? " section' date=' it speaks about the relative wind speed.

 

What is that??[/quote']

 

The wind speed as measured by the boat. If the wind speed measured on the dock is 20 mph, and the boat is moving 20 mph in the direction of the wind, the relative wind speed will be zero - on the boat you will stop feeling the wind. If the boat is moving 15 mph, the relative wind speed will be 5 mph.

[albertlee]

Hmm.... so, at what speed is the boat moving then??

[J.C.MacSwell]

Hmm.... so, at what speed is the boat moving then??

 

deduct the apparent wind vector from the true wind vector and you get the boat speed (velocity) vector.

[Martin]

http://www.phys.unsw.edu.au/~jw/sailing.html

 

under the "How can boats sail faster than the wind? " section' date=' it speaks about the relative wind speed.

 

What is that??[/quote']

 

Tom Swanson and MacSwell have already answered this completely, especially if you are familiar with adding and subtracting vectors. But maybe we can have some more fun with this. the link you gave links to a site about "18 footer skiff" sailboats. These, it is claimed, often go faster than the wind. Perhaps more surprisingly, when a skipper wants to sail "downwind" he may find he can go faster sailing a zig-zag course instead of going straight in the same direction as the wind.

 

This is explained pretty well at the site you linked to, albertlee. But to understand basic sailing physics you must understand vector arithmetic so maybe we should look at one or two examples.

 

imagine an xy plane with unit vector (1,0) pointing east and

(0,1) pointing north.

so if the wind vector, for landsman, is (10, -10) that means wind blows 10 mph east and 10 mph south

I think you see that it blows from the NW at 14 miles per hour?

that is, 14.14 the square root of 200?

 

now suppose the boatspeed vector is (10,0)-----it is sailing east at 10 mph.

then the wind that the boatman feels must be (10, -10) - (10,0) = (0, -10)

So the boatman feels a 10mph north wind.

 

these 18foot skiff boats have very little drag, so imagine that the boat speeds up some. It is now going (20,0)-----20 mph to the east.

the wind that the boatman feels, and that the sails feel, is now

(10, -10) - (20, 0) = (-10, -10)

So the boatman feels a 14mph North East wind.

 

if that is all obvious to you, please do not be offended (I don't know if you are familiar with vector arithmetic or not)

 

this link albert gave is pretty good:

http://www.phys.unsw.edu.au/~jw/sailing.html

it shows these interesting high-performance skiffs

http://www.18footer.org/

[albertlee]

So, why the final boat velocity isn't wind velocity + initial boat velocity??

 

Isn't that more logical??

[YT2095]

the same way the tip of a Whip breaks the speed of sound with a simple arm movement

[J.C.MacSwell]

So' date=' why the final boat velocity isn't wind velocity + initial boat velocity??

 

Isn't that more logical??[/quote']

 

Why would it be?

 

Nothing in the above is an attempt to analyse the forces, accelerations final velocity etc. The relative wind speed can change, of course, with any change in the true wind or boat velocity. (speed and direction)

[albertlee]

Ok, so, what sort of conclusion does that section say??

 

Albert

[Martin]

... analyse the forces, accelerations final velocity etc...

 

albert listen to what macswell is saying. many basic physics problems, including the sailboat, boil down to newton 2nd and 3rd laws----"F = ma" and "equal and opposite" force law.

 

JC mentions the words "force" and "acceleration". he wants you to use F=ma.

He uses the phrase "analyze the forces"----he wants you to draw a vector picture in the xy plane.

 

the boatman could make the boat go without sails if he had a box of rocks and he would throw the rocks one by one in a backwards direction

(to make the boat go east, he would throw the rocks west, off the back of the boat)

 

to use the wind, the boatman needs only to deflect the wind to a more backwards direction. it is like "throwing air" off the back of the boat, instead of rocks.

 

deflecting the wind, from a more sideways to a more backwards direction, generates a forwards force (also sideways force but the keel of the boat opposes sideways motion) and as long as there is any forwards force

(beyond the "drag" forces like e.g. friction of the boat hull in the water) the boat will accelerate.

 

all that is necessary is to have very small "drag" force, resisting motion, and to get some small forwards force by deflecting the air in a backwards direction.

 

have you ever heard of ICE-BOATS? these boats sail on frozen lakes in winter. they have been able to go more than 100 mph. even a moderate wind like 15 mph can make these boats go 50, 60 mph. they travel on SKATE-BLADES so the drag is very small.

 

So' date=' why the final boat velocity isn't wind velocity + initial boat velocity??

 

Isn't that more logical??[/quote']

 

if you want to calculate numbers you must learn to use F = ma and draw force-diagrams with vectors in the xy plane. you cannot reason it only with words. verbal logic is not enough. you need trigonometry

 

to use F=ma you must estimate the mass of the air being deflected, and the backwards component of acceleration being given to the air by the sail.

naturally angles are involved.

it would be convenient to work in the coordinates of the boatman (less complicated) and so you will draw the wind vector with the direction which the boatman sees

[albertlee]

The most confusion, is, why the ICE-BOAT can travel even faster than wind??

[J.C.MacSwell]

The most confusion, is, why the ICE-BOAT can travel even faster than wind??

 

It basically comes down to it's thrust to drag efficiency. It reaches a speed where the drag equals the thrust. Since the apparent wind is shifting further and further forward it is harder and harder to gain thrust to match the increasing drag (mostly aerodynamic drag on an iceboat) inspite of the increase in available energy.

Due to the efficiency of the iceboat this can still come at over twice the wind speed.

[albertlee]

you mean the wind force exerts on the ice-boat?? and its velocity accelerates??

 

Secondly, how do you calculate the mass of the air exerting on the boat's sail?? since the volume is indeterminable eventhough you know its density.

 

Thirdly, again, what is that section in my website talking about relative wind velocity trying to conclude, stress on?? It only says, if the boat goes against wind, when the boat velocity increases, the relative wind increases. Whereas if the boat goes in the direction of the wind, when the boat velocity increases, the relative wind decreases.

 

Albert

[Martin]

 

Secondly' date=' how do you calculate the mass of the air exerting on the boat's sail??

 

Albert[/quote']

 

good question albert.

imagine the boat goes east and the boatman feels wind come from north at 10 meter per second. Suppose he puts sail at 45 degree angle so as to deflect wind backwards. (simplified picture)

 

Now do the analysis for each square meter of sail.

Each sq. meter of sail only shows 0.707 sq. meter cross-section to the wind (because sail is not FACING wind, it is angled 45 degrees)

 

the wind is coming at 10 meters per second, so each square meter of sail deflects 7 cubic meters of air per second

 

and one cubic meter of air is about one kilogram mass

 

therefore each square meter of sail deflects 7 kilograms of mass per second.

 

the momentum of a 7 kilogram mass moving 10 meters per second is 70 kilogram meters per second. force equals time rate of change of momentum.

 

so each second, this area of sail gives to the air a backwards momentum of 70 kilogram meter per second

 

by Newton Third "equal opposite" Law this area of sail experiences a 70 Newton forwards force (air is pushed backwards towards back of boat, sail is pushed forwards towards front). Backwards push on air must equal forwards push on boat.

 

Suppose the sail is 20 square meters, then the total forwards force is 1400 Newtons

and until this force is balanced by "drag" (aerodynamic hydrodynamic "friction") this forwards force will accelerate the boat.

 

the drag increases with speed, so the boat will go faster and faster until the drag is enough to balance the forwards force, then the boat will stop accelerating and that will be the boat's top speed at the conditions of this example.

 

To describe the example in vector notation, suppose the landsman sees the wind go (10, -10)

and the boat go (10,0) meters per second

so the boatman sees the wind go (0, -10) meters per second.

Boat going east, sees wind coming from north.

To find the sail force, what matters is what the boatman sees. He sees the wind coming at 90 degrees from his course so he puts the sail at 45 degree angle. this is extremely oversimplified. please excuse any careless haste

[albertlee]

does the mementum, 70kgm/s, neccessarily mean a force of 70N??

[Martin]

does the mementum, 70kgm/s, neccessarily mean a force of 70N??

 

so each second, this area of sail gives to the air a backwards momentum of 70 kilogram meter per second

 

think of the sail giving backwards momentum to the air at a certain rate

 

this rate is 70 kilogram meter per second per second

 

each second, the sail gives to the air an amount of momentum which is 70 kg m/s

 

it does this by deflecting the air and diverting some sideways momentum that the air had into backwards directed momentum

 

OK, now what, by definition, is 70 kg m/s² ?

 

Albert, what physical quantity in what units is

70 kilogram meter per second per second?

[albertlee]

70n

 

[Martin]

70n