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Old Nov 21st 2010, 09:26 AM   #1
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VMG greater than the speed of the wind

An often quoted statement about sailing is causing me some puzzlement, there are even pages on the web claiming data that back the statement explained below. I would be interested to know what others think from a physics standpoint. I think it has the makings of a very interesting debate.

It is a well known fact that a sailing boat can achieve a velocity in excess of the speed of the wind. Ice boats can achieve a velocity multiple times that of the wind speed. The velocity made good (VMG) is usually measured in the direction against the wind or with the wind. Optimal boat speed is gained at somewhere between 25 degrees and 40 degrees to the apparent wind which is the vector sum of the headwind created by boat speed and the true wind.

If a boat starts from zero speed, at a direction of 45 degrees down wind, the direction of the apparent wind will be from 135 degrees to the front of the boat. Once the boat is traveling at wind speed, the wind will be coming from 67.5 degrees from the front of the boat. Since the boat only sees the apparent wind and not the true wind, and considering that the idealised situation of no friction is closely approximated in the case of and ice boat. Is is possible as is often claimed that an ice boat is able to travel faster than the wind. Since the boats can control there position by tacking, it is equivalent to stating that a balloon is let go to travel freely with the wind and that the ice boat will arrive at a designated point directly downwind before the balloon does.
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Old Nov 22nd 2010, 10:40 PM   #2
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Boat and Windspeed

If the area of the sail is large enough, the initial gust of wind might be able to transfer a large amount of momentum to the boat. This could lead to a high initial acceleration which might cause the boat to momentarily move faster than the wind. But this cannot continue to happen if the wind continues to blow at the same speed because if the boat tries to move faster in the wind than the wind itself, the wind will effectively be moving backwards compared to the boat which in effect will cause the boat to slow down. Thus the boat could achieve wind speed after which there is no more force exerted on the sail, and the boat continues in its state of........(Newton takes over). I however don't know anything about the angles involved.
In a car etc this problem will not arise as the propulsion system there is different. Here,the propulsion is by the wind itself.
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Old Nov 24th 2010, 05:25 AM   #3
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There is a lot of references in this article:
Sailing faster than the wind - Wikipedia, the free encyclopedia
Especially
http://www.nalsa.org/Articles/Cetus/...ance-Cetus.pdf

Originally Posted by YellowPeril View Post
I would be interested to know what others think from a physics standpoint.
No problem with physics. See vector diagram below.



And this animation which also contains a force diagram:

YouTube - DOWNWIND VMG GRATER THAN WINDSPEED (STREAM LINE)

The squeezed wedge analogy:

YouTube - DOWNWIND VMG GRATER THAN WINDSPEED (WEDGE)

It gets even more baffling if you go DIRECTLY downwind, faster than the wind. Impossible with conventional sail craft but possible with this:



See this for more information:
DowindRecord

Last edited by A.T.; Nov 24th 2010 at 05:35 AM.
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Old Nov 25th 2010, 03:06 PM   #4
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Vmg

Those ice boats are pretty amazing.

After giving the matter some thought, I am convinced that an iceboat at least can have a vmg faster than the wind and that this is easy to show graphically. A sail boat should have significantly more drag making this feat more difficult

In the same way the vehicle pictured by AT must turn the propellor backward to maintain a constant apparent wind (think of a corkscrew). What puzzles me is how the vehicle overcomes the torque that this would obviously generate. A sailboat overcomes this problem by using a heavy keel.
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Old Nov 25th 2010, 10:27 PM   #5
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Originally Posted by YellowPeril View Post
Those ice boats are pretty amazing.
After giving the matter some thought, I am convinced that an iceboat at least can have a vmg faster than the wind and that this is easy to show graphically. A sail boat should have significantly more drag making this feat more difficult
Some yachts achieve a downwind VMG of 2 x true wind on water. From wiki:
During the first race of the 2010 America's Cup, the winning yacht USA sailed 20 nautical miles (37 km) to windward in 1 hour 29 minutes, in winds of 5 to 10 knots (19 km/h). Thus its velocity made good upwind was about 1.8 times windspeed, consistent with being able to sail about 13 degrees off the apparent wind when sailing upwind. It sailed 20 nautical miles (37 km) downwind in 1 hour 3 minutes, so its velocity made good downwind was about 2.5 times windspeed, consistent with being able to sail about 14 degrees off the apparent wind when sailing downwind.
Originally Posted by YellowPeril View Post
In the same way the vehicle pictured by AT must turn the propellor backward to maintain a constant apparent wind (think of a corkscrew).
Once you understand how downwind VMG greater than true wind works, it is easy to see how you can go directly downwind faster than the wind wind. The trick is to have only the airfoils tacking (what the propeller blades do), while the hull goes directly downwind. This animation shows that transition:

YouTube - Sail_to_Prop.mov

Originally Posted by YellowPeril View Post
What puzzles me is how the vehicle overcomes the torque that this would obviously generate. A sailboat overcomes this problem by using a heavy keel.
They have an asymmetric axis.



On water you would need a multi-hull design for stability. A displacement hull with a heavy keel would have too much water drag to go DDWFTTW.

See Ride Like the Wind (only faster) for more information.
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Old Nov 28th 2010, 12:45 PM   #6
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Dwfttw

A.T.

Could you elaborate on how the asymmetric axis design deals with the torque generated by the turning blades.
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Old Nov 29th 2010, 02:42 AM   #7
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Originally Posted by YellowPeril View Post
A.T.
Could you elaborate on how the asymmetric axis design deals with the torque generated by the turning blades.
The torque that turns the propeller, has an opposite reaction torque that tries to turn the cart the other way (roll it on the side). That reaction torque is countered by the forces on the back wheels. The asymmetric axis distributes the load on the back wheels more evenly.

Or did you mean the torque generated by the thrust of the propeller (and the breaking force on the wheels)? That torque pushes the nose of cart down and is simply countered by the increasing force on the front wheel.

On water all this stabilizing would be much more difficult. You would need automatically controlled hydro-foils or something like this.
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Old Nov 29th 2010, 11:47 AM   #8
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Vmg

A.T.

I was actually unsure what the greater force was, the propulsion force or the breaking force caused by the torque on the wheels (due to moving the blades). However after posting this post, I realised that vector diagram number 4 (with the red sail force arrow) can be used to show graphically that the forward propulsive force is about twice as strong as the breaking force generated by the torque of the DWFTTW vehicle.

Thanks for your help, the vector diagrams really did help a lot bring the point home to me.
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Old Nov 29th 2010, 01:52 PM   #9
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Originally Posted by YellowPeril View Post
A.T.
I was actually unsure what the greater force was, the propulsion force or the breaking force caused by the torque on the wheels (due to moving the blades).
Obviously, to accelerate above windspeed the thrust of the propeller must be greater, than the braking force at the wheels needed to turn the propeller. This is only possible, if the air moves slower relative to the cart than the ground. Like a gear-box the cart converts a low braking force coming in fast at the wheels, into a large thrust force coming out slow at the propeller.

Originally Posted by YellowPeril View Post
However after posting this post, I realised that vector diagram number 4 (with the red sail force arrow) can be used to show graphically that the forward propulsive force is about twice as strong as the breaking force generated by the torque of the DWFTTW vehicle.
The ratio depends on the configuration of the vehicle and it's current speed. In the following animation at about 0:50 there is an example vector diagram showing that the braking force is less than the thrust, above wind-speed:

YouTube - Directly Down & Up Wind Faster than the Wind
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