Merged DDWFTTW in a “Real” Wind

[humber]

That is not a cite, and it is not a land yacht. And you provided no information about the wind.

Now again, a land yacht that can beat the wind while staying within 20 degrees of DDW in a <= 15 knot wind.

Look up any data on DN class ice-boats. They readily do that.

Right off the bat, the terms are not defined. But so far I have to guess that K = density * Cd * Frontal area. And if so, okay so far.... except the lack of definitions.

Aww...
Fd = 1/2*Cd*A*p*V^2
Fd = (1/2*Cd*A*p)V^2
=k*V^2

Power = force*velocity.
Pcart = Fd*Vcart

Vhead = Vwind - Vcart*.

Now try again

F_Drag = K*Vhead^2
Power_Drag = K*Vhead^3

Pcart = K*Vhead^2*Vcart.

ETA:*
http://en.wikipedia.org/wiki/Apparent_wind#Instruments
Definition of apparent wind
The Apparent wind is the wind experienced by an observer in motion and is the relative velocity of the wind in relation to the observer.
Apparent wind velocity is the vector sum of the true wind and the headwind an object would experience in still air. The headwind velocity in still air is inverse of the object's velocity, therefore the apparent wind can also be defined as a vector subtraction: the Velocity of the wind minus the Velocity of the object.

 

[Llyricist]

Look up any data on DN class ice-boats. They readily do that.

Once again, an ice boat is not a land yacht. And you fail to support your claim. Again.

Aww...
Fd = 1/2*Cd*A*p*V^2
Fd = (1/2*Cd*A*p)V^2
=k*V^2

Damn, forgot the 1/2

Power = force*velocity.
Pcart = Fd*Vcart

Wrong, as I already showed. Pcart is useless, what you have computed is Power_drag in an FOR where the ground is stationary. Which is irrelevant to a force analysis, F_Drag is F_Drag.

ETA:*
http://en.wikipedia.org/wiki/Apparent_wind#Instruments
Definition of apparent wind
The Apparent wind is the wind experienced by an observer in motion and is the relative velocity of the wind in relation to the observer.
Apparent wind velocity is the vector sum of the true wind and the headwind an object would experience in still air. The headwind velocity in still air is inverse of the object's velocity, therefore the apparent wind can also be defined as a vector subtraction: the Velocity of the wind minus the Velocity of the object.

I know what apparent wind is, but I bet you don't.

 

[Subduction Zone]

Look up any data on DN class ice-boats. They readily do that.

Still not a cite humber. Wrong, try again.

 

[humber]

Once again, an ice boat is not a land yacht. And you fail to support your claim. Again.

The Detroit News and DN class ice-boats in general have a Cd of around .02.
and edges to increase that for lateral forces.
You have .008, so what excuse do you have this time?

Wrong, as I already showed. Pcart is useless, what you have computed is Power_drag in an FOR where the ground is stationary. Which is irrelevant to a force analysis, F_Drag is F_Drag.

You have shown nothing. You have no particular expertise in FOR, that can allow you to simply state your opinion.

Frr = Vcart*m*g*Crr,
Power = Frr*Vcart^2

Frr is a force opposing the cart, just as Fd is, and must be met by force at the wheel. Therefore, Pcart, the power necessary to maintain motion against the drag is:

F_Drag = K*Vhead^2
Pcart = K*Vhead^2*Vcart

 

[Llyricist]

The Detroit News and DN class ice-boats in general have a Cd of around .02.
and edges to increase that for lateral forces.
You have .008, so what excuse do you have this time?

Still waiting for evidence of land yachts beating the wind while staying within 20 degrees of DDW in <= 15 knot winds.

I had a Cd of .28 btw, not that it is relevant.

You have shown nothing. You have no particular expertise in FOR, that can allow you to simply state your opinion.

I don't need any expertise, I'm either right or not, in this case, I am right.

Frr = Vcart*m*g*Crr,

Wrong, there is no V in that formula.
So this is wrong too:

Power = Frr*Vcart^2

No squared there.

Frr is a force opposing the cart, just as Fd is, and must be met by force at the wheel.

Wrong, it must be met by an opposing force, period.

Therefore, Pcart, the power necessary to maintain motion against the drag is:

F_Drag = K*Vhead^2
Pcart = K*Vhead^2*Vcart

You are adding an unnecessary element, and leaving one out. There is no need to convert any forces to power unless your aim is to obfuscate.

The force needed to oppose the resistance is equal to the sum of the forces of resistance. Period.

 

[humber]

Still waiting for evidence of land yachts beating the wind while staying within 20 degrees of DDW in <= 15 knot winds.
I had a Cd of .28 btw, not that it is relevant.

The Cd applies as was clear from the context, to the coefficient of friction to the ice, and so corresponds to Frr.
The cart data is readily available, and it's existence or not, does nothing to say that the cart works ddw. It is never for one second, ddw.
The speeds obtained are due to winds that can exceed 20mph, and change rapidly over very short time scales.
Steve Smith again:

One other possibility is worth discussing — that it could be the time-variation of the wind direction, rather than simply the fact that the wind is not DDW.

The trial meets has both criterion. You still like to ignore the fact that the prop - the device that you say powers the cart - spends 50% of its time in winds that are above the line where both windspeed and direction are recorded .

I don't need any expertise, I'm either right or not, in this case, I am right.

And your credentials are?

Frr = Vcart*m*g*Crr,
Power = Frr*Vcart^2

Wrong, there is no V in that formula.
Power = Frr*Vcart^2

You are dying on your feet.

Frr = Vcart*m*g*Crr,
P = F*V
P = Frr*Vcart
P = m*g*Crr*Vcart^2

You are adding an unnecessary element, and leaving one out. There is no need to convert any forces to power unless your aim is to obfuscate.
The force needed to oppose the resistance is equal to the sum of the forces of resistance. Period.

Is the power.

http://en.wikipedia.org/wiki/Drag_(physics)#Power
Power
The power required to overcome the aerodynamic drag is given by:

[latex]P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
[/latex]

F_Drag = K*Vhead^2
The force generated on the cart wrt the wind.

The cart must oppose that force, but is traveling at Vcart.
Power = F_drag*Vcart
The power required by the cart wrt the ground.

 

[Llyricist]

The Cd applies as was clear from the context, to the coefficient of friction to the ice, and so corresponds to Frr.
The cart data is readily available, and it's existence or not, does nothing to say that the cart works ddw. It is never for one second, ddw.
The speeds obtained are due to winds that can exceed 20mph, and change rapidly over very short time scales.
Steve Smith again:

One other possibility is worth discussing — that it could be the time-variation of the wind direction, rather than simply the fact that the wind is not DDW.

The trial meets has both criterion. You still like to ignore the fact that the prop - the device that you say powers the cart - spends 50% of its time in winds that are above the line where both windspeed and direction are recorded .


And your credentials are?


You are dying on your feet.

Frr = Vcart*m*g*Crr,
P = F*V
P = Frr*Vcart
P = m*g*Crr*Vcart^2



Is the power.

http://en.wikipedia.org/wiki/Drag_(physics)#Power
Power
The power required to overcome the aerodynamic drag is given by:

[latex]P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
[/latex]

F_Drag = K*Vhead^2
The force generated on the cart wrt the wind.

The cart must oppose that force, but is traveling at Vcart.
Power = F_drag*Vcart
The power required by the cart wrt the ground.

Still waiting for evidence of land yachts beating the wind while staying within 20 degrees of DDW in <= 15 knot winds.

 

[humber]

Still waiting for evidence of land yachts beating the wind while staying within 20 degrees of DDW in <= 15 knot winds.

Then you will have to do your research.

There is still this outstanding.

Frr = Vcart*m*g*Crr,
P = F*V
P = Frr*Vcart
P = m*g*Crr*Vcart^2

Is the power.

http://en.wikipedia.org/wiki/Drag_(physics)#Power
Power
The power required to overcome the aerodynamic drag is given by:

[latex]P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d<br />[/latex]

F_Drag = K*Vhead^2
The force generated on the cart wrt the wind.

The cart must oppose that force, but is traveling at Vcart.
Power = F_drag*Vcart

An iceboat can achieve 2WS at 30 degrees. Check any polar chart

But, you have no claim that may be comparable, because the prop spends 50% of its time in fasterand constantly accelerating wind.

That is more than enough to drive the very slow and pushed cart to 1.8WS.

It does that in winds that are at least 20 degrees RMS, and at least 15mph.
The windspeed is also wrong, because it is not an RMS calculation, and does not take into account the wind above the hub.

The cart is a monumental under-achiever.
The idea is ancient folk-lore, the TM is plagiarized, and does not represent a cart in the wind. The small cart is a recorded failure, the math and theory are wrong, and the trial a sham.

 

[Llyricist]

Then you will have to do your research.

I can't prove a negative. I have looked and looked, and not found the evidence. You have the positive claim, that it CAN be done, you have to support it.

 

[humber]

I can't prove a negative. I have looked and looked, and not found the evidence. You have the positive claim, that it CAN be done, you have to support it.

No, I am saying that the cart did not achieve what an iceboat can.

This cart is run in winds that are higher than measured, and angles that are greater than measured, so there really is nothing to explain.

If you want to make comparisons with a sail boat, then you will need to support that argument, and then still face the differences.
To propose that comparison, is to admit that the cart is not traveling ddw. Why should anyone agree that is acceptable? Test the ddw cart, ddw, and not a great deal from it, and then challenge skeptics to prove otherwise.
That also leaves the quality of the measurements in the hands of those so incompetent, as not to be able to arrange a ddw trial, or get drunk in a brewery.

One reason the cart cannot be a ddw prop cart, is that it fails a power analysis.

One reason is this:

Cat got your tongue?

Frr = Vcart*m*g*Crr,
P = F*V
P = Frr*Vcart
P = m*g*Crr*Vcart^2

Is the power to overcome the rolling losses.


http://en.wikipedia.org/wiki/Drag_(physics)#Power
Power
The power required to overcome the aerodynamic drag is given by:

[latex]P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d<br />[/latex]

v^3 is When v = ground speed = windspeed, but the cart moves wrt respect to both the wind and the ground, so V cannot be the same for both frames.

F_Drag = K*Vhead^2 ( Vhead = Vwind -Vcart)
The retarding force generated by the wind on the cart, wrt the wind.

The cart must oppose that force, but is traveling at Vcart so
Power = F_drag*Vcart
Power = K*Vhead^2*Vcart
The power required to overcome the aerodynamic drag relative to the ground.

 

[Llyricist]

An iceboat can achieve 2WS at 30 degrees. Check any polar chart

An Iceboat is not a land yacht, 30 degrees is greater than 20

It does that in winds that are at least 20 degrees RMS,

Wrong, the RMS deviation is 11 degrees, and irrelevant to boot. The average was 9 degrees.

and at least 15mph.

at MOST 15 mph, you like to talk about the higher winds above the hub, but always ignore the much lower winds below.

The windspeed is also wrong, because it is not an RMS calculation,

You know what, RMS just MIGHT apply to wind speed, but I have to think about how that might actually work in this circumstance, it isn't as cut and dried as drag force being quadratically related to the wind speed I don't think. But it might be. And I guarantee that you would not like the result.

and does not take into account the wind above the hub.

It doesn't take into account the much lower speeds below the hub either.

 

[Llyricist]

No, I am saying that the cart did not achieve what an iceboat can.

Oh you were lying then. Got it.

There is no conventional Land yacht that can beat the wind within 20 degrees of DDW in a <= 15 knot wind.

Yes there is.

 

[profpangloss]

Only true in an FOR where the air is stationary. That is not an example of rotational power which would be invariant.

Yes indeed rotational power is frame invariant.

The treadmill belt is rotating, therefore the power from that belt is frame invariant, unlike the power from the wind.

The cart on a treadmill is not the same as a cart in the wind.

Oh, I know you are going to say "extend the top of the belt out to infinity and you can attach an inertial frame to it" or some other such nonsense. You can't attach an inertial frame to an imaginary projection!

Just mark an "X" on the belt and find a frame where that "X" is stationary. Such a frame will need to go around both ends of the treadmill at a high rate of angular velocity and undergo angular acceleration. Such a frame will not be an inertial frame moving in uniform motion.

There is no treadmill equivalency to a cart in the wind.

Goodbye treadmill.

 

[Llyricist]

Yes indeed rotational power is frame invariant.

The treadmill belt is rotating, therefore the power from that belt is frame invariant, unlike the power from the wind.

Wrong, the power from the motor is frame invariant. The power of the belt is entirely dependent on FOR.

The cart on a treadmill is not the same as a cart in the wind.

Yes it is.

Oh, I know you are going to say "extend the top of the belt out to infinity and you can attach an inertial frame to it" or some other such nonsense. You can't attach an inertial frame to an imaginary projection!

Well no, I wouldn't say that or any other such nonsense, IFOR's don't get attached to anything. Particularly not to a surface that goes through accelerations.

An IFOR that is based on the motion of the upper belt surface, will continue to move in that direction even as the belt changes direction.

Just mark an "X" on the belt and find a frame where that "X" is stationary. Such a frame will need to go around both ends of the treadmill at a high rate of angular velocity and undergo angular acceleration. Such a frame will not be an inertial frame moving in uniform motion.

Which is why IFOR's don't get attached to anything.

 

[profpangloss]

An IFOR that is based on the motion of the upper belt surface, will continue to move in that direction even as the belt changes direction.

Good. You agree that in an IFOR based on a projection of the upper belt's motion, the belt will be seen to be changing direction. In fact, it will be seen to be rotating and the power available from that belt is frame invariant. This is not the same as the power from the wind.

 

[humber]

Oh you were lying then. Got it.
There is no conventional Land yacht that can beat the wind within 20 degrees of DDW in a <= 15 knot wind.

An iceboat can achieve 2WS at 30 degrees, and have been recorded at least 5WS, and perhaps 7WS in 10mph winds.

The cart is not an iceboat, and the posted wind speed does not represent the actual wind, because the prop spends half of its time in unmeasured wind.
You would first need to conduct a trial that could be comparable, and where the rules are not amended on the day of the trial, in order to accommodate the competitor and his specific wants and needs.
Iceboat records are held in genuine competition with other boats piloted by adults.

The cart does not need to be compared to a refined and fast machine such as an iceboat, because there is nothing to explain. the BB manages to limp at a blistering pace of 0.1ms-2, all the way to 1.8WS (28mph) after being pushed to WS (11mph) in winds of unknown speed, but certainly greater than measured, 15mph, and not in the least ddw.
An ice boat is a fast and rapidly accelerating vehicle, and can reliably self-start, and does not need a nanny to apologize for it.

What are you asking to be explained? That it did not sink? No pirates?

For some reason, you have left this on your plate.

One reason the cart cannot be a ddw prop cart, is that it fails a power analysis.

One reason is this:

Cat got your tongue?

Frr = Vcart*m*g*Crr,
P = F*V
P = Frr*Vcart
P = m*g*Crr*Vcart^2

Is the power to overcome the rolling losses.


http://en.wikipedia.org/wiki/Drag_(physics)#Power
Power
The power required to overcome the aerodynamic drag is given by:

[latex]P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d<br />[/latex]

v^3 is When v = ground speed = windspeed, but the cart moves wrt respect to both the wind and the ground, so V cannot be the same for both frames.

F_Drag = K*Vhead^2 ( Vhead = Vwind -Vcart)
The retarding force generated by the wind on the cart, wrt the wind.

The cart must oppose that force, but is traveling at Vcart so
Power = F_drag*Vcart
Power = K*Vhead^2*Vcart
The power required to overcome the aerodynamic drag relative to the ground.

 

[Llyricist]

Good. You agree that in an IFOR based on a projection of the upper belt's motion, the belt will be seen to be changing direction. In fact, it will be seen to be rotating and the power available from that belt is frame invariant. This is not the same as the power from the wind.

No, I don't agree with that. The upper surface of the belt will not have any power at any time. The other parts of the belt will, but not the part that is actually in contact with the cart. Meanwhile, in that IFOR, the air will have lots of power.

 

[profpangloss]

No, I don't agree with that. The upper surface of the belt will not have any power at any time. The other parts of the belt will, but not the part that is actually in contact with the cart. Meanwhile, in that IFOR, the air will have lots of power.

In fact, the treadmill will not even stay in such a frame! You would need to continuously recreate the frame over and over again as "X" marked on the belt moves out of sight, because as you stated: "An IFOR that is based on the motion of the upper belt surface, will continue to move in that direction even as the belt changes direction." So your IFOR, based on a projection, continues to move as the belt goes round and round and the belt cannot remain in such a frame at all!

 

[Llyricist]

Humber, you are too far off the rails at this point to bother, and it's 2:30 am here.

All that crap about ice boats is of your own making, I didn't bring them up, you did. I made this claim:

There is no conventional Land yacht that can beat the wind within 20 degrees of DDW in a <= 15 knot wind.

And you said this:

Yes there is.

And you refused to back it up. Instead you started throwing red herrings left and right, including goading me to do yet another power analysis of the cart, of which I do several hundred in a matter of seconds.

It won't change by converting force to power and back, or using torque * AV, or any other gimmicks you try to throw. It's old hat, everyone understands it. It always boils down to the simple equation I posted.

 

[Llyricist]

In fact, the treadmill will not even stay in such a frame! You would need to continuously recreate the frame over and over again as "X" marked on the belt moves out of sight, because as you stated: "An IFOR that is based on the motion of the upper belt surface, will continue to move in that direction even as the belt changes direction." So your IFOR, based on a projection, continues to move as the belt goes round and round and the belt cannot remain in such a frame at all!

Frames of reference have no intrinsic boundaries.