Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147300 times)

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Offline Brumby

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There is nothing you need to do to that equation.  The correct one is the one I presented.
Wrong.

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The extra bit that is missing the the stored energy. The propeller was powered by wind power when well below wind speed then as the wind power decreased the stored energy started to provide most of the power to vehicle.
Wow.  Just, wow.  That is just so wrong.  There is no energy storage as you maintain.  There can't be.

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The way that stored energy is calculated is a bit complicated but it can be done as I already did in the spreadsheet calculator.
Say 60% of wind power is sent to propeller (from the wheel) then the other 40% gets to accelerate the vehicle increasing the vehicle kinetic energy and obviously speed.
The 60% that gets to propeller will be contributing to increased pressure differential (of course propeller may be 50% efficient so half of that 60% ends up as stored energy) You can see this increase in pressure differential about the same way as if natural wind speed has increased.
Again, you are aware of the phenomenon, but you continually misrepresent it!  The "pressure differential" is thrust.  This thrust is not stored energy - it is continuously created moment by moment.  Disconnect the drive shaft and it will cease as soon as the rotational momentum of the propeller is spent.

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In the calculator all that energy put in the propeller increases the potential wind energy and with such a mechanism 2x even 3x wind speed is not a problem but since this is stored energy as soon as it is all used up vehicle will slow down below wind speed.
You need to get over this "stored energy" thing.  It's stopping you from seeing the correct picture.

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All that it will need to be done is to run the experiment fully not stop the experiment just before vehicle will start to slow down.
Now this is a problem I have with you.  This is an assertion on your part.  Aside from your own, self-serving claims, there is (as far as I know) NO evidence that this will happen.  You can't say "I've proven it" because you are biased.  Your opinion does not count.  What is needed is an INDEPENDENT, authoritative source.

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It is a shame that this problem is wrongly presented in some schools and I wish I could do something about it.  I was much more optimistic about being able to explain this but it seems so far I'm failing to do so.
Oh - you are doing an EXCELLENT job in helping people understand what is going on.  By presenting an absolutely flawed explanation - one that has been disproven experimentally - you have engaged curious minds in a process of examination of the problem.  As a result, they are much more aware of how the mechanism actually works.

And for that, I thank you.
 

Online Kleinstein

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Besides, I use the word "thrust" here simply as a descriptive term.  If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square. If you want the power, than one has to multiply by a speed, that may depend on the reference system used and this often is not the same as the one used for the thrust.
The is still the point when the force changes direction - this is not included in the expression.


The calculation done to show that the blackbird can work are usually steady state, so they don't need to care about possibly energy storrage. The idea of energy storrage is only introduced to cast some doubt on the experiments.  A closer look however does not support it: there can be some energy in the compressible air, but this would be only very short time (and thus to little) and there is no control for it and thus the energy in the air going up when the vehicle goes faster. So the energy in the air behaves more like the kinetic energy in the vehicle, working against the vehicle going faster.
 

Offline Brumby

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Besides, I use the word "thrust" here simply as a descriptive term.  If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square.

Again - I used the term "thrust" in a conversational sense, not an engineering one, which is why I offered the verbose form.
« Last Edit: December 22, 2021, 11:08:51 am by Brumby »
 

Online Kleinstein

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Besides, I use the word "thrust" here simply as a descriptive term.  If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square.

Again - I used the term "thrust" in a conversational sense, not an engineering one, which is why I offered the verbose form.
Even in the verbose form it is still not allways right to use a simple speed difference to the 3rd power.  There is a 2nd power for the force and than an often different speed for the speed to make is a power.
 

Offline Brumby

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Then you are saying there is a fundamental problem with our friend's original formula?

I would have characterised the bit I added as completing the net air speed component.  It's not a separate entity, but one part in the determination of effective air speed.
« Last Edit: December 22, 2021, 02:33:58 pm by Brumby »
 

Offline fourfathom

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Again, you are aware of the phenomenon, but you continually misrepresent it!  The "pressure differential" is thrust.  This thrust is not stored energy - it is continuously created moment by moment.  Disconnect the drive shaft and it will cease as soon as the rotational momentum of the propeller is spent.

In the electrodacus theory, this pressure differential is stored and accumulated while the vehicle is below wind speed (sort of like inflating a balloon behind the vehicle).  As the vehicle reaches wind speed, the balloon begins to deflate, blowing the vehicle above windspeed for a while.  Once the balloon is deflated the vehicle slows down until it is again under windspeed.

I don't buy it.  There is no stored pressure differential.
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Offline IanB

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Then you are saying there is a fundamental problem with our friend's original formula?

You can see in post #1012 where I showed the more correct formula for wind power on a simple sail going downwind. Surprisingly, it turns out the maximum efficiency is only about 15%.

With a wind turbine, the common calculation is to consider the total rate of wind kinetic energy flowing through the swept area of the turbine blades (the energy flux), and then consider how much of that energy can be captured by the blades, which leads to the Betz limit.

With a sail, no kinetic energy can flow "through" the sail, so the calculation instead needs to consider the wind force on the sail multiplied by the speed that the sail is moving (power = force x velocity).

With a sail, the transferred power is zero when the sail is stationary (no velocity), and it is zero when the sail is moving at the same speed as the wind (no force). Maximum power transfer appears to occur somewhere around one third of the wind velocity.
 

Offline fourfathom

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An irrelevant question that has no bearing on the problem:
  Is the "sail" in that equation just a planar surface?  Actual sailboat sails, even those used for DDW sailing, are actually cut to create a three-dimensional airfoil, and are trimmed (tensioned with ropes) to optimize airflow and create lift.  So the efficiency numbers may be different than the Bernoulli numbers.
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Online Kleinstein

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Then you are saying there is a fundamental problem with our friend's original formula?

The problem is in what the formula stands for:  The (w-v)³ type formular is for the power available to a moving wind turbine (ignoring the power needed of vailable form the movement of the turbine). It is the the same as the normal ergy availabel to a wind turbine, just with a different wind speed.
This is different from it is totally different from the power from a sail on the vehicle: that power is force times vehichle velocity as also shown in the calculation of IanB. So the 2 parts together may make more sense.


Anyway the equation for the power a sail vehicle could use is not even relevant for the calculation that shows that the Backbird vehicle can work.
With only using a formula for a sail based setup on can not calculate the prop driven vehincle, as this is something different. When you do the calculation for the simple (e.g. 1 D world directly downwind) sail driven vehicle the result is that it can not go faster than the wind. This still does not say anything about a sail driven vehicle going zig-zag. This is known to be able to go faster.

I only took a more detailed look with the case of zero and low speed, to show that the  0.5 * air density * area * (wind speed - vehicle speed)^3 form is obviously wrong for a sail based vehicle.
For some reason electrodacus has difficulties in accepting that he can be wrong with his equation. It is not the only point he is wrong, but I had hope he would recognize the error in a equation that is not really related (at least I don't see a good way to use it one way or the other) to the main question.
With a failure to recognize even such a simple point, I think I will give up on convincing electrodacus. Not sure if he does not know better or is acting as a troll just to keep the discussion running in circles, repeating his wrong claims over and over again.

The sail in the calculation is just the planar surface with wind perpendicular (down wind). So not sophisticated areodynamics included.
 

Offline IanB

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An irrelevant question that has no bearing on the problem:
  Is the "sail" in that equation just a planar surface?  Actual sailboat sails, even those used for DDW sailing, are actually cut to create a three-dimensional airfoil, and are trimmed (tensioned with ropes) to optimize airflow and create lift.  So the efficiency numbers may be different than the Bernoulli numbers.

Yes, that derivation is for a "square" sail, perpendicular to the wind direction, sailing directly downwind. That is not, of course, representative of real sails on real boats with wind angles and keels and all sorts of other cleverness. It is just a theoretical exercise.

Having said that, all analysis of real sails will involve force triangles and resultant vectors, and real and apparent wind velocities. In fact, not much different from the engineering analysis of the blades on a wind turbine.
« Last Edit: December 22, 2021, 05:21:29 pm by IanB »
 

Offline IanB

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With a failure to recognize even such a simple point, I think I will give up on convincing electrodacus. Not sure if he does not know better or is acting as a troll just to keep the discussion running in circles, repeating his wrong claims over and over again.

This is, apparently, the unfortunate truth. When I did the analysis of the cart on the belts a few posts back, and showed how the cart can move against the direction of the belts with appropriate gearing, this analysis was simply rejected as contrary to the belief that such a result is impossible.
 

Offline electrodacus

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Wow.  Just, wow.  That is just so wrong.  There is no energy storage as you maintain.  There can't be.

Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.
I will no longer post the graph but you can check that here https://en.wikipedia.org/wiki/Axial_fan_design
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.
The fact that you (all here) do not understand what energy storage is (among many other things) dose not make what I say invalid.
The equations I presented and the explanation fully matches what happens in all tests that where done.
On the other had you multiple wrong equation do not predict what is observed from all experimental test's.
Take just Derek's formula using (vehicle speed - wind speed) in the video I critiqued and that formula will predict deceleration when below wind speed (never observed in any experiment not to mention ridiculous) and predicts ever increased rate of acceleration when vehicle is above wind speed again never seen in any test and ridiculous to even claim.

Not being able to provide the most important equation of wind power available to vehicle for a wind power only vehicle means you (all) have no understanding of the subject. Seems to be more of a cultural thing that you never answer with I do not know and trow in guesses no matter how unlikely they are to be correct.

Correct formula for a direct down wind vehicle in therms of available wind power is this: 0.5 * air density * area * (wind speed - vehicle speed)^3
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
It is as simple as that if you (all) where to understand physics.

It was a waste of time to discus the issue with you (all) but I did try.
Hopefully someone with more patience and better teaching skills will give it a try.
For now I have better things to do and if I can think of a simpler way to explain this to you (all) I will maybe attempt that in the future.

Offline electrodacus

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The problem is in what the formula stands for:  The (w-v)³ type formular is for the power available to a moving wind turbine (ignoring the power needed of vailable form the movement of the turbine). It is the the same as the normal ergy availabel to a wind turbine, just with a different wind speed.
This is different from it is totally different from the power from a sail on the vehicle: that power is force times vehichle velocity as also shown in the calculation of IanB. So the 2 parts together may make more sense.


Anyway the equation for the power a sail vehicle could use is not even relevant for the calculation that shows that the Backbird vehicle can work.
With only using a formula for a sail based setup on can not calculate the prop driven vehincle, as this is something different. When you do the calculation for the simple (e.g. 1 D world directly downwind) sail driven vehicle the result is that it can not go faster than the wind. This still does not say anything about a sail driven vehicle going zig-zag. This is known to be able to go faster.

I only took a more detailed look with the case of zero and low speed, to show that the  0.5 * air density * area * (wind speed - vehicle speed)^3 form is obviously wrong for a sail based vehicle.
For some reason electrodacus has difficulties in accepting that he can be wrong with his equation. It is not the only point he is wrong, but I had hope he would recognize the error in a equation that is not really related (at least I don't see a good way to use it one way or the other) to the main question.
With a failure to recognize even such a simple point, I think I will give up on convincing electrodacus. Not sure if he does not know better or is acting as a troll just to keep the discussion running in circles, repeating his wrong claims over and over again.

The sail in the calculation is just the planar surface with wind perpendicular (down wind). So not sophisticated areodynamics included.

The equation 0.5 * air density * area * (wind speed - vehicle speed)^3  perfectly applies to any direct down wind vehicle no matter if it is a cart with a sail one with a wind turbine or one with a propeller/fan.
The difference on the propeller/fan is that energy can be stored as pressure differential.  Maybe pressure differential energy storage is a bit complicated to understand  but all you should need to know is that the above mentioned equation is correct and if you can understand that you will see why energy storage is needed else wind speed can not be exceeded direct down wind and no speed can be achieved direct upwind.
The type of energy storage is very different for direct down wind (pressure differential) and direct upwind (small elastic or gravitational storage and stick slip hysteresis).

Online Kleinstein

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Correct formula for a direct down wind vehicle in therms of available wind power is this: 0.5 * air density * area * (wind speed - vehicle speed)^3
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.

Were do you get that magic formula from ?  It is not supported by the wind power PDF you linked and I have not seen it in any books. The standard formular is just the 3rd power of the wind speed, without the vehicle moving. With the idealization of the Betz limit this is OK. The moving wind turbine is a rather exotic problem and thus usually no read made tabulated solution for this problem and one has to derive it from the general principles.
 

Offline electrodacus

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Correct formula for a direct down wind vehicle in therms of available wind power is this: 0.5 * air density * area * (wind speed - vehicle speed)^3
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.

Were do you get that magic formula from ?  It is not supported by the wind power PDF you linked and I have not seen it in any books. The standard formular is just the 3rd power of the wind speed, without the vehicle moving. With the idealization of the Betz limit this is OK. The moving wind turbine is a rather exotic problem and thus usually no read made tabulated solution for this problem and one has to derive it from the general principles.

The formula of course contains just wind speed for a stationary device like a ground mount wind turbine but since this refers to a vehicle moving directly down wind the wind speed relative to vehicle is what counts and that is wind speed minus vehicle speed.
And this same formula will also apply to a wind turbine if the wind turbine is installed on a vehicle moving directly down wind.
And yes is not a usual thing to install the wind turbine on a moving vehicle (it will not make sense since if you need to move the vehicle a sail will be more efficient and if you need electricity you just get that at the wheel in any wind powered vehicle).
The formula is for the ideal case but applies to all wind powered vehicles and in the form I shown applies to a directly down wind vehicle or a directly upwind .
Since for direct upwind the speed direction is negative (wind speed - (-vehicle speed)) you get (wind speed + vehicle speed) so you always have access to wind power but you need energy storage as direction of the wind opposes vehicle direction.

You can find different ways to calculate but result can not be different since this perfectly matches what is observed in real tests.

And also this formula is used if you want to know how what is the minimum power your vehicle engine or motor requires in order to overcome the drag due to wind.

Most small passenger vehicles have around 2.5m^2 of frontal area and if they are aerodynamic the Coefficient of Drag may be around 0.3 thus effective frontal area 2.5m^2 * 0.3 = 0.75m^2
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km

Offline Labrat101

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This problem is impossible to solve without the correct information .
Firstly the Prop . being variable pitch it also has centre clutch assembly . The length of the
blades are a very important . How many blades it has . 3 or 5  are common 6 & 8 are normally shorter . (example) Most cars use 6 or 8 on the engine fan these are also variable pitch operated by temperature . Just to give you some idea what's involved . your car fan will spin free when cold & or if the speed is below set point of the magnetic clutch .
Also Its not a turbine blade these are totally different .
The maths and details of these Props are well documented and cover over 100 pages of heavy
reading .
Without having the propeller specs at hand . You can make up any formulae you like . It will not
work  . Its just guess work , finding a cool load of numbers that look good to you .
 But its BS  .
Get all the Data sheets & Specs . Then try again .
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Offline fourfathom

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This problem is impossible to solve without the correct information .
Firstly the Prop . being variable pitch it also has centre clutch assembly [...]

Yes, an exact solution requires data and details not used in these simple formulae.  But basic principles can still be discussed and insights achieved.  For example, the simple treadmill experiments used fixed-pitch props.  They did effectively start at windspeed, and then proceeded to accelerate from there.  The simple gear-coupled wheels on moving belts (or one set on a belt, the coupled set on the ground) also demonstrate the principle.  So does the spool of thread example.  Better data will let you determine drag and energy requirements, etc, but using 100% efficiency, zero-friction models is still instructive.

About the electrodacus theory of stored energy, I had suggested an inflated balloon, but perhaps a better model would be a spring between the wind-face (moving at windspeed) and the rear of the vehicle.  While the vehicle speed is less than windspeed the spring is being compressed.  When the vehicle reaches windspeed, the spring continues to push the vehicle above windspeed for a while.

I suppose this makes sense to electrodacus, but I've sailed thousands of miles downwind at less than windspeed.  There is no significant energy storage, you use it or lose it.
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Offline Labrat101

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I suppose this makes sense to electrodacus, but I've sailed thousands of miles downwind at less than windspeed.  There is no significant energy storage, you use it or lose it.
Yes your correct .
 Quote " There is no significant energy storage, you use it or lose it. "
True But,  electrodacus  would argue that it could be stored in a Paper bag and squeezed out .
With a formulae  found in the Dummies Book of wind Power (for beginners )

& for the tread mill . A hamster in its wheel would be more appropriate
« Last Edit: December 22, 2021, 09:37:54 pm by Labrat101 »
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Online PlainName

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Quote
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km

When the vehicle is at wind speed there is no air drag. Nothing. 0. At faster than wind speed there is a small amount (assuming just slightly faster), increasing as the vehicle exceeds wind speed more. The increased drag prevents infinite acceleration, but at just over wind speed there is no drag to speak of, hence the power needed to combat drag could  be as small as you can arrange to measure.
 

Offline IanB

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Quote
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km

When the vehicle is at wind speed there is no air drag. Nothing. 0. At faster than wind speed there is a small amount (assuming just slightly faster), increasing as the vehicle exceeds wind speed more. The increased drag prevents infinite acceleration, but at just over wind speed there is no drag to speak of, hence the power needed to combat drag could  be as small as you can arrange to measure.

Bear in mind there is no wind in the problem statement, so the vehicle is pushing through still air at 33 m/s. The mistake made is in the power formula. The power required to overcome drag is equal to drag resistance times vehicle speed. In still air the vehicle speed and the apparent wind speed are the same, but if there is some wind present the vehicle speed and the apparent wind speed are different. One could be more or less than the other.
 

Offline fourfathom

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And also this formula is used if you want to know how what is the minimum power your vehicle engine or motor requires in order to overcome the drag due to wind.

Most small passenger vehicles have around 2.5m^2 of frontal area and if they are aerodynamic the Coefficient of Drag may be around 0.3 thus effective frontal area 2.5m^2 * 0.3 = 0.75m^2
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km

To be fair, while I initially was thinking "but DDW at windspeed there's no drag at all", in further reflection I read this as an interesting digression from the DDWFTTW arguments, and an illustration of how drag affects automobiles at speed.  I don't think that electrodacus is claiming that a vehicle driving 120 km/h downwind in a 120 km/h wind will see 16.2 kW of drag.
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Offline IanB

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To be fair, while I initially was thinking "but DDW at windspeed there's no drag at all", in further reflection I read this as an interesting digression from the DDWFTTW arguments, and an illustration of how drag affects automobiles at speed.  I don't think that electrodacus is claiming that a vehicle driving 120 km/h downwind in a 120 km/h wind will see 16.2 kW of drag.

However, drag is a force, with SI units of Newtons. So "16.2 kW of drag" is not a statement that can be made. What you can say is that the vehicle will require 16.2 kW to overcome the drag resistance. This may seem picky, but it matters when the wind speed and the vehicle speed are different.

Power comes into the equation involving force and speed, hence:

   (power) = (force) x (velocity)

Wherein it must be understood that force is the magnitude of a vector pointing in the opposite direction of velocity.

The drag force comes from the apparent wind velocity impeding the vehicle, and the velocity itself is the velocity of the vehicle over the ground. Since these can be different, it is important to keep the separate in the equation.
« Last Edit: December 23, 2021, 01:08:20 am by IanB »
 

Offline Brumby

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In the electrodacus theory, this pressure differential is stored and accumulated while the vehicle is below wind speed (sort of like inflating a balloon behind the vehicle).  As the vehicle reaches wind speed, the balloon begins to deflate, blowing the vehicle above windspeed for a while.  Once the balloon is deflated the vehicle slows down until it is again under windspeed.

My apologies for being so dense as to not see this.  Such a concept is so absurd that I never pictured it this way.

Quote
I don't buy it.  There is no stored pressure differential.

Absolutely.



Wow.  Just, wow.  That is just so wrong.  There is no energy storage as you maintain.  There can't be.

Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.

Which causes air movement, since there is no containment.  Air will flow in any available direction, dissipating any energy contained into the surrounding environment.

Quote
...
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.

A compressed spring contains energy within that spring - but only while the compression is held.  That is, the energy is contained.  There is no containment in an unbounded air mass, therefore energy cannot be stored.  A balloon can store energy, as the air under compression is bounded by the balloon, but once ruptured, that energy is dissipated within milliseconds.

If you disagree, please illustrate in a diagram how energy can be stored as you claim.
 

Offline IanB

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Quote
...
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.

A compressed spring contains energy within that spring - but only while the compression is held.  That is, the energy is contained.  There is no containment in an unbounded air mass, therefore energy cannot be stored.  A balloon can store energy, as the air under compression is bounded by the balloon, but once ruptured, that energy is dissipated within milliseconds.

Well, obviously you can compress a spring by pressing on only one end of it, while leaving the other end free...  :-//
 
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Offline Brumby

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Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.

Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.

You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released.  Include any necessary description and formulae that would be required for it to stand up to examination.
« Last Edit: December 23, 2021, 02:07:32 am by Brumby »
 


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