Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[Subductionzon]

From my understanding the students quickly lost interest in the project. Too bad for them, they could have been part of a truly fascinating build.

 

[rcgldr]

The force in the chain is creating a clockwise torque at the prop axis, but also an anti clockwise on the wheel axis.

I got that, but take the case where the right side of the prop sprocket is directly above the front of wheel axis sprocket, so that the chain under tension takes a vertical run (just twisting 90 degrees) between the sprockets. The upwards vertical force to the right of the center of the wheel axis generates a counter clockwise torque, but the tension force in the chain will be countered by the compression forces in the framework, so the linear forces cancel.

To compensate for the offset tension (upwards force on the wheel axis), I'm thinking that the compression force in the right post will be greater than the left post, which makes sense in that this would generate a lateral force to the left on the hub, corresponding to the counter-clockwise torque on the vehicle.

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

 

[OmCheeto]

I got that, but take the case where the right side of the prop sprocket is directly above the front of wheel axis sprocket, so that the chain under tension takes a vertical run (just twisting 90 degrees) between the sprockets. The upwards vertical force to the right of the center of the wheel axis generates a counter clockwise torque, but the tension force in the chain will be countered by the compression forces in the framework, so the linear forces cancel.

To compensate for the offset tension (upwards force on the wheel axis), I'm thinking that the compression force in the right post will be greater than the left post, which makes sense in that this would generate a lateral force to the left on the hub, corresponding to the counter-clockwise torque on the vehicle.

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

I will probably replace the chain with a set of gears in my analysis. Chains are too much like noodles, and I have no idea how to create a free body diagram with a twisted noodle.

 

[ThinAirDesign]

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

Of course if we had extended the right side the same length we extended the left side we would have had the same roll protection (actual marginally more because of the added weight of the longer right side). Problem is we would have no longer been able to trailer the Blackbird legally down the road as it would been too wide.

Extending only the left side gave us 99% of the increase (as compared to both sides) while not adding the disadvantages. Extending the right side would have essentially been a waste of time.

JB

 

[rcgldr]

Of course if we had extended the right side the same length we extended the left side we would have had the same roll protection (actual marginally more because of the added weight of the longer right side). Problem is we would have no longer been able to trailer the Blackbird legally down the road as it would been too wide.

That's what I thought about the roll protection, but I wasn't aware of the trailering issue. Thanks for the response.

 

[Llyricist]

Okay, someone wanted math?

I'll take a cart that is very much like the actual cart, weighs 650 lbs (295 kg) w/pilot;

Has fairly efficient wheels giving a coefficient of rolling resistance (crr) of .013;

The fairings and such give it a relatively sleek coefficient of drag of .26 and it has a frontal area of 20 Sq Ft, or 1.858 m²;

It uses a twisted chain drive transmission with a 23 tooth sprocket on the axle, and a 65 tooth sprocket on the propeller shaft, and the twist will take the normal 98% efficiency of a chain drive down to 92%;

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

And one more specification is that the wheels have an outer diameter of 27 inches (.686 m).


Now I'm just going to analyse the main forces acting on the cart when it is travilling at 24 MPH (10.72 m/s) with a 12 MPH (5.36 m/s) tailwind. That's twice wind speed. Additionally I will use the conditions at El Mirage Lake Bed on July 3, 2010 to derive an air density of 1.063 kg/m³. All the caluculations will be done using SI units for convenience.

First, there will be a rolling resistance acting on the cart of (crr * mass * 1G) 0.013 * 295 kg * 9.81 = 37.6 N

Next there will be aerodynamic drag, using the Rayleigh formula: (F_D = 0.5 * density * v² * coeeficient of Drag * frontal area);

0.5 * 1.063 kg/m³ * 5.36 m/s (airspeed: 10.72 m/s - 5.36 m/s) * .26 * 1.858 m² = 7.39 N

Next, using the gearing and wheel size specified above, I determine that while the wheels are rolling along the ground at 10.72 m/s the prop shaft will be turning at 105.7 rpm.

Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Transmitting that torque through the gears to the axle will result in an opposing torque on the axle of 270.38 * (23 / 65) / .92 (drive efficiency) = 104 N-m

Dividing that by the radius of the wheel will result in an opposing force of 104 N-m / .343 m = 303.2 N.

For total opposing Forces of: 37.6 N + 7.39 N + 303.2 N = 348.2 N

Javaprop also gives a thrust in those conditions of 376.6 N, For a net acceleration force of 376.6 N - 348.2 N = 28.4 N;

Which will accelerate the cart at a rate of (a = F / m) 28.4 N / 295 kg = 0.09 m/s/s, well in line with the observed and documented performance of the actual cart.

Next I will quantify how the energy balances in a frame of reference where the ground can be considered stationary.

 

[OmCheeto]

Okay, someone wanted math?

I'll take a cart that is very much like the actual cart, weighs 650 lbs (295 kg) w/pilot;

Has fairly efficient wheels giving a coefficient of rolling resistance (crr) of .013;

The fairings and such give it a relatively sleek coefficient of drag of .26 and it has a frontal area of 20 Sq Ft, or 1.858 m²;

It uses a twisted chain drive transmission with a 23 tooth sprocket on the axle, and a 65 tooth sprocket on the propeller shaft, and the twist will take the normal 98% efficiency of a chain drive down to 92%;

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

And one more specification is that the wheels have an outer diameter of 27 inches (.686 m).


Now I'm just going to analyse the main forces acting on the cart when it is travilling at 24 MPH (10.72 m/s) with a 12 MPH (5.36 m/s) tailwind. That's twice wind speed. Additionally I will use the conditions at El Mirage Lake Bed on July 3, 2010 to derive an air density of 1.063 kg/m³. All the caluculations will be done using SI units for convenience.

First, there will be a rolling resistance acting on the cart of (crr * mass * 1G) 0.013 * 295 kg * 9.81 = 37.6 N

Next there will be aerodynamic drag, using the Rayleigh formula: (F_D = 0.5 * density * v² * coeeficient of Drag * frontal area);

0.5 * 1.063 kg/m³ * 5.36 m/s (airspeed: 10.72 m/s - 5.36 m/s) * .26 * 1.858 m² = 7.39 N

Next, using the gearing and wheel size specified above, I determine that while the wheels are rolling along the ground at 10.72 m/s the prop shaft will be turning at 105.7 rpm.

Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Transmitting that torque through the gears to the axle will result in an opposing torque on the axle of 270.38 * (23 / 65) / .92 (drive efficiency) = 104 N-m

Dividing that by the radius of the wheel will result in an opposing force of 104 N-m / .343 m = 303.2 N.

For total opposing Forces of: 37.6 N + 7.39 N + 303.2 N = 348.2 N

Javaprop also gives a thrust in those conditions of 376.6 N, For a net acceleration force of 376.6 N - 348.2 N = 28.4 N;

Which will accelerate the cart at a rate of (a = F / m) 28.4 N / 295 kg = 0.09 m/s/s, well in line with the observed and documented performance of the actual cart.

Next I will quantify how the energy balances in a frame of reference where the ground can be considered stationary.

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

...


Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.



I know the truth is out there...

https://www.youtube.com/watch?v=JDZBgHBHQT8

 

[Llyricist]

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:


Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

Here's the numbers I used:

r(ft) chord(ft) incidence(deg)
0.93983 0.18983 80.1249
1.06718 0.31718 76.89058
1.21091 0.46091 73.75922
1.36252 0.61252 70.74371
1.51452 0.76452 67.8535
1.66077 0.91077 65.09468
1.7967 1.0467 62.47044
1.91922 1.16922 59.98132
2.02652 1.27652 57.62588
2.11789 1.36789 55.40093
2.1934 1.4434 53.30211
2.25373 1.50373 51.32415
2.29988 1.54988 49.46123
2.33312 1.58312 47.70721
2.35476 1.60476 46.05583
2.36616 1.61616 44.50084
2.3686 1.6186 43.03614
2.3633 1.6133 41.65586
2.35135 1.60135 40.35435
2.33373 1.58373 39.12632
2.31133 1.56133 37.96673
2.28491 1.53491 36.87091
2.25513 1.50513 35.83446
2.22255 1.47255 34.85332
2.18766 1.43766 33.92371
2.15088 1.40088 33.04211
2.11254 1.36254 32.20531
2.07293 1.32293 31.41029
2.03228 1.28228 30.65429
1.99078 1.24078 29.93474
1.94858 1.19858 29.24929
1.90579 1.15579 28.59575
1.86249 1.11249 27.97211
1.81875 1.06875 27.37649
1.77457 1.02457 26.80716
1.72997 0.97997 26.26254
1.68491 0.93491 25.74113
1.63934 0.88934 25.24157
1.59319 0.84319 24.76258
1.54632 0.79632 24.30298
1.49857 0.74857 23.86167
1.44974 0.69974 23.43764
1.39951 0.64951 23.02992
1.34748 0.59748 22.63764
1.29305 0.54305 22.25996
1.23533 0.48533 21.89613
1.1729 0.4229 21.54542
1.10312 0.35312 21.20715
1.02023 0.27023 20.8807
0.90414 0.15414 20.56549
1.10312 0.35312 21.20715
1.02023 0.27023 20.8807
0.90414 0.15414 20.56549

You can copy just the numbers to your clipboard and when you bring up the import geometry dialog in Javaprop, it will paste them automatically.

Also, the airfoils used are not included in javaprop, I had to follow the directions and used the same guy's JavaFoil program to create the necessary data files for the NACA 6412.


Also also, I learned java and accessed the "propeller object" directly from my own program to get more specific results than are provided by the program the way it is.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.

was it this site?

http://www.mh-aerotools.de/airfoils/javaprop.htm"

If so, that is surprising, I never had any propblem

 

[Llyricist]

The first thing about the energy balance to understand is that the external input of energy to the cart system only has to equal the losses in the system plus any gain in KE of the system.

The losses of the cart system (again, all in a frame of reference where the ground can be considered stationary) are the aerodynamic drag, the rolling resistance, the drivetrain losses, and the aerodynamic losses at the propeller. It is important to note here that the negative work of the prop on the cart is not a loss, since much of it is put right back into the system, just the part lost in the drivetrain and aerodynamically are losses.

And the cart will gain KE as it accelerates from 10.72 to 10.81 m/s over the course of a second.


So starting with the aerodynamic drag, that is a force of 7.39 N acting over 10.72 m/s, for a loss to the system of 10.72 * 7.39 = 79.2 J/s.

Next the rolling resistance, a force of 37.6 N acting over 10.72 m/s gives a loss of 403.1 J/s.

Now it starts to get more complicated. The torque demand at the propeller shaft equates to a power demand of 2993.5 J/s (Watts), While the negative work on the cart from that demand equals 303.2 N * 10.72 m/s = 3250.3 J/s, giving a drivetrain loss of 3250.3 - 2993.5 = 256.8 J/s.

So far, that is 79.2 + 403.1 + 256.8 = 739.1 J/s in losses.

Skipping the aerdynamic losses at the prop for the moment, since the same calculations will be used to determine the energy input, The ke gain of the cart system can be calculated from the difference in KE after and before.

KE = 0.5 * m * v²

before = 0.5 * 295 kg * 10.72 m/s² = 16950.46 J
after = 0.5 * 295 kg * 10.81 m/s² = 17236.27 J

Difference = 285.8 J

So far, an energy requirement of 739.1 + 285.8 = 1024.9 J

Now for the really complicated part.

From http://www.mh-aerotools.de/airfoils/propuls4.htm" , which is the same site that Javaprop comes from, there is a generic formula for propeller thrust:

Where;

T = thrust [N]
D = propeller diameter [m]
v = velocity of incoming flow [m/s]
Δv = additional velocity, acceleration by propeller [m/s]
ρ = density of fluid [kg/m³]

Since I already have everything except Δv, I have to solve for that, which results in a quadratic equation that is too messy to post unless absolutely necessary. But I need that value both for the KE loss of the air and the prop's aerodynamic losses.

Long story short, I get a Δv of 2.417 m/s.

With that in hand I move to the "Engine Power" formula on that same page, to get the minimum power, assuming no losses at all, for a propeller to generate that thrust in those conditions.

Which results in a power of 2473.6 W, which is far less than the 2993.5 W that is actually at the shaft. In fact, the difference is the aerodynamic losses at the prop.

So those losses are 2993.5 - 2473.6 = 519.9 J/s

For total losses and KE gain of the cart system of 519.9 + 1024.9 = 1544.8 J/s


Now, for the KE loss of the air that is actually transferred to the cart, I need the mass of the air that is decelerated, and how fast it was going before and after.

The thrust formula above is actually F = ma, where everything on the right side except for the last Δv is the mass (it calculates the volume of the cylinder of air and multiplies by density), and the last Δv is the acceleration.

So using that part to calculate the mass results in a mass of 155.79 kg.

The KE of the air before the cart moves through it is 0.5 * 155.79 kg * 5.36² = 2238 J

The air is decelerated by 2.417 m/s, so its velocity after the cart moves through it is 5.36 - 2.417 = 2.94 m/s;

so its KE will now be 0.5 * 155.79 kg * 2.94² = 673.3 J

For an energy input to the cart system of 2238 - 673.3 = 1564.7 J/s


So we have 1564.7 J/s into the system, and 1544.8 either lost, or gained in KE, by the cart system. Close enough considering all the rounding I did, and the lack of using calculus for really precise answers.

 

[RonL]

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:


Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.



I know the truth is out there...

https://www.youtube.com/watch?v=JDZBgHBHQT8

"Trust no one"

 

[ThinAirDesign]

Cool. I thought I was going to have to do all the maths.

No sure why you would think that when repeatedly you've been given more 'maths' than you could ever know what to do with.

https://www.physicsforums.com/showpost.php?p=3423521&postcount=176

JB

 

[OmCheeto]

No sure why you would think that when repeatedly you've been given more 'maths' than you could ever know what to do with.

https://www.physicsforums.com/showpost.php?p=3423521&postcount=176

JB

Those maths seemed fishy to me.

ddw2.pdf was a boat design, with a ~"and land vehicles work the same" paragraph thrown in at the end.

2009_28.pdf was about a car that travels against the wind, with a "ddw" thrown in. But the author was correct, I did find it amusing.

Despite the obvious lack of commercial exploitability of wind turbine cars, the presentation will hopefully illuminate surprise and amuse the readers.

...

The issue with such a vehicle is of course that it needs to be pushed up to velocities above the wind velocity for the energy conversion process to take over since a propeller does not work when the flow comes from “the wrong side”

This seems to contradict how the original ddw vehicle worked.

Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf struck me as amateurish(he listed omega as propeller tip velocity , rather than radians per second), and I stopped reading the paper after the 2nd equation, as it was simply wrong.

I think I'd rather start from scratch.

I hear...I forget
I see...and I remember
I do...and I understand

"Trust no one"

Welcome back Ron!
I trust you.
Even though you're a nut.
(Takes one to know one. )

 

[A.T.]

ddw2.pdf was a boat design, with a ~"and land vehicles work the same" paragraph thrown in at the end.

And that makes the math fishy? Because he talks about both, water and land ddwfttw vehicles? Of course they work on the same principle, just with different efficiencies.

2009_28.pdf was about a car that travels against the wind, with a "ddw" thrown in.

And that makes the math fishy? Because he talks about both directions, upwind and downwind? Of course they work on the same principle, just with reversed power flow.

The issue with such a vehicle is of course that it needs to be pushed up to velocities above the wind velocity for the energy conversion process to take over since a propeller does not work when the flow comes from “the wrong side”

This seems to contradict how the original ddw vehicle worked.

No, it doesn't. The authors simply didn't analyze the self-start in propeller mode. They are correct to say that the energy conversion that they described assumes being above wind speed, and they didn't want to speculate beyond that.

Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf struck me as amateurish(he listed omega as propeller tip velocity , rather than radians per second)

Where exactly did he do this? As far I can see, he defines omega as "propeller rotational speed" and he calls "propeller tip velocity" Vc.

, and I stopped reading the paper after the 2nd equation, as it was simply wrong.

How can equation 2 be wrong? It is a pure definition of symbols.

I think I'd rather start from scratch.

I think you should try to understand those 3 papers first.

 

[spork]

I think you should try to understand those 3 papers first.

Absolutely not. I think it will be far more entertaining to watch him start from scratch.

 

[OmCheeto]

Where exactly did he do this? As far I can see, he defines omega as "propeller rotational speed" and he calls "propeller tip velocity" Vc.

I think you are correct on this point. Damn morning brain!

How can equation 2 be wrong? It is a pure definition of symbols.

μW = W(μ₁ + μ₂ + μ₃

W = weight of the vehicle
μW = force on the vehicle
μ₁ = mechanical rolling friction
μ₂ = aerodynamic friction
μ₃W = force as the result of acceleration...

Since when did the weight of a vehicle affect it's aerodynamics?

I think you should try to understand those 3 papers first.

But that's the problem. I don't understand them. Hence why I have to start from scratch.

Llyricist's numbers look interesting. I think I'll start there.

But GEEZ! I've already wasted 6 hours today on this problem. Oh well, only 9 more weekends to go.

 

[A.T.]

Since when did the weight of a vehicle affect it's aerodynamics?

It doesn't, and nobody claimed it does. Defining a dimensionless parameter as the ratio of two forces, doesn't imply that the forces affect each other.

 

[OmCheeto]

Since when did the weight of a vehicle affect it's aerodynamics?

It doesn't, and nobody claimed it does. Defining a dimensionless parameter as the ratio of two forces, doesn't imply that the forces affect each other.

bolding mine.

Andrew B Bauer, the author of the paper, did.

I see distinctly where the author multiplied the weight of the vehicle by it's aerodynamic 'friction' to come up with a force component.

The paper was sited as a reference that I should study and learn from. Since the reference is wrong, I don't see that I can learn anything from it.

I thought you kids were experts in aerodynamics? This is so basic, even wiki get's it right:

Where is a reference to weight, or even mass in that equation?

Wait! OMG. I found it.

ρ = mass/volume

 

[spork]

Sunlight will never cost $5/gal-GP

Perhaps not, but the power you get from it costs more than that currently.

 

[A.T.]

I see distinctly where the author multiplied the weight of the vehicle by it's aerodynamic 'friction' to come up with a force component.

Nope. The author simply defined a dimensionless parameter as the ratio of aerodynamic drag force and weight. Multiplying the weight with that parameter naturally gives you the aerodynamic drag force.

The paper was sited as a reference that I should study and learn from. Since the reference is wrong, I don't see that I can learn anything from it.

Nothing wrong with the paper. But your claim that a definition is wrong is not even wrong, just nonsensical.

I thought you kids were experts in aerodynamics? This is so basic, even wiki get's it right:

Where is a reference to weight, or even mass in that equation?

Your understanding problem is indeed very basic, and has nothing to do with aerodynamics. Nobody claims that aero drag depends on weight. But there is nothing wrong with expressing aero drag as a fraction of weight.

 

[OmCheeto]

Your understanding problem is indeed very basic, and has nothing to do with aerodynamics.

Your command of the English language appears to be "indeed very basic". Is English a second language for you?

Nobody claims that aero drag depends on weight. But there is nothing wrong with expressing aero drag as a fraction of weight.

Also, your command of aerodynamics, not to mention logic, seems to be somewhat lacking. What is your primary area of expertise?

How I interpret your last sentence: "F_d does not depend on weight, but it's ok if you throw weight into the equation."

ps. Is this what happened last week when ZZ threatened to shut us down? I missed the whole thing.

 

[A.T.]

How I interpret your last sentence: "F_d does not depend on weight, but it's ok if you throw weight into the equation."

Instead of reinterpreting everything you should read what is actually written. It is perfecty ok to define a parameter x as:

x = drag / weight

so that:

drag = x * weight

This doesn't imply in any way that drag depends on weight.

 

[OmCheeto]

drag = x * weight

This doesn't imply in any way that drag depends on weight.

Alrighty then. I can now understand fully why ZZ threatened to shut this thread down.

Taxi! Oh Taxi! Can you take me away from crazyland please.

 

[A.T.]

Instead of reinterpreting everything you should read what is actually written. It is perfecty ok to define a parameter x as:

x = drag / weight

so that:

drag = x * weight

This doesn't imply in any way that drag depends on weight.

g_mars = 0.376 * g_earth

Does the statement above imply that Earth's gravity affects Mars' gravity?

 

[OmCheeto]

g_mars = 0.376 * g_earth

Does the statement above imply that the Mars gravity depends on Earth's gravity?

0.376 does not have dimensions.

weight does.

your logic is flawed.

Is this why ddwfttw threads die and get resurrected all the time? Because no one ever wants to solve the problem?

hmmm... This thread appears to be spiraling into a black hole of smoke and mirrors. Perhaps I should just write my own paper.

 

[Llyricist]

Testing one two three

 

[Llyricist]

I posted 2 posts last night, one in reply to omcheeto, and both got lost somewhere.

 

[RonL]

It's difficult for me to visualize how and where this prop torque is transmitted to the vehicle. Is it at the prop hub or at the axle hub? I don't remember studying the physics of chains on chain driven vehicles. I suppose I'll have to back up and figure out how a bicycle works first.

But as I said, I don't have time to work on this except on weekends.

Thanks for the welcome back, from my self imposed ban
In answer to your above question, my thought goes back to early day flight, when a pilot tried to make a turn and the reaction to the plane was a nose dive into the ground. In this thread, weight and spread of wheel from center of propeller axis is important for stability. (just my thoughts).
I'll not say more, as it would be easy for me to get this thread locked.
I feel like a recovering alcoholic resisting his first temptation for a drink. You do remember spaghetti physics ? I have gone through the entire thread and at post 153 rcgldr made comments about the real answers of what is going on. Where there is motion there is temperature change.
That's all for me. Back to my ban and studies.

Ron

 

[Llyricist]

Trying again, here is the geometry data I used for the propeller, I had to shift the radius stations from the original that spork posted since it was originally designed for a 16 foot diameter, but they extended the spars to make it 17.5 feet.

Edit: something went very wrong, what I originally posted was messed up somehow.

 

[Llyricist]

I have a feeling it was because I used links in the posts that got waylayed. I guess I need to read the rules about that.

 

[Llyricist]

Let me try that data again:

Radius(ft) Chord(ft) Angle of incidence (deg)
1.325 0.18983 80.1249
1.475 0.31718 76.89058
1.625 0.46091 73.75922
1.775 0.61252 70.74371
1.925 0.76452 67.8535
2.075 0.91077 65.09468
2.225 1.0467 62.47044
2.375 1.16922 59.98132
2.525 1.27652 57.62588
2.675 1.36789 55.40093
2.825 1.4434 53.30211
2.975 1.50373 51.32415
3.125 1.54988 49.46123
3.275 1.58312 47.70721
3.425 1.60476 46.05583
3.575 1.61616 44.50084
3.725 1.6186 43.03614
3.875 1.6133 41.65586
4.025 1.60135 40.35435
4.175 1.58373 39.12632
4.325 1.56133 37.96673
4.475 1.53491 36.87091
4.625 1.50513 35.83446
4.775 1.47255 34.85332
4.925 1.43766 33.92371
5.075 1.40088 33.04211
5.225 1.36254 32.20531
5.375 1.32293 31.41029
5.525 1.28228 30.65429
5.675 1.24078 29.93474
5.825 1.19858 29.24929
5.975 1.15579 28.59575
6.125 1.11249 27.97211
6.275 1.06875 27.37649
6.425 1.02457 26.80716
6.575 0.97997 26.26254
6.725 0.93491 25.74113
6.875 0.88934 25.24157
7.025 0.84319 24.76258
7.175 0.79632 24.30298
7.325 0.74857 23.86167
7.475 0.69974 23.43764
7.625 0.64951 23.02992
7.775 0.59748 22.63764
7.925 0.54305 22.25996
8.075 0.48533 21.89613
8.225 0.4229 21.54542
8.375 0.35312 21.20715
8.525 0.27023 20.8807
8.675 0.15414 20.56549

 

[A.T.]

0.376 does not have dimensions.

The parameter \mu_{2} in Bauer's paper is also dimensionless. It relates two forces, just like 0.376 relates two accelerations. In neither case it is meant to imply that the two quantities affect each other.

 

[Redbelly98]

I posted 2 posts last night, one in reply to omcheeto, and both got lost somewhere.

The two posts were automatically put in moderation, meaning they were not visible except to forum Mentors -- we're not sure why this happened.

I have made the posts visible again, here are direct links:

Post # 208
https://www.physicsforums.com/showthread.php?p=3427707"

 

[RonL]

Just can't help myself
If all the smoke and mirrors, brain twisting mechanics, and mathematical mysteries are ignored, this reduces to a simple open air, compressor system. The surroundings are what would be considered a container.

Compression and expansion with heat being converted to work.

Llyricist did a beautiful job with his posts.



Ron

 

[spork]

Just can't help myself
If all the smoke and mirrors, brain twisting mechanics, and mathematical mysteries are ignored, this reduces to a simple open air, compressor system. The surroundings are what would be considered a container.

Compression and expansion with heat being converted to work.

I designed and built the thing, and you've lost me. What heat is being converted to work?

Llyricist did a beautiful job with his posts.

Yes he did. He's also done some great analysis and some really nice animations.

 

[A.T.]

heat being converted to work.

I think you have it backwards. The work done by the air on the cart (in the ground frame) is converted into heat (once cart is at constant speed). To convert heat into work you would need a temperature difference. But this is powered by the wind, which is a velocity difference.

 

[OmCheeto]

Thanks for the welcome back, from my self imposed ban
In answer to your above question, my thought goes back to early day flight, when a pilot tried to make a turn and the reaction to the plane was a nose dive into the ground. In this thread, weight and spread of wheel from center of propeller axis is important for stability. (just my thoughts).
I'll not say more, as it would be easy for me to get this thread locked.
I feel like a recovering alcoholic resisting his first temptation for a drink. You do remember spaghetti physics ? I have gone through the entire thread and at post 153 rcgldr made comments about the real answers of what is going on. Where there is motion there is temperature change.
That's all for me. Back to my ban and studies.

Ron

And thank you for pointing out inconsistencies:

Post # 153

A zero loss cart could achieve 5x wind speed with these numbers, but we also know it's actual speed in one of it's runs was "only" 2.8x wind speed. With a 10 mph tailwind, that means 28 mph ground speed.

It is seen from Equation (9), that the velocity of this type of vehicle tend to infinity for a given velocity difference between the media (= wind speed) as the product of the efficiencies tend to unity.

all bolding mine

5 and infinity are not quite the same.

Consistency people! Consistency!

Reference #2 shot down.

But just imagine if equation #9 were correct!

Equation #9
(Vehicle Velocity) / (Velocity Difference between wind and ground) = 1/((1/efficiency)-1)

solving for Vehicle Velocity and a cart efficiency of 95% we get:
95% efficiency --> 19x multiplier

So with a 40.42 mph wind, we have a theoretical vehicle velocity of 768mph = mach 1



Where have I seen https://www.physicsforums.com/showthread.php?t=450364" before?

Oh shoot. That was for a cart going against the wind. Maybe they should turn the Blackbird around. I would really like to see it break the sound barrier in a 41 mph headwind.



And I would try and analyze Llyricist's numbers, but I have to go to work today. So until next weekend, I can do little more than exchange a multitude of friendly barbs.

 

[spork]

But just imagine if equation #9 were correct!

It is correct.

Equation #9
(Vehicle Velocity) / (Velocity Difference between wind and ground) = 1/((1/efficiency)-1)

solving for Vehicle Velocity and a cart efficiency of 95% we get:
95% efficiency --> 19x multiplier

So with a 40.42 mph wind, we have a theoretical vehicle velocity of 768mph = mach 1

It says the cart could theoretically achieve any multiple of wind speed. It doesn't say it could achieve any speed. Compressiblity effects will reduce the cart's efficiency by definition as the prop tips start getting into the 250 mph range.

Don't read too much into a single (correct) equation.

 

[Subductionzon]

Where have I seen https://www.physicsforums.com/showthread.php?t=450364" before?

Oh shoot. That was for a cart going against the wind. Maybe they should turn the Blackbird around. I would really like to see it break the sound barrier in a 41 mph headwind.

Probably not this year, but very probably next year:
http://www.fasterthanthewind.org/"

 

[OmCheeto]

It is correct.

It says the cart could theoretically achieve any multiple of wind speed. It doesn't say it could achieve any speed.

Yes it did.

the velocity of this type of vehicle tend to infinity for a given velocity difference

bolding mine again

even just approaching infinity is pretty fast in my book.

I would do the math of how efficient your vehicle would have to be, if powered only by a http://www.petsnails.co.uk/faq.html#q120", to reach mach 1, but that would be silly.

Compressiblity effects will reduce the cart's efficiency by definition as the prop tips start getting into the 250 mph range.

Don't read too much into a single (correct) equation.

Agreed.

And just in case my references to https://www.physicsforums.com/showthread.php?p=3423521#post3423521" get confusing:
http://www.boatdesign.net/forums/attachments/propulsion/28167d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddw2.pdf"
http://orbit.dtu.dk/getResource?recordId=241183&objectId=2&versionId=1"
http://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf"

 

[spork]

Yes it did.

Yes - the author seems to have been a little too fast and loose with his wording. I stand corrected.

I will say however that one could say "its speed will tend to infinity in theory". That's the classic catch-all that means "I don't want to consider all real-world effects like compressibility - so I'll use an over-simplified theory".

 

[A.T.]

5 and infinity are not quite the same.

"5 x windspeed" referred to a cart with a certain effective gearing (transmission ratio & pitch)

"tend to infinity" simply means that there is no upper bound on the windspeed multiple if effective gearing is not fixed.

Consistency people! Consistency!

Reference #2 shot down.

Not quite. You have quoted rcgldr out of context, to compare apples & oranges, so that you can falsely claim "inconsistency".

And even if some forum post would contradict the reference, how is that "inconsistency" that shoots down the reference?

But just imagine if equation #9 were correct!

It is correct

Equation #9
(Vehicle Velocity) / (Velocity Difference between wind and ground) = 1/((1/efficiency)-1)

solving for Vehicle Velocity and a cart efficiency of 95% we get:
95% efficiency --> 19x multiplier

So with a 40.42 mph wind, we have a theoretical vehicle velocity of 768mph = mach 1

Yes, IF you can achieve 95% overall efficiency at mach 1. But that's a big IF. The equation doesn't say anything about what overall efficiency is possible at which speed. It just says that there is no upper bound on windspeed multiple IF there is no lower bound (> 0) on losses.

 

[rcgldr]

One thing I don't remember seeing is an explanation of the offset rear wheels which they claim is to counter the effect of prop torque.

To summarize the explanation that ended up in a few posts spread out through this thread:

Note that if a vehicle is at rest in a no-wind condition on a level surface, offset wheels do not produce any net torque. The BB just needed a wider wheelbase to prevent rolling over due to the torque at the propeller, but the BB was too close to the legal width limitation for a towed vehicle to extend the wheelbase equally, so they only extended the left axle.

But just imagine if equation #9 were correct! Equation #9:
(Vehicle Velocity) / (Velocity Difference between wind and ground) = 1/((1/efficiency)-1)

From a previous post:

Assuming a sailcraft holds a constant heading θ relative to the true wind (θ = zero means in the direction of the true wind), then the apparent wind can be split into two components:

apparent_crosswind = wind_speed x sin(θ)
apparent_headwind = sailcraft_speed - (wind_speed x cos(θ))

Note the crosswind component is constant, regardless of the speed of the sailcraft. Only the apparent headwind is related to the sailcraft speed.

All of the thrust from the sail on a sailcraft is due to diversion of the apparent crosswind.

In the case of vehicles that get their power from two media moving respect to each other, power input is ideally independent of the vehicles speed, and with some finite amount of power input, and power consumption that approaches zero as a vehicle's efficiency goes to 100% (no losses, no drag), then it makes sense that there is no mathematical limitation on speed for a theorectial vehicle in a theoretical situation (one where supersonic speeds don't affect the outcome).

 

[ThinAirDesign]

... the BB was too close to the legal width limitation for a towed vehicle to extend the wheelbase equally, so they only extended the left axle.

Not really -- we didn't extend both sides because it was only productive to extend one side. Extending the other side would have been a waste of time, money, materials.

JB

 

[OmCheeto]

Not really -- we didn't extend both sides because it was only productive to extend one side. Extending the other side would have been a waste of time, money, materials.

JB

Very smart indeed.

But I am still trying to calculate the torque on the vehicle, and the assumed angle of attack of the prop that I ran in my simulation were pointed out to be wrong. I've gone back and tried to figure out what the effective AOA of the Blackbirds prop is but I seem to get conflicting data now.

From A.T.'s images on https://www.physicsforums.com/showpost.php?p=3352297&postcount=51", there is the following graph:

It shows a blade_angle(deg) of from ~17 to ~26 degrees
It also lists a blade_AOA(deg) of from ~9 to ~4 degrees

Not being an aeronautical engineer, I do not understand the difference.

Also, Llyricist listed two sets of prop data:

Post #208 seemed to have some very strange data. The radius went up, and then the radius went down, whilst the incidence went down.
Post #230 seemed to make more sense, as the angle of incidence minimized at ~20.6 degrees. More in line with A.T.'s "blade angle".

As I recall, I simplified this problem in my simulation to have a single AOA for the prop.

I don't know if that was proper, but it made for much easier maths, given I'd never heard of Javaprop at the time.

The image of the prop on the Discover Channel (post #75) makes it look as there is almost no twist at all, which seems to contradict Llyricist's numbers of from 80 to 20 degrees from axis to tip end.

I've decided that there is a torque on the vehicle due to the prop, (Yes, there is hope for me yet.), just not the value.

btw, did I mention this was a very fun problem?

 

[Llyricist]

The first numbers I posted were indeed messed up, like I said.

The prop IS twisted, like all well designed props, and it is twisted to keep the angle of attack more or less constant at design speed.

If you go to fasterthanthewind.org, and scroll down on the right to get to the older blogs, for the end of 2009 and the beginning of 2010, you can see the construction process of the propeller blades and get a better idea of the twist.

 

[spork]

To first order approximation we run about 5 degrees angle of attack at all stations of the blade at our design point.

 

[Llyricist]

The two posts were automatically put in moderation, meaning they were not visible except to forum Mentors -- we're not sure why this happened.

I have made the posts visible again, here are direct links:

Post # 208
https://www.physicsforums.com/showthread.php?p=3427707"

Thank you.

 

[Llyricist]

Llyricist did a beautiful job with his posts.



Ron

Thanks, I have to admit, I had a tough learning curve for the energy accounting in a frame of reference where the ground is stationary. I knew it worked, and could do it in other frames, but that negative work thing gave me a hard time.

 

[Llyricist]

Yes he did. He's also done some great analysis and some really nice animations.

And if anyone is interested, and you have flash player installed, they are animations of all the runs they did on July 3, 2010, showing the wind direction and relative wind direction, cart speed, wind speed, apparent wind speed, and wind speed multiple. And they can be viewed here:

Runs 1 and 2: http://megaswf.com/serve/1146662/"

Runs 3 and 4: http://megaswf.com/serve/1146664/"

Runs 5 and 6: http://megaswf.com/serve/1146723/"

Runs 7 and 8: http://megaswf.com/serve/1147000/"

I had to split them up because of Flash's Total frames limitation.

 

[A.T.]

From A.T.'s images on https://www.physicsforums.com/showpost.php?p=3352297&postcount=51", there is the following graph:

It shows a blade_angle(deg) of from ~17 to ~26 degrees
It also lists a blade_AOA(deg) of from ~9 to ~4 degrees

This numbers came from JavaProp when I automatically chose the best prop pitch stetting for every given speed, and estimated efficiencies.

blade_angle : the angle between airfoil cord and propeller disc at 75% propeller radius
blade_AOA : is the angle between airfoil cord and the relative airflow at 75% propeller radius

Post #208 seemed to have some very strange data. The radius went up, and then the radius went down, whilst the incidence went down.
Post #230 seemed to make more sense, as the angle of incidence minimized at ~20.6 degrees. More in line with A.T.'s "blade angle".

Yes, in the propeller geometry the angle of incidence (deg) at 0.75 radius corresponds to my blade_angle. But the variable propeller pitch can offset the angle of incidence along the blade by some number.

The image of the prop on the Discover Channel (post #75) makes it look as there is almost no twist at all, which seems to contradict Llyricist's numbers of from 80 to 20 degrees from axis to tip end.

Better perspective:
http://www.fasterthanthewind.org/2009/12/checking-alignment.html