Sailing downwind faster than the wind

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spork

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origin":2bd0sj33 said:
Fine, how did I change the scenario you presented?

In my scenario there is a generator driven by the wheels, the prop is powered by an electric motor, and the electric motor is powered by the generator. This would appear to most to suggest perpetual motion - and indeed it would be perpetual motion if it could maintain any speed at all in such a configuration if it could do so without an external power source. But I showed that the wind acts as the external power source that powers that feedback loop.

The energy harnessed by the wheels in this case can clearly exceed the energy required by the prop to maintain DDWFTTW.
 
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origin

Guest
Here was your scenario:

spork":17r6bu95 said:
Let's say our cart is moving along at 55 ft/sec (in the ballpark of 40 mph). And let's say it's taking 10 lbs of force at the road surface to turn the wheels. Without worrying just yet about the fact that this would slow our cart down, wouldn't you agree that I'm putting exactly 1 horsepower into the cart (550 ft-lbs/sec)? In other words, let's assume for now that I'm towing the thing behind my car. I could hook up a generator to the cart's wheels and that generator could put out 1 h.p. of electricity (if it were 100% efficient). And yes - that energy would have to come from my car towing the cart.

I agreed with this. You even acknowledge that the car would have to put out more power because of the breaking affect of the generator. If the power transfer from the wheels to the generator was 100% efficient (I am being generous) you would now have to put into 2 hp to maintain your speed. I have changed nothing I have only put a number on your correct claim that it would take more energy to hook up the generator.

You then said:
spork":17r6bu95 said:
OK, we're on the same page so far. You agree we can pull the cart with a motor and generate 1 H.P. with a generator attached to the wheels, or we can put that motor on the cart and do the same thing. *presumably* that motor will have to put out MORE than 1 H.P. to accomplish this - but we'll get back to that.

You then finished with this:
spork":17r6bu95 said:
Great. So now let's see what it actually takes to turn that prop. Our cart is going 55 ft/sec. Let's say we have a 22.5 ft/sec tailwind (to keep the math simple). So this means there's 22.5 ft/sec headwind as far as the cart is concerned.

OK, our prop is producing 10 lbs of thrust, and it's using that thrust to advance through a 22.5 ft/sec headwind. Using the definition of prop efficiency, we can see that it would require exactly 1/2 H.P. input to that prop if the prop were 100% efficient (10 lbs x 22.5 ft/sec). But our prop is not 100% efficient. Let's make it 85% efficient. That means it's going to take about 0.6 H.P. to turn our prop.

This means we still have an excess of 0.4 H.P. from the generator that's being turned by our wheels if the prop pushes us with 10 lbs of thrust. And the reason this is possible is that the wheels are getting 10 lbs at 55 ft/sec, while the prop is delivering 10 lbs at only 22.5 ft/sec.

Of course we're going to need some of that 0.4 excess H.P. to overcome the inefficiency of the generator on the wheels, rolling resistance, and aerodynamic drag - but with careful engineering, we can easily cover that with our 0.4 H.P.

This is where you changed your own scenario. You said “And let's say it's taking 10 lbs of force at the road surface to turn the wheels [at 55 ft/sec]” So clearly you are going to need more than 0.4 hp to maintain your speed since 0.4 hp is less than 1 hp, you will need more than twice that amount of power available to maintain your speed.

You keep forgetting that the you said you start out with 1hp loss due to the wheels and then you add another 1 hp loss due to the CEMF of the generator for a total of 2 hp loss. This is your scenario
 
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spork

Guest
origin":1qjjo51l said:
You keep forgetting that the you said you start out with 1hp loss due to the wheels and then you add another 1 hp loss due to the CEMF of the generator for a total of 2 hp loss. This is your scenario

I'm forgetting nothing. I did not say there was a 1 H.P. loss from the wheels. I said it would take 1 H.P. to turn the generator. We were not yet looking at losses. So that is NOT my scenario. Honestly, who do you think is more likely to interpret my scenario wrong - you or me?

But once again, you don't want me to ignore real-world losses? Fine - give me the numbers and I'll show you a scenario where I can use YOUR numbers and go DDWFTTW. Give me:

Rolling resistance (our cart is about 0.01)
Coefficient of drag (our cart is probably about 0.3 now)
Frontal area (we're maybe 10 sq-ft)
Transmission efficiency (I'm pretty sure we're at 90% or better)
Prop efficiency (also in the 90% or better range)

So let's quit arguing about whose scenario we're using. We'll use YOUR scenario. Give me your real-world numbers and I'll show you how they add up to DDWFTTW - just as our real-world cart does in the real-world.
 
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SubductionZone

Guest
origin, Drela's boat works on the same principle as spork's cart. They both reduce the relative speed of one medium with respect to the other.

Think of it this way, is not the wind blowing over the land a huge potential energy source? Since you are not here right now I will answer for you, it is obvious, yes.


If you slowed down a bit of that wind would not the potential energy of the wind also be lowered?

Again the answer is yes.

Third question, where do you think the energy went? In spork's case it went into his cart to power it faster than the wind. As long as it can slow down the wind it interacts with with respect to the ground it has a source of energy. The same goes for Drela's boat. It lowers the difference in speed between the two mediums that it interacts with. Now you know the where the energy comes from. The how is a bit more difficult.
 
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spork

Guest
SubductionZone":1mn9lxpr said:
Now you know the where the energy comes from. The how is a bit more difficult.

If he'll give me his real world numbers we can walk through the how without arguing about whether I'm using ludicrously optimistic numbers or leaving anything out. Fingers crossed...
 
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origin

Guest
spork":wf79gt25 said:
origin":wf79gt25 said:
You keep forgetting that the you said you start out with 1hp loss due to the wheels and then you add another 1 hp loss due to the CEMF of the generator for a total of 2 hp loss. This is your scenario

I'm forgetting nothing. I did not say there was a 1 H.P. loss from the wheels. I said it would take 1 H.P. to turn the generator. We were not yet looking at losses. So that is NOT my scenario. Honestly, who do you think is more likely to interpret my scenario wrong - you or me?

This is what you said:
Let's say our cart is moving along at 55 ft/sec (in the ballpark of 40 mph). And let's say it's taking 10 lbs of force at the road surface to turn the wheels. Without worrying just yet about the fact that this would slow our cart down, wouldn't you agree that I'm putting exactly 1 horsepower into the cart (550 ft-lbs/sec)? In other words, let's assume for now that I'm towing the thing behind my car.

That is 1hp.

I could hook up a generator to the cart's wheels and that generator could put out 1 h.p. of electricity (if it were 100% efficient).

That is another 1 hp

edited to fix the format
 
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SubductionZone

Guest
Wrong origin, that would be a different 1 hp. spork is not proposing both directly driving the propeller from the wheels and driving a generator. He is proposing doing one or the other.
 
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spork

Guest
origin":3b3vzu7l said:
This is what you said...

I know what I said - it's still here in black and white. You're claiming my analysis is wrong because I've left certain losses out - but you refuse to give me numbers to work with. I have to assume that's because you realize I'll plug YOUR numbers in my equation and show you that I can make it work - just like the real cart in the real world.

I thought you agreed above to have an honest discussion. I have no interest in discussing whether you think I understand my own scenario. Give me some real world numbers with real losses.
 
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ThinAirDesigns

Guest
origin":3dt0jx7k said:
That is 1hp.

That is correct orgin ... there is 1hp going *INTO* the vehicle through the tow rope from the car. The car is applying 10lbs of force at 55ft per sec.

That is another 1 hp

Yes orgin, that is 1hp taken OUT of the vehicle through the drive axle.

And your question was?

JB
 
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origin

Guest
ThinAirDesigns":qodyvwtj said:
origin":qodyvwtj said:
That is 1hp.

That is correct orgin ... there is 1hp going *INTO* the vehicle through the tow rope from the car. The car is applying 10lbs of force at 55ft per sec.

That is another 1 hp

Yes orgin, that is 1hp taken OUT of the vehicle through the drive axle.

And your question was?

JB

I have no question, you at least seem to understand. I takes 1 hp going *into* the vehicle to tow the vehicle. Now we hook up a generator (according to the scenario) to the wheels which takes 1 hp OUT through the drive axle. Now obviously to maintain the speed you must REPLACE the 1 hp taken OUT of the vehicle - so it will take 2 hp of towing power.

You guys should really relax - I am on the verge of agreeing with the ability of a craft to go faster than the wind going directly down wind. Not through any of your explanations though, all you guys do is confuse the issue. I have read Dr. Drela's paper and have been corresponding with him. I was amazed he would take the time answer me especially at the end of the spring semester. I am not sure if he has time to answer my last question but I think I will be able to infer the answer if he can't, so freaking chill out.
 
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spork

Guest
origin":1f80xmfr said:
You guys should really relax - I am on the verge of agreeing with the ability of a craft to go faster than the wind going directly down wind. Not through any of your explanations though, all you guys do is confuse the issue.

I'm sorry that our very simple analysis is confusing to you. But I assure you if you didn't INSIST on misinterpreting our words it would seem less so. I can also assure you that if you were willing to have an honest and responsive discussion as you claimed you would, you'd have an easier time understanding. As near as I can tell, you simply choose to be a troll on this issue.

I have read Dr. Drela's paper and have been corresponding with him.

Let me know when you come to the realization that it says exactly what we've been telling you - or do you insist on changing what's in his paper as well?

...so freaking chill out.

Quit trolling. Complaining about my numbers, while refusing to offer your own (or even acknowledge the request), gives you away.

Very soon you will have achieved step 2 of the process - you will admit DDWFTTW works, but claim we explained it all wrong. Soon after, you will achieve step 3 - you will explain it to us "correctly" (but sadly you will get it completely wrong).
 
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wick07

Guest
Okay, I just jumped into this post today, and Origin took up my sword of skepticism for me. But I think I made some sense out of it, let me know if I'm on the right track.

The thrust generated by a propeller is related to the mass times the change in velocity across the propeller (very simplified but it should work for our purposes since I don't care what this force actually is, but rather the relationship between the tailwind and headwind are important). So we shall us the equation

Thrust = m * (Vi - Vo)

Where:
m=mass of airflow across the propeller (constant)
Vi = air velocity at the input of the propeller
Vo = Velocity of air at the output of the propeller

In the static case, the thrust will be equal to the force required to turn the propeller Assuming a static case:

F1 = m * (V1 -V2)

or in other words with 'F1' power I can accelerate a mass of air 'm' from V1 to V2.

Now, lets assume Vt is a wind blowing AGAINST the propeller output. Our Vo = V2 + Vt in this case. So now our equation looks like:

F2 = M * (V1 – (V2 + Vt))

F2 < F1, and F1 is required to maintain our static case. Since there is an excess of power the system will begin to accelerate.

Am I understanding this correctly?

Origin - Would you buy-off on my math? Have I erred?
 
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spork

Guest
wick07":384mpst0 said:
Okay, I just jumped into this post today, and Origin took up my sword of skepticism for me. But I think I made some sense out of it, let me know if I'm on the right track.

The thrust generated by a propeller is related to the mass times the change in velocity across the propeller (very simplified but it should work for our purposes since I don't care what this force actually is, but rather the relationship between the tailwind and headwind are important). So we shall us the equation

Thrust = m * (Vi - Vo)

Where:
m=mass of airflow across the propeller (constant)
Vi = air velocity at the input of the propeller
Vo = Velocity of air at the output of the propeller

Yup. This is pretty good. m is mass flow rate of course.

In the static case, the thrust will be equal to the force required to turn the propeller Assuming a static case:

F1 = m * (V1 -V2)

or in other words with 'F1' power I can accelerate a mass of air 'm' from V1 to V2.

Hmmm... you've lost me a bit here. The force required to turn the prop would be expressed as a torque. That torque multiplied by the rotation rate would give power_in. Power_out / power_in = prop efficiency. Power_out is also equal to thrust * free stream velocity. That's how we could compute thrust from the torque required to turn the prop.

Now, lets assume Vt is a wind blowing AGAINST the propeller output. Our Vo = V2 + Vt in this case. So now our equation looks like:

F2 = M * (V1 – (V2 + Vt))...

I'm getting further lost. What is Vo? Also, except for the case of the cart going less than wind speed, there is no wind blowing against the prop output. The prop is immersed in a fluid that's moving (the wind). Props are typically analyzed down to the static condition. It's hard to find prop data for the reverse flow condition (i.e. our cart below windspeed).
 
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origin

Guest
spork":27k6yum7 said:
origin":27k6yum7 said:
You guys should really relax - I am on the verge of agreeing with the ability of a craft to go faster than the wind going directly down wind. Not through any of your explanations though, all you guys do is confuse the issue.

I'm sorry that our very simple analysis is confusing to you. But I assure you if you didn't INSIST on misinterpreting our words it would seem less so. I can also assure you that if you were willing to have an honest and responsive discussion as you claimed you would, you'd have an easier time understanding. As near as I can tell, you simply choose to be a troll on this issue.

Please tell me specifically where I misrepresented what you wrote.

I have read Dr. Drela's paper and have been corresponding with him.

Let me know when you come to the realization that it says exactly what we've been telling you - or do you insist on changing what's in his paper as well?

Your whole discussion about sailboats and iceboats are not germane to the subject. The reason it may be possible to go directly down wind faster than the wind has nothing to do with sailboats. Because you guy suck so badly at physics when you try to discuss generator and motors you make goofy errors and then refuse to admitt it.

You talk about Dr. Drelas math like it is complicated, it is not, it basically high school algebra. I was not farmiliar with some of the terms so I had to do some research but the math is simple.

...so freaking chill out.

Quit trolling. Complaining about my numbers, while refusing to offer your own (or even acknowledge the request), gives you away.

Very soon you will have achieved step 2 of the process - you will admit DDWFTTW works, but claim we explained it all wrong. Soon after, you will achieve step 3 - you will explain it to us "correctly" (but sadly you will get it completely wrong).

Actually, I will not explain it to you incorrectly or any other way. Talking to you guys is like talking to petulant children. I am done wasting my time.
 
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ThinAirDesigns

Guest
origin":1ybvx2vi said:
I have no question, you at least seem to understand.

Not so fast:

It takes 1 hp going *into* the vehicle to tow the vehicle.

Not without the generator dragging it doesn't -- remember for the moment we are ignoring all drags and inefficiencies (we account for them later)

Now we hook up a generator (according to the scenario) to the wheels which takes 1 hp OUT through the drive axle. Now obviously to maintain the speed you must REPLACE the 1 hp taken OUT of the vehicle - so it will take 2 hp of towing power.

Nope -- you're adding the drag of the generator *twice*. Remember, at the initial stages of this scenario, the vehicle is *100% efficient* -- it takes *no* HP to drag the vehicle at 55ft/sec ... all the 1HP going in from the towing car is being pulled out of the drive axle by the generator.

Once again, if we pull our 'no drag', 'no resistance', '100% efficient' vehicle with 10lbs of force at 55ft/sec, we can draw 1HP out of our 100% efficient generator at the drive axle ... 10lbs @55ft/sec IN and 1HP out.

Agreed?

JB
 
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spork

Guest
origin":398esfqa said:
Please tell me specifically where I misrepresented what you wrote.

Nope. That's just further trolling on your part. I've told you several times that we should use YOUR scenario and quit arguing about how you changed mine. But you refuse to give me yours because you can see where it leads.

Your whole discussion about sailboats and iceboats are not germane to the subject. The reason it may be possible to go directly down wind faster than the wind has nothing to do with sailboats.

You're absolutely wrong.

Because you guy suck so badly at physics when you try to discuss generator and motors you make goofy errors and then refuse to admitt it.

And yet our analysis accurately predicts our real-world results - while YOUR analysis explains that it's impossible.

You talk about Dr. Drelas math like it is complicated, it is not, it basically high school algebra. I was not farmiliar with some of the terms so I had to do some research but the math is simple.

I never said it was complicated. I did exactly that analysis before Drela ever considered it. There are plenty of references to that on the net. I didn't claim it was complicated - I claimed that you couldn't understand it - and that is the case.

Actually, I will not explain it to you incorrectly or any other way. Talking to you guys is like talking to petulant children. I am done wasting my time.

And thus the trolling ends (or so I wish).
 
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origin

Guest
spork":kommkdv9 said:
origin":kommkdv9 said:
Please tell me specifically where I misrepresented what you wrote.

Nope. That's just further trolling on your part. I've told you several times that we should use YOUR scenario and quit arguing about how you changed mine. But you refuse to give me yours because you can see where it leads.

Your whole discussion about sailboats and iceboats are not germane to the subject. The reason it may be possible to go directly down wind faster than the wind has nothing to do with sailboats.

You're absolutely wrong.

Because you guy suck so badly at physics when you try to discuss generator and motors you make goofy errors and then refuse to admitt it.

And yet our analysis accurately predicts our real-world results - while YOUR analysis explains that it's impossible.

You talk about Dr. Drelas math like it is complicated, it is not, it basically high school algebra. I was not farmiliar with some of the terms so I had to do some research but the math is simple.

I never said it was complicated. I did exactly that analysis before Drela ever considered it. There are plenty of references to that on the net. I didn't claim it was complicated - I claimed that you couldn't understand it - and that is the case.

Actually, I will not explain it to you incorrectly or any other way. Talking to you guys is like talking to petulant children. I am done wasting my time.

And thus the trolling ends (or so I wish).

I don't like being accused of trolling, especially when I was trying to engage you in some real communication but due to your complete ignorance of physics and your adolesent attitude that has become impossible.
 
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ThinAirDesigns

Guest
Origin, I'm very happy that someone is helping you correct your knowledge of basic physics, but when you have been insisting on adding in the simple drag of the generator *twice* when of course there is only one generator, it's a bit hard for you to convince people that it's *us* who have been making the simple errors.

JB
 
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spork

Guest
origin":39kwmrm9 said:
I don't like being accused of trolling, especially when I was trying to engage you in some real communication but due to your complete ignorance of physics and your adolesent attitude that has become impossible.

Sadly, the trolling has not ended (as I suspected). Curious that I concieved of, designed, and built (along with JB) a wind powered vehicle that goes downwind faster than the wind when I'm completely ignorant of physcs. Yet you, who knows so much about physics, claims it can't work. :lol:

And the record shows you refused to enage in an actual responsive discussion. You claimed I ignored factors, but refused (and still refuse) to provide them so I can show where you're wrong. That's pretty much the definition of trolling.
 
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ThinAirDesigns

Guest
origin":171o5uf2 said:
... due to your complete ignorance of physics ...

Yes ... lets look back at the evidence, shall we?:

A: we, from first physics principles, independently conceived of, designed, tested and built a vehicle that has approached 3 times the speed of the wind, directly downwind.

B: you, have insisted that according to the long established laws of phyics what we did is "impossible".

You might wish to reconsider who has demonstrated the better grasp of physics.

ETA: I see spork alreadly beat me to this point.

JB
 
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wick07

Guest
spork":2qomg78k said:
wick07":2qomg78k said:
In the static case, the thrust will be equal to the force required to turn the propeller Assuming a static case:

F1 = m * (V1 -V2)

or in other words with 'F1' power I can accelerate a mass of air 'm' from V1 to V2.

Hmmm... you've lost me a bit here. The force required to turn the prop would be expressed as a torque. That torque multiplied by the rotation rate would give power_in. Power_out / power_in = prop efficiency. Power_out is also equal to thrust * free stream velocity. That's how we could compute thrust from the torque required to turn the prop.

This was part of my simplification. To keep the math easier I took some shortcuts:

1) I assumed a 100% efficient system (unrealistic, but I'm just trying to establish if this is even possible; if it can't be done in an ideal system it will be impossible in a realistic one). Therefore I concluded that all power created by the wheels would have to be turned into thrust by the prop.

2) Since the mass of air I am moving is constant the trust will be proportional to the difference between the apparent air velocity at the input of the system and the apparent air velocity at the output of the system.

Spork":2qomg78k said:
Wick07":2qomg78k said:
Now, lets assume Vt is a wind blowing AGAINST the propeller output. Our Vo = V2 + Vt in this case. So now our equation looks like:

F2 = M * (V1 – (V2 + Vt))...

I'm getting further lost. What is Vo? Also, except for the case of the cart going less than wind speed, there is no wind blowing against the prop output. The prop is immersed in a fluid that's moving (the wind). Props are typically analyzed down to the static condition. It's hard to find prop data for the reverse flow condition (i.e. our cart below windspeed).

That's probably because I messed up my signs. Let's do this over again shall we. Here are the variables I am going to use:

Pa = Power available from the generator attached to the wheels. This is constant for a given speed.
F1 = The thrust created when there is NO TAILWIND.
F2 = The thrust created when their is A TAILWIND.
Vi = Air velocity on the input side of the prop system.
Vo = Air velocity at the output side of the prop system.
Vt = True tailwind velocity.
Assuming an ideal prop, the thrust created by the prop will be equivalent to the power used to turn the prop. No loss.

First the NO TAILWIND case:

F1 = m * (Vi - Vo)
Pa = F1

There will be no net forces on the system, at best we can achieve equilibrium.

Second the TRUE TAILWIND case:

Vi = Vi (The prop blocks the tailwind from interfering in the input airflow)
Vo = Vo - Vt (The true Tailwind is blowing into the prop, decreasing the output air velocity)
F2 = m * (Vi - (Vo - Vt)) (The thrust created by the prop)
F2 > F1

In the TAILWIND case the thrust created is greater than in the NO TAILWIND Case.

The power available however, has nothing to due with true or relative winds. The power available is related to the speed of the cart (which hasn't changed), not the wind speed.

If I have increased thrust because of a tailwind, that means I need less input power to the prop to achieve the same thrust as was achieved in the NO TAILWIND case. But I still have the same power available.

This means I have excess power in the TAILWIND case! I will end up accelerating!

My top speed will now be limited by my ability to utilize this excess power. The more efficient I make my machine, the higher the possible speed. This has become an engineering problem. Note that this is not perpetual motion. As we get faster and faster, the tailwind effect will decrease. At some point the internal resistance of the system will overwhelm it and we will have reached an effective no-tailwind condition. At that point the system will be in equilibrium again.

Despite this derivation, my gut tells me this is inherently wrong (I know, you did this in the real world so it must work). Which is why I want someone to double-check my logic.
 
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eyytee

Guest
origin":1imchdyp said:
I looked at Drela's analysis and it is very strange to say the least. (...)
To me this presents a problem. The turbine powers the propeller, which moves the boat, which powers the turbine, which powers the propeller, which powers the turbine, which...
The only problem here is: You are analysing a feedback-loop in terms of a linear cause-effect chain. With this naive reasoning you will get the same strange result with the ruler cart:

[youtube]http://www.youtube.com/watch?v=k-trDF8Yldc[/youtube]

The reels power the blue gear, which moves the cart, which powers the reels, which power the blue gear...

Hint: In both examples I indicated in red the effect that has two causes, not just the one mentioned.

origin":1imchdyp said:
Your whole discussion about sailboats and iceboats are not germane to the subject. The reason it may be possible to go directly down wind faster than the wind has nothing to do with sailboats.
You either don't understand sailboats or this vehicle, or neither of them. This might help you:

[youtube]http://www.youtube.com/watch?v=UGRFb8yNtBo[/youtube]

origin":1imchdyp said:
You talk about Dr. Drelas math like it is complicated, it is not, it basically high school algebra.
That "non complicated" math was given to you on the 1st page of this thread, and yet you went for 10+ pages arguing against the result of it.
 
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spork

Guest
wick07":i9efm7fy said:
In the static case, the thrust will be equal to the force required to turn the propeller

One of the main problems I'm having with the derivation is that you're mixing units. The thrust from the propeller will be a simple force. The "force" required to turn the prop will be a force at a specified distance (e.g. 10 ft-lbs).

Assuming a static case:

F1 = m * (V1 -V2)

In this case "m" has to be "m-dot" or mass flow rate (i.e. mass/time).

in other words with 'F1' power I can accelerate a mass of air 'm' from V1 to V2.

Again, F1 is force and power is force x velocity.

1) I assumed a 100% efficient system (unrealistic, but I'm just trying to establish if this is even possible; if it can't be done in an ideal system it will be impossible in a realistic one). Therefore I concluded that all power created by the wheels would have to be turned into thrust by the prop.

2) Since the mass of air I am moving is constant the trust will be proportional to the difference between the apparent air velocity at the input of the system and the apparent air velocity at the output of the system.

Both are good assumptions.

Assuming an ideal prop, the thrust created by the prop will be equivalent to the power used to turn the prop. No loss.

Again, we have a problem. Thrust is force and power is energy/time (or force times velocity)

First the NO TAILWIND case:
...
There will be no net forces on the system, at best we can achieve equilibrium.

Agreed.

Second the TRUE TAILWIND case:

Vi = Vi (The prop blocks the tailwind from interfering in the input airflow)

We can't look at this as the prop "blocking" the tailwind. The prop is immersed in the fluid and operating in the normal way a prop does in air.

In the TAILWIND case the thrust created is greater than in the NO TAILWIND Case.

This is true. The reason for this can be looked at in at least two ways.

With a tailwind, the prop is operating closer to the static case (think of a plane sitting still on the runway and revving its engine). This gives a greater AOA (angle of attack) of the blades on the air vs. blades trying to get a bite into air that's already flowing rapidly through the prop disk. Thus greater thrust.

The second way to look at it is this... power out = thrust x free_stream_velocity. The power out can never be more than the power in. With a tail-wind, the free_stream_velocity is the vehicle_velocity - the speed of the tail wind. So a given power can give me more thrust in this case.


The power available however, has nothing to due with true or relative winds. The power available is related to the speed of the cart (which hasn't changed), not the wind speed.

Yes, the power available is equal to the speed of the cart times the retarding force on the wheels from "turning the generator".

If I have increased thrust because of a tailwind, that means I need less input power to the prop to achieve the same thrust as was achieved in the NO TAILWIND case. But I still have the same power available.

Exactly!

This means I have excess power in the TAILWIND case! I will end up accelerating!

Right again.

My top speed will now be limited by my ability to utilize this excess power. The more efficient I make my machine, the higher the possible speed. This has become an engineering problem. Note that this is not perpetual motion.

Right on all counts.

As we get faster and faster, the tailwind effect will decrease.

I don't think this is true - but perhaps in a relative sense.

At some point the internal resistance of the system will overwhelm it and we will have reached an effective no-tailwind condition. At that point the system will be in equilibrium again.

This is right except for the "effective no-tailwind".
 
T

ThinAirDesigns

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origin":3ffe0vmk said:
Your whole discussion about sailboats and iceboats are not germane to the subject. The reason it may be possible to go directly down wind faster than the wind has nothing to do with sailboats.

eyytee":3ffe0vmk said:
You either don't understand sailboats or this vehicle, or neither of them.

Yes eyytee, I alway love the "it's got nothing to do with sailboats" display of ignorance. As that animation (and all the math and vector analysis) so easily shows, the only difference between the airfoil on the reaching boat and the airfoil on the cart is the diameter of the circle.

JB
 
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