Sailing downwind faster than the wind

[origin]

The ground moves backwards faster than the air relative to you, so you can apply a greater force to the air than you apply to the ground, while using less power than you get from the ground: power = force * velocity

Sorry that is not how reality works. If this true (which it is not) then all I have to do is get the cart moving in any direction and as long as the cart is moving faster than the air I will be producing power.

Nope this is impossible - perhaps you misspoke?

 

[eyytee]

The ground moves backwards faster than the air relative to you, so you can apply a greater force to the air than you apply to the ground, while using less power than you get from the ground: power = force * velocity

Sorry that is not how reality works.

Sorry that is exactly how reality works.

Every gear box does this:
low force at high velocity comes in
high force at low velocity comes out

If this true (which it is not) then all I have to do is get the cart moving in any direction and as long as the cart is moving faster than the air I will be producing power.

No the condition is:

The ground moves backwards faster than the air, relative to the cart

This never happens without true tailwind (air moving in the same direction as the cart, relative to the ground)

 

[spork]

Let's take this in steps. Do you agree that the prop blades would be feeling a relative wind while spinning on the cart that's going directly downwind at windspeed?

They would feel exactly the same relative wind speed as a stationary cart with no wind - sure.

O.K. we agree so far. Next step...
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.

Do you buy all of this?

 

[origin]

Let's take this in steps. Do you agree that the prop blades would be feeling a relative wind while spinning on the cart that's going directly downwind at windspeed?

They would feel exactly the same relative wind speed as a stationary cart with no wind - sure.

O.K. we agree so far. Next step...
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.

Do you buy all of this?

OK - I think it makes more sense to say you have to have a motor on the cart that is delivering 1 Hp to maintain the velocity - but it is the same either way.

 

[spork]

OK - I think it makes more sense to say you have to have a motor on the cart that is delivering 1 Hp to maintain the velocity - but it is the same either way.

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.

So let's go with your suggestion. We'll put that motor on the cart. We'll have it turn the prop. So for now, let's not worry about how big that motor has to be, but would you agree that it can turn the prop to produce the same 10 lbs of thrust to be equivalent to my towing the cart with 10 lbs of force?

 

[origin]

OK - I think it makes more sense to say you have to have a motor on the cart that is delivering 1 Hp to maintain the velocity - but it is the same either way.

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.

So let's go with your suggestion. We'll put that motor on the cart. We'll have it turn the prop. So for now, let's not worry about how big that motor has to be, but would you agree that it can turn the prop to produce the same 10 lbs of thrust to be equivalent to my towing the cart with 10 lbs of force?

Sure

 

[spork]

Sure

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.

 

[origin]

Sure

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.

No no no. You have changed the problem and gone off on a tangent!!

Here 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)?

So if you were not taking the wind into account then the headwind would require the cart to have more power not less.

If you were taking a head wind into account then it would still take 1 Hp.

If you were assuming still air and now you have a tail wind then of course the cart would increase in speed because the headwind is less and there is a freaking motor turning the propeller!!

 

[spork]

No no no. You have changed the problem and gone off on a tangent!!

No no no. I haven't. I have a cart that we agreed:
- Can be pulled by my cart - or by an engine on the cart itself.
- When I pull that cart at 55 ft/sec, with a 10 lb retarding force on the tires, I can generate 1 H.P. by connecting a generator to those tires. Wind never comes into this in any way yet. I have no prop or windage. We can worry later about the coefficient of drag of the chassis.
- I can take the 0.6 H.P from the 1 H.P. generated by my generator to turn my prop to create 10 lbs of thrust in a 22.5 ft/sec wind. That thrust is just as good at the 10 lbs my car is pulling the cart with, so I can cut the tow line and I'm good to go.
- This leaves me with 0.4 excess H.P. to handle the real-world drag, losses, and innefficiencies.

My analysis above matches the results that we get in the real world. Your analysis tells us why our real-world results aren't possible.

 

[ThinAirDesigns]

And as to the few pages of previously posted analysis from a preeminent MIT aero Professor -- have you found the flaw in that short set of equations Origin?

JB

 

[ThinAirDesigns]

Origin, I don't understand why you think spork is off on a "tangent". Put a generator on the wheels and a motor on the prop and do the very simple math:

It's easy to calculate the horsepower that can be generated from the wheels at 55ft sec when pulled by the car with 10lbs of force (1 H.P.)

It's easy to calculate the net headwind speed on the vehicle with a 22.5ft sec tailwind (55 - 22.5 = 22.5ft sec)

It's easy to calculate the HP required for an 85% efficient propeller to generate the same 10lbs of thrust that the car is providing at 22.5ft sec (0.6 H.P.)

Now, it also easy to subtract 0.6 from 1.0, release the tow rope from the car and see that at 2x windspeed, there is still 0.4 H.P. left over from the tailwind to spread around between the other ineffiencies in the system ... transmission losses, rolling resistance, aero drag on the chassis, etc.

Do the same calcs at the same vehicle speed but with no tailwind blowing (in other words try to get 10lbs of thrust in a 55ft sec headwind), and you soon see that your 85% efficient prop has made it impossible even without the other losses. This of course is because TINFL. The power of the wind does not constitute a free lunch.

JB

 

[origin]

No no no. You have changed the problem and gone off on a tangent!!

No no no. I haven't. I have a cart that we agreed:
- Can be pulled by my cart - or by an engine on the cart itself.
- When I pull that cart at 55 ft/sec, with a 10 lb retarding force on the tires, I can generate 1 H.P. by connecting a generator to those tires. Wind never comes into this in any way yet. I have no prop or windage. We can worry later about the coefficient of drag of the chassis.
- I can take the 0.6 H.P from the 1 H.P. generated by my generator to turn my prop to create 10 lbs of thrust in a 22.5 ft/sec wind. That thrust is just as good at the 10 lbs my car is pulling the cart with, so I can cut the tow line and I'm good to go.
- This leaves me with 0.4 excess H.P. to handle the real-world drag, losses, and innefficiencies.

My analysis above matches the results that we get in the real world. Your analysis tells us why our real-world results aren't possible.

So let me see if I have the scenario correct.

I have the cart with a motor hooked up to the propeller such that the propeller is developing a thrust that equates to 1 Hp, so that the cart moves at a steady 40 mph. Now a 20 mph tail wind comes and the cart can either go faster of if I had a potentiometer I could dial back the power output of the motor and maintain my speed.

Is this right? If so I agree.

 

[spork]

So let me see if I have the scenario correct.

I have the cart with a motor hooked up to the propeller such that the propeller is developing a thrust that equates to 1 Hp, so that the cart moves at a steady 40 mph. Now a 20 mph tail wind comes and the cart can either go faster of if I had a potentiometer I could dial back the power output of the motor and maintain my speed.

Is this right? If so I agree.

That's not precisely my scenario, but it embodies the principle of my scenario, and yes - it's correct.

 

[Mee_n_Mac]

No no no. You have changed the problem and gone off on a tangent!!

No no no. I haven't. I have a cart that we agreed:
- Can be pulled by my cart - or by an engine on the cart itself.
- When I pull that cart at 55 ft/sec, with a 10 lb retarding force on the tires, I can generate 1 H.P. by connecting a generator to those tires. Wind never comes into this in any way yet. I have no prop or windage. We can worry later about the coefficient of drag of the chassis.
- I can take the 0.6 H.P from the 1 H.P. generated by my generator to turn my prop to create 10 lbs of thrust in a 22.5 ft/sec wind. That thrust is just as good at the 10 lbs my car is pulling the cart with, so I can cut the tow line and I'm good to go.
- This leaves me with 0.4 excess H.P. to handle the real-world drag, losses, and innefficiencies.

My analysis above matches the results that we get in the real world. Your analysis tells us why our real-world results aren't possible.

So let me see if I have the scenario correct.

I have the cart with a motor hooked up to the propeller such that the propeller is developing a thrust that equates to 1 Hp, so that the cart moves at a steady 40 mph. Now a 20 mph tail wind comes and the cart can either go faster of if I had a potentiometer I could dial back the power output of the motor and maintain my speed.

Is this right? If so I agree.

This is pretty much the way the MIT analysis goes about looking at the issue. Given this example go and read the 1'st one linked to. He uses a boat and a turbine suspended in the water inlieu of a vehicle and wheels but the idea is the same as just described.

 

[origin]

So let me see if I have the scenario correct.

I have the cart with a motor hooked up to the propeller such that the propeller is developing a thrust that equates to 1 Hp, so that the cart moves at a steady 40 mph. Now a 20 mph tail wind comes and the cart can either go faster of if I had a potentiometer I could dial back the power output of the motor and maintain my speed.

Is this right? If so I agree.

That's not precisely my scenario, but it embodies the principle of my scenario, and yes - it's correct.

OK - So what's your point?

 

[origin]

This is pretty much the way the MIT analysis goes about looking at the issue. Given this example go and read the 1'st one linked to. He uses a boat and a turbine suspended in the water inlieu of a vehicle and wheels but the idea is the same as just described.

I read it but there were several effeciency terms that I am not farmiliar with so I want to do a little research to make sure I understand it. But it is low on my priority list...

 

[ThinAirDesigns]

OK - So what's your point?

His point is pretty damn clear (and doesn't require learning any new terms) ... the simple math he walked you through clearly and easily shows that in a tailwind, one can get more power from the wheels than is required by the prop.

JB

 

[origin]

OK - So what's your point?

His point is pretty damn clear (and doesn't require learning any new terms) ... the simple math he walked you through clearly and easily shows that in a tailwind, one can get more power from the wheels than is required by the prop.

JB

No he didn't. He showed that a tail wind produces a force - no kidding. If you cart has a motor turning the prop then it will go faster than the wind. That is all that was proven.

 

[Mee_n_Mac]

OK - So what's your point?

His point is pretty damn clear (and doesn't require learning any new terms) ... the simple math he walked you through clearly and easily shows that in a tailwind, one can get more power from the wheels than is required by the prop.

JB

No he didn't. He showed that a tail wind produces a force - no kidding. If you cart has a motor turning the prop then it will go faster than the wind. That is all that was proven.

How is using a generator driven by the wheels and a motor to drive the prop any different from doing it with gears ? The former is an electro-mechanical transmission, the latter purely mechanical. The effect is the same.

FWIW: the MIT analysis lacks the kind of simple force diagram the link I provided earlier had. Had it I might hope all would be clear.

 

[ThinAirDesigns]

OK - So what's your point?

His point is pretty damn clear (and doesn't require learning any new terms) ... the simple math he walked you through clearly and easily shows that in a tailwind, one can get more power from the wheels than is required by the prop.

JB

No he didn't. He showed that a tail wind produces a force - no kidding. If you cart has a motor turning the prop then it will go faster than the wind. That is all that was proven.

Please tell me which of the following you disagree with:

A: A force of 10lbs, pulling forwards on our theoretical vehicle at 55ft sec, can be used to generate 1 H.P. at the drive axle (ignoring generation losses, etc. for the moment).

B: A vehicle traveling at 55ft sec, in a 22.5ft sec tailing will be experiencing a 22.5ft sec. headwind.

C: An 85% efficient propeller can turn 0.6 H.P. into 10lbs of thrust in a 22.5ft sec. headwind.

D: 1 > 0.6

Thanks

JB

 

[spork]

OK - So what's your point?

No point. I can't take you any further with known means. I'm inclined to go with the math, physics, and real-world results. You're free to base your "understanding" on whatever you like.

 

[origin]

No he didn't. He showed that a tail wind produces a force - no kidding. If you cart has a motor turning the prop then it will go faster than the wind. That is all that was proven.

Please tell me which of the following you disagree with:

A: A force of 10lbs, pulling forwards on our theoretical vehicle at 55ft sec, can be used to generate 1 H.P. at the drive axle (ignoring generation losses, etc. for the moment).

B: A vehicle traveling at 55ft sec, in a 22.5ft sec tailing will be experiencing a 22.5ft sec. headwind.

C: An 85% efficient propeller can turn 0.6 H.P. into 10lbs of thrust in a 22.5ft sec. headwind.

D: 1 > 0.6

Thanks

JB

Lets look at the scenario again:

I have the cart with a motor hooked up to the propeller such that the propeller is developing a thrust that equates to 1 Hp, so that the cart moves at a steady 40 mph. Now a 20 mph tail wind comes and the cart can either go faster of if I had a potentiometer I could dial back the power output of the motor and maintain my speed.

Spork agreed that this is what he was trying to say. So I am maintaining a steady velocity of 40 mph, this means I am overcoming the losses in the friction in the wheels and gearing and I am overcoming the drag of an effective 40 mph headwind. So if there is a tail wind of 20 mph there will be an effective headwind of only 20 mph and it will take less power to move the cart at the same speed.

This is not very earth shattering, pretty obvious. What have we shown is that the wind has power, so in theory a sail boat should work.

Have we shown that you can travel faster than the wind going directly down wind - clearly not, this scenario soesn't even directly address it.

 

[eyytee]

Have we shown that you can travel faster than the wind going directly down wind - clearly not, this scenario soesn't even directly address it.

Well, it's your scenario. Why don't you address spork's original scenario, with a generator at the wheels that powers the prop? It is easy to show that the prop can produce more forward force than the breaking force at the wheels needed to generate the power, if the ground moves faster backwards than the air relative to the cart .

 

[spork]

Lets look at the scenario again:

Sure - but let's use the actual scenario I presented, not the new one you made up.

Spork agreed that this is what he was trying to say.

No - spork said "That's not precisely my scenario, but it embodies the principle of my scenario, and yes - it's correct."

That in no way suggests that your scenario is "what I was trying to say". Listen origin, you're free to choose not to accept math, physics, or reality - but I prefer that you not put words in my mouth.

Have we shown that you can travel faster than the wind going directly down wind - clearly not, this scenario soesn't even directly address it.

Indeed THAT scenario does not directly address DDWFTTW. I presume this is precisely why you chose it over the scenario that I laid out quite clearly.

 

[Mee_n_Mac]

Have we shown that you can travel faster than the wind going directly down wind - clearly not, this scenario soesn't even directly address it.

I'd have said the doing 40 mph directly downwind with a 20 mph wind coming directly from behind is exactly the scenario in question, 40 is going faster than the 20 mph wind. Let's recall that all the power being supplied to the prop and motor is coming from a generator driven off the wheels.

Let me rephrase the scenario a bit. We have a vehicle doing 40 mph and the power generated by the generator is exactly 1 HP (assuming the gen is 100% efficient). This is independant of any wind. It depends solely on the vehicles ability to overcome whatever drag forces there are and do 40 mph. With this 1 HP I to drive a motor and prop. Will I be able to do 40 mph all things left the same ? The answer is no because I need 1 HP, output from the prop, to go 40 mph and in real life nothing is 100% efficient. The generator won't be, the motor won't be, the prop won't be, etc, etc. Now along comes a 20 mph tailwind. This reduces the work that needs to be done by the motor and prop. The only question now is if that reduction is large enough to allow for real world in-efficiencies to be overcome, to have enough power coming from the real life gen (less than 1 HP) to drive an inefficient motor and prop to overcome the now reduced load.