Power available to a wind powered vehicle traveling directly downwind

[electrodacus]

I will like the equation describing the amount of wind power available to a wind powered vehicle driving directly downwind.
I'm interested in the ideal case so max theoretical power available to a wind only powered vehicle driving directly downwind.

I do know what the equation is but I like to hear it from others so that we can agree on this first.

 

[Henryk]

It depend on how big the sail is, really.

 

[electrodacus]

It depend on how big the sail is, really.

Yes but that area will be part of the equation. It also depends on wind speed and vehicle speed.
I'm interested in the equation not a numeric value.

 

[phinds]

Why don't you tell us what you DO know instead of being coy about it !

 

[electrodacus]

Why don't you tell us what you DO know instead of being coy about it !

I wanted to be from one of you but since you asked here it is.
0.5 * air density * area * (wind speed - vehicle speed)³

 

[erobz]

I wanted to be from one of you but since you asked here it is.
0.5 * air density * area * (wind speed - vehicle speed)³

Thats not quite what I get.

Do you know the definition of power $P$ in terms of force and velocity?

 

[electrodacus]

Thats not quite what I get.

Do you know the definition of power $P$ in terms of force and velocity?

Please feel free to post your equation especially as you think mine is wrong.

 

[phinds]

I wanted to be from one of you but since you asked here it is.
0.5 * air density * area * (wind speed - vehicle speed)³

Do the units of that equation come out to be the units of power?

 

[erobz]

Please feel free to post your equation especially as you think mine is wrong.

Use the definition of Power $P = F \cdot v$, you will see what I mean. ( oh and you forgot to include the coefficient of drag as well in the force of drag )

 

[electrodacus]

Do the units of that equation come out to be the units of power?

Yes. There result of that equation will be Watt

 

[electrodacus]

Use the definition of Power $P = F \cdot v$, you will see what I mean. ( oh and you forgot to include the coefficient of drag as well in the force of drag )

Like I mentioned post the wind power equation (ideal case) for a wind powered vehicle traveling directly downwind.

 

[erobz]

Like I mentioned post the wind power equation (ideal case) for a wind powered vehicle traveling directly downwind.

You should be able to figure it out yourself. You are pretty close.

Ask yourself carefully. What is the force the wind will apply, and what is the velocity at which that force is applied.

Also, you should refrain from making demands to people who are trying to help you. It shows poor character.

 

[electrodacus]

You should be able to figure it out yourself. You are pretty close.

Ask yourself carefully. What is the force the wind will apply, and what is the velocity at which that force is being applied.

:)
OK let say the sail area is 1m² the wind speed is 10m/s and the vehicle speed is 5m/s directly downwind so what is the wind power available to vehicle ideal case.

Using the equation I provided
0.5 * 1.2 * 1 * (10-5)³ = 75W

What is your numerical result for the problem and how did you made the calculation ?

 

[erobz]

:)
OK let say the sail area is 1m² the wind speed is 10m/s and the vehicle speed is 5m/s directly downwind so what is the wind power available to vehicle ideal case.

Using the equation I provided
0.5 * 1.2 * 1 * (10-5)³ = 75W

What is your numerical result for the problem and how did you made the calculation ?

Two things.

1) You are leaving out the drag coeficient. https://en.wikipedia.org/wiki/Drag_coefficient

2) You are improperly applying the definition of power

 

[electrodacus]

Two things.

1) You are leaving out the drag coeficient. https://en.wikipedia.org/wiki/Drag_coefficient

2) You are improperly applying the definition of power

1) consider the drag coefficient as 1
2) How so ?

There is something else that I want to explain but this first step is important.
We need to agree on what the wind power available to a direct downwind vehicle is.

 

[erobz]

1) consider the drag coefficient as 1

You can, doesn't make it true

2) How so ?

What is the velocity of the of the applied force?

$$ P = F \cdot v $$

 

[electrodacus]

You can, doesn't make it true

What is the velocity of the of the applied force?

$$ P = F \cdot v $$

Is my problem incomplete ? Is it something else that you will need to find the amount of wind power available to this vehicle ?
You have 1m² sail with coefficient of drag of 1
Wind speed 10m/s
Vehicle speed 5m/s directly downwind.
air density of 1.2kg/m³

I do not care about the force but of course you can use the generic equation you provided earlier to find the force from power and speed (witch is the speed delta between wind and vehicle speed or put in a different way the wind speed relative to vehicle that is (10-5)).

 

[erobz]

Is my problem incomplete ? Is it something else that you will need to find the amount of wind power available to this vehicle ?
You have 1m² sail with coefficient of drag of 1
Wind speed 10m/s
Vehicle speed 5m/s directly downwind.
air density of 1.2kg/m³

I do not care about the force but of course you can use the generic equation you provided earlier to find the force from power and speed (witch is the speed delta between wind and vehicle speed or put in a different way the wind speed relative to vehicle that is (10-5)).

There are two things you need to consider.

There is the force that is applied to the sail . That will involve the relative velocity $(w-v)$

Then, there is the velocity at which that force is applied to the vehicle...which is?

 

[electrodacus]

There are two things you need to consider.

There is the force that is applied to the sail . That will involve the relative velocity $(w-v)$

Then, there is the velocity at which that force is applied to the vehicle...which is?

I'm going to ask again.
Do you think the data I provided is not enough for you to calculate the available wind power to vehicle ?
You do not need to know the force in order to calculate the wind power available to vehicle.
You already have air density and vehicle area interacting with air.

 

[erobz]

I'm going to ask again.
Do you think the data I provided is not enough for you to calculate the available wind power to vehicle ?
You do not need to know the force in order to calculate the wind power available to vehicle.
You already have air density and vehicle area interacting with air.

The data is missing the drag coefficient, it’s important whether you like it or not.

Otherwise the formulation is still incorrect because you refuse to answer my questions that are leading you to it. The old adage “You can lead a horse to water, but you can’t make it drink” comes to mind.

1. What is the force acting on the cart?
2. What is the velocity of the cart to which it is applied?

The power is calculated from answering the questions and multiplying them.

 

[electrodacus]

The data is missing the drag coefficient, it’s important whether you like it or not.

Otherwise the formulation is still incorrect because you refuse to answer my questions that are leading you to it. The old adage “You can lead a horse to water, but you can’t make it drink” comes to mind.

1. What is the force acting on the cart?
2. What is the velocity of the cart to which it is applied?

The power is calculated from answering the questions and multiplying them.

I think I already mentioned that coefficient of drag is 1.
1. The force acting on the cart can be calculated from the data I already provided.
2. In the example I use cart velocity of 5m/s but you can take any other value you prefer.

Maybe 5m/s for cart speed was not the best example due to symmetry as 10m/s wind speed - 5m/s cart speed will also be 5m/s so we can use 4m/s for the cart to eliminate the possible confusions due to symmetry.So for:
wind speed = 10m/s
cart speed = 4m/s
area = 1m² with a coefficient of drag of 1
air density 1.2kg/m³

Wind power available to the cart will be

0.5 * air density * area * coefficient of drag * (wind speed - cart speed)³
0.5 * 1.2 * 1 * 1 * (10-4)³ = 129.6W

Since I now now the available wind power and wind speed relative to vehicle I can calculate the force but that is irrelevant as it is not something I'm interested in.

The point I want to make will be related to conservation of energy but before I can do that we need to agree on the equation describing the amount of wind power available to a vehicle traveling directly downwind.

 

[erobz]

I think I already mentioned that coefficient of drag is 1.
1. The force acting on the cart can be calculated from the data I already provided.
2. In the example I use cart velocity of 5m/s but you can take any other value you prefer.

Maybe 5m/s for cart speed was not the best example due to symmetry as 10m/s wind speed - 5m/s cart speed will also be 5m/s so we can use 4m/s for the cart to eliminate the possible confusions due to symmetry.So for:
wind speed = 10m/s
cart speed = 4m/s
area = 1m² with a coefficient of drag of 1
air density 1.2kg/m³

Wind power available to the cart will be

0.5 * air density * area * coefficient of drag * (wind speed - cart speed)³
0.5 * 1.2 * 1 * 1 * (10-4)³ = 129.6W

Since I now now the available wind power and wind speed relative to vehicle I can calculate the force but that is irrelevant as it is not something I'm interested in.

The point I want to make will be related to conservation of energy but before I can do that we need to agree on the equation describing the amount of wind power available to a vehicle traveling directly downwind.

Lets back up:

What is force of drag acting on the cart? Just the force. NOT the power. Leave it in all variables.

 

[electrodacus]

Lets back up:

What is force of drag acting on the cart? Just the force. NOT the power. Leave it in all variables.

You have an obsession with force :)
I on the other hand have an obsession with power and I do not care about the force.

You can continue the example I provided and find out the force but I do not care about force at all.
Calculated wind power was 129.6W and speed delta was 10m/s - 4m/s = 6m/s
Thus force will be 129.6W / 6m/s = 21.6N

Air is made out of small particles that collide with the vehicle and thus transfer their kinetic energy to vehicle.

 

[erobz]

You have an obsession with force :)
I on the other hand have an obsession with power and I do not care about the force.

You can continue the example I provided and find out the force but I do not care about force at all.
Calculated wind power was 129.6W and speed delta was 10m/s - 4m/s = 6m/s
Thus force will be 129.6W / 6m/s = 21.6N

Air is made out of small particles that collide with the vehicle and thus transfer their kinetic energy to vehicle.

Power equals Force times Velocity.

In this case $P = F_D v$

In order to correctly determine the power tell me what the force of drag is $F_D$ is! Your equation for Power is wrong.

 

[electrodacus]

Power equals Force times Velocity.

In this case $P = F_D v$

In order to correctly determine the power tell me what the force of drag is $F_D$ is! Your equation for Power is wrong.

Please provide me with what you think is the correct equation for wind power available to cart.
You have all the data you need and force is not needed to calculate this.

Are you saying that the below data is insufficient for you to calculate the wind power available to cart ?
"
wind speed = 10m/s
cart speed = 4m/s
area = 1m2 with a coefficient of drag of 1
air density 1.2kg/m3
vehicle travels directly downwind.
"

 

[erobz]

Please provide me with what you think is the correct equation for wind power available to cart.
You have all the data you need and force is not needed to calculate this.

Are you saying that the below data is insufficient for you to calculate the wind power available to cart ?
"
wind speed = 10m/s
cart speed = 4m/s
area = 1m2 with a coefficient of drag of 1
air density 1.2kg/m3
vehicle travels directly downwind.

No, I can find the power easily with that data.

You however cannot properly determine that power because of your lack of cooperation. In the future, don't bother asking for help if you are unwilling to entertain the idea you might be incorrect.

 

[electrodacus]

No, I can find the power easily with that data.

You however cannot properly determine that power because of your lack of cooperation. In the future, don't bother asking for help if you are unwilling to entertain the idea you might be incorrect.

I was not asking for help with the equation. I was asking for confirmation that equation is correct (witch I know it is).
I want to explain something else but for me to do so we need to agree on equation describing available wind power to a wind powered only vehicle traveling directly downwind.

So as you say you can find the power easily with that data please provide the equation (as you seem to claim the one I provided is wrong).
Once you do so we can compare the prediction the equation make and compare to reality.

 

[erobz]

I was not asking for help with the equation. I was asking for confirmation that equation is correct (witch I know it is).
I want to explain something else but for me to do so we need to agree on equation describing available wind power to a wind powered only vehicle traveling directly downwind.

So as you say you can find the power easily with that data please provide the equation (as you seem to claim the one I provided is wrong).
Once you do so we can compare the prediction the equation make and compare to reality.

Maybe someone else will give you the proper result...I'm not wasting any more time.

 

[electrodacus]

Maybe someone else will give you the proper result...I'm not wasting any more time.

You do not seems to have the answer and you are also unwilling to learn.

My question is super simple
What is the equation describing wind power available to vehicle assuming the known data above.

 

[A.T.]

I'm interested in the ideal case so max theoretical power available to a wind only powered vehicle driving directly downwind.

That limit irrelevant to DDWFTTW vehicles, for the reason stated here:

This limit is not relevant for the DDWFTTW cart, because the goal is not to extract as much energy per air volume as possible. The goal is to extract just enough energy to offset the losses and continue to accelerate. The goal is to be efficient, and for propellers that means: Accelerate a lot of air by a little bit. So the optimum is not to stop the air relative to the ground, but the opposite: disturb it as little as possible.

There is no hard theoretical limit on how efficient the cart can be made, and how many wind-speed multiples it can achieve.