Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147435 times)

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Offline Kleinstein

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Going to 0.5 m/s in only 2 ms  sounds like a lot of acceleration - actually 250 m/s² and thus about 25 G, which for a 10 kg vehicle would need 2500 N of force. Something in this calculation does not sound right !  :-//
 

Online bdunham7

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No motion of the vehicle means zero power can be generated at the wheel.

Let's stick with that for a moment.

So you say that for any non-zero speed, the power is (whatever your formula is) or about 60N * (wind speed - vehicle speed), but right at zero the power goes to zero?

So if I'm sitting there and the vehicle is stopped with brakes, there's no energy being dissipated by the brake, but if I allow the vehicle to creep forward at 1mm/s, now the brakes are dissipating almost 600W?

A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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No motion of the vehicle means zero power can be generated at the wheel.

Let's stick with that for a moment.

So you say that for any non-zero speed, the power is (whatever your formula is) or about 60N * (wind speed - vehicle speed), but right at zero the power goes to zero?

So if I'm sitting there and the vehicle is stopped with brakes, there's no energy being dissipated by the brake, but if I allow the vehicle to creep forward at 1mm/s, now the brakes are dissipating almost 600W?

Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W
In real world there is elasticity and plasticity if needed as everything is made up of atoms.
There is also a stick slip hysteresis that will prevent you to move to slow it will be a fairly sudden transition from moving to stopped.
A wind turbine is only about 40% efficient and one fixed to ground and about 1m^2 swept area will produce around 240W in 10m/s wind.
Sails are just so much more efficient but you do not usually want your wind turbine to move.
For accelerating a vehicle you can not beat a sail so even if you want to use energy storage you will still select a sail and not a propeller.
The propeller/fan just has that effect that it can store energy in the compressible medium and use that to accelerate latter.
Nice thing about that is that stored energy being in material will not take any space/weight on the vehicle.

Similar thing will be a vehicle with wheels only having one wheel on a solid road and one wheel on a very long moving conveyor belt but that conveyor surface will be some super stretchy rubbery material so if you are stopped a lot of potential energy is stored in the long rubbery surface of the conveyor belt then if that wheel was just with brake on and not connected to other wheel it will be like a sail vehicle impossible to exceed conveyor speed.
But if the wheel on the solid road is connected with some gear ratio to the wheel on the flexible conveyor belt taking some of the power from accelerating the vehicle and putting it back in stretching that rubbery material even more than vehicle can exceed conveyor speed for some limited amount of time.   

 
   

Online bdunham7

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Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W

One of us has a fundamental misunderstanding of a very basic principle of physics.  I think the brakes will dissipate approximately 60mW under those conditions. 
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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Going to 0.5 m/s in only 2 ms  sounds like a lot of acceleration - actually 250 m/s² and thus about 25 G, which for a 10 kg vehicle would need 2500 N of force. Something in this calculation does not sound right !  :-//

See my video at minute 2 https://youtu.be/4Hol57vTIkE?t=121  that graph there where you can see both the speed (blue) and power (green) calculated with exactly this formula and similar 1m^2 sail just lower wind speed 6m/s and a 10kg vehicle and the result perfectly match reality if you where to also add the small friction losses as that example is for an ideal vehicle with no losses.

And so for that ideal vehicle it takes 154ms to get to a quarter of the wind speed 1.5m/s then 1.39seconds to get to half wind speed 3m/s and 12.5 seconds to get to 3 quarters of the wind speed 4.5m/s and of course it will take forever to get to wind speed.

Offline electrodacus

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Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W

One of us has a fundamental misunderstanding of a very basic principle of physics.  I think the brakes will dissipate approximately 60mW under those conditions.

If you apply only 60W of breaking power will get in about 1 second to around half the wind speed so 5m/s maybe just slightly more I will need to calculate exactly.
We can test this and see that a real test will exactly be predicted by my theory (is not my theory but the one I presented to you).

Online bdunham7

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If you apply only 60W of breaking power

You need just under 60N of braking force, not 'braking power'.  Then the power would be 60N X 0.001m/s = 60mW.  Power is force x speed. 
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Online IanB

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If you apply only 60W of breaking power will get in about 1 second to around half the wind speed so 5m/s maybe just slightly more I will need to calculate exactly.
We can test this and see that a real test will exactly be predicted by my theory (is not my theory but the one I presented to you).

Work = Force x Distance (for example: joules = Newtons x meters)

Power = Force x Speed (for example: watts = Newtons x meters / second)

If the vehicle speed is zero the power is zero because Power = Force x zero.

If the vehicle speed is nearly zero the power is nearly zero because Power = Force x (very small number).

If you calculate power vs speed for a simple sail driven vehicle, you should get power = zero when the speed is zero, and power = zero when the vehicle is travelling at the same speed as the wind (if the vehicle speed and the wind speed are the same, no power can be transferred to the sail).

When you plot this on a graph, you should find that the maximum power transfer occurs somewhere in the middle, when the vehicle speed is approximately half the wind speed.

This has some similarity with the power transfer theorem in electronics, where maximum power transfer occurs when source and sink impedances are equal.
 

Offline electrodacus

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If you apply only 60W of breaking power

You need just under 60N of braking force, not 'braking power'.  Then the power would be 60N X 0.001m/s = 60mW.  Power is force x speed.

Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s)
So you are thinking about slowing down 10m/s wind on a 1m^2 sail to 9.999m/s

Online IanB

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Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?

Yes!
 

Offline electrodacus

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    • electrodacus

If the vehicle speed is nearly zero the power is nearly zero because Power = Force x (very small number).


You are looking at this from the opposite direction. You just think the vehicle is powered from an internal on (board) energy source and there is no wind.
The highest power is available from wind when vehicle speed is lowest relative to the ground.
The lowest power available is when vehicle speed is almost the same as wind speed while vehicle travels directly downwind.
 

Offline electrodacus

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Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?

Yes!

:) That is a very big Yes for a wrong answer.

Online bdunham7

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Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s)

I'm not sure if you can say you are 'slowing the wind down' to the vehicle speed, but since you said it, lets use that.

If I can slow the wind down from 10m/s to ZERO using by dissipating 0 power, why is it so difficult to believe that I can slow it down to 0.001m/s with dissipating 60mW?
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s)

I'm not sure if you can say you are 'slowing the wind down' to the vehicle speed, but since you said it, lets use that.

If I can slow the wind down from 10m/s to ZERO using by dissipating 0 power, why is it so difficult to believe that I can slow it down to 0.001m/s with dissipating 60mW?


You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction). 

Online bdunham7

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You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).

I think perhaps you don't realize that "the ground deals with that" in all of the situations, not just the zero speed one?

Yes, if the sail is producing about 60N force, then 60mW will get me upwind at 1mm/s, provided there are no losses.
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Online IanB

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Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).

Yes, of course.

The force acting on the sail is 60 N. The vehicle speed required is 0.001 m/s. Therefore the power required from the motor is 60 N x 0.001 m/s = 60 mW.
 

Offline electrodacus

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You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).

I think perhaps you don't realize that "the ground deals with that" in all of the situations, not just the zero speed one?

Yes, if the sail is producing about 60N force, then 60mW will get me upwind at 1mm/s, provided there are no losses.

You forgot about the 10m/s headwind.

Offline electrodacus

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Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).

Yes, of course.

The force acting on the sail is 60 N. The vehicle speed required is 0.001 m/s. Therefore the power required from the motor is 60 N x 0.001 m/s = 60 mW.

Where is the 10m/s headwind in all this ?

Online IanB

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Where is the 10m/s headwind in all this ?

It produces a force on the sail of 60 N.
 

Offline electrodacus

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Where is the 10m/s headwind in all this ?

It produces a force on the sail of 60 N.

You need to consider your speed relative to air not relative to ground so 9.999m/s not 0.001m/s which is just the speed relative to the ground.

Offline electrodacus

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I will try to add more to the example maybe that will help.

Vehicle wants to drive against wind direction so upwind at 0.001m/s

Power needed is 60N * (10+0.001) = 600.06W

Vehicle wants to drive down wind it can waste the excess as heat in brake pads or generate

60N * (10-0.001) = 599.94W

This Is just simplified as I kept the force the same 60N but force will be slightly different just did not wanted to use the more complex looking formula as this is a good enough approximation. 

Offline Kleinstein

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One needs to include the speed of the wind in the power, when you calculate the power the wind is loosing, but not in the energy the vehicle with the sail is gaining.  A sail is not 100% efficient and quite some of the power taken from the wind is just converted to heat, warming up the wind.
 

Offline fourfathom

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I don't see much point arguing about efficiencies, when the electrodacus model only applies to drag-propulsion (a sail DDW), and his equation insists that available force is a function of AWS (Apparent Wind Speed), or  (wind speed - vehicle speed)^3.  This is not the case with the spinning propeller.  The model with a generator being driven by AWS does not represent the wheel/propeller vehicle.  Of course electrodacus can "prove" that the vehicle can't indefinitely exceed windspeed DDW, but his proof doesn't apply to the actual vehicles in question.
We'll search out every place a sick, twisted, solitary misfit might run to! -- I'll start with Radio Shack.
 

Offline tom66

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No motion of the vehicle means zero power can be generated at the wheel.

Let's stick with that for a moment.

So you say that for any non-zero speed, the power is (whatever your formula is) or about 60N * (wind speed - vehicle speed), but right at zero the power goes to zero?

So if I'm sitting there and the vehicle is stopped with brakes, there's no energy being dissipated by the brake, but if I allow the vehicle to creep forward at 1mm/s, now the brakes are dissipating almost 600W?

Same idea as a short circuited solar panel dissipating almost no power as the voltage across the panel is effectively zero.  Or open circuit where current is effectively zero.
 

Online CatalinaWOW

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Electrodacus is in love with his model.  A common problem with simulation people.  Simulations are wonderful things.  They can provide insight that is hard to obtain with actual physical tests, allow measurements that are literally impossible with physical tests and often save tremendous amounts of time and money.

But simulations have two fundamental flaws, that are often difficult to recognize.  First, they are all approximations and do not provide comprehensive information on when the omissions are important.  Second, usually a small problem but huge here, is that simulations no matter how large and wonderful do not inherently match the problem being simulated.  The math is all correct but doesn't represent the physics of the situation. 

In Electrodacus case there is a pretty obvious problem which he is overlooking.  He is not intrinsically wrong with his (10m/s -0.001m/s) formulation.  That is one way of presenting the problem.  But he is overlooking the fact that 0.001 m/s second of incremental velocity requires only a trivial amount of power.  Regardless of source.  In his own thought process there is zero drag in the condition of interest since the vehicle is moving at wind speed.  So the only consumer of power is the increase of momentum due to making the vehicle move faster.  The flawed thinking is ignoring the power required to maintain the initial 10 m/s velocity.

Having said all of that, it has been regularly asserted that a sailboat can tack downwind faster than a balloon blown down wind.  If this is indeed true the advantage must be small and dependent on perfect execution and perhaps on boat configurations which perform poorly in other conditions.  I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster.  These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
 


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