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Mess with your minds: A wind powered craft going faster than a tail wind speed.
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electrodacus:

--- Quote from: IanB on December 17, 2021, 01:08:17 am ---
--- Quote from: electrodacus on December 17, 2021, 12:53:28 am ---Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.
--- End quote ---

If you want to get 437 W from a vehicle moving at 1 m/s then you can use the formula: power = force x velocity

You have:

437 W = force x 1 m/s

So the force on the sail would have to be 437 N.

From the sail equation:

Force = 0.5 x (air density) x (apparent wind velocity)^2 x (area of sail)

437 N = 0.5 x 1.2 kg/m3 x va^2 x 1

va^2 = 437 / (0.5 x 1.2) = 728

va = 27 m/s

This tells that the wind speed would have to be 27 + 1 = 28 m/s.

--- End quote ---

I feel like you see the air something less relevant than the road. Maybe just because it is invisible and low density.

If you apply no brakes at all then those 437W will all be available to accelerate the vehicle. Sort of seems you agree with that then not sure why if you agree with that you do not agree with the fact that you need to apply 437W of braking in order to maintain speed.
Ideal vehicle needs no power to maintain speed so you need to take all that available wind power and convert it into heat through friction brakes or use a generator to take advantage of that in some other ways.

On the other side you can also look at a vehicle driving at 9m/s with no wind so there is a 9m/s apparent wind and that 9m/s apparent wind requires you to provide 437W in order to be able to maintain vehicle speed at 9m/s
If you had a 9m/s vehicle and a 9m/s wind from the back then 0m/s apparent wind you need 0W so no power to maintain vehicle speed.
I know something seems maybe too good to be true but it is the reality and can be tested.
bdunham7:

--- Quote from: electrodacus on December 17, 2021, 01:22:13 am ---If you apply no brakes at all then those 437W will all be available to accelerate the vehicle.

--- End quote ---

No, it won't.  You don't accelerate a vehicle with power, you do it with force.  At 1m/s with a 10m/s wind, you have your 46.8 Newtons available to accelerate the vehicle, which at 1m/s represents 46.8 watts of power.  The remainder of the wind's power will be dissipated.  Of course that number changes as the speed change.  And b/t/w, from an earlier post, 46.8 * 9 = 421.2, not 437.4.  Not a big issue, but it was making something not look right.
IanB:

--- Quote from: bdunham7 on December 17, 2021, 01:15:40 am ---What I meant is that there is another term in the full equation, the laminar drag.  This is the low-speed drag component dependent on speed, the viscosity of the medium and constant based on the shape and size of the object. the viscosity of the medium.

--- End quote ---

As the "drag coefficient", I think the Cd value includes that. The Cd value may not exactly be a constant, and may vary with wind speed, but within the region it applies, it is a kind of "catch all" for all the indirect friction effects.

The viscosity of the medium would get included in the Reynolds number, and the Cd may then be correlated against the Reynolds number.
electrodacus:

--- Quote from: bdunham7 on December 17, 2021, 01:42:32 am ---
No, it won't.  You don't accelerate a vehicle with power, you do it with force.  At 1m/s with a 10m/s wind, you have your 46.8 Newtons available to accelerate the vehicle, which at 1m/s represents 46.8 watts of power.  The remainder of the wind's power will be dissipated.  Of course that number changes as the speed change.  And b/t/w, from an earlier post, 46.8 * 9 = 421.2, not 437.4.  Not a big issue, but it was making something not look right.

--- End quote ---

If you look back at earlier example I stated that it takes around 100ms to get to 1m/s and the vehicle kinetic energy at 1m/s is 50Ws
So according to your theory it will take more like one full second for this vehicle to get to 1m/s instead of just 100ms.

Sorry is 48.6N it was just a typo that I will correct. Thanks for noticing.
Brumby:
I think I'm gonna regret this.....


--- Quote from: electrodacus on December 17, 2021, 12:14:44 am ---.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or a external energy source.

--- End quote ---
Taking the above literally - which is what you have been doing all along - and you will never be able to see the full mechanism.  The above is the "obvious" bit - but it's not the full story.

If I were to change your statement to something more correct, it would look like this:

--- Quote ---.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or an additional energy source.

--- End quote ---

Here is the key - there IS an additional energy source, but it is not as obvious as wind on a sail (which is why "intuition" is a really bad influence - really bad.).  It is, however, fully contained within the system consisting of only the Blackbird, the wind and the ground.  It is not external to that system and it is not an energy storage mechanism.

If you cannot OPENLY consider such a possibility and work through the relevant calculations, you will never understand.

Such closed-mindedness would have kept the study of physics from progressing any further than Newton.
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