General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
electrodacus:
--- Quote from: Brumby on December 23, 2021, 02:05:18 am ---
--- Quote from: electrodacus on December 22, 2021, 06:14:14 pm ---Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
--- End quote ---
Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.
You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released. Include any necessary description and formulae that would be required for it to stand up to examination.
--- End quote ---
I will make an attempt to explain pressure differential.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.
Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.
If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.
All this is possible because air is a compressible fluid and it is much heavier than you actually imagine.
This is also the reason I insist that none of the wheels only vehicles can represent direct down wind faster than wind and all of them represent direct upwind faster than wind as there pressure differential energy storage is not used.
I saw some comments about my example with vehicle drag.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
I was making a point of the fact that the equation I insist on is one of the most important equations in anything that has to deal with air from wind turbines to any type of wind powered vehicles and to any aerodynamic drag calculation.
And yes using power to define drag is absolutely the right type of unit. When you design a vehicle you need to know the drag power as that is the most significant part affecting vehicle consumption.
I see people comparing Tesla model 3 a very small super aerodynamic vehicle with a large SUV in therms of consumption at highway driving speeds and they think that the SUV was just badly designed with a very inefficient drive-train where the reality is that the larger air drag is by far the largest factor in an EV consumption.
bdunham7:
--- Quote from: electrodacus on December 23, 2021, 07:06:37 am ---If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
--- End quote ---
Nope. It will need 1/6 of the power.
electrodacus:
--- Quote from: bdunham7 on December 23, 2021, 07:10:37 am ---
--- Quote from: electrodacus on December 23, 2021, 07:06:37 am ---If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
--- End quote ---
Nope. It will need 1/6 of the power.
--- End quote ---
You have no idea what air is.
IanB:
--- Quote from: electrodacus on December 23, 2021, 07:06:37 am ---That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
--- End quote ---
Why do you keep going back to old things that we have proved to you are not true?
A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.
You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage
If you keep going back and re-stating wrong things that have been corrected earlier, then this thread is going round in circles and cannot make progress. That is why we think you are trolling.
Insisting on something doesn't make you right, no matter how often you repeat it. In order to demonstrate the correctness of what you are saying, you have to be able to prove it with appropriate equations and logic, which is something you consistently fail to do. Moreover, you keep ignoring inconvenient facts when they go against your preconceptions.
IanB:
--- Quote from: electrodacus on December 22, 2021, 06:14:14 pm ---Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.
I will no longer post the graph but you can check that here https://en.wikipedia.org/wiki/Axial_fan_design
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.
--- End quote ---
If air is compressible, why does your power formula only have a single, constant density in it?
If air is compressed like a spring, then the density will vary with pressure and your equation for power cannot be correct.
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