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Mess with your minds: A wind powered craft going faster than a tail wind speed.
Labrat101:
This problem is impossible to solve without the correct information .
Firstly the Prop . being variable pitch it also has centre clutch assembly . The length of the
blades are a very important . How many blades it has . 3 or 5 are common 6 & 8 are normally shorter . (example) Most cars use 6 or 8 on the engine fan these are also variable pitch operated by temperature . Just to give you some idea what's involved . your car fan will spin free when cold & or if the speed is below set point of the magnetic clutch .
Also Its not a turbine blade these are totally different .
The maths and details of these Props are well documented and cover over 100 pages of heavy
reading .
Without having the propeller specs at hand . You can make up any formulae you like . It will not
work . Its just guess work , finding a cool load of numbers that look good to you .
But its BS .
Get all the Data sheets & Specs . Then try again .
fourfathom:
--- Quote from: Labrat101 on December 22, 2021, 09:04:34 pm ---This problem is impossible to solve without the correct information .
Firstly the Prop . being variable pitch it also has centre clutch assembly [...]
--- End quote ---
Yes, an exact solution requires data and details not used in these simple formulae. But basic principles can still be discussed and insights achieved. For example, the simple treadmill experiments used fixed-pitch props. They did effectively start at windspeed, and then proceeded to accelerate from there. The simple gear-coupled wheels on moving belts (or one set on a belt, the coupled set on the ground) also demonstrate the principle. So does the spool of thread example. Better data will let you determine drag and energy requirements, etc, but using 100% efficiency, zero-friction models is still instructive.
About the electrodacus theory of stored energy, I had suggested an inflated balloon, but perhaps a better model would be a spring between the wind-face (moving at windspeed) and the rear of the vehicle. While the vehicle speed is less than windspeed the spring is being compressed. When the vehicle reaches windspeed, the spring continues to push the vehicle above windspeed for a while.
I suppose this makes sense to electrodacus, but I've sailed thousands of miles downwind at less than windspeed. There is no significant energy storage, you use it or lose it.
Labrat101:
--- Quote from: fourfathom on December 22, 2021, 09:20:55 pm ---I suppose this makes sense to electrodacus, but I've sailed thousands of miles downwind at less than windspeed. There is no significant energy storage, you use it or lose it.
--- End quote ---
Yes your correct .
Quote " There is no significant energy storage, you use it or lose it. "
True But, electrodacus would argue that it could be stored in a Paper bag and squeezed out .
With a formulae found in the Dummies Book of wind Power (for beginners )
& for the tread mill . A hamster in its wheel would be more appropriate
PlainName:
--- Quote ---So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km
--- End quote ---
When the vehicle is at wind speed there is no air drag. Nothing. 0. At faster than wind speed there is a small amount (assuming just slightly faster), increasing as the vehicle exceeds wind speed more. The increased drag prevents infinite acceleration, but at just over wind speed there is no drag to speak of, hence the power needed to combat drag could be as small as you can arrange to measure.
IanB:
--- Quote from: dunkemhigh on December 22, 2021, 11:16:51 pm ---
--- Quote ---So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km
--- End quote ---
When the vehicle is at wind speed there is no air drag. Nothing. 0. At faster than wind speed there is a small amount (assuming just slightly faster), increasing as the vehicle exceeds wind speed more. The increased drag prevents infinite acceleration, but at just over wind speed there is no drag to speak of, hence the power needed to combat drag could be as small as you can arrange to measure.
--- End quote ---
Bear in mind there is no wind in the problem statement, so the vehicle is pushing through still air at 33 m/s. The mistake made is in the power formula. The power required to overcome drag is equal to drag resistance times vehicle speed. In still air the vehicle speed and the apparent wind speed are the same, but if there is some wind present the vehicle speed and the apparent wind speed are different. One could be more or less than the other.
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