General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
bdunham7:
--- Quote from: IanB on December 17, 2021, 12:24:10 am ---Just a quick summary on the calculations and units:
The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)
--- End quote ---
That omits the laminar drag portion, but I'm guessing perhaps that is negligible in this case? (I don't actually know)
And that does appear to be the equation, more or less, that he is using. I didn't recognize it because of the way he was expressing it, but it works out--as I showed above!
electrodacus:
--- Quote from: IanB on December 17, 2021, 12:39:42 am ---
--- Quote from: electrodacus on December 17, 2021, 12:33:32 am ---Then 46.8N * 9m/s = 437.4W so same exact result I got.
--- End quote ---
But the 46.8 N is not moving at 9 m/s. If you make the sail move at 9 m/s, then the apparent wind velocity is going to be 1 m/s.
In that case: Force on sail = 0.5 * 1.2 * (10-9)^2 * 1 = 0.6 N
Power = 0.6 N * 9 m/s = 5.4 W
You cannot have a high wind pressure on the sail, and also make the sail move at the speed of the wind. That is double accounting, and it does not work.
--- End quote ---
That will be if you completely ignored the wind. So if there was no wind and vehicle was driving at 1m/s then this will be the power it will need to supply to his motor to counteract drag 0.5 * 1.2 * 1 * 1^3 = 0.6W
0.6W / 1N = 0.6N
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.
In the other direction if there is no wind and this vehicle with 1m^2 frontal area needs to drive at 9m/s then it needs 437.4W to maintain that speed just to counter the drag. So that is how much an electric bike will use at this speed if frontal area will be 1m^2 (usually is less than half that but this is just an example with round numbers).
IanB:
--- Quote from: bdunham7 on December 17, 2021, 12:45:45 am ---
--- Quote from: IanB on December 17, 2021, 12:24:10 am ---Just a quick summary on the calculations and units:
The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)
--- End quote ---
That omits the laminar drag portion, but I'm guessing perhaps that is negligible in this case? (I don't actually know)
And that does appear to be the equation, more or less, that he is using. I didn't recognize it because of the way he was expressing it, but it works out--as I showed above!
--- End quote ---
The equation works in general for any shape of object. The area is calculated in some standard way (usually the cross section facing the wind), and the drag coefficient accounts for the shape of the object. Basically, all the complex geometry, skin friction, turbulent effects and everything else is wrapped up in the Cd value. It can vary a lot, from as little as 0.01 to as much as 2.
IanB:
--- 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.
bdunham7:
--- Quote from: IanB on December 17, 2021, 12:59:28 am ---The equation works in general for any shape of object. The area is calculated in some standard way (usually the cross section facing the wind), and the drag coefficient accounts for the shape of the object. Basically, all the complex geometry, skin friction, turbulent effects and everything else is wrapped up in the Cd value. It can vary a lot, from as little as 0.01 to as much as 2.
--- End quote ---
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.
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