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| Newton's third law problem. |
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| wraper:
Fixed the topic for you. |
| electrodacus:
--- Quote from: Nominal Animal on November 24, 2022, 05:16:14 pm ---Modeling the macro-scale properties of compressible fluids as colliding solid particles is just not tenable. I know, because my proper field is molecular dynamics simulations. --- End quote --- That is because of our limited compute capability. That will not change the fact that air molecules collide with the vehicle body and that is how energy is transferred. The equation we have and I provided many times is doing a great approximation of the wind power available without calculating energy provided by each air molecule. We just select an air density typical 1.2kg/m3 and the area and coefficient of drag of the solid surface it interacts with. --- Quote from: Nominal Animal on November 24, 2022, 05:16:14 pm ---It depends on the vehicle. For a ground vehicle, what matters is that it depends not on the vehicle airspeed (speed of vehicle relative to the surrounding air), but on the velocity and pressure of air with respect to ground. This is because the vehicle spends its energy in an effort to accelerate with respect to ground, not with respect to wind; to compensate for the drag and friction (i.e., losses) the vehicle experiences. Yes, aeroplanes and ground vehicles differ significantly in this. An easy to understand vehicle that can extract energy from air when their relative speeds are the same, is a cart with a vertical axis airfoil wind turbine. The airfoils do not "catch" wind, they act like aeroplane wings, generating torque from the pressure differential on different sides of the blade. During any motion of the vehicle, the VAAWT is rotating. When the vehicle travels downwind at the same speed as the wind, the blades of the VAAWT can still extract energy from the wind in all phases except when the blade is in its extremum position with respect to the travel direction. It is obvious, since the blade is moving wirth respect to air surrounding it. Because of the pressure differential principle, the direction of the wind is easily compensated by changing the blade orientation/attack angle. The only limiting factor is how efficient you can make your vehicle, how small you can make the losses in its mechanisms. --- End quote --- A lot of words and no equations. I demonstrated that I can predict exactly what happens using very few simple and well known equations. While all I get from you is "It depends on the vehicle". Facts: - Yes a vehicle with no energy storage like a simple sail vehicle can drive only using available wind power thus up to wind speed in wind direction and no speed directly upwind. - A vehicle that uses some form of energy storage can drive both upwind and for a limited amount of time proportional with amount of stored energy directly downwind. Understanding the difference between power and energy may be critical in understanding the above two facts. So all you need to now are this: a) P = 0.5 * air density * area * coefficient of drag * (wind speed - vehicle speed)3 to get ideal so best case scenario for wind power available. b) KE = 0.5 * mass * (vehicle speed)2 c) details about the vehicle energy storage if it has any. Knowing the above 3 things you can predict exactly what happens and so understand how all this vehicle work. |
| electrodacus:
--- Quote from: Nominal Animal on November 24, 2022, 05:33:35 pm ---Direct upwind is trivial to demonstrate: just put an efficient horizontal turbine on a cart, and use a very low gearing to the wheels (i.e. turbine turns many tens of times per one rotation of the wheels). Make it a VAWT and the cart will move regardless of the wind direction. This matches my Lego trike I demonstrated earlier, and whose exact behaviour I described mathematically (in terms of velocities or displacements, and gear ratio); instead of wind, it used a spool of thread, or a Lego chain/track. If one were to build a model where the ground wheel is actually a pulley with a wire looped around it and fixed to ground at both ends, it shows that no slip is required at all for it to function as described. Indeed, if the wire is looped many times around the pulley, and fixed to the pulley at the midpoint, then no slip is even possible if one uses e.g. thin steel cable. --- End quote --- I will assume you have basic electricity knowledge. Vehicle has a wind turbine connected to an electric generator. The electric generator is connected to an electric motor that drives the wheels. Based on wind speed and wind turbine swept area 1000W are available from the wind. That means wind pushes against the vehicle with 1000W so if all 1000W output from the wind turbine are applied to an ideal 100% efficient electric motor the best that the vehicle can do is stay in the same place it is. So the problem of you not understanding can come from a) Not understanding that wind wants to accelerate the vehicle in the direction of the wind and so if your wind turbine output 1000W electrical of mechanical power you need 1001W at the wheels in order to begin to accelerate the vehicle mass against wind direction with 1W b) Not understanding what power is and that if you include time you deal with energy So you can wait for 2 seconds and store (in some energy storage device) 2000Ws worth of energy then you can use that stored energy to accelerate for say 1 second at 2000W or for 200ms at 10kW. No energy storage means you can not accelerate the vehicle directly upwind. |
| electrodacus:
--- Quote from: wraper on November 24, 2022, 05:45:46 pm ---Fixed the topic for you. --- End quote --- What ? |
| Nominal Animal:
--- Quote from: electrodacus on November 24, 2022, 05:50:43 pm ---A lot of words and no equations. --- End quote --- Well, it is better than unphysical nonsense and equations unrelated to the physical phenomena at hand. You seem to believe that if an equation applies in some specific situation, it must apply always, and that just isn't so. Like I explained, if a vehicle uses the fixed ground instead of surrounding air in its propulsion mechanism, the amount of energy it can obtain from wind is not relative to the vehicle speed, it is relative to the ground. So no subtracting vehicle speed from the wind speed relative to ground. Extracting the energy is only an engineering problem. Because of the Venturi effect, the exact wind speed just isn't an issue wrt. harvesting energy from it. You insist on using static surfaces, ignoring all pressure-related effects, as if the vehicle is and has to be a simple rectangular box, with no internal mechanisms and only interacting with its environment via elastic collisions. It is silly, and quite annoying. You need to move past your preconceptions. For example, you could start gently, and consider a vehicle with a very aerodynamic shape, say saucer-shape with a sharp edge, with upper and lower curvatures different (to balance out any vertical forces, including lift, due to the pressure differential), and with a Gorlov helical turbine poking up on top. It has minimal drag, and the turbine efficiency is about 35%. It typically operates at tip speed ratio over 1, which means that the induced flow rate due to rotation of the turbine is greater than the flow rate with respect to the axis. Go read Gorlov's 2001 paper on it, but do recall that it only considers the properties of a stationary turbine. And before you assert it, no, a moving turbine is not the same thing as stationary turbine in zero-wind situation. It only matches if we assume the turbine is not moving in either case, and that's not going to happen with a moving turbine. Because of the tip speed ratio being over 1, even when the wind speed approaches zero relative to the turbine axis, the turbine isn't going to stop (unless it has bad bearings or other significant losses). Because it can keep accelerating the vehicle all the way, it will be less efficient at that particular point compared to speeds immediately above and below, but it isn't zero even there. What makes that seemingly simple construction interesting, is the fact that airflow speed varies as a distance from static surfaces like the ground, below the boundary layer. This means that along the vertical turbine axis, there is always a range of wind speeds relative to the axis, and because the wind direction does not affect the turbine, it can extract energy from the wind even when the vehicle is traveling downwind at the nominal wind speed. This applies even to wind tunnels (and in those, it is most noticeable, because they have laminar flow patterns by design; in nature, wind tends to be a bit turbulent). That sort of a vehicle does not even need wheels, it can even use a propeller for the propulsion, as long as it keeps close to a static surface (ground), and still achieve faster than nominal wind speed downwind or upwind. When you wrap your mind around that, go look at aerodynamics, propulsion, and so on, and expand your understanding from there. Reconsider the models and practical examples I've shown earlier in this thread. Don't get stuck in your preconceived notions just because you're emotionally heavily invested in them. Learn to entertain ideas and theories and models that you do not believe are true, and learn how to rationally examine why your beliefs and the idea/theory/model do not agree. |
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