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| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
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| gnuarm:
--- Quote from: electrodacus on December 28, 2021, 10:29:21 pm ---OK lets compare the output of the two equations air density 1.2kg/m3 area 1m2 Wind speed 20m/s Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s 0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed) a) 4800W b) 600W c) 0W 0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed) a) 0W (how will your wind powered vehicle ever start) b) 600W c) 0W Any comments ? Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed. --- End quote --- The first equation is the power available in the wind. Not relevant to the discussion of how much power it takes to move the vehicle. The second equation is the power required to move the vehicle at that speed against the wind. So when the vehicle is not moving, it takes no power since there is no motion. P = F * V, if V = 0, P = 0. See ma? No hands!!! You even admitted this in a previous message saying the bike rider can simply keep his foot on the pedal to hold the bike against the wind with no power being exerted. Now you wish to retract that and claim a vehicle at "anchor" is a special case somehow. Whatever. Your arguments get lamer every time you post. |
| gnuarm:
--- Quote from: electrodacus on December 28, 2021, 11:47:06 pm --- --- Quote from: IanB on December 28, 2021, 11:10:01 pm --- --- Quote from: Alex Eisenhut on December 28, 2021, 10:31:05 pm ---https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf --- End quote --- Good find. Thank you! That will take some time to digest. --- End quote --- Do not waste your time it is the same rubbish from another person that has no understanding of power and energy. --- End quote --- LOL!!!! By virtue of the fact that the author comes to a conclusion that does not agree with ED the author must not understand power or energy. If that is the case, clearly ED is not qualified. He has already shown he doesn't understand the nature of a "rate" being related to time such as power. He claims it is energy that is related to time. His arguments remind me of an "engineer" who couldn't apply basic math. They were doing a survey which required calculating the area of a room. He wanted to divide length by width rather than multiply. When someone asked him to explain he said, "How many quarters in a dollar? How many quarters in two dollars? See???" That's ED. Nobody understands quarters like ED does. |
| IanB:
--- Quote from: gnuarm on December 29, 2021, 12:40:59 am ---Is that clear? --- End quote --- Yes, all clear and correct. In the case of a wind turbine, the equation with wind speed cubed gives the total kinetic energy of the wind flowing through the swept area of the turbine. The turbine blades are able to capture some, not all, of this energy. This capture ratio can be expressed as an efficiency factor, which has a theoretical maximum at the Betz limit of 59.3%. For simplified design purposes when installing a wind turbine, you can just make use of a quoted efficiency factor as given by the designer of the turbine, which will typically be presented in a chart looking something like this one: |
| electrodacus:
--- Quote from: gnuarm on December 29, 2021, 12:49:52 am --- --- Quote from: electrodacus on December 28, 2021, 10:29:21 pm ---OK lets compare the output of the two equations air density 1.2kg/m3 area 1m2 Wind speed 20m/s Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s 0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed) a) 4800W b) 600W c) 0W 0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed) a) 0W (how will your wind powered vehicle ever start) b) 600W c) 0W Any comments ? Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed. --- End quote --- The first equation is the power available in the wind. Not relevant to the discussion of how much power it takes to move the vehicle. The second equation is the power required to move the vehicle at that speed against the wind. So when the vehicle is not moving, it takes no power since there is no motion. --- End quote --- The first equation is the wind power available to any wind powered vehicle. Second equation is not representing anything as it is just the wrong version of the first one. The first equation (the correct one is valid for all cases). When vehicle wants to travel direct upwind the vehicle speed will be negative so equation will look like this 0.5 * air density * area * (wind speed - (-vehicle speed))2 * (wind speed - (-vehicle speed)) = 0.5 * air density * area * (wind speed + vehicle speed)3 And yes for any vehicle speed direct down wind you need the wind power available plus some more meaning it is impossible for any wind powered vehicle to drive directly upwind without energy storage or external energy source. |
| gnuarm:
--- Quote from: electrodacus on December 29, 2021, 01:16:02 am --- --- Quote from: gnuarm on December 29, 2021, 12:49:52 am --- --- Quote from: electrodacus on December 28, 2021, 10:29:21 pm ---OK lets compare the output of the two equations air density 1.2kg/m3 area 1m2 Wind speed 20m/s Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s 0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed) a) 4800W b) 600W c) 0W 0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed) a) 0W (how will your wind powered vehicle ever start) b) 600W c) 0W Any comments ? Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed. --- End quote --- The first equation is the power available in the wind. Not relevant to the discussion of how much power it takes to move the vehicle. The second equation is the power required to move the vehicle at that speed against the wind. So when the vehicle is not moving, it takes no power since there is no motion. --- End quote --- The first equation is the wind power available to any wind powered vehicle. Second equation is not representing anything as it is just the wrong version of the first one. The first equation (the correct one is valid for all cases). When vehicle wants to travel direct upwind the vehicle speed will be negative so equation will look like this 0.5 * air density * area * (wind speed - (-vehicle speed))2 * (wind speed - (-vehicle speed)) = 0.5 * air density * area * (wind speed + vehicle speed)3 And yes for any vehicle speed direct down wind you need the wind power available plus some more meaning it is impossible for any wind powered vehicle to drive directly upwind without energy storage or external energy source. --- End quote --- Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind? Lets simplify the messy bits and assume the force generated by the wind on the vehicle is Fw = Kv * (wind speed + vehicle speed)2 Where Kv is an accumulation of all the factors that relate the force to the relative wind speed. So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed? Now, forget wind. Lets say there is a drag on the vehicle defined by Fdrag = Kdrag * (vehicle speed)2 What is the power required to maintain the speed of this vehicle? I bet $50 he won't answer this and ducks the question. I suppose there's a chance he comes up with something amazingly convoluted. Probably starts talking about driving the vehicle on a river bottom. I really can't believe this guy designs anything. He must be pulling our legs. |
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