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| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
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| Brumby:
--- Quote from: electrodacus on December 18, 2021, 03:40:16 am ---The crazy explanation that vehicle when above wind speed directly down wind will take power from the wheel to supply the propeller is exactly like saying a motor and generator connected together with a belt are a free energy generator. --- End quote --- No, it's not. Your motor and generator are operating at the same speed - but the the two sides of the Blackbird mechanism are not. This is clearly apparent in the following statement: --- Quote ---That is because when you take energy from the wheel that is directly taken out of the vehicle kinetic energy Ws or Joules and since propeller is maybe at best 70% efficient can only put back 70% of that back in to kinetic energy and so vehicle will have lower speed. --- End quote --- Yes, there will be a reduced efficiency in the power extracted from the wheels through to the propeller - but the propeller is pushing against the wind. With the blackbird travelling at wind speed, the wind will appear as a stationary wall of air - and THIS is what the propeller is pushing against. <== IMPORTANT POINT! Warning: The following uses terms like power and energy in a conversational style - NOT an engineering one! Change the words as you see fit - but follow the intent. - Let's say wind speed is 20 km/h - Power is extracted from the wheels which are rotating at a tangential speed of 20km/h - This power is transferred through a gearbox to rotate a propeller. (The ratio of the gearing is important here - which includes propeller characteristics.) - The propeller then pushes against a wall of air that appears to be stationary ( 0 km/h ) from the perspective of the vehicle. In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference. The extracted energy will cause a drop in speed, but the additional thrust will cause an increase. Finding the equilibrium point is the trick. Is is below wind speed? Is it above? Is it equal to wind speed? |
| Brumby:
--- Quote from: electrodacus on December 18, 2021, 03:40:16 am ---What you can see if you take a video from the side is the decrease acceleration rate --- End quote --- It doesn't matter even if the acceleration rate drops to zero. If the resulting "terminal velocity" is higher than wind speed, then we have found the equilibrium point which supports Blackbird's achievement. --- Quote ---and that will not correspond to Derke's or anyone else's prediction. --- End quote --- You say "will not". That sounds like an intuitive belief rather than a mathematical proof. |
| Brumby:
After re-reading, I felt this statement was worth emphasising. --- Quote from: Brumby on December 18, 2021, 12:50:58 pm --- In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference. --- End quote --- |
| fourfathom:
[I just can't help myself!] One more way to consider the situation (and this has been said here before, by myself and others) is that the vehicle is located between and mechanically coupled to the interface of the ground and the moving air. It really doesn't matter if the vehicle is moving or stationary, the difference in velocity is always there and can be used to generate power. But you need to have an appropriate mechanism to extract this power and drive the vehicle DDWFTTW. A rock can't do it. The Blackbird can. |
| electrodacus:
--- Quote from: Kleinstein on December 18, 2021, 08:37:34 am --- And that is wrong and there is no sense in repeating this all over !. :horse: :horse: :horse: :horse: --- End quote --- You need to explain it not repeat something you memorised. --- Quote from: Kleinstein on December 18, 2021, 08:37:34 am ---Very confident is has to be seen with a grain of salt: The calculator is for a different problem: the force a wall has to withstand, when close to an edge (kind of clif if you want). The Zones are not the place where the pressure in the air is measured, but the distance of the wall from the edge. Close to the edge there is some concentration of the wind - so the real value to compare is the one far away from the edge. The calculated force is not that impressive to support the claimed high power from a sail. So I am glad you are confident in a calculator that contradicts your claims. Better have more confidence in the calculator for the bicycle power that showed that it is possible to drive with relatively low power against an head wind. --- End quote --- The calculator is for exactly this type of problem. You may not like it as it supports my theory and calculations for a compressible fluid are much more complex. Look at the type of teren you can select and no there is no close to the edge of a clif. Please look more closely at that calculator and input some data. The calculation will perfectly apply to a sail. The pressure is what is shown there and is the average for a certain volume. Use the following data Terrain category 0 Wind speed 10m/s Wall Length 10m Wall height 2m Distance from base say 1m (so wall is not touching the ground) length return corner 0 Solidity ratio 1 Ortogonal factor 1 Air density 1.2 Then you will see that average pressure for 0.3*h so 0.6m behind the wall/sail you have 407N/m^2 Then from that point 60cm to 4m the average pressure is 253N/m^2 You can see how average pressure increases as you get closer to the wall/sail and it makes sense as this is a compressible fluid so way more air molecules will be squeezed just right next to the wall so much higher air density in those regions. Same was shown for a propeller with the difference that propeller can increase this pressure differential even more while vehicle is below wind speed and it has access to wind power. |
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