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| electrodacus:
--- Quote from: SiliconWizard on July 07, 2022, 08:20:48 pm ---So. Have we revolutionized EE yet? :popcorn: --- End quote --- There is nothing to revolutionize. Conservation of energy is nothing new is just people seems to not understand it is a law and nobody has ever proved a case where it is not valid. Wind power is the only input to this vehicle and it is a positive value only when vehicle speed is below wind speed for direct downwind case. When vehicle finds himself above wind speed for any reason taking energy from the wheel will reduce the vehicle kinetic energy by that exact amount and that means reducing vehicle speed unless there is some energy storage that was charged when vehicle was below wind speed. Do you have an opinion :) are are you just a spectator ? |
| PlainName:
--- Quote from: electrodacus on July 07, 2022, 08:29:36 pm --- --- Quote from: dunkemhigh on July 07, 2022, 08:16:00 pm --- --- Quote ---b) Vehicle can be accelerated by 60W from wind + 40W you take at the wheels and put all in propeller still 100W ideal case. --- End quote --- Wrong, sorry. It takes almost nothing to maintain speed. It doesn't take 60W to accelerate - it takes whatever is available over the practically nothing that maintains speed. There is no drag the normal speed vs power curves you might want to consult are not relevant. And you don't take 40W or whatever from the wheel - just a bit more than the practically nothing that maintains speed would be fine. So we are looking at maybe 5% being taken via the wheel, and whatever slower speed relates to 5% less power from the direct wind. If you want 100% to be 100W I'm sure you can do the maths. (Actually, it could be anything from 0.001% to 99.9% - solving the equation would tell you what value, if there is a working value, would do the biz.) And that is it. The question is whether that 5% of power can produce thrust that at least equals the lost speed due to 5% less direct wind power. But recall that most of the power is used in accelerating up to speed, so the reduction may not be very much at all. Forget your 100/60/40 splits - that might be appropriate when the thing is stationary but not when it is in full flight. --- End quote --- Thanks for answering. Ideal case you need no energy to maintain speed. But to accelerate the vehicle you need energy. You have some wind power available to vehicle depending on wind speed relative to vehicle (wind speed - vehicle speed) and equivalent surface area of the vehicle on witch wind interacts with. I just gave an example of 100W available from wind to vehicle. All this 100W will accelerate the vehicle. If you take anything at the wheel I gave 40W as an example then only 60W are available for vehicle acceleration so acceleration rate will be lower. If you then use the 40W at the propeller ideal case you get back to 100W of acceleration power but nothing more. I will like to know exactly what part you disagree with but to me it seems that you think taking 40W at the wheel will not reduce the available acceleration by exactly that amount. --- End quote --- I disagree with taking 40% for the wheel, mainly because I think the power required to product thrust increases proportionally to the speed of the thrust - drag is involved there. So low thrust is better, and it isn't going to need anywhere near 40%. But... let's say it did take 40%. How much slower will the vehicle be? We don't know; we only know that the acceleration is affected so the question is whether it could accelerate to a high enough speed for the thrust to be in with a shout. Maybe it could, maybe it couldn't. I referred previously to the feathered Blackbird blades. In all the demos I've seen, none of them (except blackbird) started from zero speed. That is, none of them accelerated up to wind speed - all of them started at effective wind speed. Blackbird was the exception but it used feathered blades (which wouldn't suck much power). In the other demos I would just assume that it's difficult to start from zero, but the fact that Blackbird had to have feathered blades may be significant. |
| electrodacus:
--- Quote from: dunkemhigh on July 07, 2022, 08:54:49 pm ---I disagree with taking 40% for the wheel, mainly because I think the power required to product thrust increases proportionally to the speed of the thrust - drag is involved there. So low thrust is better, and it isn't going to need anywhere near 40%. But... let's say it did take 40%. How much slower will the vehicle be? We don't know; we only know that the acceleration is affected so the question is whether it could accelerate to a high enough speed for the thrust to be in with a shout. Maybe it could, maybe it couldn't. I referred previously to the feathered Blackbird blades. In all the demos I've seen, none of them (except blackbird) started from zero speed. That is, none of them accelerated up to wind speed - all of them started at effective wind speed. Blackbird was the exception but it used feathered blades (which wouldn't suck much power). In the other demos I would just assume that it's difficult to start from zero, but the fact that Blackbird had to have feathered blades may be significant. --- End quote --- 40% was just a random example. If you provide a vehicle 100W of power to accelerate for 10ms that means you provide the vehicle with 100W * 0.01s = 1Ws = 1 Joule of energy. It at that time vehicle was at 3m/s and had a weight of 10kg his kinetic energy initially was 0.5 * 10kg * (3m/s)2 = 45Ws So due to 1Ws of energy applied (100W for 10ms) the vehicle new kinetic energy will be 45Ws + 1Ws = 46Ws That means vehicle new speed will be sqrt(46/(0.5*10)) = 3.033ms If you instead apply a brake of 100W the kinetic energy of the moving vehicle will drop meaning the speed will drop. So in my example the choice was 100W of acceleration power from the wind directly or ideal case 100W with extra complications like taking some power at the wheels and putting in to a propeller or another wheel at 100% efficiency. If you have 100W of wind power than that is it. And if you are at wind speed and have 0W of available power to accelerate that is it and above wind speed power available is negative meaning deceleration. The only way for any wind powered vehicle to exceed wind speed directly downwind is to have an energy storage device (other than the vehicle kinetic energy). It is theoretically impossible for any wind only powered vehicle to exceed wind speed directly downwind or travel at any speed directly upwind. |
| rfeecs:
electrodacus, here is a paper that analyzes the problem from an energy point of view: "Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion" https://backend.orbit.dtu.dk/ws/portalfiles/portal/3748519/2009_28.pdf They conclude: --- Quote ---By applying basic concepts from mechanics and fluid mechanics we have derived the equations applicable for the performance of vehicles “powered” by a difference in velocity between two media. A myriad of intriguing results emerge from these relations. • It is theoretically and practically possible to build a wind driven car that can go directly upwind (using a generator/wind turbine in the air). ... • It is theoretically possible to build a wind driven car that can go in the downwind direction faster than the free stream wind speed (using a propeller in the air) --- End quote --- This idea has been around a long time. It's been demonstrated. You have been proven wrong. Despite your hundreds of posts you have failed to convince anyone by mindlessly repeating your flawed theory. |
| electrodacus:
--- Quote from: rfeecs on July 07, 2022, 09:34:29 pm ---electrodacus, here is a paper that analyzes the problem from an energy point of view: "Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion" https://backend.orbit.dtu.dk/ws/portalfiles/portal/3748519/2009_28.pdf They conclude: --- Quote ---By applying basic concepts from mechanics and fluid mechanics we have derived the equations applicable for the performance of vehicles “powered” by a difference in velocity between two media. A myriad of intriguing results emerge from these relations. • It is theoretically and practically possible to build a wind driven car that can go directly upwind (using a generator/wind turbine in the air). ... • It is theoretically possible to build a wind driven car that can go in the downwind direction faster than the free stream wind speed (using a propeller in the air) --- End quote --- This idea has been around a long time. It's been demonstrated. You have been proven wrong. Despite your hundreds of posts you have failed to convince anyone by mindlessly repeating your flawed theory. --- End quote --- The paper seems to concentrate on direct upwind witch uses different type of energy storage and can be debunked even easier. All you need to do is take a high speed video of any direct upwind device and see how it moves in discrete steps with time to charge then discharge the energy storage device tens of charge discharge cycles per second depending on design. Here is my video proof for direct upwind is just a 20 second slow motion video https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8 Input power is provided here by the middle paper the one that moves relative to camera frame and the vehicle advances against the direction that paper is moving only because it first stores energy then it releases the energy to move a bit forward. Wheels in front the one on moving paper are the generator wheels (same as the wind turbine on a direct upwind vehicle) and the back wheels are the motor wheels powered by the energy stored in the rubber belt. If there was no energy storage then all times Pin = Pout and vehicle will not move just sit in place or be dragged back. But since the belt is charged (elastic potential energy) and then when the back wheel spins because it slips there is power available at the back wheel from the belt that is now discharging and the process repeats. Edit: Looking a bit more at that paper and that Figure 1 contradicts what the paper is trying to prove. Vg is wind speed and direction is left to right Fg and Fp are correct same value and opposite directions And then Vp is the same direction as wind direction meaning the vehicle is being dragged directly downwind at significantly less than wind speed. They try to explain how direct upwind will work and the only diagram they have in there correctly predicts that without energy storage the vehicle will either not move or be slowly dragged in the wind direction. |
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