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#61
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| If the prop is producing 9N at 4m/s, it's energy demands are 36W, not 15W. If you allow for a prop efficiency of 80%, the demands are closer to 45W. |
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#62
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If you apply your logic to an airplane with 100kts airspeed in a tail wind of 100kts it would require to double its power to achieve 200kts over the ground. The propeller on the cart is not reacting against the ground. It is reacting against the air. This is exactly the same condition as an aircraft flying at 10,000m. It does not matter if it is in the air or on the ground. Increasing ground speed with a tail wind does not increase the power required by the prop. So if you can prove that an aircraft increases power to maintain its airspeed in a tail wind you may be able to mount a plausible argument against DDWFTTW. Rick W |
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#63
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| Your diagram clearly shows that 9N are required to push the cart 4 m/s. Therefore, 36W are required. It doesn't matter what the airspeed thru the prop is. You indicated that there is 9N of drag. So 36W of power are required to overcome it at 4 m/s. FYI: I happen to be a pilot and I do understand indicated airspeed vs. ground speed. I've actually flown a plane backwards in level flight according to the GPS. |
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#64
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That fact has nothing to do with how much electrical power is needed to rotate the prop. You are mixing too very different things. In craft IRF the craft doesn't move at all meanig speed is zero, and power is 9N*0m/s = 0 Watts. That doesn't tell anything about needed shaft power. Quote:
Or do you claim your power consumption was negative while going backwards related to the ground ? If not, why don't you just say so. |
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#65
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If I was to replace the propeller with a spinnaker in 3m/s tail wind I could get maybe 2m/s over the ground. The power input from the wind in this case is 18W. My cart is capable of generating 9N of thrust with an airspeed of 1m/s. If I give it a tail wind of 3m/s then it will generate 9N with an airspeed of 1m/s and a ground speed of 4m/s. The 9N is enough to maintain the ground speed of 4m/s. Go back to your aeroplane. If it takes 100HP for level flight at 100kts you do not double horsepower to achieve 200kts over the ground in a 100kt tail wind. You only need the same 100HP. Exactly the same with my cart. The conditions for the prop have not changed. It sees a headwind of 1m/s. It is not bothered by what it happening at ground level. The prop operates in air. Rick W |
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#66
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| You are pushing against the ground and surrounding environment with 9N of force at 4 m/s. That's 36W. You can't escape that by arbitrarily juggling the frame of reference until you find one that makes your energy demands go away. As far as the airplane goes, its frame of reference is the free air stream. Ground speed is meaningless in flight. And before you say that cart's frame of reference should be the free airstream like the airplane, let me remind you that the cart's wheels are still attached to the ground. Even if we allow for zero aerodynamic drag in spite the fact that the cart has a headwind, you are still in need of 20W of power just to turn the wheels against the ground which has busted your budget of 15W. |
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#67
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When it is operating in a tail wind of 3m/s and doing 4m/s across the ground the wind provides the extra power. I have not busted the budget because the power demanded by the prop has not changed. It sees 1m/s wind and is producing 9N. The wind provides the rest of the power. You have not responded to how a spinnaker could actually provide energy to the cart doing 2m/s in a 3m/s tail wind. How much power is the spinnaker generating. By your analysis you are saying the spinnaker needs to produce at least 18W. What is the energy input to the spinnaker. Where is its battery connected????? I also disagree on your point that ground speed is not important to flight. If you can get a 100kt tail wind in a plane capable of 100kts then you get from point A to point B in half the time using half the fuel. The wind has saved you all that energy. My cart is simply exploiting the tail wind the same way. If it makes life easier for you assume there is no wheel drag. Will it speed up in this case with a tail wind? Rick W |
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#68
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| Thanks for the warning. We'll try not to use any complicated aero terms or formulas. Quote:
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#69
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Wait, weren't the wheels providing power to the prop? How can the wheel drag go to zero then? |
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#70
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#71
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You dodged my point. What is driving the prop? Where is the energy coming from? Is the prop driving the air to produce the thrust needed to exceed the wind speed? |
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#72
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I have an electric powered vehicle as shown in these slides. I am only considering battery power here. Do you agree with what I am saying in Slide 1 and Slide 2 as depicted? If not tell me where my analysis is flawed. Rick W |
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#73
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What's driving the prop? The prop shaft. What's driving the prop shaft? The axle (through a 90 degree bevel gear set). What's driving the axle? The wheels. What's driving the wheels? Well, that depends on how you look at it. We can say it's the road (which isn't the most pleasing answer) or the propeller thrust (which SEEMS like it would constitute perpetual motion, but it does not for reasons we can easily explain). Quote:
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#74
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I notice that you're mixing rolling resistance with rolling power. Slide 1 shows the energy required at this point is 1.75W and that you have a surplus of power available from a battery. Cool. Slide 2 shows that the power needed to roll the cart is now 3W. So far so good. |
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#75
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Do you agree with my analysis here? Rick W. |
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