General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
iMo:
I've been still trying to decouple the concept of the propeller and the wheels here.
The Zeppelin has no contact with ground.
Let us imagine an Iceboat - the same construction as the Blackbird (the same body and propeller construction), but running on 3 runners made of steel with sharpened edges, the blades having pretty low forward resistance.
Will the Iceboat travel faster than the tail wind speed?
Kleinstein:
To travel faster than the wind (in the same direction) you still need grip to the ground. To generate power you need to transfer force to the ground, so simple low friction sliding does not work.
The power harmessed from the wind is force transferred to the ground times wind speed minus losses from the prop from generating vortices and similar distortions in the wind field. So with no force to the ground there is no power to start with.
With an ice sail going sideways, the runners can transfer force (sideways) to the ice easy and thus plenty of power.
Labrat101:
100 years ago we all thought that the Plus & Minus on a battery that electricity flowed . From Plus to Minus
Every one was happy with this .. But it was found to be WRONG it goes from Minus to Plus .. :wtf: :blah:
Most people now except this as this is what you are told in school / Uni or what ever.
weather right or wrong the world maybe Flat . Depends on your perspective thinking . What you see is not always what is .
Mother Nature does not have to play by the rules of mea Humans . So what when you see something that you Don't understand your Brain
will try to fit some thing in to stop it blowing a fuse ..
A computer if it does not understand will return ERROR . No Data . because the Human did not enter it .
Somethings are as is . Appling maths formulators to do quick fix may solve your problem but not be Right solution
Murphy's Law is always true .. Humans will Always Get it round the wrong way.
cbutlera:
--- Quote from: electrodacus on September 03, 2021, 04:56:32 am ---
--- Quote from: bdunham7 on September 03, 2021, 03:14:19 am ---
--- Quote from: electrodacus on September 03, 2021, 02:01:17 am ---I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.
Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%
So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s
Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.
--- End quote ---
OK, that is easy enough. Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car. I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W. Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather 2.422.24m/s, the airspeed that the propeller sees. So the force generated will be (3.816W * 0.7 efficiency / 2.422.24m/s) = 1.1041.1925N. The net force (propeller force minus wheel force) will then be 0.11925N and the vehicle will accelerate at 0.11925m/s2. Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed. In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) = 2.028 2.385W of additional power for a total of 4.6992 5.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W. And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.
But I think I've made an additional error, so wait a bit.
Edit: Should be all good now. If not, someone point out the errors.
--- End quote ---
Good effort but you forgot something super important. What is the kinetic energy of the vehicle before and after you did the break ?
Say you maintain that 3.816W break for one second now the vehicle kinetic energy is reduce by 3.816Ws
Now you apply 3.816W * 0.7 = 2.671W for one second and you can see you end up with lower kinetic energy than you started with.
And is irrelevant if you do that for 1 full second or a 1ms or even 1us the end result is that vehicle speed will be reduced.
The wind can not provide anything unless you refer to stored energy. As long as wind speed is lower than vehicle speed there is nothing to gain from that.
Forces have no role here since all you care is vehicle kinetic energy change. If you end up with lower kinetic energy then vehicle has slowed down.
Do you see a difference in kinetic energy gain between a vehicle traveling at 2.24m/s with no wind and one traveling at 4.24m/s and 2m/s wind when power is applied to the propeller ?
--- End quote ---
No, it is you who have forgotten something super important - Newton's third law. Something that I thought that you had finally grasped from my thought experiment with me walking in an aeroplane. Remember, the energy that I expended pushing against the aeroplane floor was 35 joules. The aeroplane floor was pushing right back against me (Newton's third law), while moving at 200 m/s. Which is how it contributed the additional 14,000 joules to my total gain in kinetic energy of 14,035 joules.
In my thought experiment, that 14,000 joule contribution was so great, compared to the 35 joules that my muscles had contributed, that you were unable to pretend that it didn't exist. Clearly my muscles couldn't have contributed the entire 14,035 joules of my gain in kinetic energy.
Now that much lower speeds are being discussed, you have gone right back to pretending that the 2.385W contribution to the total power from the wind in bdunham7's calculation doesn't exist. As bdunham7 has shown, once you include that additional power, there is a net gain in kinetic energy, not a net loss as you appear to stubbornly believe.
Alex Eisenhut:
--- Quote from: electrodacus on September 03, 2021, 05:27:53 am ---
If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power. Breaking power and propulsion power. As long as breaking power is more than propulsion power vehicle will slow down unless there is something else that can offer propulsion.
--- End quote ---
Teaching you the difference between breaking and braking is also kind of pointless, I guess. |O
But since you like special cases, here's a breaker that brakes, or a brake that breaks.
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