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The engines don't directly cause the airflow. The engines push the airplane, whose movement through the air causes the airflow over the wings.
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So there will be an airflow, and the plane will lift into the air.
The airflow is a consequence of the engines pushing the plane into speed, exactly as at a "standard" take off. The only difference is that the free running wheels will be spinning twice as fast when the plane leaves the ground, but the speed of the plane - relative to the surrounding air and the solid ground - will be exactly as for a normal take off. The air flow in the same.
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Not quite accurate. The airflow over the wings is only part of the lift needed to fly. Modern aircraft are too heavy to use the Bernoulli Principle to fly. Instead they use the redirected air flow from the belly of the fuselage. Watch an aircraft in flight - the nose is always higher than the tail and the plane is staying aloft from Newton's 3rd Law of motion. The force keeping the plane in the air is the air being deflected down by the slope of the fuselage.
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Yes it can, the wheels spin freely and have nothing to do with propulsion, even in normal takeoffs. The Mythbusters even did a show about it.
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The answer is not without wind.
An airplane is lifted off the ground, not because of the speed of the plane, per se.
But because of the speed of the air moving above and below the wing.
The shape of the wing leverages the Bernoulli affect. (High Pressure below the wing, lower pressure above),
giving the plane "lift".
In fact, during a strong wind storm. Planes that are stored OUTSIDE, and TIED DOWN. WILL Lift into the air, and pull against the ropes. Being in Florida, I have witnessed this first hand. It's wild. (And it only works if the plane is facing the wind! The other planes get pushed "down/away" as their wings are "reversed", or they get turned.)
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Obviously yes I would say
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Not if its speed over the ground is zero.
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I’d argue that it’s the airspeed that matters. If hurricane force winds start blowing during the experiment, the airplane might take off. However the original question didn’t mention anything like that and, under normal conditions, airspeed and ground speed are roughly equal.
Mircea
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It is airspeed that matters, but how fast the wheels are spinning is irrelevant, they aren't what propels the airplane forward.
If you think 'goto' is evil, try writing an Assembly program without JMP.
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Of course you are right. I realized (late) that engines push the air back and, hence, the airplane forward irrespective of wheels moving or not (or even not existing at all as in the case of seaplanes).
Seems my brain was taking a day off yesterday . Luckily it was a weekend day.
Mircea
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I hear you. My brain takes frequent breaks, and not just on weekends. I just happened to have argued this same scenario a few years ago (and was on the wrong side at first) and recalled the facts.
If you think 'goto' is evil, try writing an Assembly program without JMP.
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Thanks for an interesting link.
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Other thought experiment: a toy car is on a conveyor, you push it forward and the conveyor goes backward "at the same speed" (whatever that means, which is not quite clear). Can you push it forward?
Whatever the answer, an airplane would do the same thing, because its thrust is applied in the reference frame of the air around it. The wheels are not driven, they spin freely except when the brake is applied.
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Quote: The wheels are not driven, they spin freely except when the brake is applied. ... and that is an important point
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harold aptroot wrote: a toy car is on a conveyor, you push it forward and the conveyor goes backward "at the same speed" (whatever that means, which is not quite clear). Can you push it forward? Yes, but you might want to rethink about the problem. Derek Muller was able to show that multiple physics professors at prestigious universities didn't even understand the basic underlying principles.
A Physics Prof Bet Me $10,000 I'm Wrong
It's not exactly the same problem, in the video the propeller will generate a greater force and the vehicle will move forward. Increasing the speed of the conveyor belt will result in an even greater vehicle speed.
The wheels are irrelevant. So are university degrees.
Best Wishes,
-David Delaune
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Funny you would mention that. As a teen, I did an (I think) related experiment, which is easier to think about.
The setup is like this. Part A: a small lego (technic) car has a gear (which will act as pinion gear) on one of its axles, and you hold a long rack so that it meshes with the gear on the top. Clearly you can push the little car by pushing the rack, it will go at half the speed that you push it.
Part B: if you make a little sliding mount for the rack under the gear, can you still push the car? It turns out that you can, and it makes the car go very fast. That is a sort of similar situation as with that propeller car, except instead of wind there is a rack, and instead of a propeller there is a pinion gear that meshes with the rack.
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Hmmm,
I don't know much about this one.
I was curious as to what you're talking about. I looked it over and the article is referenced from another source:
Sullivan W. (1996). The Secret of the Incas: Myth, Astronomy, and the War Against Time. Three Rivers Press, New York.
Using my ever-improving Wikipedia editor skills, I checked that book beginning on page 41. The reference is based on references from yet another source and I quote 'from a discussion with Gary Urton'. Looks like Gary Urton is guy to contact on this. He probably has alot of free time[^].
Btw that guy has great credibility, you should check to see if there is something you missed.
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Sorry, I must not have been clear. Sullivan's book outlined how the fox's tail had became invisible due to the precession of the equinoxes on the morning of winter solstice in 650 AD. The pic I uploaded shows the sky with a sun depression of 20° (I believe) at that time. The 'fox' was a Milky Way 'object,' as it wasn't a constellation in those terms (if memory serves me correctly). The Milky Way disappears from the sky slightly before astronomic dawn, so the Milky Way would have started disappearing shortly after that 20° depression, if not at that 20° depression.
Benfer, in his article, was arguing that the Milky Way and the 'fox's tail' was clearly visible in the sky, because he didn't take twilight into account. In other words, he was saying that the sky instantly goes from dark enough to see the stars, to instantly light once the sun breaches the horizon. Therefore the 'fox' could be seen all the way up to sunrise. Which is utter nonsense.
Does it make sense that way?
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David O'Neil wrote: Does it make sense that way? Yes, I fully understood your position on the subject. Not sure if I want to get involved in your academic debate with that guy.
I would be willing to play the straw man with you:
Straw Man wrote: Is the altitude at sealevel in your astronomy software? The Andes are at nearly 7000m at the peak. A quick calculation tells me that you would be able to see approximately 300km over the horizon from the Andean mountain peaks at that elevation.
Not my argument, just trying to generate a position for you to defend.
Best Wishes,
-David Delaune
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The only way to find out is to go there and see for ourselves when the Milky Way disappears before sunrise. Or have a very, very good model of atmospheric absorption vs temp at that time of year, yada, yada, yada. Even at 7000m, I know it won't be visible ten minutes before sunrise, which is what he was saying. Go to street view. Here is Machu Picchu: Google Maps. The horizon is just more mountains.
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David O'Neil wrote: The only way to find out is to go there and see for ourselves Looks like Dr. Bromberg from the University of Toronto has a neat little tool that will calculate civil, nautical, and astronomical twilight values for arbitrary dates.
The Kalendis Calendar Calculator[^]
Plug in your dates/coordinates and let's have a look at the values.
Best Wishes,
-David Delaune
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I wrote a script for Stellarium to do that. You can view (an old version of?) it here: Stellarium Scripting. The picture I posted is at 20° sun depression I believe, which is 2° greater than astronomic dawn. If you want my most recent version (which may be exactly the same as that - I haven't looked at it in a while), I'll post it.
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