Engineering question
-
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.
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?
Our Forgotten Astronomy | Object Oriented Programming with C++
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
Thanks everyone for your replies, the links and the videos. When I posted this question, little did I imagine that it would raise such an intense discussion. Thanks once again.
-
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?
Our Forgotten Astronomy | Object Oriented Programming with C++
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
-
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
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](https://www.google.com/maps/@-13.163738,-72.5448503,2a,75y,81.9h,112.19t/data=!3m6!1e1!3m4!1sr5yCHd2XEJCzg0Lh3RVF2Q!2e0!7i13312!8i6656). The horizon is just more mountains.
Our Forgotten Astronomy | Object Oriented Programming with C++
-
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](https://www.google.com/maps/@-13.163738,-72.5448503,2a,75y,81.9h,112.19t/data=!3m6!1e1!3m4!1sr5yCHd2XEJCzg0Lh3RVF2Q!2e0!7i13312!8i6656). The horizon is just more mountains.
Our Forgotten Astronomy | Object Oriented Programming with C++
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
-
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
I wrote a script for Stellarium to do that. You can view (an old version of?) it here: [Stellarium Scripting](https://github.com/Stellarium/stellarium/issues/766). 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.
Our Forgotten Astronomy | Object Oriented Programming with C++
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
The speed of the wheels is irrelevant - they aren't powered or driven in any way. The speed of the airflow over the wings is the source of lift, along with the wings angle of incidence. Assuming the conveyor is moving at the same speed (but in the opposite direction) as the aircraft would be on a normal runway then the aircraft would actually be stationery and the airflow over the wings would effectively be zero and thus not generating any lift?
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
The airplane does or does not lift off owing to the upward airflow forces on the wings. If the air does not move relative to the airplane (or vice versa) the plane will stay on the ground. And the speed of the conveyorbelt relative to the air close by will cause some drag, therefore some lift, but it is likely to be way too little, unless you add a quite signifant ventilator to help. That should be pretty obvious, but I miss the joke - if there is one?
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
Only if you have a really big fan blowing at a speed at least that of the aircrafts V2 (take off) speed. :laugh:
-
Only if you have a really big fan blowing at a speed at least that of the aircrafts V2 (take off) speed. :laugh:
Then the aircraft will go and hit that fan :-)
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
-
It obviously can since propulsion for airplanes is provided by i/e a jet engine. Airplane doesn't accelerate using wheels, like i/e a car. Conveyor belt would only make the wheels spin faster, and that's it.
Actually ... no. The conveyor will have an effect and can increase lift - slightly - since it will also drag the air along with it and if that interacts with the wings it effectively act like the plane is going faster. (That's why aircraft carriers always launch steaming full ahead into the wind when launching and recovering aircraft - it adds a few MPH to their airspeed and reduces the chances of a stall.)
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt AntiTwitter: @DalekDave is now a follower!
-
Actually ... no. The conveyor will have an effect and can increase lift - slightly - since it will also drag the air along with it and if that interacts with the wings it effectively act like the plane is going faster. (That's why aircraft carriers always launch steaming full ahead into the wind when launching and recovering aircraft - it adds a few MPH to their airspeed and reduces the chances of a stall.)
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt AntiTwitter: @DalekDave is now a follower!
Your answer is confusing. First, you say: actually no, then continue with a reasonable explanation on why some extra lift would be achieved. So, airplane WOULD take off, which I assume was the point of the q. Whether it'd have more lift or no due to conveyor belt is a minor detail. There's also the matter of how long the conveyor belt is. If it's as long as the runway, then yea, but if it's as long as the airplane, airplane could fall off and probably crash in front of it.
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
The ONLY way the airplane can take off is if the speed of air respect of the airplane's speed is equal to the minimum speed the airplane needs to take off when the wind is absolutely calm. That is, because the conveyor makes the plane to be static respect to the ground, the only way the plane will take off is if there is a really hard hurricane that accelerates de wind to the plane's take off speed.
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
What a plane needs to take off is lift which is created by the pressure difference between the top and bottom sides of the airfoils (wings). To generate that difference, we use high-speed air flow aimed at the leading edge of the wing. The shape of the wing (an airfoil) makes air pile up in a high-pressure zone under the wing and zip over the top to create a low-pressure zone above the wing. This crazy magic lifts the plane with all of its weight into the air. To generate the air flow, we generally use the easiest thing at hand, the velocity of the plane itself. On a conveyor belt, it won't move against the wind, but if it's facing into a gale strong enough, it could theoretically lift into the air and its jets would then be sufficient to make it go so long as the gale persists long enough for the jets to achieve enough speed through the air to maintain lift.
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
Has anyone posted the Mythbusters episode about the question?
I’ve given up trying to be calm. However, I am open to feeling slightly less agitated.
-
If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ... Can the airplane take off?
Yes. An airplane is not propelled forward by the wheels, but by the propeller(s) or jet engines. How fast the wheels are spinning is irrelevant. It may be a little trickier to steer but it'll take off.
If you think 'goto' is evil, try writing an Assembly program without JMP.
-
Not a joke. A question asked by my friend, for which I am not aware of the answer.
Degree in physics, as if it mattered: No. The lift depends on airflow over the airplane wings. There will be none.
-
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
