no, it could never stay in a straight line
[quote=“BluBalls,post:300,topic:37377"”]
For the people that think it is not going to take off… Answer me this… could a plane take off on ice? If so, why?
[/quote]
Yes, and they do ALL THE TIME in winter… stopping is a little harder though
[quote=“boardjnky4,post:301,topic:37377"”]
no, it could never stay in a straight line
[/quote]
wait, really, did you really just say this?
:mad:
the wheels may slip on ice, less friction, but, there is still some friction, so the plane can move forward and eventually take off. ice is not a moving surface… a treadmill is effectively keeping the plane in the same location by going in the opposite direction, and in the perfect (impossible example), at the same speed to offset the thrust.
[quote=“badazzss,post:305,topic:37377"”]
the wheels may slip on ice, less friction, but, there is still some friction, so the plane can move forward and eventually take off. ice is not a moving surface… a treadmill is effectively keeping the plane in the same location by going in the opposite direction, and in the perfect (impossible example), at the same speed to offset the thrust.
[/quote]
do you realize that the thrust is completely independent of the wheels right? i have not taken physics nor have i taken higher level math…but logic tells me the wheels will spin twice as fast in the opposite direction when independent thrust (like maybe a jet engine) is applied.
[quote=“newman,post:281,topic:37377"”]
at an infinitely small increment after the plane has changed forward velocity to V=0+, then wheelspeed is (V=0+) and thus treadmill speed is -(V=0+).
So i think the function is just V(treadmill)=2*V(plane)
but some part of me wants to tend this thing to infinity…
[/quote]
Since i know you will listen, I’ll post a bit here.
If the question states that the treadmill will match the speed in the oposite direction, why would it trend to infinity? It does not say it will exceed the speed, only match.
So V(treadmill) = -V(plane). At any instant in time, depending on what point on the wheel you are at, your V(wheel,X) will obviously vary. At the top, the instantaneous linear velocity of the wheel will =2V(plane) and at the bottom, it will = -2V(plane). At 90 degrees, it would be V(wheel,X)=V(plane).
As you power up your retardo boosters (engines), the plane will start to move forward. The treadmill will accelerate in the opposite direction, helping create faster spinning wheels. However were you to maintain thrust, so that you cruised down the taxi-way at 10mph. The belt would be at a constant -10mph as well. If you decided to punch it so you didn’t get rear-ended by the moron behind you, you would accelerate, having the belt accelerate at the same time. Eventually you would reach take off velocity and the experiment ends. To reach an infinite belt speed, you would need infinite acceleration, which would assume you kept going, and never took off.
The reason you won’t reach infinite belt speed, is the same reason the plane will take off. The plane is going to accelerate on its own, thanks to the thrust of the engines, the belt will spin, increasing the frictional losses to twice what you would experience at that same gven V(plane) relative to the ground. Even in the real world, this would still not create some type of overwhelming experiment ending force.
[quote=“jrod0187,post:236,topic:37377"”]
In all seriousness unlike yesterday when I was posting, the plane will move as Newman stated but it will not create enough lift under the Airfoil. For all you idiots (Walter) instead of posting my knowledge in avionics, I’ll just refer to a few examples that are somewhat “dumbed down” (again this is for people like Walter). Taken for Wikipedia:
where x is the location at which induced velocity is produced, x’ is the location of the vortex element producing the velocity “X” doesnt exist because there is no induced velocity produced. Why? Glad you asked. Again from Wikipedia since they explain things so well:
A symmetric airfoil must have a positive angle of attack to generate positive lift. At a zero angle of attack, no lift is generated. At a negative angle of attack, negative lift is generated
This right here is why I said no… the plane will not take off. That being said IN MY OPINION, I think that the plane will not create enough lift around the surrounding airfoil. The plane will move with resistance but depending on how far the conveyor belt is (which I’m sure it wont be long at all) I dont think it will get up to speed to fly. Now all you who said yes it will fly and supported a diagram that doesn’t represent the proper magnitude of the question is quite appalling. Maybe someone smarter than you said “yes it will fly” and then you jumped on the band wagon. :tup: to all those who agreed or disagreed and provided some what intellagent explanations.
[/quote]
There is so much wrong with this, I don’t know where to start, but I’ll give you an easy one.
Did you know, that outside of aerobatic aircraft, there are no planes that utilize a symmetric airfoil?
That said, symmetric airfoil aircraft do exist, and all start on flat horizontal ground, similar to our treadmill, yet still manage to take off. Care to explain?
I still want to see someone try to dyno a plane… you know… strap it down, with the wheels on the dyno then go full thrust.
why? obviously nothing will happen… we realize that. lol
same thing.
[quote=“badazzss,post:310,topic:37377"”]
why? obviously nothing will happen… we realize that. lol
[/quote]
Nothing will happen?
I envision the Dyno being dragged down the runway in a HUGE shower of sparks
The wheels of a plane on a conveyor belt do about as much as the wheels on a sea plane.
[quote=“BluBalls,post:309,topic:37377"”]
I still want to see someone try to dyno a plane… you know… strap it down, with the wheels on the dyno then go full thrust.
[/quote]
it will make as much torque as an RX-8. :mamoru:
[quote=“Dex,post:314,topic:37377"”]
it will make as much torque as an RX-8. :mamoru:
[/quote]
LOL
[quote=“badazzss,post:310,topic:37377"”]
why? obviously nothing will happen… we realize that. lol
[/quote]
its the same concept behind a treadmill runway
oh, i didnt know the rollers on a dyno actively moved? yeah…
[quote=“66impalass,post:318,topic:37377"”]
its the same concept behind a treadmill runway
[/quote]
why would it matter?