Well, that's really what this comes down to, isn't it?
If the Chinook can drag the plane fast enough to generate lift, the tires will rub on the surface of the conveyor belt (just as the tires of little Billy's skateboard will rub on the treadmill).
If the jet engines are powerful enough, we will have the same exact effect as being dragged by the Chinook.
If it can take off being dragged by the Chinook, it can be take off with the engines.
Correct.If the conveyor is going backwards and the plane isn’t moving relative to the belt (engines off) which way does the plane go?
Backwards.
Correct, again.So the plane needs forward motion relative to the belt to remain in the same place relative to the ground.
No, this is incorrect. The opening post stated the conveyor belt speed matched the wheel speed, not the plane speed. In effect, the wheel speed would be twice the rotational speed compared with the wheel speed of a plane on a non-moving belt.As the plane speed increases the conveyor speed increases (as stated in the OP).
The relative motion through the air will be zero.
Yes, you are technically correct, I just used the example of pushing against the air to illustrate that the motion and lift of the plane is due to the engine/air interaction + wing/air interaction, not friction from the tires like the treadmill,etc., examples given by some. Also, air speed, and not ground speed, is the important parameter for lift.Yeah?
No. It’s the air being thrown backwards by the power plant (action-reaction) that provides thrust.
No, this is incorrect. The opening post stated the conveyor belt speed matched the wheel speed, not the plane speed. In effect, the wheel speed would be twice the rotational speed compared with the wheel speed of a plane on a non-moving belt.
This is the opening post. It is actually poorly worded as to how 'speed' is actually measured. A small airplane wheel cannot rotate at the 'same' RPM as the much larger conveyor belt, obviously (think small and large gears). The top of the conveyor belt would have a 'speed' relative to the ground, of course. Assume the top of the conveyor belt is moving at 120 mph relative to the ground. What speed would the wheels need to exactly match this speed of 120 mph? Not '0'. The wheels would need a forward speed of 120 mph relative to the ground, the same as the conveyor belt, but in the opposite direction. Understand where I am comming from now?Imagine a plane is sat on the beginning of a massive conveyor belt/travelator type arrangement, as wide and as long as a runway, and intends to take off. The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.
“ If the conveyor is going backwards and the plane isn’t moving relative to the belt (engines off) which way does the plane go?
Backwards. ”
Correct.
Draw a force diagram, If you can show me how a moving road in a perfect system (thus the wheel is completely free spinning, no friction of it on the axel, and infinit friction between it and the road) Then show me.
Not in the least! You are adding speeds when you should be subtracting - thus the major flaw in your vision of the problem.
<----------- attempted motion of plane
------------> countering motion of conveyor
Each are equal in value.
When they are added algebraically, the net result is zero.
oli,
when are you going to deal with post 97?
http://sciforums.com/showpost.php?p=1440199&postcount=97
yes, different indeed. so much so that the plane will indeed take off. agreed?Umm, post #99 addressed it somewhat. And then post #108 clarified my assumptions.
If the bearings ARE frictionless then it's a different ballgame.
so, how much friction will be needed to keep the plane on the ground assuming a 200 horsepower engine? what will that friction do?In 99 I said I wasn't considering frictionless bearings.
Just to be bloody-minded I'm still thinking about that. Although the inertia of the plane itself would certainly suggest so.yes, different indeed. so much so that the plane will indeed take off. agreed?
Oh that's a fun one. Especially since HP has little to do with the plane's actual speed or acceleration - I'd need the plane's mass, prop rotational speed and diameter among other things...so, how much friction will be needed to keep the plane on the ground assuming a 200 horsepower engine? what will that friction do?
Of such things are discussions born...i would imagine the OP was/ is a hypothetical scenario.
holy crap you guys burnt midnight oil on this?? lmao
oli,
i would love to meet you in a bar and see how much money i could weasel from you before you realized you been had.
google the phrase "plane on a conveyor belt" and get back with me.
what say you now oli?Well, I was gonna post and say that the plane would not be able to take off... that was until i did the experiment with a treadmill and a RC plane. I was actually very shocked to see that the plane moved forward at the same velocity and speed than if the treadmill was turned off. Im quite amazed.
what say you now oli?
let me guess, something about the above setup isn't like the OP, right?