Airplane Question that Drove Me NUTS!!! - Electroboom

[exxos]

I must say I don't really get this. I mean surely a aeroplane has to be actually physically moving in relation to "open air" in order for the wings to generate lift. If there was a conveyor belt running at "take off speed" it would be irrelevant because there is no airflow over the wings in open-air. All it would be doing is turning the wheel surely ?! Even if the engines were running it would just mean the conveyor belt would have to run faster in order for the plane to be still stationary. If the plane's engines were generating more forward speeds and the conveyor belt counter acting it, and the plane would effectively be taking off at half speed or something.

[mrbombermillzy]

I didn't think it was too much of a puzzle really...but then again, I did recently read one of my children's books called 'rocket science for babies' (no joke!).

If you *just* have thrust, it's possible to achieve takeoff if the thrust is sufficient and positioned correctly (think of a Harrier VTOL jet).

You can also achieve the task with less thrust, but then there must be uplift (from the wings) which is only possible from a high enough air speed.

Anyhow, I'm going to say 'no' to this, as the takeoff of this particular craft is designed to have both of these variables and it won't have any uplift on that conveyor.

[mfro]

Frankly, I don't understand what the problem is.

If there is a conveyor with no friction, it doesn't even matter what speed it's running at or if its there or not.

The only two variables that decide whether the bird will fly or not is the thrust of its engines and the resulting relative airspeed at the wings generating lift. Both couldn't care less on what's happening under the plane's wheels.

[rubber_jonnie]

Well, no, it can't take off is the simple answer, because regardless of the thrust involved, no forward motion means no lift which means the plane cannot become airborne, and unless you can vector the thrust like in a Harrier or other similar VSTOL aircraft (Lift=thrust in these cases) you won't go anywhere.

Wheel speed is irrelevant to air speed.

However, if you replace the 747 with a B29 with 4 propellors it changes things a little, as due to the location of the engines and propellors, they generate thrust but also airflow over the wings.

I doubt that propellors alone would generate enough airflow to get a B29 airborne, but they may generate some lift that wouldn't be present in the 747 due to the engine type and location.

However if you substituted a STOL aircraft with very high lift wings and a propellor, then it might just work.

Or of course you could just say, as Medhi did, it's sitting there doing nothing, so nothing will happen...

[chronicthehedgehog]

I didn't watch past 1 minute - didn't want to spoil it. What if the airplane was facing into wind and windspeed was very high. That would generate lift wouldn't it? Enough to lift the nose and reduce contact with the ground perhaps.

[coonsgm]

Having worked in engineering and commercial roles for two of the largest jet engine manufacturers (the US based ones, not RR) over a 20 year period, I can safely say that no, the airplane will not have lift from the wings and will thus not take off.

There are very few engines / aircraft that have a thrust to weight ratio of greater than 1. This is what is required in order to get off the ground without any lift from the wings.

If anyone is interested, I'd be happy to draw a force diagram explaining this. (didn't watch the youtube video to conclusion, so maybe he already did that).

[alexh]

Everyone GCSE physics student knows that the aeroplane requires airflow over the wings to produce lift.

The question is given the scenario can there be air flow over the wings "if the conveyor belt can match the speed of the wheels"?

You automatically think not, that all the energy of the engines would go to turn the plane's wheels and in-turn the conveyor belt and the air speed would remain zero.

But as demonstrated in that thread it is not true. The aeroplane moves forwards, creates the normal amount of airflow across it's wings and takes off having reached the same air speed it normally requires to take off.

[mfro]

The description says the conveyor would be as long as a normal runway, the picture has it just a little longer than the plane.

In the former case it will just take off and don't care about the belt, in the latter case it will roll on slowly as normal and then fall off :lol:

[dad664npc]

No takeoff - there is no forward motion. No forward motion = no lift. The aircraft is on a giant treadmill. If you are running as fast as you can on a treadmill, do you get a breeze in your face?

[Darklord]

I think some points have to be very clear.

The conveyor belt, if only as long as the picture shows, would result in no take off. It's not a VERTOL so there's just no way it could lift off in that space.

If the conveyor belt is as long as a normal runway, things get a little more interesting. If the conveyor belt runs as fast as the wheels *only* and the thrust from the jet engines exceeds that, then the plane can take off, assuming it reaches minimal lift off speeds at V1 and V2.

However, if the conveyor belt can match the plane's speed, including engine thrust, then it's still not going to go anywhere, and no takeoff. The key difference being whether we are talking about the conveyor belt matching the planes wheels (and not taking engine thrust into account) or the conveyor belt matching the plane's speed, including engine thrust.

All of this is, of course, "IMHO". (and I hope they have an in-flight movie and drinks/meals 'cause I think this flight might be on that conveyor belt for a very long time!). :lol: