Neddy Bate
Valued Senior Member
Maybe you should have deleted it this time as well, because it sounds wrong. Large enough square concrete bocks (instead of wheels) would perilously impede takeoff, surely???
Aircraft carriers - ramps - makes less than zero sense to me
- Thread starter Jonathan Ainsley Bain
- Start date
Maybe you should have deleted it this time as well, because it sounds wrong. Large enough square concrete bocks (instead of wheels) would perilously impede takeoff, surely???
Also, unfortunately, seems to be a low understanding of rhetorical questions and "what if, but it would never occur, scenarios (aircraft with concrete blocks for wheels) posed only as THOUGHT
experiment"
What about if a
Tomcat on a aircraft carrier had skis not wheels and the carrier had a ramp with a conveyor belt going sideways on the ramp?
Please show your working
foghorn
Valued Senior Member
Probably, but if the thrust is great enough the plane would rip free and fly. It does sound odd, but it makes the point that it is a question of thrust. I think??
DaveC426913
Valued Senior Member
Exactly. The point it was supposed to make is that an aircraft takes off based on its thrust acting on the air, not on its wheels on the ground. But it makes that point poorly.
I only brought it up because Sarkus had a similar thought in post 37.
DaveC426913
Valued Senior Member
Sure but that still obfuscates the issue for reader trying to grapple with it because the river provides friction that slows the plane down. Naive readers can then surmise a river that gets faster and faster in the wrong direction, and that would impede thrust.
Yazata
Valued Senior Member
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DaveC426913
Valued Senior Member
Rockin' those JATOs.
The engine thrust moves the plane forward. The weight of the plane is transferred via the wheels to the conveyor, pushing it backward. Unlike the ground, the conveyor offers no resistance. It moves backward a distance equal to the point of contact of the tire moving forward. Tires do not rotate on the rims.
The mass of the plane exerts a downward force=mg. There is more than thrust at work.
The mass of the plane exerts a downward force=mg. There is more than thrust at work.
Not thrust per se, but airspeed over the aerofoils (i.e. wings - the things that create the lift). An "aircraft" can have all the thrust in the world but if the airspeed (for some reason) remains zero, or below the level sufficient to create the necessary lift, the aircraft will not fly. Similarly, an aircraft may not need any engine or thrust at all yet still achieve flight: point it into a headwind that results in sufficient airspeed over the wings, and Bob's your Auntie's brother! Flight.
I was going to post something like that as an example!
Correct, there is indeed more than thrust at work.
1. The aircraft turns on its engines.
2. The aircraft will only start to move forward once it has generated sufficient Thrust to overcome both the inertia and friction that its mass is generating.
3. But it's not quite that simple, as the faster the aircraft moves forward, the more Drag it creates, and the Thrust needs to also overcome this. Drag is a function of V^2, the aircraft cross-sectional shape (and other parameters that make up the "Coefficient of Drag"), air density etc.
4. The excess Thrust (Engine thrust less friction, less drag) will be translated into an acceleration, with velocity being a function of acceleration and time.
5. Lift is similar to drag in that it is a function of V^2. Instead of a Coefficient of Drag (CD) you have a Coefficient of Lift (CL) which is a function of the aerofoil parameters (shape, size etc). In order to achieve take-off, the aircraft must generate Lift in excess of its Weight.
6. You thus hope that the engines produce sufficient Thrust to give you sufficient Velocity to generate sufficient Lift before you run out of runway. If you do, happy flying. If you don't... oops.
Velocity in the above is the airspeed - i.e. the speed of the air over the aerofoils - and not the aircraft's forward velocity relative to the ground.
Note: the inertia and friction, at least in normal operations (i.e. no concrete block tyres) can almost be ignored. All concrete block "tyres" do is increase the friction and inertia element of the equation, making them not insignificant.
Being on a conveyor belt really doesn't change the above analysis in any way, because forward motion is not determined by the rotation of the tyres (unlike in a car).
Hope this helps.
DaveC426913
Valued Senior Member
Sorry, you assert that the wheels push the conveyor backward? How do you figure?
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DaveC426913
Valued Senior Member
Air drag, at below takeoff speeds, is pretty negligible.
Unless you are including runway friction under drag. But it's still pretty small.
Certainly for a powerplant that is capable of getting the plane up to 150mph.
Yeah, that pretty much sums it up.
Well, for a reasonably designed aircraft it is.
. At uni, we were given a wing CL with which to design an aircraft, and one of our teams managed to design one with an horrendous L/D ratio.They did so deliberately, mind you, just to see what you'd have to come up with. Imagine a double decker bus with tiny wings and huuuge engines! I think they ended up with an L/D of between 1 and 2! (Typically you're looking for an L/D of 30+ for recreational and 20+ for commercial.)
Anyhoo, yes, it is pretty negligible at low speed, you're correct.
DaveC426913
Valued Senior Member
Yeah, like these acrobatic stunt planes they fly at airshows.
With a big enough engine, all you need is enough control surfaces to point it where you want to go.
Once you manage to lift the concrete blocks that are your tyres, of course.
Yazata
Valued Senior Member
Paging Mr. Ramp or Mr. Catapult!
DaveC426913
Valued Senior Member
Yeah. Using a Harrier is kinda cheating.
forum;
It was back to the drawing board.
The goal is to prevent an airplane from taking off (to the right) from a moving conveyor belt.
Initially the wheel of the plane is at conveyor position 0 (red line) which is also ground position 0.
The conveyor drive system is only capable of motion to the left.
One method of control could use a photo sensor on the belt to detect the presence of the wheel.
With synchronized motions, the tire moves a distance dw to the right while the conveyor belt moves a distance db to the left, with a net movement of 0.
It was back to the drawing board.
The goal is to prevent an airplane from taking off (to the right) from a moving conveyor belt.
Initially the wheel of the plane is at conveyor position 0 (red line) which is also ground position 0.
The conveyor drive system is only capable of motion to the left.
One method of control could use a photo sensor on the belt to detect the presence of the wheel.
With synchronized motions, the tire moves a distance dw to the right while the conveyor belt moves a distance db to the left, with a net movement of 0.
DaveC426913
Valued Senior Member
I think perhaps you are kidding around, but in case you are not, your setup still fails to prevent the aircraft from taking off.
The conveyor could accelerate to its max speed to where it risks flying apart - and the wheels could be spinning madly and start smoking - but the plane will still accelerate to the right and eventually take off.
The conveyor's movement to the left - no matter how fast it goes - is unable to prevent the wheel from moving to the right. Because the wheel is attached to the airplane, and the airplane is being moved by its propeller.
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