Could a 747 Take Off Using Engine Thrust Alone with No Wheels?

The thrust to weight ratio is too low to move the airplane without wheels...only about .3 thrust to weight.....You would need to exceed 1 to 1...

https://www.grc.nasa.gov/WWW/K-12/BGP/Donna/t_w_ratio_answers.htm

https://en.wikipedia.org/wiki/Thrust-to-weight_ratio

If the aluminum-to-concrete coefficient of friction is less than the thrust-to-weight ratio, then the plane could take off.

Thanks for the quick reply and the included link. During our conversation thrust to weight ratios did come up, so I am glad we were on the right track. So.. in reviewing the link noting fighter aircraft 1 to 1 or better ratio; would a F-16 theoretically be able to overcome the friction of the runway on aluminum to reach take off speed in the previous scenerio? Not sure if there are aluminium to runway friction tables out there or not.

The heat would build from the friction of the aluminum dragging against the runway surface producing a spectacular shower of sparks until the entire plane exploded in a huge fireball of red hot aluminum shrapnel...but I'm just guessing...

Like when you see some idiot driving a car on the rims?

could A 747 take off on rims?

Probably have less rolling resistance...

it's been done... partially

no thanks!

The engine thrust is only helpful in opposing gravity if the trust can be directed downward.

Not so, the thrust moves the plane forward, as the speed increases so does the lift generated by the wing surface, lift is in direct opposition to gravity....

SE, we tend to the antagonist side of relationships but seriously, You should really work on reading and not translating or interpreting. Red was very clear in what he said using far fewer words than you did to express the same thought.

How so? I don't agree with any of that...

LoL Nice. You get a GA from me.

Gear up landing....

http://www.historyofpia.com/forums/viewtopic.php?t=15552

If the angle of incidence is positive not negative. I thought it is usually negative, so this makes the angle of attack negative so the plane moving forward would make it unable to rotate to a positive angle of attack. -- JHF

Teflon is ~0.04...So if the runway was coated with teflon, and you had some way to keep the wings straight and level, and some way to steer the plane, possibly with magnets...it still wouldn't work.... You still need more thrust!

..."The friction between aircraft and runway will be proportional to the normal force exerted by the aircraft on the runway. The normal force will be the difference between Weight of aircraft and Lift, the friction coefficient will be typically of a magnitude of 0.02 for a standard tarmac runway."...

Ref...

Where does the 0.02 COF come from? More likely would be around 0.20.

You might be right, this seems more like the rolling resistance coefficient...

Agreed; that makes sense as rolling resistance.

Teflon? Why not something that occurs in nature, freezing rain (liquid water on top of ice).

Sure...

"Water ice at temperatures not far below 0 °C is remarkable for low coefficients of around 0.05 for static friction and 0.04–0.02 for dynamic friction, but these figures increase as the temperature diminishes."

Ref

Maybe with a set of ski's and an ice covered runway...

"The Air National Guard's LC-130 planes carry heavy loads to the south pole and other National Science Foundation bases in Antarctica, plus the high, cold summit of the Greenland ice cap."

"The aircraft are equipped with retractable skis that allow the aircraft to land on snow and ice as well as on conventional runways. The aircraft have provisions for using rocket-assisted-takeoff (RATO) rockets, four on each side of the aircraft, that are installed and used when the LC-130 operates from rough, unprepared snow surfaces, sticky snow or when shorter takeoff runs are needed. Originally the expended rocket bottles were jettisonable, but due to several accidents which occurred when a bottle detached from the aircraft during takeoff, the mounting provisions were changed so that the bottles could not be released in the air."

https://www.popularmechanics.com/flight/g634/how-the-lc-130-flies-to-greenland-and-antarctica/

https://en.wikipedia.org/wiki/Lockheed_LC-130

Wow new turboprops....what a machine...notice they lift the ski's off the ground when stopped to keep them from freezing to the ground....

https://www.thedrive.com/the-war-zone/4142/fat-albert-c-130-gives-you-a-buzz-cut-in-this-crazy-head-on-takeoff-video

You would need enough thrust to overcome static friction, or stiction, to get the plane to move...and then you would need enough thrust to overcome wind resistance....So you might need well in excess of 1 to 1 thrust to weight ratio, perhaps even 2 to 1 or more....

Wouldn't the resting surface area of the belly be a factor as increasing the static friction to be overcome by the thrust as well? The actual surface area of the tire to runway very small compared to the belly of the aircraft.

Believe it or not, the area is not a factor in sliding friction.

For example, if you have a 1" x 2" x 4" block on an inclined plane, and you raise the plane until it starts sliding, it makes no difference which surface of the block is in contact with the slide.

http://zebu.uoregon.edu/disted/ph101/lf.html

The reason is that the frictional force is proportional to the area times pressure, where pressure is the force perpendicular to the plane, N, divided by area, A. So F = μAN/A, and the Area, A, cancels out.

You are overlooking several implied assumptions. If the contact area stays uniform during only sliding and no debris provide either roller lubricating or galling action then the contact area will not matter. In this case as whole layers of sheet metal slide off the COF will drop and as exposed superstructure attempt to dig into the sturdy runway the COF will climb. (What if the cement runway has expansion joints that exposed superstructure can catch upon?)

Remember those marvelous Physics demonstrators (i.e. Sargent-Welch) are meticulously engineered to prevent other obfuscating phenomena and forces from interfering from the presented phenomena. Attempting to lift-off any airplane in a crash landing scenario is far from an ideal scenario.

Item "B" better be pretty good or the engine nacelles will likely scrape--I don't have the detailed dimensions to determine degree of control to counter any tendency to roll (bank for you non-pilots).

we saw the results of this approach with the recent Pakistani air crash, the engines don't like being sanded off at the bottom.

Yes, given your extra variables.

On a normal runway, no, because the friction would not be overcome within the length of the runway. However, you gave an infinitely long runway, so friction can be overcome.

You also gave control and a belly that will not be destroyed. Thus, it is possible to gain enough speed to fly the plane.

However, the speed necessary would be higher than that needed with wheels because the plane is now closer to the ground, cannot lift the front up, or rotate, I believe it is called, to increase the lift ratio on the wings, so herefore the lift must be generated by a flatter wing surface.

Unfortunately no.

Because (as is obvious if you look at a picture of a 747), if the aircraft is resting on its belly you have at the very least destroyed the inboard pair of engines, as they protrude lower to the ground than the bottom of the fuselage and therefore the bottoms of these engines would be crushed by the weight of the aircraft.

Could it take off under these circumstances on the remaining two engines?

If not heavily loaded, a 747 (on wheels) can take off with just one engine. This is not normally advised, but it has been done.

Maybe ... with a long greasy downward sloping runway with a headwind of 150mph.....

A Hungarian aircraft took of without landing gears in 2000, though not from a full stop:

Malév Flight 262

The Melev scenario is about the closest to my original question. how fast (continuous) momentum-energy the plane skidding down the runway verses could it be done from complete stop? I would concede, with enough thrust... you can fly anything with wings..

the question remains does 747 have enough thrust to overcome the static friction at gross weight to reach a rotation speed? Even with unlimited distance; I think it would not but I don't know how to prove this mathematically.

The thrust to weight is all you need. Even with it being the smallest 747 version and using the same engines as Air Force One and having those engines rotatable like a Harrier there still isn't enough thrust to lift off vertically.

Basically you can simply use thrust versus weight like a Rocket. once you realize there isn't enough thrust you stop. Sliding down a runway would only increase the thrust needed which is already not enough. What more do you need?

You are trying to generate lift with the wings moving through the air. with the wings that close to the ground lift will not form at any speed.

Can it be done, by lifting the aircraft a bit off the ground, la hovercraft till we establish enough speed and then take off. I am not sure about the calculations, but a thrust vectoring kind of thing could work.

Just an idea! :)

Pass the Aspirin, please?