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Hello Forum,

I understand that an airplane can fly when its overall weight $W_{plane}$ (force pointing down) is perfectly balanced by an upward directed force called lift $F_{lift}$. If the lift is larger than the weight, the plane will rise in height.

When the airplane is empty and starting from rest on the runway, the plane eventually reaches the right speed $v_{max}$ to generate lift $F_{lift}$ and make the airplane fly. The lift is larger than the weight when the plane takes off: $F_{lift} > W_{plane}$. Is that correct? If weight and lift were immediately equal, the plane would not gain altitude. Once in the air, the pilot can either increase or reduce the lift force compared to the weight (which stays constant).

What if the plane had a load so that its total weight was $W_{total} = W_{plane}+W_{load}$? The airplane max speed $v_{max}$ that it can reached on the runway is still the same and dictated by the plane's engine and its fixed power. With load or without load, the pilot always pushes the engines to the maximum. My understanding is that the lift force, always bigger than the total weight, will still be sufficiently large (up to a certain maximum load) to lift the airplane.

If correct, how much larger is the lift force compared to the weight of the airplane when the airplane is empty? 20% bigger? If so, if the airplane weighted $1000 N$, the overall lift force would be $F_{lift}= 1200 N$ the max loading capacity must be less than $200 N$ so that the lift force remains larger than the overall weight...

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256bits
Gold Member
With load or without load, the pilot always pushes the engines to the maximum
That is not true. Reduced engine power for a lighter load.
Perhaps maximum thrust for a military aircraft that wants to get the aircraft in the air and gain altitude as quickly as possible in a combat situation but for normal flights reduced power saves fuel and engine wear.
the engine power utilized can be also dependant upon runway length, wind speed ( which affects the air speed for head or tail winds, but not the ground speed ), temperature.

The airplane max speed vmaxvmaxv_{max} that it can reached on the runway is still the same
also not true. A heavier plane needs a greater air speed for nose up.
Actual maximum ground speed is related to the tires and their carrying capacity, above which tire failure is a possibility.

If correct, how much larger is the lift force compared to the weight of the airplane when the airplane is empty? 20% bigger? If so, if the airplane weighted 1000N'>1000N1000N 1000 N , the overall lift force would be Flift=1200N'>Flift=1200NFlift=1200N F_{lift}= 1200 N the max loading capacity must be less than 200N'>200N200N 200 N so that the lift force remains larger than the overall weight...
In that particular case approximately 20% larger loaded than when empty. Of course, the rate of climb will be somewhat different, meaning the vertical distance covered to the horizontal distance covered.

berkeman
Mentor
When the airplane is empty and starting from rest on the runway, the plane eventually reaches the right speed vmaxv_{max} to generate lift FliftF_{lift} and make the airplane fly. The lift is larger than the weight when the plane takes off: Flift>WplaneF_{lift} > W_{plane}. Is that correct?
No. Can you please post a link to where you are reading something like this? Vmax for fighter jets is Mach 1+. Are you saying that fighter jets take off from runways supersonic all the time?

Maybe try using Google to learn about the "rotation" speed at aircraft takeoff from conventional runways...?

CWatters
Homework Helper
Gold Member
Hello Forum,

I understand that an airplane can fly when its overall weight $W_{plane}$ (force pointing down) is perfectly balanced by an upward directed force called lift $F_{lift}$. If the lift is larger than the weight, the plane will rise in height.
If lift is greater than weight it will accelerate upwards.

CWatters
Homework Helper
Gold Member
When the airplane is empty and starting from rest on the runway, the plane eventually reaches the right speed $v_{max}$ to generate lift $F_{lift}$ and make the airplane fly. The lift is larger than the weight when the plane takes off: $F_{lift} > W_{plane}$. Is that correct? If weight and lift were immediately equal, the plane would not gain altitude.
.
No.

The plane accelerates up to speed and then the pilot rotates the aircraft so the wings make more lift.

If the lift is greater than weight the aircraft accelerates vertically. If lift stayed greater than weight it would climb faster and faster.

In practice lift eventually reduces or is reduced until it equals weight and the aircraft continues to climb at a constant velocity.

CWatters
Homework Helper
Gold Member
What if the plane had a load so that its total weight was $W_{total} = W_{plane}+W_{load}$? The airplane max speed $v_{max}$ that it can reached on the runway is still the same and dictated by the plane's engine and its fixed power. With load or without load, the pilot always pushes the engines to the maximum. My understanding is that the lift force, always bigger than the total weight, will still be sufficiently large (up to a certain maximum load) to lift the airplane.
Lift (and drag and the power needed) can also be changed by altering the angle of attack of the wing and by using flaps.

It's not always necessary to use maximum power, at least not these days. It was for many early aircraft as they were under powered.

CWatters
Homework Helper
Gold Member
If correct, how much larger is the lift force compared to the weight of the airplane when the airplane is empty? 20% bigger? If so, if the airplane weighted $1000 N$, the overall lift force would be $F_{lift}= 1200 N$ the max loading capacity must be less than $200 N$ so that the lift force remains larger than the overall weight...
If you gradually increase the weight of an aircraft the amount of lift needed to maintain level flight increases. This can be achieved by flying faster or by changing the angle of attack of the wing (flying nose high) or by lowering flaps. All three options increase drag so more power is required.

Eventually your aircraft can't carry any more weight. Your engine has limited power and/or your wing can't provide any more lift. For example if you increase the angle of attract too much the flow of air over the wing becomes turbulent, the wing stalls and off reduces, sometimes suddenly.

Thanks for all the reply and sorry for my sloppy description. I did some homework last night and this is what i found:

a) For the same runway length and the same engine power, a heavier airplane will need to reach a larger speed to be able to takeoff compared to a lighter airplane (no cargo, no people). The heavier plane will reach that required speed over a longer distance by using all the engine power. I read about derated takeoff. Airplanes don't run their engine at max power to take off (to save fuel and wear the engine less).

b) When an airplane flies horizontally, the lift force must be exactly equal to total weight. But when the plane has a certain nonzero pitch angle (nose up) the lift force magnitude in the vertical direction is less than the overall weight. Vectorially, the thrust force adds the missing vertical force component to balance the weight.
Another example: when an airplane flies vertically (like jets), the lift force is almost completely horizontal. The lift generated by the fuselage (not by the wings) + engine's thrust force will match the downward weight. Another example: when an airplane banks using its ailerons, the lift force generated by the wings is not vertical anymore but inclined. The banking allow the plane to turn left or right. Its vertical component is less than the downward weight. Even in that case, I think thrust will compensate for the missing upward vertical force.

Thanks!

CWatters