I just really don't understand how an airplane can fly with greater weight than lift.

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    $\begingroup$ I'm not sure I understand what you're asking here, but your question as it's phrased is the equivalent of asking "If I can lift 400 pounds how can I lift this 500 pound box?" (The answer should be self-evident.) $\endgroup$ – voretaq7 Aug 7 '15 at 15:32
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    $\begingroup$ I don't understand your question fully. Can you give some actual numbers, or a link to some article that explains what you're asking about? $\endgroup$ – Pondlife Aug 7 '15 at 15:33
  • $\begingroup$ weight 5:4 lift $\endgroup$ – Ethan Aug 7 '15 at 15:35
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    $\begingroup$ As described by Peter Kampf's answer in the question @fooot linked, I believe that if the weight of the plane is slightly greater than the lift generated by the wing, it's possible to compensate by increasing the AoA and using excess thrust to generate a lift component. However, it's going to be an expensive flight (fuel consumption) and most likely illegal, since you'd be significantly over certified MTOM for the plane. $\endgroup$ – FreeMan Aug 7 '15 at 16:09
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    $\begingroup$ @reirab: Thanks. Agreed, but it gave me a strong headache. Negative mass: Rudiments for flying saucer science :-), hence would be plagiarism. $\endgroup$ – mins Aug 28 '15 at 5:56

There are many situations where the weight is not equal to the lift.

Lift is, per definition, the aerodynamic force perpendicular to the undisturbed incoming airflow.

The weight of the aircraft is due to gravity, hence pointing down.

An example of a steady situation in which the lift is not equal to the weight is the climb phase. During climb the lift is actually less than the weight.

In your comment you mention a weight-to-lift ratio of 5 : 4. If that would be during an unaccelerated climb, the flight path angle would be 37 degrees and the thrust minus drag would have to be 60% of the weight.

The image below shows the weight vector $W$, lift vector $L$, and excess thrust vector $T$ in a ratio 5:4:3.

The combination of the lift and thrust vectors is equal is size and opposite to the weight vector. It is denoted $W'$ in the image.

For simplicity I left drag out of the image and assumed thrust is purely acting in the direction of travel.

enter image description here

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