I'm working on a species of large bird, and trying to determine whether they will be able to fly or not.

I know that this equation likely applies:

$$ A = \cfrac{L}{\cfrac{1}{2} v^2 \rho C_L} $$

where $A$ is (wing) surface area, $L$ is lift force, $v$ is speed, $\rho$ is air density, and $C_L$ is coefficient of lift.

For an Earth-like planet, specifically one with Earth-like gravity, applying this would be relatively straight forward, as already illustrated in the linked answer. However, my planet has a different gravity! Specifically, the surface gravitational acceleration is a shade over 12.2 m/s², corresponding to about 1.25 G, which (unfortunately in this case, but deliberately chosen) amounts to a rather significant difference.

The atmospheric pressure on the surface is 1930 millibar, compared to Earth's just over 1013 millibar in a standard atmosphere. These birds fly at such altitudes that, as an approximation, their atmosphere can be assumed to be uniformly at that pressure.

The birds don't need to be fast flyers -- I'd be happy with 15-20 km/h in level flight, though I don't mind it at all if they can be faster.

However, they are big. My current design has the wing span of a large one at up to 300 cm tip to tip (compared to e.g. a bald eagle's typical up to 230 cm), 80 cm from the beak to the tail feathers, and a mass (not weight) of 13 kg (up to three times that of a harpy eagle). On Earth, that corresponds to a shade over 127 N (13 kgf); with the higher gravity, that'd be 159 N (about 16.2 kgf?) of force directed toward ground. By simplifying (in the manner of the spherical cows of physics), I estimated the total wing surface area of a large bird to be about 15000 cm².

  • How do I apply the above equation to a non-unity gravity? (What is the actual unit of $L$?)
  • If the above equation doesn't apply in non-unity gravity, what formula do I need?
  • Will my birds likely be able to fly in the specified environment?
  • $\begingroup$ I can't quite decide whether I want this to be [hard-science] or just [science-based]. It's starting out as [hard-science] and I may relax that later if I don't get any good answers. $\endgroup$
    – user
    Commented Aug 8, 2017 at 17:02
  • $\begingroup$ It looks like you're wanting a yes or no with the math to back it up. Hard-science looks like the right tag. $\endgroup$
    – sphennings
    Commented Aug 8, 2017 at 17:23
  • $\begingroup$ @sphennings Yes, that's about it. In this case, I even think I have the math in the question, I just don't know how to apply it to my particular problem. $\endgroup$
    – user
    Commented Aug 8, 2017 at 17:28
  • $\begingroup$ You're concerned about gravity's impact due to having a different value than Earth's surface, but so does the question you linked. The equation may implicitly include gravity via the atmospheric density and lift force values, so it would cancel out. $\endgroup$
    – Frostfyre
    Commented Aug 8, 2017 at 17:36
  • 1
    $\begingroup$ I think this quote is appropriate here: "Get your facts first, and then you can distort them as much as you please." - Mark Twain $\endgroup$
    – JYelton
    Commented Aug 8, 2017 at 20:49

3 Answers 3


To know if an object can fly in a given environment upward force must be greater than the downward force.

If we're assuming that the only forces that matter are gravity and lift, we want the downward force of gravity upon the object (159N) to be less than the force of lift.

To calculate the minimum flight speed we need to find the speed where L = Fg. This can be done by setting L = 159N and solving for v.

  • $\begingroup$ This is exactly what the OP is looking for, I think. OP should now consider whether the bird can actually move that fast considering the increased drag from huge wings and thicker air. $\endgroup$
    – bendl
    Commented Aug 9, 2017 at 13:39

The extinct Haast's Eagle (Harpagornis moorei) flew in Earth's atmosphere in New Zealand until the 15th century. Big females weighed up to 15 kg, and had a wingspan of 2.6 to 3 meters, says Wikipedia. Since your world's gravitational acceleration is only about 25% higher than Earth's but the atmosphere is about 100% more dense, I don't see why it couldn't fly there too, lift being proportional to the density of the atmosphere.

  • $\begingroup$ Was about to link Haast's Eagle also. Fairly close to what Michael describes. Also worth mentioning is Teratornis: en.wikipedia.org/wiki/Teratornis which was actually even larger. (Up to 4 meters wingtip to wingtip.) So given that it has worked here it should work at a slightly higher G with a denser atmosphere. $\endgroup$
    – Doomfrost
    Commented Aug 9, 2017 at 14:45

This can be resolved by considering the forces involved in avian flight.

Looking at flight from a physics point of view, there are four main forces that you have to worry about. Weight is a force produced by gravity in the downward direction, and every flyer has to produce lift in order to counteract weight. Anything moving through air also experiences drag, which slows it down, so there must be a forward-moving force, called thrust, to oppose the force of drag. These two pairs of forces – weight and lift, drag and thrust – have to be roughly balanced in order for a bird or plane to fly.

Source: The Flight of Birds by Joanna Tong & Adele Schwab.

Lift can be calculated as the weight of the bird in the specific gravity field. Force equals mass multiplied by the planet's force of gravity.

Additional information about bird flight can be found here. This goes further than simply answering the querent's question, but which may prove useful.

ADDENDUM: The above link to The Flight of Birds is broken, despite the fact it can be accessed via a Google search, this seems to be due to MIT reorganizing their courseware files, however, this link should take people to a list of courseware containing this item. Scroll down the list until find the relevant PDF document.

  • $\begingroup$ Your link to "The Flight of Birds" is broken. $\endgroup$
    – user
    Commented Aug 9, 2017 at 10:55
  • $\begingroup$ @MichaelKjörling I have fixed a link in my addendum. While it's pretty dodgy, I have tested it and it works for me. YMMV. Thanks for advising the link was broken. $\endgroup$
    – a4android
    Commented Aug 10, 2017 at 1:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .