This is for a story I'm writing. I can't find any information on how far various parrot species can travel without needing to land-- the closest I could find is this page saying that a macaw flies up to 15 miles looking for food. Intuitively I would think that larger birds, like macaws and African greys, would be able to fly farther than smaller ones due to having stronger wings, but the nonstop-flight record-holder is about the size of a robin so I guess that's not necessarily true.

Can anyone tell me how far various parrots can fly in a stretch, or at least the farthest that any parrot species can fly?

  • $\begingroup$ related biology.stackexchange.com/q/23530/3340 $\endgroup$
    Jun 27, 2019 at 10:20
  • $\begingroup$ @David. This site is open to anybody that wishes to use it. The OP clearly is asking a biological question which is on-topic here. It doesn't matter what their end-use of this information is. Please review our on-topic guidelines and our Code of Conduct. Most importantly, be nice to new users! $\endgroup$ Jun 27, 2019 at 14:02
  • $\begingroup$ @theforestecologist — OK, then it’s off-topic because he should have done his own research. I know nothing about parrots (other than you’re not supposed to shoot them in Australia) but was able to find an answer in a few minutes googling (at parrot.org). The site is supposed to be for serious students of biology and I think this sort of question is too like a Guinness Book of Records question. $\endgroup$
    – David
    Jun 27, 2019 at 16:35
  • $\begingroup$ @David Could you provide a link? I haven't been able to find an answer to this, and parrot.org doesn't seem to be at all related to my question. $\endgroup$ Jun 27, 2019 at 19:21
  • $\begingroup$ The page I found was parrots.org/ask-an-expert/…. It is a bit iffy in that some of the figures are miles per day (presumably landing in between) but others are non-stop between islands. Probably not as much detail as you would like, but a start. I searched for "range of flight of parrots". Another problem is that there is a drone with the name "parrot" so best to use plural. $\endgroup$
    – David
    Jun 27, 2019 at 19:57

1 Answer 1


Birds of flight were the original inspiration for the design of a machine that could fly and carry a person aloft, hence it's not surprising that the aerodynamics of avian flight and aircraft have much in common. Specifically, they both consume mass as the source of energy to maintain flight; jet fuel or gasoline in the case of airplanes, and stored body fat in birds, and they both have wings that provide aerodynamic lift as the air moves over them during flight. In addition, both share another characteristic of flight, the ability to glide, to continue flight without providing any of their own energy to maintain that flight. This energy is provided by the atmosphere itself in the form of rising air currents caused by a difference in temperature of a local 'pocket' of air; a pocket of air that is warmer than the surrounding air will rise because it has lower density, the Archimedes Principle in action. A similar process occurs when a parcel of moist air is surrounded by dry air at the same temperature as the moist air hence less dense than dry air. The third source of rising air is due to the local topography; the air on the windward side of a ridge or mountain is forced upward and is frequently used by birds as a source of lift.

Any discussion of gliding flight will unavoidably involve some aspects of atmospheric physics (a.k.a., weather), it's no different here. As stated above, a parcel of moist air surrounded by dry(er) air at the same temperature will rise. As long as that temperature is above the saturation temperature (the dew point) for that parcel of air, the water will remain in vapor form. We all know that as we go higher in the atmosphere the temperature drops; it's cooler at the top of a mountain than at its base. Therefore, as our parcel of moist air rises its temperature will drop, and eventually that temperature is the same as the dew point in that parcel leading to the condensation of that moisture, i.e., a cloud forms. As a surface of constant temperature in the atmosphere is nearly a level surface, we see clouds in the sky whose bases are all at the same level, the level where this condensation begins. Now, for a little of thermodynamics; when we boil water by adding heat (that is energy) to it, we are turning liquid water into a vapor (steam). Here's the thing, when we cool that steam down to the dew point it will condense back into liquid water, and in doing so, we get the heat (that was put in to make it boil) back again! That recovered heat shows up as an increase in temperature of the air that just gave up the water vapor. This increase in temperature causes that air to continue to rise, now due to a temperature difference with the surrounding air rather than a water vapor pressure difference; the cloud continues to grow upwards. This is the source of the cumulonimbus clouds we see in the sky that may eventually form thunderstorms. This discussion hi-lights a key fact about weather that relates directly to our discussion about gliding flight; if there are no updrafts, there are no clouds. That's correct, for a cloud to form, there must be updrafts containing moist air. No clouds indicates no updrafts. If there are no updrafts, there is no gliding flight. However, we note that really dry air is very hard to find; there still might be thermals around, but not likely, and those not very strong. The take away from this discussion is this: if we want to include increases in maximum range resulting from gliding flight, we need to be able to predict the weather (which hasn't happened yet, and I say this as one who has spent years as an undergraduate and graduate student active in atmospheric research.). Hence, long distance gliding flight will not be addressed any further here.

We begin our analysis of powered flight by considering a specific airplane, say a Boeing 787 passenger jet. To find its maximum range, the aircraft would be completely fueled up, take off and fly a level, constant speed flight path, as any accelerations (by changing altitude or going faster) would waist fuel. When the fuel tank goes dry, you have reached the maximum range of powered flight (assuming no head or tail winds of course).

From an analytic point of view, the fuel carried by the 787 is the source of energy, $E_s$, that powers its engines. These engines produce the thrust force, $\mathbf{T} = T\mathbf{\hat{T}}$ directed horizontally, parallel to the 787's longitudinal axis and to the flight path, that counteracts the effect of the atmospheric drag force, $\mathbf{D} = D\mathbf{\hat{D}}$ that opposes the 787's motion along its flight path. Under steady flight conditions (constant speed and altitude), the net horizontal forces on the 787 is zero so that $\mathbf{T} + \mathbf{D} = \mathbf{0}$, or $\mathbf{D}=-\mathbf{T}$. Taking the magnitude of both sides of this expression we find that $D=T$ so that $\mathbf{\hat{D}}=-\mathbf{\hat{T}}$. We find that the thrust generated by the engines has the same magnitude as, but directed opposite to the atmospheric drag.

Under the same flight conditions, we find a similar relationship for the vertical components of force acting on the 787, its weight, $\mathbf{F}_w = F_w\mathbf{\hat{F}}_w$ is balanced by the lift $\mathbf{L} = L\mathbf{\hat{L}}$ generated by the wings so that $F_w = m_p g = L$ and $\mathbf{\hat{L}}=-\mathbf{\hat{F}}_w$ where $m_p$ is the instantaneous mass (= takeoff mass of the plane, $m_{p_0}$, less the mass of fuel expended so far generating thrust) of the 787 and $g=9.8\, \text{m/s}^2$ is the standard gravitational acceleration on the surface of the Earth. We note here, that under these flight conditions, both $\mathbf{L}$ and $\mathbf{F}_w$ are perpendicular to $\mathbf{T}$ and $\mathbf{D}$.

If the thrust is removed so that $\mathbf{T}=\mathbf{0}$, then the drag force will no longer be opposed and will slow the plane down, reducing the speed of air flowing over the wing, which in turn will cause the wing to generate less lift, thus initiating the plane's descent (its weight is greater than the lift produced by the wings). If the plane is then 'nosed down' by an angle $\alpha$ from the horizontal, the projection of the plane's weight vector, $\mathbf{F}_w$ onto the plane's longitudinal axis will no longer be zero, but will instead be $\mathbf{F}_w\sin\alpha$ directed forward opposing the drag force. If $\alpha$ is chosen so that the sum of this projection and the drag vector is zero, then the plane will descend at a constant rate and the magnitude of the drag is given by $D = F_w\sin\alpha$. The projection of the weight vector onto the axis perpendicular to the plane's longitudinal axis, $\mathbf{F}_w\cos\alpha$, is balanced by the equal magnitude but oppositely directed lift vector, whose magnitude now becomes $L = F_w\cos\alpha$. If we form the ratio $D/L$ we find \begin{equation} \frac{D}{L} = \frac{F_w\sin\alpha}{F_w\cos\alpha} = \tan{\alpha} \tag{1}\label{1} \end{equation} The inverse of this ratio, $L_D = L/D = (\tan\alpha)^{-1}$, is known in aerodynamics as the lift to drag ratio while the angle $\alpha$ is called the glide slope angle. These two parameters are important in the overall characterization of the aerodynamics of an air-frame. Once this ratio is known, it can be used to estimate the drag in level flight. But in level flight, the lift is equal in magnitude to the plane's weight, $L=F_w=m_p g$. Substituting this expression into Eq.~$\eqref{1}$ and solving for the drag \begin{equation} D = L\tan\alpha = F_w\tan\alpha = m_p g\tan\alpha \tag{2}\label{2} \end{equation}

We have reached the point in our analyses that we need to address the mass/energy budget for the plane's flight. It will be useful to separate the mass of the plane into its empty (no fuel) mass, $m_{p_e}$, and the mass of available fuel, $m_f$, with the initial, takeoff mass of fuel given by $m_{f_0}$. With these quantities defined, the initial takeoff mass of the plane is given by $m_{p_0} = m_{p_e} + m_{f_0}$ while the instantaneous mass is given by $m_p = m_{p_e} + m_f$. During flight, the mass of the fuel available, $m_f$, varies such that $m_{f_0} \ge m_f \ge 0$ while the mass of the plane, $m_p$, varies as $m_{p_0} \ge m_p \ge m_{p_e}$.

There are two additional constants required to determine the net effective energy available to do work against the drag force when consuming the (differential) quantity $\delta m_f$ of fuel while flying the (differential) distance $\delta \mathbf{r}$. The first of these, $\kappa$, determines the total (differential) energy, $\delta E$, available from the combustion of the quantity $\delta m_f$ of fuel \begin{equation} \delta E = \kappa \delta m_f \tag{3}\label{3} \end{equation} For an American airplane such as the 787, $\kappa$ will have units something like BTU per pound of fuel expended. The second, $\eta$, specifies the efficiency of converting the available energy into actual work, $\delta W$, generating thrust that counteracts the drag \begin{equation} \delta W = \eta \delta E = \eta \kappa \delta m_f = - \mathbf{T}\cdot\delta\mathbf{r} = - m_p g\tan\alpha \delta r \tag{4}\label{4} \end{equation} where $\delta\mathbf{r} = \delta r \mathbf{\hat{T}}$ is a differential displacement vector along the flight path during constant velocity, horizontal motion and the minus sign accounts for the fact that the plane's energy stores are consumed as that energy is used to counteract drag (a fundamentally dissipative process).

Letting the $\delta$'s become derivatives, dividing through by $m_p$ and using $m_p = m_{p_e} + m_f$ and replacing the integrated variables by primed quantities, , Eq.~$\eqref{4}$ can be rewritten in the integral form \begin{equation} \eta \kappa \int_{m_{f_0}}^{m_f} \frac{dm'}{m_{p_0} + m'} = - g\tan\alpha \int_0^r dr' \tag{5}\label{5} \end{equation} with the limits of integration evaluated at takeoff and the current downrange position a distance $r$ from takeoff.

Performing the integrations indicated in Eq.~$\eqref{5}$ and simplifying, we have the result \begin{equation} m_p = m_{p_0} e^{-\frac{g\tan\alpha}{\kappa\eta}r} \tag{6}\label{6} \end{equation} We find that the mass of the plane, $m_p$, is an exponentially decreasing function of the distance flown, $r$. Letting $r=r_m$ be the maximum range of plane where all fuel has been expended (when $m_f=0$ so that $m_p = m_{p_e}$), Eq.~$\eqref{6}$ becomes \begin{equation} m_{p_e} = m_{p_0} e^{-\frac{g\tan\alpha}{\kappa\eta}r_m} \tag{7}\label{7} \end{equation} We note the similarity of this expression to that of the Tsiolkovsky rocket equation.

Eq.~$\eqref{7}$ can be solved for the maximum range $r_m$ \begin{equation} r_m = \frac{\kappa\eta}{g\tan\alpha}\ln\left(\frac{m_{p_0}}{m_{p_e}}\right) \tag{8}\label{8} \end{equation} an amazingly simple result, all things considered! This result remains valid for any aerodynamic system that obtains its lift via forward motion through the air provided by a propulsion system that consumes mass to produce thrust. It could be applied to a Cessna 172, or even a nitro powered radio controlled (RC) model of a 172. It could not be applied to an electrically (battery) powered model of the 172 because there is no mass loss from a battery, or to any type of glider (no thrust or mass loss). And it can, however, be applied to any bird of flight, including our parrot!

For the parrot, the energy source is the fat stored in its body. This mass is consumed through metabolic processes that convert it into $\text{CO}_2$ and water vapor that is expelled during respiration, and as sweat and urine as the parrot flies (the parrot's "exhaust" as it were!). The energy content of body fat ($\kappa$ as defined in Eq.~$\eqref{3}$) is 9 (food) Calories per gram. One food Calorie is equal to one kilocalorie which is in turn equal to 4184 Joules in SI units, see the Wikipedia article Food energy.

The efficiency of converting stored energy in the human body into mechanical work has been estimated to be $18\%$ -- $26\%$ (see Wikipedia page Muscle). One would expect similar numbers for other warm blooded vertebrates, so that, to one significant figure, we take $\eta = 20\% = 0.2$ (a dimensionless quantity).

There seems to be a very broad range for the percent of body mass that is fat. Some migratory birds have up to $70\%$ (see Obese super athletes: fat-fueled migration in birds and bats, however the parrot is not generally considered a migratory bird. The web page Comparison of flight mileage for various wild parrot species states a migration distance of 320 km for Thick-billed Parrots for example. Hence the $70\%$ number is likely far too large. At the other extreme, ground beef is considered lean if it contains $10\%$ fat, but more generally it is closer to $20\%$. We will select a value somewhat below the median of these extremes, say $35\%$.

A typical mass for a parrot is another difficult number to ascertain, as there is a very large difference in body mass for the various members of the parrot family. For example, the web page Average Bird Weights of Common Parrot Species gives data for 52 parrot species with links to four other species, each with several entries. These vary from 10 grams for the Zebra finch to 1530 grams for the Green-winged Macaw covering a mass range of over two orders of magnitude! Upshot: there is no such thing as a "typical" parrot! We will choose the Thick-billed Parrot as we have some long distance data to compare our result to. The Wikipedia page Thick-billed parrot gives its mass range as 315-370 grams, we shall use 370 grams so that $m_{p_0} = 0.37 \, \text{kg}$, $35\%$ of which is to be considered fuel so that $m_{f_0} = 0.16 \, \text{kg}$ leaving the parrot's 'empty mass' at $m_{p_e} = 0.24 \, \text{kg}$.

We have one remaining parameter to estimate, that being the glide slope angle, $\alpha$, used to find the lift to drag ratio above. Consider the order of magnitude estimates of $\alpha = 10^0 = 1\, \text{radian} \approx 60^o$, $\alpha = 10^{-1} = 0.1\, \text{radian} \approx 6^o$ or $\alpha = 10^{-2} = 0.01\, \text{radian} \approx 0.6^o$. Clearly $60^o$ is far too steep and $0.6^o$ is much too shallow, leaving $6^o$ as the only acceptable order of magnitude choice, hence we set $\alpha=10^{-1}$ radian, a number valid for most birds of flight.

Repeating Eq.~$\eqref{8}$ above, $$ r_m = \frac{\kappa\eta}{g\tan\alpha}\ln\left(\frac{m_{p_0}}{m_{p_e}}\right) $$ and substituting the parrot's values from above (including unit conversion factors)

$$ r_m= \frac{\left[\left(\frac{4184\,\text{J}}{\text{gm}}\right) \left(\frac{1000\,\text{gm}}{\text{kg}}\right) \left(\frac{\text{kg m}^2}{\text{J s}^2}\right) \right] \left(0.2\right)} {\left(\frac{9.8\,\text{m}}{\text{s}^2}\right) \left(\tan\left(0.1\right)\right)} \ln\left(\frac{0.37\,\text{kg}}{0.24\,\text{kg}}\right) \approx 370 \text{ km } $$

we find the answer to the question, "How Far Can a Parrot Fly [under power] in a Single Day?" to be

$$ \boxed{r_m\approx 370\, \text{km}} $$

a number which is in close agreement with the (limited) data available which gave an actual (vs maximum) daily migration range of 320 km.

It's interesting to note that this maximum range for powered flight can be viewed as the minimum range when gliding flight is included. Under ideal weather conditions, the actual maximum range could be extended considerably if the parrot were to capitalize on any available thermals it encountered during its flight.


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