# Is this calculation for the 'free' climbing speed when flying in a headwind with a gradient correct?

From energy considerations, it can be found that a plane flying straight and level, if it enters a headwind with a positive gradient, will climb, extracting energy from the wind gradient.

The extra energy involved will be

$E_{EXTRA} = mg(h_f – h) = \frac{1}{2} m (vfg^2– vg^2)$

where $m$ is the mass of the plane, $h$ the initial altitude, $h_f$ the final altitude after the climb, $vfg$ the speed of the plane (when starting the climb) in relation to the mass of air present at the altitude where the climb will end, and $vg$ the TAS of the plane when starting the climb.

Now, the question is what the extra power of that airplane will be when climbing under those conditions. It’s clear that there is an extra power, since energy is being taken from the wind gradient at a given rate…

As that rate is related to the magnitude of the gradient, we may say that, for a given wind gradient $ß$, the extra power $P_{EXTRA}$ will be:

$P_{EXTRA} = E_{EXTRA} · ß$

$P_{EXTRA} = \frac{1}{2} m (vfg^2 – vg^2) ß$

Knowing the power, we can now find also the extra vertical speed w of the plane of mass m due only to the gradient:

$w = \frac{1}{2} m (vfg^2 – vg^2) ß/mg = (vfg^2 – vg^2) ß/2g$

In order to 'test' the formula, we may think that the difference between the wind speeds at the top and the bottom of a 'block of wind' is $12 m/s$, that we fly with a TAS of $vg = 60 m/s$ and that $vfg = 72 m/s$. We set the gradient at $10 m/s/100m = 0.1 s^{-1}$. Thus, the height of the ‘wind block’ will be 120 m.

Then,

$w = (722 - 602) · 0.1/2 · 9.8 = 8.08 m/s = 1591 feet/min$

Not implausible, I think…

I propose to look at it in a different way. Let's assume a constant airspeed climb.

We have two scenarios: one where there is no headwind gradient, and one with a headwind gradient. We can calculate the achieved climb speed in both scenarios and then compare them to see what the effect of the wind is. By selecting a constant airspeed, the drag is constant in both scenarios so they can be easily compared.

In both scenarios the aircraft will have the same constant thrust, such that the thrust exceeds the drag. The excess thrust is equal to 5% of the weight.

For the purpose this answer, I do not look at calibrated airspeed but only at true airspeed. Basically I neglect effects of changing density with altitude.

Also, the angle of climb (flight path angle $\gamma$) is sufficiently small to be in the linear domain (such that $\cos\gamma \approx1$ and $\sin\gamma \approx \gamma$)

## Scenario 1: climbing in a constant wind field

In the first scenario the aircraft is climbing in a constant wind field. This can be no wind at all, or a constant headwind or a constant tailwind.

For the aircraft to perform a constant airspeed climb in this constant wind field, all forces acting on the aircraft (thrust, drag, lift, weight) need to be in equilibrium.

• Thrust = green
• Drag = red
• Lift = blue
• Weight = black

$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W} - \sin \gamma\right) = 0$

With a thrust excess ($T-D$) of 5% of the weight, the aircraft need to climb at a flight path angle of 0.05 radians which is approximately 2.86 degrees.

$\frac{T - D}{W} = \sin \gamma = 0.05$

The vertical speed $w$ is then equal to:

$w = V\cdot\sin\gamma$

So in the first scenario the aircraft will achieve a climb speed that is equal to 5% of the true airspeed. Taking 60 m/s from your calculation for the airspeed, the aircraft will climb at 3 m/s = 591 ft/min.

## Scenario 2: climbing in a wind gradient

In our second scenario, the same aircraft climbs with the same velocity and same thrust setting, but now it encounters a headwind gradient of 10 m/s per 100 meter of climb. This equates to a wind gradient of 0.1/s:

$\tau = \frac{10 \textrm{ m/s} }{100 \textrm {m}} = 0.1 \frac{1}{\textrm{s}}$.

A climb of 3 m/s would result in an effective wind increase of $\tau \cdot w$ = 0.3 m/s2 which is effectively an increase of true airspeed. Therefore, this isn't a constant speed climb anymore.

To make sure the climb is performed at a constant air speed, the aircraft needs to climb faster.

$\frac{dV}{dt} = \frac{T - D - W sin \gamma}{m} = g\left( \frac{T - D}{W} - \sin \gamma\right) + \tau V \sin \gamma = 0$

$\sin \gamma = \frac{g}{g-\tau V}\cdot\frac{T-D}{W}= \frac{9.81}{9.81-0.1\cdot 60}\cdot0.05 = 0.129$

$w=V\sin\gamma =$ 7.72 m/s = 1521 ft/min

In the wind gradient the climb speed of 1521 ft/min is considerably higher than the 591 ft/min achieved in a constant wind field with the same power setting. The difference is a factor of 2.57 which can be attributed to the wind.

• Thanks for your valuable comments. However, and in 'scenario 2', I think that the climb is 'unintended'... I mean that the pilot is flying s/l, in non-moving air, then suddenly the wind conditions change to a headwind with gradient, and then the aircraft starts to climb automatically, and 'entirely free'... I mean that no fuel is spent on the climb, all the energy gained in the climb coming from the wind gradient... – xxavier Mar 31 '17 at 7:27
• @xxavier I'll think about that scenario and see if I can add it. The problem is that as soon a you start climbing, the lift vector is tilted backwards so the aircraft will decelerate. Of course kinetic energy can always be traded for height, but I am not sure if I can find a good way of expressing the energy transfer from the wind to the aircraft. Another problem is that with non-constant airspeed, the drag will change and hence there will be energy transfer independent of the gradient. – DeltaLima Mar 31 '17 at 7:47
• IMHO, the airspeed is constant during the climb... But it's just an intuition... – xxavier Mar 31 '17 at 7:49
• The air speed is only constant if you manage to balance all forces with the change in wind speed. Without having any excess thrust that is only possible at one particular airspeed which depends on the wind gradient. And if you fly at that airspeed, the rate of climb can be anything you like. You just harvest the energy needed for climbing from the wind. – DeltaLima Mar 31 '17 at 7:52
• Imagine that you are flying s/l with no wind. Now, things change abruptly and the plane is in a headwind with a positive gradient. You do nothing... The plane will keep a constant airspeed while climbing through that gradient, with no need for extra gas, since the necessary energy will be provided by the gradient, which will 'suffer' accordingly. The way I have tried to demonstrate it is, always in my opinion, OK, but there is a step which has only intuitive (and dimensional) support, and is where I introduce the gradient ß as the factor of time, converting energy into power... – xxavier Mar 31 '17 at 11:06