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I was reading a discussion about the "Downwind Myth" and a doubt arose:

If we are taking off with a 50knots headwind, that wind component will be part of our dynamic pressure, and so in that case part of our IAS.

Of course turning 180° after take off will not cause any change in IAS, because from that moment the aircraft will be part of that airmass. The only change will be on the GS.

Taking off with the same wind intensity, but this time from the opposite direction, we will not have that advantage in IAS, and dynamic pressure.

Is that increase in dynamic pressure only possible when on the ground? (excluding headwind shear). What is the physics behind that?

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    $\begingroup$ I'm not sure that this question is going to "fly". It's very broad, and it appears that it may contain a misconception.. Of course it's always true that you can decompose the total dynamic pressure into a component attributable to wind and a component attributable to groundspeed. You just can't make a logical step from this starting point, to a "conclusion" that the airspeed (and the total dynamic pressure) will tend to change when you rapidly turn 180 degrees while flying into or against the wind. As to why-- well that's a big can of worms. Is that what you are really asking? $\endgroup$ Jan 19 at 14:41
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    $\begingroup$ Oh dear god I always get a slight headache whenever the concept of downwind turn pops up here 😏 $\endgroup$
    – Jpe61
    Jan 19 at 17:41
  • $\begingroup$ This is discussed in this question and perhaps this one. Do either of them help you? $\endgroup$
    – Pondlife
    Jan 19 at 18:21
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It's always true that you can decompose the total dynamic pressure into a component attributable to headwind and a component attributable to groundspeed. You just can't make a logical step from this starting point, to a "conclusion" that the airspeed (and the total dynamic pressure) will tend to change when you rapidly turn 180 degrees while flying into or against the wind.

As to "why"-- well that's really a separate question. The short answer is that when an aircraft makes a coordinated turn, all the acceleration (whether seen from the airmass reference frame or the ground reference frame) is attributable to the net force generated by the banked wing, which acts purely in the centripetal direction in the airmass reference frame. So there is no change in airspeed or total dynamic pressure, even though there is a change in groundspeed, and in the component of total dynamic pressure that can be attributed to groundspeed, and in the headwind component, and in the component of the total dynamic pressure that can be attributed to the headwind. Since the force from the banked wing is not purely centripetal in the ground reference frame-- i.e. not purely perpendicular to the direction of the ground track-- the groundspeed does not remain constant. The wing-- not the wind-- is the thing that provides the force that is required to change (accelerate) the groundspeed, as the aircraft turns.

Now you may ask-- obviously if you rapidly turn 180 degrees while stationary on the ground, the head-on component of the total dynamic pressure-- which is fully attributable to the wind-- will change. What is causing the difference between the ground case and the airborne case? The thing that is different between the ground case and actual flight, is that when you are stationary on the ground on a windy day, when you rotate to change the heading by 180 degrees, the force exerted on the aircraft by the ground (through the tires) changes direction in the aircraft's reference frame. The situation is completely different from actual flight.

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Lift is created when there's a relative airflow over the wing. In order to produce lift, this airflow has to be from the front to the rear (leading edge to trailing edge). The Pitot Tube points forward, to measure the pressure due to the impact of air molecules from that direction, ie the relative airflow component flowing from the direction to which it points

The Pitot of a stationary airplane on a runway facing a 50kt headwind will register a dynamic pressure to show an Indicated Airspeed of 50KIAS. If the airplane requires a take off (rotation) speed of 65KIAS, it needs to accelerate to a speed of 15kts relative to the runway as would be measured by an automobile speedometer based on wheel/tyre rpm. At this 15kts, the IAS would read 65KIAS enabling the airplane to get airborne.

Facing the opposite direction with the 50kts blowing from tail to nose, the stationary airplane would not indicate any IAS till it achieves 50kts relative to the ground as measured by the hypothetical automobile speedometer. Thereafter, accelerating further, the Pitot would start registering dynamic pressure and it would start registering IAS therefore. So the automobile speedometer would read 50 + 65 = 115kts for the IAS to be at 65KIAS.

The difference between the 2 take-offs above is in the distance required for the take-offs and the automobile speedometer represents the same as forward Ground Speed.

Once airborne, in line with what you stated, the airplane loses it's 'tethering' to the ground and shear conditions excepted, the direction of flight relative to the wind will not cause problems such as stalling the airplane.

(Shear conditions are defined by rapid changes to the flow of the air with relation to the air mass itself. Such conditions exist due to terrain, jet-streams, Fronts, storm conditions man made obstacles and related phenomena.)

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Taking off with a 50 knot tailwind means your airspeed will stay at zero until your groundspeed reaches 50 knots, then your airspeed will begin rising (hopefully it will rise to liftoff speed before you run out of runway.)

The airspeed indicator is not able to show negative numbers therefore the dynamic pressure is sucking instead of blowing at the beginning off your takeoff roll. (Please excuse my less than proper terminology, I have been awake 25 hours.)

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  • $\begingroup$ Negative airspeed can be shown, but only in simulators :) $\endgroup$ Jan 19 at 16:26
  • $\begingroup$ I guess that's a tongue in cheek statement you're making @Camille_Goudeseune. $\endgroup$
    – skipper44
    Jan 19 at 18:32

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