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For a school project, I'm trying to make an automated flight planner, considering mainly the differences in flight paths according to wind.

Now I've heard that flying with/against the wind affects airspeed. But I really don't know any specific(even empirical) laws on determining by how much it does so.

What's more, I've also heard about gas savings through these effects. That I would have pretty much no idea of how to calculate but that's less important.

Basically if anyone can point me in the right direction to figure out how to figure out solid speed gains, I'd be grateful, even moreso if I can find out the theoretical origins of such laws.

(The only thing I found that I think is close to what I want is this, but I have no clue what the laws derive from)

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what level of calculus you wanna obtain? i mean, do u want, for example, to implement also parameter such as the cross section of the plane against the wind? or is something more happy go lucky like i travel at 300knots against a 30knots wind so my effective speed is 270knots? – Steve Nov 21 '10 at 18:19
    
I think some "complicated" physics would be necessary (I'm no stranger to working with differentials and surface integrals in physics, and this should need a bit of complicated physics) – Dasuraga Nov 21 '10 at 19:02
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some useful equations at en.wikipedia.org/wiki/E6B – Chris Burt-Brown Mar 16 '11 at 14:52
    
@Dasurga: there is nothing complicated, the plane is moving at a constant speed relative to the air, as if the air was stationary. Its completely Galilean invariance. There is no calculus. There is no cross sections. Only vector addition, and the answer is exact, and correct for all purposes. – Ron Maimon Nov 21 '11 at 17:20
    
@Ron Dasuraga was last seen here one Year ago! – Georg Nov 21 '11 at 22:00

Well, the formulae in the Wikipedia page you reference are simple consequences of this model:

  • Your aircraft instruments measure speed relative to the surrounding air (air speed)
  • The surrounding air moves with respect to the ground (wind speed)
  • The speed of the aircraft vs the ground will be the vector sum of both these speeds.

To first order (and I'm pretty sure that should suffice for your school project) the resulting fuel savings ensue through the fact that you only need to expend fuel to maintain air speed - basically, if the wind carries the surrounding air in your direction of travel, you have to travel shorter distances "through the air".

The mathematics underlying this consists of vector addition in the simple case and maybe addition of vector fields/line integrals in these vector fields for more complex cases.

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For your purposes, you can simply assume that the airplane travels at a constant velocity through the air mass. "Wind" simply means that this air mass is moving with respect to the ground.

If the airplane's velocity through the air is (a vector) v, and the air velocity with respect to the ground is a vector w, then the airplane's velocity with respect to the ground is simply v+w.

It's the same problem as swimming in a river with a current. If you are swimming directly upstream (or downstream), then the velocity of river water simply subtracts (or adds) to your velocity. If, on the other hand, you want to swim directly across, without floating downstream with the current, you have to angle your swim slightly upstream so that the component of your velocity in the direction of the river flow cancels out the river velocity.

In flying, this correction is called the "wind correction angle"; it corrects for the crosswind component of the wind, in order to fly a straight path (relative to the ground) from point A to point B. If you google for "cross-country flight planning" you will find lots of materials about how this is done in practice. The old-fashioned way is to use the E6B circular slide-rule, which you mentioned. The math is actually easier than that device/page might have you believe. The advantage of the E6B is that you can use it with one hand while you are flying an airplane.

If you assume that the wind is constant everywhere, then the problem is fairly trivial. If you assume that the wind is different from place to place and at different altitudes, and that the airplane's engine's efficiency is different at different altitudes, then the problem is more complex/interesting.

Where the wind really matters is when the airplane is transitioning from the ground to the air (and thus their relative velocities really matter!). Search youtube for "cross-wind landing" for some dramatic examples.

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Can you please not say "for your purposes", because this might lead people to believe this is an approximation to something more difficult. This is not a simplification, it is just the answer. – Ron Maimon Nov 21 '11 at 17:20
    
Hm, I'll have to think of a good re-wording. It is a simplification in the sense that it ignores the possibilities of turbulence, wind sheer, etc. – nibot Nov 21 '11 at 20:55

Wind speed affects ground speed, not air speed. The airplanes fly at a specified IAS (indicated air speed). Add the IAS with the wind speed in the direction of travel and you get the ground speed. Using simple algebra

$$ u_{ground} = u_{IAS} + u_{wind} $$

So to travel a distance $S$ with head wind takes $t_{A\rightarrow B}=\frac{S}{u_{IAS}-u_{wind}}$, but to return with tail wind $t_{B\rightarrow A}=\frac{S}{u_{IAS}+u_{wind}}$. Obviously $t_{A\rightarrow B} > t_{B\rightarrow A}$

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+1 on the notion that ground speed is the straightforward vector addition of air speed and wind speed.

Manual flight computers such as the classic Dalton E6B allow you to do this vector math graphically. From a starting point in the middle of your plastic plate you draw a line in the direction of the wind speed, with a length proportional to the wind speed. From there you draw a line in the direction of your air speed (which is your aircraft's heading), with a length proportional to your air speed. Then the vector from start to finish is your ground speed and your ground track (direction you are travelling over the ground).

Concerning the extra fuel you need to burn to fly into a headwind or the fuel you save by flying with a tailwind, a most practical consideration to note is that if you fly from pt. A to pt. B into a headwind, and then turn around and fly from B back to A with a tailwind of the same strength, you lose more flying into the headwind than you gain from flying with the tailwind.

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Get yourself an E6B. It has two sides. One side is a simple circular slide rule, for doing all kinds of multiplication and division, for speed-time-fuel-flow problems.

You want the other side, which features a sliding card and a transparent disk overlay on it, on which you can make pencil marks. What it does is solve the "wind triangle", which is just a vector sum.

If the aircraft is heading North, say, at a speed of 100 knots, and the wind is from the southwest at a speed of 30 knots, meaning the air containing the aircraft is moving Northeast at 30 knots, those two vectors are added to get the aircraft's actual motion vector over the ground, which is roughly to the East of North, and roughly 120 knots.

(A knot is one nautical mile per hour. A nautical mile is about 15% larger than a statute mile, and it is defined as 1 minute, i.e. 1/60 of a degree, of latitude. It is used because it makes navigation easier, because charts have latitude/longitude lines.)

What pilots do with that is determine the correction angle they need to use to fly in the direction they want to go, and the ground speed tells them how long it will take to get there.

It's a little more complicated than that, of course. They have to account for magnetic compass variation, and the fact that wind speeds generally increase with altitude. Now with GPS gear, it takes a lot of the guesswork out of enroute navigation, but in terms of flight planning, they still need to deal with wind, time-of-flight, fuel, etc.

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This answer, my friend, is blowing in the wind! Check for the questions date and when Dasuraga was seen last! – Georg Nov 21 '11 at 14:17
    
@Georg: I know, but other people might be interested, and it's a jubject I like to talk about :) – Mike Dunlavey Nov 21 '11 at 14:21

Wind just carries the aircraft with it. For the aircraft, everything feels the same, just the ground passes by at the sum of wind and airspeed.

The tricky part of the question is about the best airspeed for maximum range with wind. Best range means you cover the most distance while the wind is carrying the plane with it. If you have a headwind, the longer you stay aloft, the more you are carried back, so you better hurry up. With a tailwind, it helps to slow down because now the wind is helping you to cover even more distance.

Graphical solution

But how much? We need to pick that particular speed where the change in fuel consumption just balances the change in speed over ground. I always found that easier to explain with gliders, and there you can really observe which polar point is best. Just picture yourself as an observer on the ground who sees the plane fly by in the distance. If you plot a line with the combination of positions and altitude, there must be one flight speed that produces a line where the flight path angle is the shallowest. This is the desired optimum. That has only little to do with optimum L/D - this is just one other point you can find with a sink speed polar. And it happens to be the point for best glide in still air. But there is so much more that humble polar will tell you, if you look at it the right way.

With powered aircraft you need to pick the polar point where your fuel flow is lowest for the given speed over ground. Basically, you fly like the glider and add enough power to stay at the same altitude. That is all the difference. Please look at the plot below which shows airspeed on the X axis and sink speed on the Y axis.

Optimum Speed for Best Range

The solution is graphical: You start on the X-axis at the point which corresponds to the wind speed and put a tangent on the sink speed graph. Where the tangent touches the sink polar (blue line), the plane flies at the best L/D for that given wind speed. Move the starting point to positive speeds for headwind and to negative speeds (not shown here) for tailwind. If the term "best L/D" is already reserved in your mind, please read this as the "best polar point". It is really the same.

Analytical solution

For powered flight things become more tricky, because thrust changes with speed. To simplify things, we can say that thrust changes over speed in proportion to the expression $v^{n_v}$ where $n_v$ is a constant which depends on engine type. Piston aircraft have constant power output, and thrust is inverse with speed over the speed range of acceptable propeller efficiencies, hence $n_v$ becomes -1 for piston aircraft. Turboprops make some use of ram pressure, so they profit a little from flying faster, but not much. Their $n_v$ is -0.8 to -0.6. Turbofans are better in utilizing ram pressure, and their $n_v$ is -0.5 to -0.2. The higher the bypass ratio, the more negative their $n_v$ becomes. Jets (think J-79 or even the old Jumo-004) have constant thrust over speed, at least in subsonic flow. Their $n_v$ is approximately 0. Positive values of $n_v$ can be found with ramjets - they develop more thrust the faster they move through the air.

Now for fuel flow: This goes up and down with the power output of the engine. Again a simplification, but it helps to get to grips with the problem and gives useful results. This lets us re-formulate the problem as: At what airspeed do I have the best ratio between power and ground speed?

Mathematically, we want to fly with $\frac{P}{v_w+v}$ at the lowest possible value. $P$ is the power, $v_w$ is the wind speed and $v$ the airspeed. To express the thrust behavior over speed, I break P up into a product of a constant $K_S$, the throttle setting $\delta$ and the speed like this: $K_S\cdot\delta\cdot v\cdot v^{n_v}$. Here is the general solution, pasted as a PNG to save me all the typing in the editor:

formulas

Please note that implicitly the lift coefficient is on both sides of the equation. To solve it, you need to do it recursively, until speed and lift coefficient match. I took this form because of the similarity to the general form at still wind which can be found in many performance books.

Now I have put the results into a plot. In order to eliminate the aircraft-specific parameters, it shows the ratio of $c_L$ with wind over $c_L$ without wind. The plot is metric, but will work for all units if you use the same units for wind speed and air speed.

best cl over speed

To give an example for the application of the correction factor: If you are flying in a 20 m/s headwind and your best range speed at still wind is 50 m/s (approx 97 kts), the $c_L$ needs to be 70% of the $c_L$ at still wind for piston-powered aircraft. This makes your corrected airspeed 60 m/s (v is proportional to $\sqrt{c_L}$), and now the recursive nature of the formula rears its ugly head. At 60 m/s, the correction is only 77.5%, so we need to do a few loops until we arrive at a point where the airspeed and the correction factor match. In this example, this would be 57 m/s or 110 kts in case of a piston-powered aircraft.

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Basically what you can precisely calculate are just simple effects based on the fact that the velocity of a plane in ground reference frame is a vector sum of plane's velocity in air reference frame and the velocity of air in ground's reference frame (=wind) -- this lead to some corrections in direction (very handy in GPS era, though), speed and time of flight (and so fuel issues), but this is kindergarten physics. Of course wind modifies the flight dynamics too, but this on the other hand cannot be precisely calculated because hydrodynamics is just too complex. This gap is partially filled by some phenomenological laws, the rest is guided by "piloting skills" of the pilot.

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-1: Wind doesn't modify anything, and the "kindergarten physics" answer is the exact solution. This is a symmetry of nature. – Ron Maimon Nov 21 '11 at 17:23

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