Effect of wind on flightpath.

[Aviator]

I am a bit of a hobby programmer and went about coding up some tools to speed up my flight planning.

What i wanted was to be able to input my desired track and True airspeed, the wind direction and speed and the distance to destination.

Then be returned a heading to counteract the wind drift and ground speed as well as time to destination as a result.

So went under the assumption to calculate this that if an aircraft flew for 1 hour in a straight line the wind will move the aircraft downwind by exactly the windspeed which would reduce the problem to simple trigonometry.
For example if the windspeed was 20kt then could i expect to be exactly downwind by 20nm from where i would be in nil wind after 1 hour? or is this approach flawed.

(Mathmatically of course, wind will obviously vary considerably throughout a flight so im just asking this from a planning perspective)

 

[Urwumpe]

That approach is good enough for government work. generally, you would also need to include the effects of Earths rotation and curvature for longer distances (> 350 NM)

 

[icedown]

In aviation it is called a crab. I still use my E6-B(a type of circular slide rule) to calculate it when I am planning my flight. In these calculations you also have to account for head/tail wind components as they will change your final flight path slightly too.

To start out wind information is given in from context, so you have flip it 180 degs so

WindDir = ReportWindDir + 180

Next calculate angle of wind relative to your desired course

 WTA = DesCourse - WindDir

Then finally calculate your wind correction angle

SinWCA = windspeed * sin(WTA) / airspeed
WCA = arcSin(SinWCA)

Then use that to find your desired heading

 Heading = DesCourse + WCA

And your groundspeed

 GS = airspeed * cos(WCA) + windspeed * cos (WTA)

Example Problem:
Course: 90
Airspeed: 90m/s
ReportWinDir = 330
Windspeed = 7m/s

WinDir = 150 = 330 - 180
WTA = -60 = 90 - 150
SinWCA = -0.0673575  = 7 * sin(-60) / 90
WCA = -3.86 = asin(7 * sin(-60) / 90 )
Heading = 86.1 = 90 -3.86
GS = 93.3m/s = 90 * cos(-3.86) + 7 * cos(-60)

 

[Aviator]

Thanks for that, yes this is the approach i had been taking.

As for the earth's curvature that shouldn't be an issue.. this is mostly for local flights no more than 150nm from departure.

About accounting for rotation however.. yes i can see how that would affect space craft where the earth rotates beneath you - but for an aircraft moving within a mass of air rotating with the earth wouldn't that just manifest itself as a change in wind direction? (coriolis effect) meaning i should just be able to use my forcasts?

 

[Urwumpe]

About accounting for rotation however.. yes i can see how that would affect space craft where the earth rotates beneath you - but for an aircraft moving within a mass of air rotating with the earth wouldn't that just manifest itself as a change in wind direction? (coriolis effect) meaning i should just be able to use my forcasts?

It affects you more than it affects the wind, simply put. If you look at the flight paths of aircraft over longer distances, you can see that they don't just follow a simple great circle, but actually have a tiny bias to compensate for coriolis effect as well, additionally to the wind. You can for example see that in opposing latitudes. The great circles would be pointing away from the equator, wind strengths and directions are generally equal, but the rotation bias moves the apex of the trajectory further to the east, the higher the latitude gets.

Not compensating coriolis effect actually means ~150 km error already for a ICBM.