When I was starting the nav lessons my Instructor said to plan a flight to overhead nearby disused airfield but to ignore the wind when plotting track. He just wanted me to fully experience drift. We duly set off and before long I found myself way North of the waypoint. A great demonstration Statistics: Posted by T6Harvard — Fri May 03, 2024 17:53 ..read more

Mz Hedy wrote: Orion1210 wrote:... I think I understand that concept; you spend less time enjoying the tailwind and more time battling the headwind vs flying the same two legs without wind therefore you burn more fuel overall. An amusing apparent paradox is that the same applies when the wind is at 90º to your track on both legs. Yes. Another way to think about this is to imagine what happens if you ignore the wind. Say you know in no wind it’ll take you an hour to get to your destination, so you fly for an hour on the no wind course. When you’ve done that you aren’t in the right ..read more

Think of a 2x2 square. If you measure from corner to opposite corner it will be 2.82 (it is shorter to cross from corner to corner than traverse two sides). So if you apply that the other way round, if from corner to corner is 2, you need sides of 1.4 to get to the opposite corner. Have a play with https://www.calculator.net/triangle-cal ... =Calculate Statistics: Posted by riverrock — Fri May 03, 2024 11:20 ..read more

Orion1210 wrote: ... I think I understand that concept; you spend less time enjoying the tailwind and more time battling the headwind vs flying the same two legs without wind therefore you burn more fuel overall. An amusing apparent paradox is that the same applies when the wind is at 90º to your track on both legs. Statistics: Posted by Mz Hedy — Fri May 03, 2024 11:05 ..read more

Thanks for your insights on this, very helpful. lobstaboy wrote: Here’s another counter intuitive thing about wind. Imagine you fly from A to B with a tailwind exactly from behind. There’s no sideways drift but you’ll get to B sooner than you might expect because of the wind carrying you along from behind. For the return journey you’ll have a headwind and cover the ground more slowly by the same amount. You’d think that in terms of fuel consumption the two journeys would balance, the fuel saved on the outward journey will match the extra used on the way home. But it isn’t like that. Over ..read more

xtophe wrote: The energy linked to the wind will principally be kinetic energy and so linked to the root of the wind speed. 20^2=400 14^2 * 2 = 192 * 2 = 392 (Close enough considering we rounded the number) I know the OP used the term energy, but energy is nothing to do with the question. It’s wind speed and direction , ie a vector. Energy isn’t a vector quantity. Statistics: Posted by lobstaboy — Thu May 02, 2024 15:53 ..read more

As the others have said, you have to understand how vectors work. Drawing a wind triangle is the best way to grasp what’s going on. There’s loads on the internet to help. Here’s another counter intuitive thing about wind. Imagine you fly from A to B with a tailwind exactly from behind. There’s no sideways drift but you’ll get to B sooner than you might expect because of the wind carrying you along from behind. For the return journey you’ll have a headwind and cover the ground more slowly by the same amount. You’d think that in terms of fuel consumption the two journeys would balance, the fuel ..read more

Orion1210 wrote: For example, a 20knt wind 45 degrees off from the runway heading in my mind should give a resultant 10knt headwind and a 10knt crosswind given that 45 is half of 90, but the maths gives a 14knt crosswind and a 14knt headwind. Where did the other 8knts of apparent wind energy come from? The energy linked to the wind will principally be kinetic energy and so linked to the root of the wind speed. 20^2=400 14^2 * 2 = 192 * 2 = 392 (Close enough considering we rounded the number) Statistics: Posted by xtophe — Thu May 02, 2024 15:45 ..read more

Whilst we need to work out headwind and crosswind components, it’s not the most intuitive way to picture what’s actually happening. In your example, with a 20kt wind 45 degrees off the nose, there isn’t a certain amount of wind pushing us backwards and a certain amount pushing us sideways - it’s all pushing in the direction of the wind, which doesn’t particularly care which way the nose is pointing! As @GrahamB says, the maths we do to work out headwind and crosswind components uses a bit of trigonometry to calculate what the equivalent wind would be directly on the nose and to our 90 degree ..read more

It’s basic trigonometry demonstrated by Pythagoras’ Theorem. The square of the hypotenuse is the sum of the squares of the two other sides. If you draw it out to scale on a bit of paper it will be clearer. Statistics: Posted by GrahamB — Thu May 02, 2024 10:17 ..read more