Speed Dist Time question

[flyingpom]

Ok, So here's a silly question that should be easy but it's got me stumped...

Using 120kts TAS as a base line:

1. If I travel 60 miles @ 120kts it will take me 30 mins

2. If have a 20kt headwind across the same distance:

• I will be travelling at 100kts ( 16.67% slower)
• I will travel 60 miles in 36 mins
• Which is 6 mins longer i.e 20% longer

My confusion arises when i reverse the wind from headwind to tail wind

3. If I have a 20kts tailwind across the same distance:

• I will be travelling at 140kts ( 16.67% faster )
• I will travel 60 miles in 25.71 mins
• Which is 4.3 mins shorter than @ 120kts

WHY for a 16.67% increase in speed I am NOT seeing the same equivilant decrease in time as with the 100kts example. i.e i was expecting ( outside the mathematics ) that if 16.67% slower made me 6 mins late then 16.67% faster will make me 6 mins early......

Am missing something logical here so feel free to point out my stupidity....

[Zulu Alpha]

Cos you spend more time on the into wind leg

[BackPacker]

Yep. Headwind costs more time than a tailwind will reduce travel time.

It's really simple to explain by looking at the extreme case. If you have a tailwind equal to your airspeed, your groundspeed doubles and you will get there in half the time. If you have a headwind equal to your airspeed, your groundspeed is zero and you will never get there at all.

[Jan Olieslagers]

This is a nice exercise for those learning to use spreadsheet software. If it helps you to sleep better: you are not the only one to be bewildered here, not by a long way. My ground class teacher spent a good deal of time on it, probably from his mathematics backgrounds.

[Deeday]

flyingpom, that's not a silly question, as the answer is rather counter-intuitive.
In math terms, the reason is because the relationship between time and speed is non-linear: t = d/s (t = time; d = distance; s = speed).
For a given distance d, if you plot the time that it takes to fly it versus the ground speed, what you get is a hyperbola (x = speed; y = time; never mind the units nor the bottom-left quadrant):

Starting from any point on the curve (e.g. the one corresponding to 120 kt, wherever that is on the diagram) and moving 20 kt to the left and right, you'll see that the corresponding change in time (i.e. the how much you move on the vertical axis, in absolute terms) is bigger when you move 20 kt to the left (headwind) than when you do the same to the right (tailwind), because the slope of the curve always gets steeper going from right to left.

[Mark1234]

Love PPrune.. I understood this just fine, then along came spreadsheets and equations, and graphs - eek!

A good rule of thumb is that (for a round trip), 'any wind is a headwind' Precisely because you will spend more time pushing the headwind than you will benefiting from a tailwind.

[Deeday]

My earlier explanation was admittedly a bit confusing, with some irrelevant details that I've now removed. It should be clearer now.

[flyingpom]

Thanks All.

Makes perfect sense now. The reason this came up relates to some rules of thumb I'm using namely:

Note: My aircraft has a 120kts TAS

• For every 1 min early/late in 20nm I have 10kts of HW/TW. This then becomes 2 mins early/late in 40 and 3 in 60 etc
• So for say 20kts of HW / Tail wind I will be 2 mins late in 20, 4 in 40 etc etc
• I use 10% of the HW/TW component as the mins early or late in 20

Now from your mathematical obervations, the Speed / Time relationship is not linear which I've just realised doesn't matter for how I am applying this technique:

Lets say we are flying a 60nm Leg and have a 30kts tail wind i.e 3 mins early in 20 = 150kts GS

Initial calc is:

• At 20miles = 7mins ( Math calculated ETI = 8 mins )
• At 40miles = 14mins ( Math calculated ETI = 16 mins )
• At 16miles = 21mins ( Math calculated ETI = 24 mins )

So our ETI's are slightly out from the actual mathematical calculation T=D/S etc

-Assuming we are actually doing 150kts GS we would arrive at our first check point @ 8 mins which means we are 1 min late, which means I just update my ETI's for each of my next check points

• 2 mins late @ 40
• 3 mins late @ 60

-Updating out ETI’s we now get

• At 20miles = 7mins
• At 40miles = 16mins
• At 60miles = 24mins

And we get an accurate ETI for our destination

I am using this a technique in diversions so I work purely in time i.e no need to calc ground speed. Plus i use some small % factor for < or > 20mn intervals

[24Carrot]

Expanding on Mark1234's comment about 'spending longer pushing the headwind', you can see any wind will lose you time by assuming the opposite and looking for the contradiction.

For your 60nm round trip at TAS 120kts it will take you 60 minutes in no-wind conditions. If the time lost upwind was recovered downwind you would also make the trip in 60 mins with a wind of 20kt.

Looking at 60 min trips in general, the air moves 20 kts downwind during them, and you travel through the air at 2nm/min.

A simple 30mins out, 30 mins back takes you back to the same bit of air you took off from. A balloon launching as you took off would be there to meet you after 60 mins, and both of you would be 20 nm downwind of the airfield.

If you want to return to the airfield, you must travel (net) upwind 20 nm during your 60 minute trip. 20nm at 2nm/min is 10 mins, so the new plan is to fly upwind for 35 mins, then downwind for 25 mins. So you spend 60 mins, and get back to the airfield. So far so good.

But how far do you travel during each leg?

Upwind you fly 35 mins at 2 nm/min, which is 70nm, but the 20kt wind blows you back for more than half an hour, so you don't make the 60nm.

Downwind you fly 25 mins at 2nm/min, which is 50nm, but the 20kt wind helps you for less than half an hour, so you don't make the 60nm.

In each case you fall short by the distance travelled in 5 min at 20kt. So there is the contradiction, and so you do need extra time to fly the legs in windy conditions.