Crosswind components

That's simple - it's a vector sum. But can someone explane this to me:

Scenario One

An aicraft is flying from A to B, TAS=200 Kts, Distance AtoB=100nm, Wind
blowing from B to A with a speed=100Kts
Everyone should be able to claculate that GS(A-B)=100 Kts and GS(B-A)=300
Kts , therefore round trip time A-B-A=60min+20min=1hr20min

Scenario Two

Same as One, but remove the wind completely. The GS in both cases = 200 Kts,
therefore round trip time A-B-A = 30min+30min = 1hr

Can someone explain the difference?

"James L. Freeman" wrote in message
om...
Can someone offer a non-mathematical EXPLANATION (as opposed to
DESCRIPTION) of why the speed of headwind and crosswind components of
a wind add up to more than the speed of the wind?

Thanks.

arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

Yes, but the common logic suggest that you also spend less time in tailwind
that in head wind - and if there is no wind the two should cancell each
other... Or should they...

"Stefan" wrote in message
...
arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

"arcwi" wrote in message
...
Yes, but the common logic suggest that you also spend less time in
tailwind
that in head wind

Yes.

- and if there is no wind the two should cancell each
other...

No. Nothing makes them differ by equal amounts.

Consider the case when the headwind is equal to the TAS. Then it takes
*forever* to get to B.

Or consider a headwind that's just one knot less than the TAS. You
eventually get to B, but it takes an enormous amount of time. Even if the
trip back to A were instantaneous (which it isn't), it still couldn't cancel
out the extra time it took to get to B.

--Gary

Or should they...

"Stefan" wrote in message
...
arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

Common logic fails here, because the the commonsense explanation that the
upwind and downwind differences ought to cancel out only works if the
relationship is linear. If you do the math, the relationship between the
round trip time, round trip distance, TAS and wind speed is nonlinear:

time = distance/speed

round trip time = time out(upwind) + time back(downwind),

= D/(T - W) + D/(T + W), where D = leg distance, T = TAS, W = windspeed

= [D(T + W) + D(T - W)] / [(T - W)(T + W)], using x/y + z/w = (xw + yz)/yw

= (DT + DW + DT - DW) / (T^2 - TW + TW - W^2)

= 2DT / (T^2 - W^2)

2D is the round trip distance, so in words: round trip time = (round trip
distance x TAS) / (TAS^2 - windspeed^2)

[As a check, this reduces to: round trip time = round trip distance / TAS,
when windspeed = 0]

Hence the relationship is nonlinear with respect to wind speed. That isn't
normally so obvious because usually TAS wind speed. It's more obvious in
the original post because the poster chose a wind speed much closer to TAS.
[Work it out for windspeed = 10 kt and the other data in the original post,
and the upwind and downwind differences do almost cancel out. Then work it
out for windspeed = 199 kt!]

nuke

"arcwi" wrote in message
...
Yes, but the common logic suggest that you also spend less time in
tailwind
that in head wind - and if there is no wind the two should cancell each
other... Or should they...

"Stefan" wrote in message
...
arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

nuke,
I am standing in awe and reading your explanation
I always thought that pilots do not understand this, I mean the nonlenear
relation between the wind and the distance travelled.
I've been wrong.
My hat goes off to you!

"nuke" wrote in message
...
Common logic fails here, because the the commonsense explanation that the
upwind and downwind differences ought to cancel out only works if the
relationship is linear. If you do the math, the relationship between the
round trip time, round trip distance, TAS and wind speed is nonlinear:

time = distance/speed

round trip time = time out(upwind) + time back(downwind),

= D/(T - W) + D/(T + W), where D = leg distance, T = TAS, W = windspeed

= [D(T + W) + D(T - W)] / [(T - W)(T + W)], using x/y + z/w = (xw + yz)/yw

= (DT + DW + DT - DW) / (T^2 - TW + TW - W^2)

= 2DT / (T^2 - W^2)

2D is the round trip distance, so in words: round trip time = (round trip
distance x TAS) / (TAS^2 - windspeed^2)

[As a check, this reduces to: round trip time = round trip distance /
TAS,
when windspeed = 0]

Hence the relationship is nonlinear with respect to wind speed. That isn't
normally so obvious because usually TAS wind speed. It's more obvious
in
the original post because the poster chose a wind speed much closer to
TAS.
[Work it out for windspeed = 10 kt and the other data in the original
post,
and the upwind and downwind differences do almost cancel out. Then work
it
out for windspeed = 199 kt!]

nuke

"arcwi" wrote in message
...
Yes, but the common logic suggest that you also spend less time in
tailwind
that in head wind - and if there is no wind the two should cancell each
other... Or should they...

"Stefan" wrote in message
...
arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

Can't really say what most pilots understand. They certainly don't teach
this in ground school. I'm a low time pilot but a high time engineer :-)
nuke
"arcwi" wrote in message
...
nuke,
I am standing in awe and reading your explanation
I always thought that pilots do not understand this, I mean the nonlenear
relation between the wind and the distance travelled.
I've been wrong.
My hat goes off to you!

"nuke" wrote in message
...
Common logic fails here, because the the commonsense explanation that
the
upwind and downwind differences ought to cancel out only works if the
relationship is linear. If you do the math, the relationship between
the
round trip time, round trip distance, TAS and wind speed is nonlinear:

time = distance/speed

round trip time = time out(upwind) + time back(downwind),

= D/(T - W) + D/(T + W), where D = leg distance, T = TAS, W = windspeed

= [D(T + W) + D(T - W)] / [(T - W)(T + W)], using x/y + z/w = (xw +
yz)/yw

= (DT + DW + DT - DW) / (T^2 - TW + TW - W^2)

= 2DT / (T^2 - W^2)

2D is the round trip distance, so in words: round trip time = (round
trip
distance x TAS) / (TAS^2 - windspeed^2)

[As a check, this reduces to: round trip time = round trip distance /
TAS,
when windspeed = 0]

Hence the relationship is nonlinear with respect to wind speed. That
isn't
normally so obvious because usually TAS wind speed. It's more
obvious
in
the original post because the poster chose a wind speed much closer to
TAS.
[Work it out for windspeed = 10 kt and the other data in the original
post,
and the upwind and downwind differences do almost cancel out. Then work
it
out for windspeed = 199 kt!]

nuke

"arcwi" wrote in message
...
Yes, but the common logic suggest that you also spend less time in
tailwind
that in head wind - and if there is no wind the two should cancell
each
other... Or should they...

"Stefan" wrote in message
...
arcwi wrote:

Can someone explain the difference?

You spend more time in headwind than in tailwind.

Stefan

In article ,
arcwi wrote:
That's simple - it's a vector sum. But can someone explane this to me:

Scenario One

An aicraft is flying from A to B, TAS=200 Kts, Distance AtoB=100nm, Wind
blowing from B to A with a speed=100Kts
Everyone should be able to claculate that GS(A-B)=100 Kts and GS(B-A)=300
Kts , therefore round trip time A-B-A=60min+20min=1hr20min

Scenario Two

Same as One, but remove the wind completely. The GS in both cases = 200 Kts,
therefore round trip time A-B-A = 30min+30min = 1hr

Can someone explain the difference?

I just wrote a long reply about how vectors sort of work. Not sure if
this helps or not (as I'm not 100% sure of the question you're trying to
ask but...).
Break it down a bit more. Thing of Scenario 1 as two trips:

A-B, flying at 200kts into 100kts head win, gs = 100kts. Distance is
100kts. 100/100 = 1hr
B-A, flying at 200kts with 100kts tail wind, gs = 300kts. Distnace is
100kts. 100/300 = 20min (1/3 of an hour)
Total time 1:20

Scenario 2:
A-B, flying at 200kts no wind. Distance is 100kts. 100/200 = 30 min
B-A, flying at 200kts no wind. Distance is 100kts. 100/200 = 30 min
Total time 1hr

Here's teh take home message, wind ALWAYS slows you down (round trip).
And you spend more time flying with a headwind than a tail wind(it's not
50/50 ^_^).