Crosswind components

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