Crosswind components

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

On Thu, 26 Feb 2004 at 02:43:58 in message
, Teacherjh
wrote:
You can't, however, run before the wind faster than the wind (unless you have
one humongus current)

Yes you can. The ability to do so depends on the relative forces between
the water and the second medium. Sand and Ice yachts can readily
demonstrate a course speed downwind faster that the wind. In effect they
are tacking downwind. If you are being very precise then you cannot run
before the wind faster than the wind because running before the wind
depends only on drag and not on lift.
--
David CL Francis

Sand and Ice yachts can readily
demonstrate a course speed downwind faster that the wind.


Straight downwind? So they would be overtaking the wind? Howdeydodat?


In effect they
are tacking downwind.


Oh, then they are not "running" before the wind, they are sidestepping to some
degree.

Jose

--
(for Email, make the obvious changes in my address)

"Teacherjh" wrote in message
...

Sand and Ice yachts can readily
demonstrate a course speed downwind faster that the wind.


Straight downwind? So they would be overtaking the wind? Howdeydodat?

I suspect he's thinking about the case of holding a course a few
degrees off a direct reach. Laying down with the wind directly over the
stern, well even catching a puff from behind now and then would immediately
luff. Even if you could overcome the miniscule drag of an ice runner, the
sail becomes a HUGE air brake. Been there, done that, threw the Tee-Shirt
away.

On 24 Feb 2004 07:13:17 -0800, (James L. Freeman) wrote:

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.

Sure -- just liken it to the fact that it takes longer to walk around a
circle than straight across it.