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.

"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?

The same reason the opposite and adjacent sides of a right angle triangle
add up to more than the hypotenuse. I guess that's mathematical, but it is
an explanation...

Regards
Andrew

You shoulda stayed awake during trigonometry...

MikeM


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.

"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?

For the same reason that it's a shortcut if you can walk diagonally from one
corner of a city block to the opposite corner, rather than following the
roads around the block.

--Gary

>
> Thanks.

It's because they are separate definitions of the crosswind and are not
additive.
--

B-58 Hustler History: http://members.cox.net/dschmidt1/
-

"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.

Because wind has speed and direction. You cannot just add the numbers
to get the total. You have to do a vector sum (considering direction
and speed).



(James L. Freeman) wrote in message >...
> 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.

> (James L. Freeman) wrote in message
>...
> > 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.

"Andrew Sarangan" > wrote in message
om...
> Because wind has speed and direction. You cannot just add the numbers
> to get the total. You have to do a vector sum (considering direction
> and speed).

I don't think that's the answer. When I run a wind problem on my
Whiz-Wheel, the resultant IS a vector solution and the crosswind is only a
fraction of the wind velocity. Something makes me want to say it is the
cosine of the angle off the wing.

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
> >
>
>

In article >,
Casey Wilson > wrote:
>> (James L. Freeman) wrote in message
>...
>> > 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?

I'll give it a go (use to tutor math in college but that was a while
ago ^_^). This works best with pictures but here's teh basic outline.
Imgine you have a block on the floor. Two guys (A and B) are going to
push on the block. Since we're doing this wihtout pictures let's imgine
you're looking at the block from teh top down and it can go
North/south/east/west over teh floor.

1. A pushes on the south side of the block with 5 units (of whatever), B
pushes on the south side of the block with 5 units of force. The result
the block moves to the north with 10 units of force.

2. A pushes on teh south side of the block with 5 units, B pushes on the
north side of the block with 5 units. Result, teh block doesn't move at
all. But Both A & B are pushing with 5 units.

3. A pushes on the south side of teh block, B pushes on the east side f
teh block. Result is the block moves north-east.

That sounds ok right? But wait a mintue, in example 3 if only A was
pushing then the block would move north and only north. And if only B was
pushing the block would move east and only east. The block moves
north-east because both are pushing.
In example 2 the block didn't move at all, because A and B are pushing
in opposite directions with equal force.

But What if A pushing on the south side of rh block with 10 units and B
pushes on teh north side with 8 units? Well the block would move north,
at 2 units ofr force.
From the blocks POV having A push from teh south at 10 & B push from
teh north at 8, is THE SAME THING as just having someone push at 2 units
of force.

This is know as vector addition. You need to take teh scaler (number
of units of push) and the vector (direction) into account to do vector
addition.
So if I told you that A was pushing at 10 units and B was pushing at 10
units, and asked what direction is the block moving what would you say?
Well you 'should' say "I don't kow." Because you don't know what
direction A and B are pushing.

The above examples are easy because A nd B are pushing in opposite
direttions. It's a little harder to think of it when A and B push between
0-180 degrees. The math is easy (sin/cos stuff or use a cross wind
diagram). It's the same thing thou.

Let's say you want to land on runway 36 (to make life easy). The wind
is 045 @ 20. Well just like teh block having the win at 045 @ 20, is the
same as having two winds that are pushing the plane at the same time.
One wind is coming from 360 and the other from 090, the end result is a
single wind from 045 (yes I know wind doesn't work like that work with me
here).
The reason we 'spilt the wind up' is because the 'peice' of wind coming
from 360 doesn't concern us, but that peice from 090 does, as it'll blow
us off course when landing.
So it's the reverse of the block examples, we have one force at some
odd angle, and we spilt it into two force: head wind and cross wind.
Because these forces have strength (number of units) and direction (in
this case 045) we can't just add/subtract like number.

Just like the block example above where the block was moving forward at
2 units, wind acts the same. Landing on 360 with a wind from 360 @ 10 is
all head wind, 015 @ 10 a little cross wind, 080 @ 10 almost all cross
wind.
So the cross wind component varies with direction, and you can't just
add/subtract the units to get there. Now find any url about trig and look
up sin and/or cos and this will make a lot more sense (I hope). But you
asked why and not how to calculate it :)

This stuff is actually pretty easy to do if you have a scienctific
calculator. Sorry the above is so long ...

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 ^_^).

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
> >
>
>

"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.

I'm not sure where James's question was going, but here is an example as I
just copied it minutes ago from:

http://www.edwards.af.mil/weather/index.html
02/25/2004 16:57 LCL Rwy 4 Wind 216 Deg at 2 Kts Cross Wind 4 Kts

Note the runway crosswind is reported higher than the wind velocity. I would
be delighted if someone could explain this -specific- condition.

>>
http://www.edwards.af.mil/weather/index.html
02/25/2004 16:57 LCL Rwy 4 Wind 216 Deg at 2 Kts Cross Wind 4 Kts

Note the runway crosswind is reported higher than the wind velocity. I would
be delighted if someone could explain this -specific- condition.
<<

It's an error. It might be because the winds and the runway crosswind were
measured at different points, or at different times, or because of errors in
the math (roundoff several times?). Mathematically, the crosswind cannot be
greater than the total wind.

Jose

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

Same as a sailboat.

Can go much faster than the wind.

Karl

>>
Same as a sailboat.
Can go much faster than the wind.
<<

Actually, sailing is different because of the interaction of the keel and the
water. You can't go faster than hull speed, but ignoring that you could
theoretically go faster than the wind, if you were going at 90 degrees to it.
You can even sail INTO the wind.

You can't, however, run before the wind faster than the wind (unless you have
one humongus current)

Jose


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

"Casey Wilson" > wrote in message >...
> > (James L. Freeman) wrote in message
> >...
> > > 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.
>
> "Andrew Sarangan" > wrote in message
> om...
> > Because wind has speed and direction. You cannot just add the numbers
> > to get the total. You have to do a vector sum (considering direction
> > and speed).
>
> I don't think that's the answer. When I run a wind problem on my
> Whiz-Wheel, the resultant IS a vector solution and the crosswind is only a
> fraction of the wind velocity. Something makes me want to say it is the
> cosine of the angle off the wing.


Where you do think the cosine comes from? It is a consequence of
adding two vectors.

A 10 knot wind from the northwest has a westerly component of
10cos(45) = 7knots and a northerly component of 10cos(45) = 7knots.
Alternatively, if you have 7 knots from the west and 7 knots from the
north, you wind up with a total of 10 knots from the northwest, not 14
knots.

Try a "continuity" argument: A pushes on the (now CYLINDRICAL, with axis
vertical) block (silo), from the South, with 10 units of force. B initially
pushes from the North with 10 units. Result, 0 net force on the block. But
now let B move off to the East somewhat (to his left) and push, again (and
always) with force 10 units. Net force then has small components to the West
and North. Now let B move farther to his left. Larger components to West and
North. By the time B has moved around to the Southeast, the net force will
have large components to the North and considerable to the West. If B moves
into A's postion, the net force will be 20 units to the North. By
continuity, the net force increases continuously from 0 to 20. So at some
point that force will start exceeding 10 units (and stay that way).

John

--
John T Lowry, PhD
Flight Physics
5217 Old Spicewood Springs Rd, #312
Austin, Texas 78731
(512) 231-9391


"'Vejita' S. Cousin" > wrote in message
...
> In article >,
> Casey Wilson > wrote:
> >> (James L. Freeman) wrote in message
> >...
> >> > 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?
>
> I'll give it a go (use to tutor math in college but that was a while
> ago ^_^). This works best with pictures but here's teh basic outline.
> Imgine you have a block on the floor. Two guys (A and B) are going to
> push on the block. Since we're doing this wihtout pictures let's imgine
> you're looking at the block from teh top down and it can go
> North/south/east/west over teh floor.
>
> 1. A pushes on the south side of the block with 5 units (of whatever), B
> pushes on the south side of the block with 5 units of force. The result
> the block moves to the north with 10 units of force.
>
> 2. A pushes on teh south side of the block with 5 units, B pushes on the
> north side of the block with 5 units. Result, teh block doesn't move at
> all. But Both A & B are pushing with 5 units.
>
> 3. A pushes on the south side of teh block, B pushes on the east side f
> teh block. Result is the block moves north-east.
>
> That sounds ok right? But wait a mintue, in example 3 if only A was
> pushing then the block would move north and only north. And if only B was
> pushing the block would move east and only east. The block moves
> north-east because both are pushing.
> In example 2 the block didn't move at all, because A and B are pushing
> in opposite directions with equal force.
>
> But What if A pushing on the south side of rh block with 10 units and B
> pushes on teh north side with 8 units? Well the block would move north,
> at 2 units ofr force.
> From the blocks POV having A push from teh south at 10 & B push from
> teh north at 8, is THE SAME THING as just having someone push at 2 units
> of force.
>
> This is know as vector addition. You need to take teh scaler (number
> of units of push) and the vector (direction) into account to do vector
> addition.
> So if I told you that A was pushing at 10 units and B was pushing at 10
> units, and asked what direction is the block moving what would you say?
> Well you 'should' say "I don't kow." Because you don't know what
> direction A and B are pushing.
>
> The above examples are easy because A nd B are pushing in opposite
> direttions. It's a little harder to think of it when A and B push between
> 0-180 degrees. The math is easy (sin/cos stuff or use a cross wind
> diagram). It's the same thing thou.
>
> Let's say you want to land on runway 36 (to make life easy). The wind
> is 045 @ 20. Well just like teh block having the win at 045 @ 20, is the
> same as having two winds that are pushing the plane at the same time.
> One wind is coming from 360 and the other from 090, the end result is a
> single wind from 045 (yes I know wind doesn't work like that work with me
> here).
> The reason we 'spilt the wind up' is because the 'peice' of wind coming
> from 360 doesn't concern us, but that peice from 090 does, as it'll blow
> us off course when landing.
> So it's the reverse of the block examples, we have one force at some
> odd angle, and we spilt it into two force: head wind and cross wind.
> Because these forces have strength (number of units) and direction (in
> this case 045) we can't just add/subtract like number.
>
> Just like the block example above where the block was moving forward at
> 2 units, wind acts the same. Landing on 360 with a wind from 360 @ 10 is
> all head wind, 015 @ 10 a little cross wind, 080 @ 10 almost all cross
> wind.
> So the cross wind component varies with direction, and you can't just
> add/subtract the units to get there. Now find any url about trig and look
> up sin and/or cos and this will make a lot more sense (I hope). But you
> asked why and not how to calculate it :)
>
> This stuff is actually pretty easy to do if you have a scienctific
> calculator. Sorry the above is so long ...

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.