# Downwind Turns

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**127**
In sailing that initial reaction is called heeling. When you turn and get that first "gulp" of air . The boat is pushed into the water and takes off.

The boat is in two different media at the same time, water and air. It must correspond to both. A plane is only corresponding to one – air.

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**128**
Hmmm, actually bringing in a boat to the subject is good. They too travel on a medium that can be, like the air. It can be in motion itself.

It's part of understanding the three key ideas to grasp, in understanding the wind and how it effects the aircraft inflight.

This yet more from Stick and Rudder of course, the ideas are

The air is a soup.

Motion is relative.

Your plane is "in" the air.

The first idea: Air is a Soup.

The air is real stuff, a thick and heavy fluid, quite similar to water. To a pilot, this seems obvious; but to the laymen, and hence to many students, cadets and so on. "air" means hardly more than another word for empty space-viod.

Therefore they cannot understand what wind really is. Wind means to them merely a mysterious "force" (represented perhaps by an arrow on a blackboard diagram) that comes somehow blowing through space and hits his airplane. Or perhaps, absurdly, that wind means something that comes whistling through the air! This of course makes no sense whatsoever; it hardly bears writing down. No one would think it if he stopped to think; yet it is a fact that many a student pilot tries to base his flying on such flimsy notions.

Actually, "wind" is of course simply the fact that the air "fluid" is in flow We could see that more clearly if our vocabulary did not confuse us. the mere word wind is itself a cause of trouble. We all the stuff air when it lies still, and wind when it is in motion. That is like calling an automobile an automobile when it stands still, and then calling it a "Hickey" when it gets going. It is the same thing, moving or still. Wind is air in motion.

Ok, geez I hate typing, nuff for now but I'll touch the other ideas later.

But you get the idea, the plane travels both through, and with the air.

And is not blown, but carried along with the air's motion even though the plane still moves though it also.

When I come back I'll have some one the second idea, "Motion is relative"

And ye old "Man walking about in a moving train" comparison.

It's part of understanding the three key ideas to grasp, in understanding the wind and how it effects the aircraft inflight.

This yet more from Stick and Rudder of course, the ideas are

The air is a soup.

Motion is relative.

Your plane is "in" the air.

The first idea: Air is a Soup.

The air is real stuff, a thick and heavy fluid, quite similar to water. To a pilot, this seems obvious; but to the laymen, and hence to many students, cadets and so on. "air" means hardly more than another word for empty space-viod.

Therefore they cannot understand what wind really is. Wind means to them merely a mysterious "force" (represented perhaps by an arrow on a blackboard diagram) that comes somehow blowing through space and hits his airplane. Or perhaps, absurdly, that wind means something that comes whistling through the air! This of course makes no sense whatsoever; it hardly bears writing down. No one would think it if he stopped to think; yet it is a fact that many a student pilot tries to base his flying on such flimsy notions.

Actually, "wind" is of course simply the fact that the air "fluid" is in flow We could see that more clearly if our vocabulary did not confuse us. the mere word wind is itself a cause of trouble. We all the stuff air when it lies still, and wind when it is in motion. That is like calling an automobile an automobile when it stands still, and then calling it a "Hickey" when it gets going. It is the same thing, moving or still. Wind is air in motion.

Ok, geez I hate typing, nuff for now but I'll touch the other ideas later.

But you get the idea, the plane travels both through, and with the air.

And is not blown, but carried along with the air's motion even though the plane still moves though it also.

When I come back I'll have some one the second idea, "Motion is relative"

And ye old "Man walking about in a moving train" comparison.

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**129**Huffy01

Join Date: Dec 2008

Location: Australia

Posts: 587

I can't quite remember where the term "Heeling" came up but I was doing my ultralight pilot licence. The instructor was taking something like energy building up in the airplane and I said something about sailing causing I had my own sail boat .

If your flying towards a hill it is better to approach with a head wind because you have better control and the plane reacts faster.

If you approach with a tailwind you must compensate for the wind or you ended up in the hill!

Ye old ship's of the air!

If your flying towards a hill it is better to approach with a head wind because you have better control and the plane reacts faster.

If you approach with a tailwind you must compensate for the wind or you ended up in the hill!

Ye old ship's of the air!

#

**130**
It is because your ground reference, the hill, is changing faster downwind than upwind. The plane otherwise still reacts the same. It is like I said before, doing judged aerobatics, the reference of the judges is from the ground and any maneuver, might it be a figure 8, loop, level turn or anything in the routine, must be in the routine box. It all must be compensated for any wind by the pilot in order to get it in perfect shape as observed from the judges seats on the ground. Flying landing patterns in a controlled airspace the pilot must compensate for any wind as well to fly over the land beacons. If judged from the ground he is flying a turn with a perfect radius, but judged from the moving air as reference he is way out of shape and fly at different G forces. It is like a bird aiming for a branch to land on. Himself, the wind and the branch is moving. Compared that to a vulture flying high up, he is just moving with the wind.

Anyway, flying towards a hill I would recommend a student to fly higher from further away rather than to wait until you are close to the hill before increasing height. There are ground obstacles changing the air stream and you can not see it in advance. It might be dangerous for the inexperienced pilot.

Anyway, flying towards a hill I would recommend a student to fly higher from further away rather than to wait until you are close to the hill before increasing height. There are ground obstacles changing the air stream and you can not see it in advance. It might be dangerous for the inexperienced pilot.

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**132**
Here is the rest of the math for anyone that still believes in physics and the laws of conservation of energy.

1 kg plane flying 20 mph into 20 mph headwind

KE = 1/2 m x V^2

PE = mgh (mass * gravity * height)

g = 9.8 m/s^2 (earths gravitational force)

Plane is traveling at a ground speed of 0

KE = 1/2 x 1 x 0 = 0

Plane is traveling at a ground speed of 40 mph or 17.9 m/s

KE = 1/2 x 1 x 17.9^2 = 160 Jules

PE + KE = constant (unless energy is added or subtracted)

In order for the plane to accelerate from 0 to 40 MPH, potential energy (height) is converted to kinetic energy (speed)

KE = PE

160 joules = mgh = 1 * 9.8 * h

h = 160/9.8 = 16.3 M = 53 feet

So in this case the plane will lose approximately 50' of altitude when turning downwind.

Conversely, you can gain approximately 50' of altitude when turning back into the wind, trading KE (speed) to PE (height)

If the same plane was flying into a 10 MPH wind (which is a more reasonable case)

Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

112 - 12.4 = 99.4 Joules

99.4 joules = mgh = 1 * 9.8 * h

h = 10 M = 33 feet

And even a modest 5 MPH headwind would cause a 13' drop when turning downwind.

15 MPH = 6.71 M / S = 22.6 joules (upwind)

25 MPH = 11.2 M / S = 62.6 joules (downwind)

Change in KE = 62.6 joules

Change in PE = 63.6 / 9.8 = 4.1 M = 13'

And at 0 MPH wind speed we all agree there is no wind effect!

Note: Effect is more pronounced when flying a heavier model. For example, a 1.5 KG glider like the Radian would have 50% more loss in altitude. PE = m*g*h so change in height is directly proportional to weight.

A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with 40' of altitude would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.

**Case:**1 kg plane flying 20 mph into 20 mph headwind

KE = 1/2 m x V^2

PE = mgh (mass * gravity * height)

g = 9.8 m/s^2 (earths gravitational force)

**Upwind state:**Plane is traveling at a ground speed of 0

KE = 1/2 x 1 x 0 = 0

**Downwind state:**Plane is traveling at a ground speed of 40 mph or 17.9 m/s

KE = 1/2 x 1 x 17.9^2 = 160 Jules

PE + KE = constant (unless energy is added or subtracted)

**Change in altitude:**In order for the plane to accelerate from 0 to 40 MPH, potential energy (height) is converted to kinetic energy (speed)

KE = PE

160 joules = mgh = 1 * 9.8 * h

h = 160/9.8 = 16.3 M = 53 feet

So in this case the plane will lose approximately 50' of altitude when turning downwind.

Conversely, you can gain approximately 50' of altitude when turning back into the wind, trading KE (speed) to PE (height)

If the same plane was flying into a 10 MPH wind (which is a more reasonable case)

**Upwind state:**Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

**Downwind state:**Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

**Change in KE**112 - 12.4 = 99.4 Joules

**Change in KE = Change in PE**99.4 joules = mgh = 1 * 9.8 * h

h = 10 M = 33 feet

And even a modest 5 MPH headwind would cause a 13' drop when turning downwind.

15 MPH = 6.71 M / S = 22.6 joules (upwind)

25 MPH = 11.2 M / S = 62.6 joules (downwind)

Change in KE = 62.6 joules

Change in PE = 63.6 / 9.8 = 4.1 M = 13'

And at 0 MPH wind speed we all agree there is no wind effect!

********** EDIT - The following is incorrect. See later post for corrected version. ***********Note: Effect is more pronounced when flying a heavier model. For example, a 1.5 KG glider like the Radian would have 50% more loss in altitude. PE = m*g*h so change in height is directly proportional to weight.

**FOR THE RECORD**A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with 40' of altitude would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.

*Last edited by cbatters; 03-08-2009 at 09:04 PM.*

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**133**Here is the rest of the math for anyone that still believes in physics and the laws of conservation of energy.

1 kg plane flying 20 mph into 20 mph headwind

KE = 1/2 m x V^2

PE = mgh (mass * gravity * height)

g = 9.8 m/s^2 (earths gravitational force)

Plane is traveling at a ground speed of 0

KE = 1/2 x 1 x 0 = 0

Plane is traveling at a ground speed of 40 mph or 17.9 m/s

KE = 1/2 x 1 x 17.9^2 = 160 Jules

PE + KE = constant (unless energy is added or subtracted)

In order for the plane to accelerate from 0 to 40 MPH, potential energy (height) is converted to kinetic energy (speed)

KE = PE

160 joules = mgh = 1 * 9.8 * h

h = 160/9.8 = 16.3 M = 53 feet

So in this case the plane will lose approximately 50' of altitude when turning downwind.

Conversely, you can gain approximately 50' of altitude when turning back into the wind, trading KE (speed) to PE (height)

If the same plane was flying into a 10 MPH wind (which is a more reasonable case)

Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

112 - 12.4 = 99.4 Joules

99.4 joules = mgh = 1 * 9.8 * h

h = 10 M = 33 feet

And even a modest 5 MPH headwind would cause a 13' drop when turning downwind.

15 MPH = 6.71 M / S = 22.6 joules (upwind)

25 MPH = 11.2 M / S = 62.6 joules (downwind)

Change in KE = 62.6 joules

Change in PE = 63.6 / 9.8 = 4.1 M = 13'

And at 0 MPH wind speed we all agree there is no wind effect!

Note: Effect is more pronounced when flying a heavier model. For example, a 1.5 KG glider like the Radian would have 50% more loss in altitude. PE = m*g*h so change in height is directly proportional to weight.

A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with 40' of altitude would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.

**Case:**1 kg plane flying 20 mph into 20 mph headwind

KE = 1/2 m x V^2

PE = mgh (mass * gravity * height)

g = 9.8 m/s^2 (earths gravitational force)

**Upwind state:**Plane is traveling at a ground speed of 0

KE = 1/2 x 1 x 0 = 0

**Downwind state:**Plane is traveling at a ground speed of 40 mph or 17.9 m/s

KE = 1/2 x 1 x 17.9^2 = 160 Jules

PE + KE = constant (unless energy is added or subtracted)

**Change in altitude:**In order for the plane to accelerate from 0 to 40 MPH, potential energy (height) is converted to kinetic energy (speed)

KE = PE

160 joules = mgh = 1 * 9.8 * h

h = 160/9.8 = 16.3 M = 53 feet

So in this case the plane will lose approximately 50' of altitude when turning downwind.

Conversely, you can gain approximately 50' of altitude when turning back into the wind, trading KE (speed) to PE (height)

If the same plane was flying into a 10 MPH wind (which is a more reasonable case)

**Upwind state:**Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

**Downwind state:**Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)

KE = 1/2 m x V^2

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

**Change in KE**112 - 12.4 = 99.4 Joules

**Change in KE = Change in PE**99.4 joules = mgh = 1 * 9.8 * h

h = 10 M = 33 feet

And even a modest 5 MPH headwind would cause a 13' drop when turning downwind.

15 MPH = 6.71 M / S = 22.6 joules (upwind)

25 MPH = 11.2 M / S = 62.6 joules (downwind)

Change in KE = 62.6 joules

Change in PE = 63.6 / 9.8 = 4.1 M = 13'

And at 0 MPH wind speed we all agree there is no wind effect!

Note: Effect is more pronounced when flying a heavier model. For example, a 1.5 KG glider like the Radian would have 50% more loss in altitude. PE = m*g*h so change in height is directly proportional to weight.

**FOR THE RECORD**A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with 40' of altitude would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.

PS. This post here is all good workings.

Can you do the same math for the energy required to do an Upwind turn.

Buck Rogers knows what I am Talking about eh? Buck

When you do the Upwind turn math and place it with the Downwind math it will prove your "For the Record" statement, to be, False.

Bryce.

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**135**
All the math is interesting, but some practical things are left out. I flew with a light weight 48" Spitfire in a fairly strong wind today. As I am flying for about 35 years now, of which 29 years is RC, there are certain things I know and just do. A new RC pilot must think of his turns when flying towards him. As an experienced pilot it is like breathing, you don’t think, you just do. Sometimes it is interesting to sit and think about it again and concentrate on what is happening like a novice while flying.

As I was thinking about this thread I decided to “experiment” again. Flying into the wind I adjusted the throttle for the plane to move very slowly forward in reverence o the ground. Without adjusting the throttle, I did a gentle 180 deg turn, concentrating on what is happening and what control input there is during the turn. (Like I said, after some experience you just do without thinking about what you do.) Now, I saw that the plane after being banked with ailerons alone, gained altitude without any power increase and NO elevator input. (I fly mode 1 and moved my left thumb away from the stick to be certain not to give elevator input.) While gaining altitude, it also increased ground speed (what I observe from my position) and I levelled the wings with aileron alone. Then, flying downwind for some distance, again with my left thumb away from the elevator, I did a gentle 180 deg turn into the wind. The plane lost altitude while losing ground speed. Like I said, both turns was very gentle as not to sideslip to much so the elevator input would be needed.

What happened? While flying upwind, the wind only “see” the front profile of the plane. When the plane banked and started to turn, the wind “see” more of the plane plan form profile. The wind is hitting the plane from below, accelerating the plane downwind while also giving it more lift due to the increased sideways angle of attack. During and by the time the 180 deg turn was complete, the plane is flying at approximately the same wind speed it was flying before the turn, but it gained altitude. Some of the energy from the wind is transferred to the plane.

Now, flying downwind and doing a 180 deg turn, the opposite it happening. With the plane banked, the wind is hitting the wing from above, causing a negative sideways angle of attack. Now the plane is reducing altitude while reducing ground speed. During and after the turn the approximate wind speed is the same as before, but it lost some altitude.

I did this a few time over and over again until it got bored to see the same results over and over again.

Doing a steep and aggressive bank where the sideslip and drag is excessive you need to give lots of elevator input to counter the sideslip in order to stay at the same altitude and to complete the turn. Because of the required elevator input, there is extra drag slowing down the plane (airspeed) (I think most people know the aerodynamic theory behind this) causing it to loose altitude because of reduced horizontal lift, thus you have to increase power to hold or to regain the speed, lift and altitude.

Conclusion. Every situation and every turn is different and has to be treated and discussed on its own merits. Both sides has its valid points, (and its shortcomings) but different situations are referred to.

As I was thinking about this thread I decided to “experiment” again. Flying into the wind I adjusted the throttle for the plane to move very slowly forward in reverence o the ground. Without adjusting the throttle, I did a gentle 180 deg turn, concentrating on what is happening and what control input there is during the turn. (Like I said, after some experience you just do without thinking about what you do.) Now, I saw that the plane after being banked with ailerons alone, gained altitude without any power increase and NO elevator input. (I fly mode 1 and moved my left thumb away from the stick to be certain not to give elevator input.) While gaining altitude, it also increased ground speed (what I observe from my position) and I levelled the wings with aileron alone. Then, flying downwind for some distance, again with my left thumb away from the elevator, I did a gentle 180 deg turn into the wind. The plane lost altitude while losing ground speed. Like I said, both turns was very gentle as not to sideslip to much so the elevator input would be needed.

What happened? While flying upwind, the wind only “see” the front profile of the plane. When the plane banked and started to turn, the wind “see” more of the plane plan form profile. The wind is hitting the plane from below, accelerating the plane downwind while also giving it more lift due to the increased sideways angle of attack. During and by the time the 180 deg turn was complete, the plane is flying at approximately the same wind speed it was flying before the turn, but it gained altitude. Some of the energy from the wind is transferred to the plane.

Now, flying downwind and doing a 180 deg turn, the opposite it happening. With the plane banked, the wind is hitting the wing from above, causing a negative sideways angle of attack. Now the plane is reducing altitude while reducing ground speed. During and after the turn the approximate wind speed is the same as before, but it lost some altitude.

I did this a few time over and over again until it got bored to see the same results over and over again.

Doing a steep and aggressive bank where the sideslip and drag is excessive you need to give lots of elevator input to counter the sideslip in order to stay at the same altitude and to complete the turn. Because of the required elevator input, there is extra drag slowing down the plane (airspeed) (I think most people know the aerodynamic theory behind this) causing it to loose altitude because of reduced horizontal lift, thus you have to increase power to hold or to regain the speed, lift and altitude.

Conclusion. Every situation and every turn is different and has to be treated and discussed on its own merits. Both sides has its valid points, (and its shortcomings) but different situations are referred to.

#

**136**
For the record I was wrong on the effect being more pronounced on a heavier plane.

Went back and reviewed the math. KE is proportional to weight as is PE so the loss in height is independent of the mass of the plane.

So, you are 100% correct, the Radian turning downwind will lose 33' from a height of 40' and you will not crash. (But it will be close.)

Clint

Went back and reviewed the math. KE is proportional to weight as is PE so the loss in height is independent of the mass of the plane.

So, you are 100% correct, the Radian turning downwind will lose 33' from a height of 40' and you will not crash. (But it will be close.)

**FOR THE RECORD (2nd attempt)**

A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with**30' of altitude**would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.Clint

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**137**Went back and reviewed the math. KE is proportional to weight as is PE so the loss in height is independent of the mass of the plane.

So, you are 100% correct, the Radian turning downwind will lose 33' from a height of 40' and you will not crash. (But it will be close.)

**FOR THE RECORD (2nd attempt)**

A Parkzone Radian flying 20 MPH (airspeed) into a 10 MPH headwind with

**30' of altitude**would crash into the ground when turning downwind (unless additional throttle was added) However, the same plane flying 20 MPH (airspeed) with a 10 MPH tail wind with only 10' of altitude would have no trouble turning back upwind even with no power.

Clint

The error is in here; (let me add to it)

KE = 1/2 m x V^2

If the same plane (1Kg) was flying into a 10 MPH wind (which is a more reasonable case)

Upwind state:

Plane flying 20 MPH North into 10 MPH South headwind

Plane has a Ground speed North of 10 MPH = 4.98 M/S

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Downwind state:

Plane flying 20 MPH South with 10 MPH South tailwind

Plane has a Ground Speed South of 30 MPH = 14.9 M/S

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

Change in KE as the plane turns 180deg from north to south

12.4 +112 = 124.4 Joules

Change in KE as the plane turns 180deg from south to north

112 + 12.4 = 124.4 Joules

If the same plane (1Kg) was flying into a 10 MPH wind (which is a more reasonable case)

Upwind state:

Plane flying 20 MPH North into 10 MPH South headwind

Plane has a Ground speed North of 10 MPH = 4.98 M/S

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Downwind state:

Plane flying 20 MPH South with 10 MPH South tailwind

Plane has a Ground Speed South of 30 MPH = 14.9 M/S

KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

Change in KE as the plane turns 180deg from north to south

12.4 +112 = 124.4 Joules

Change in KE as the plane turns 180deg from south to north

112 + 12.4 = 124.4 Joules

#

**138**
Rebell - I have an open mind to consider additional forces acting on the plane when turning. Just not willing to dismiss the significant difference in KE between upwind and downwind legs.

I believe you are suggesting that the force of the wind is acting upon the bottom of the wing when banking which could provide lift for the plane and also begin acclerating the mass of the plane downwind.

Note: Even your descriptions suggest a difference when turning downwind and upwind which you are attributing to the force of the wind acting above and below the wing when banking.

Clint

I believe you are suggesting that the force of the wind is acting upon the bottom of the wing when banking which could provide lift for the plane and also begin acclerating the mass of the plane downwind.

Note: Even your descriptions suggest a difference when turning downwind and upwind which you are attributing to the force of the wind acting above and below the wing when banking.

Clint

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**139**
That is why I say there are different situations to consider and each will have different outcomes. I am not taking sides, both sides has positive and negative points. The big thing is each person wants to argue his point without giving credit to the other and he does not want to give in.

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**140**Clint

#

**141**
#

**142**2. Change in KE is based on the

**difference in velocity**as per the formulas provided - from 10 MPH going into the wind to 30 MPH when going down wind.

Note: Very little KE is lost when changing direction. (There is actually a separate thread discussing the KE losses of a glider when banking. Remember what happens when a marble bounces off of a large solid object. KE remains unchanged despite the change of direction.)

Clint

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**143**2. Change in KE is based on the

**difference in velocity**as per the formulas provided - from 10 MPH going into the wind to 30 MPH when going down wind.

Note: Very little KE is lost when changing direction. (There is actually a separate thread discussing the KE losses of a glider when banking. Remember what happens when a marble bounces off of a large solid object. KE remains unchanged despite the change of direction.)

Clint

I used north south to better illustrate that the Two KE components need to be added to each other to get the total of change.

Change in KE as the plane turns 180deg from north to south

12.4

Change in KE as the plane turns 180deg from south to north

112

12.4

**+**112 = 124.4 JoulesChange in KE as the plane turns 180deg from south to north

112

**+**12.4 = 124.4 JoulesBryce.

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**144**You need to use a direction when talking about acceleration.

I used north south to better illustrate that the Two KE components need to be added to each other to get the total of change.

Bryce.

I used north south to better illustrate that the Two KE components need to be added to each other to get the total of change.

Bryce.

Consider a car with no friction and no drag headed South at 20 MPH. You turn the wheel and change the direction so you now are headed North at 20 MPH. No change in kinetic energy of the car.

You can also consider a marble rolling at 20 MPH that impacts an infinite mass at right angle. The marble will change direction and still be going 20 MPH. (Conservation of energy)

When the plane changes from downwind to upwind, there is no change in kinetic energy until the plane velocity either increases or decreases.

The additional energy required to accelerate the plane when turning downwind or excess energy available when turning upwind is the difference of KE as per the math shown.

Clint

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**146**Or are you simply reinforcing that the velocity and corresponding KE of the plane does not change when turning from downwind to upwind (unless the altitude is changed converting KE to PE)?

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**147**I would like to politely disagree and use classic physics example to illustrate.

Consider a car with no friction and no drag headed South at 20 MPH. You turn the wheel and change the direction so you now are headed North at 20 MPH. No change in kinetic energy of the car.

You can also consider a marble rolling at 20 MPH that impacts an infinite mass at right angle. The marble will change direction and still be going 20 MPH. (Conservation of energy)

When the plane changes from downwind to upwind, there is no change in kinetic energy until the plane velocity either increases or decreases.

Clint

Consider a car with no friction and no drag headed South at 20 MPH. You turn the wheel and change the direction so you now are headed North at 20 MPH. No change in kinetic energy of the car.

You can also consider a marble rolling at 20 MPH that impacts an infinite mass at right angle. The marble will change direction and still be going 20 MPH. (Conservation of energy)

When the plane changes from downwind to upwind, there is no change in kinetic energy until the plane velocity either increases or decreases.

**The additional energy required to accelerate the plane when turning downwind or excess energy available when turning upwind is the difference of KE as per the math shown.**Clint

Your car example is not a similar example.

your Car example is not proof of you Downwind Upwind discussion. That is that there is no Downwind Upwind similarity to that example. (try the same example if the car were on a treadmill wide enough etc)

Re read the changes I made to your Math example and try to understand that if you are going to try an Use the ground as a reference to the KE between the Plane and the ground. then you need to consider the direction of the KE at start and then the Direction of the KE at the end.

Bryce.

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**148**If the math was correct, then you would have come to the same Conclusion I showed. The two forces must be added as they are in opposing directions. If you are trying to calculate the Force required to do a downwind turn V the Up wind turn.

Your car example is not a similar example.

your Car example is not proof of you Downwind Upwind discussion. That is that there is no Downwind Upwind similarity to that example. (try the same example if the car were on a treadmill wide enough etc)

Re read the changes I made to your Math example and try to understand that if you are going to try an Use the ground as a reference to the KE between the Plane and the ground. then you need to consider the direction of the KE at start and then the Direction of the KE at the end.

Bryce.

Your car example is not a similar example.

your Car example is not proof of you Downwind Upwind discussion. That is that there is no Downwind Upwind similarity to that example. (try the same example if the car were on a treadmill wide enough etc)

Re read the changes I made to your Math example and try to understand that if you are going to try an Use the ground as a reference to the KE between the Plane and the ground. then you need to consider the direction of the KE at start and then the Direction of the KE at the end.

Bryce.

#

**149**the car example is just like a plane doing a 360deg circles in no wind. there is no KE change in that example (KE relative to the Ground) and the point is. your math to try and show the amount of change in KE while doing an upwind turn V a down wind turn is Flawed.

Re Read my edit to your math. Try to understand that the Direction is a part of the Change.

Re Read the Math. Forget the car example.

or if you want to do the car example. then do it with the same Velocities weights and Headwind downwind speed (treadmill speed).

Bryce.

#

**150**
You may wish to confuse yourself and others with referring to ground speed to make your KE equations appear valid, but.... a plane circling in the air at a constant airspeed is NOT what you keep warping this idea towards.

A trimmed plane gains (or loses) NO energy when flying with no change in airspeed.

It's flying IN the air mass, not fighting it at all!

Ground speed and the KE relative to the ground is worth noting only when taking off or landing.

Takeoff speed relates to the WIND, not the ground.

That's the sole reason for paying attention to the wind direction. Ground speed is lower when flying into the wind.

With a headwind, the perceived ground speed to take off is lower. The AIRSPEED is the same.

With a tailwind, the perceived ground speed to take off is higher. The AIRSPEED is the same.

There is less KE................................. relative to the ground......... but NOT to the air, when going upwind.

There is more KE.............................. relative to the ground........... but not the air, when going downwind.

Throwing in groundspeed is nothing more than the typical red herring commonly used in debate.

.

A trimmed plane gains (or loses) NO energy when flying with no change in airspeed.

It's flying IN the air mass, not fighting it at all!

Ground speed and the KE relative to the ground is worth noting only when taking off or landing.

Takeoff speed relates to the WIND, not the ground.

That's the sole reason for paying attention to the wind direction. Ground speed is lower when flying into the wind.

With a headwind, the perceived ground speed to take off is lower. The AIRSPEED is the same.

With a tailwind, the perceived ground speed to take off is higher. The AIRSPEED is the same.

There is less KE................................. relative to the ground......... but NOT to the air, when going upwind.

There is more KE.............................. relative to the ground........... but not the air, when going downwind.

Throwing in groundspeed is nothing more than the typical red herring commonly used in debate.

.