The 1904 plane of the Wright Brothers a bit strange

[simplex1]

Something strange regarding the 1904 plane (Flyer No 2) built by the Wright Brothers
I am looking for an as simple as possible mathematical model that can explain how it really worked.

In 1904, the Wright Brothers started to test a new plane, Flyer II, somewhere near Dayton, Ohio where they managed to get permission to use a flat pasture for their experiments.
The winds were light there and, in the beginning, they had no catapult to quickly accelerate their machine and throw it into the air. They simply started the engine of the airplane which began to move along a track (a runway) while a head wind of moderate intensity was blowing and finally they got into the air and flew.

What is not quite clear (read the attached letter) is how exactly the plane took off.
W. Wright says that Flyer II lifted at 23 mph but he adds that Thrust got greater than Drag (Total Resistance) at 27 - 28 mph.
How could the plane accelerate from zero to the take off velocity if Thrust was less than Drag at speeds below 27 - 28 mph?

I made an attempt to write the equation of Flyer II as it accelerated along the track (see 1 and 2) but it is quite clear that the airplan speed gets negative unless Vw > 27 or 28 mph and such a wind speed was not available near Dayton.

1) T(Vp+Vw) - Kd * (Vp + Vw)^2 = m * dVp/dt (Thrust - Drag = ma)
2) m * g = N + Kl * (Vp + Vw)^2 (Weight = Normal reaction of the track + Lift)

where:
m = plane mass
Kd, Kl = drag and lift constants
Vp = plane speed relative to the ground
Vw = wind speed relative to the ground
g = 9.81 m/s^2
N = the normal reaction of the runway (track)

Can somebody on the forum correct my equations to make them agree with the description of W. Wright?

Fragment from a letter written by Wilbur Wright to Octave Chanute, on August 8, 1904:
"One of the Saturday flights reached 600 ft. ...
We have found great difficulty in getting sufficient initial velocity to get real starts.
While the new machine lifts at a speed of about 23 miles, it is only after the speed reaches 27 or 28 miles that the resistance falls below the thrust. We have found it practically impossible to reach a higher speed than about 24 miles on a track of available length, and and as the winds are mostly very light, and full of lulls in which the speed falls to almost nothing, we often find the relative velocity below the limit and are unable to proceed. ... It is evident that we will have to build a starting device that will render us independent of wind."
Source: http://www.loc.gov/resource/mwright.06007/?sp=52 (Library of Congress,
Page 52 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904)

 

[mfb]

W. Wright says that Flyer II lifted at 23 mph but he adds that Thrust got greater than Drag (Total Resistance) at 27 - 28 mph.

Are you sure it is not the other way round? As drag increases significantly with velocity, but thrust in general does not, initially you have more thrust than drag, until you reach the maximal velocity where both are the same.

 

[simplex1]

Are you sure it is not the other way round? As drag increases significantly with velocity, but thrust in general does not, initially you have more thrust than drag, until you reach the maximal velocity where both are the same.

You can find the original hand written letter here:
http://www.loc.gov/resource/mwright.06007/?sp=52
(Library of Congress, Page 52 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904)

You can magnify and read the text. Maybe you can discover a different interpretation of what W. Wright wrote in Aug. 1904.

Honestly, a mounted (Wright) glider can behave like this. It lifts in a 23 mph headwind assuming it is tethered like a kite but once the link to the ground is disconnected the glider can not hover and needs a speed of 4 - 5 mph relative to the ground, down the slope and against the wind, to stay in the air. A device installed on wings would measure 27-28 mph while the glider descends to the foot of the hill.

 

[simplex1]

I have found another letter addressed by W. Wright to Octave Chanute in which the elder of the two brothers again made some odd statements.

According to him, in a 17 ft/sec headwind Flyer II reached a ground speed of 42 ft/sec while, flying against a 12 ft/sec wind, the same plane averaged just 33 ft/sec, ground speed (see the Aug. 28, 1904 letter).

A 5 ft/sec increase in the wind speed induced a 9 ft/sec gain in the airplane ground speed?!

A little stronger wind might have increased the thrust of the propellers a bit but it also made the drag greater. I do not see how a + 5 ft/sec increase in the horizontal headwind speed could have made the 1904 plane fly 9 ft/sec faster.

From the Drag eq. (1) it will follow that:

T1(33+12) = Kd * (33+12)^2
and
T2(42+17) = Kd * (42+17)^2

which leads to:
T2 = 1.72 * T1 (A +5 ft/sec increase in the wind speed should have made the thrust of the propellers 1.72 times stronger !?). This is impossible.

Fragment from another letter, written by Wilbur Wright to Octave Chanute on August 28, 1904:
"Dayton, Ohio, August 28, 1904.

Dear Mr Chanute ...

... Since the first of August we have made twenty five starts with the #2 Flyer. The longest flights were 1432 ft., 1304 ft, 1296, ft. and 1260 ft. These are about as long as we can readily make on over present grounds without circling. We find that the greatest speed over the ground is attained in the flights against the stronger breezes. We find that our speed at startup is about 29 or 30 ft per second, the last 60 ft of track being covered in from 2 to 2 1/4 seconds. The acceleration toward the end being very little. When the wind averages much below 10 ft per second it is very difficult to maintain flight, because the variations of the wind are such as to reduce the relative speed so low at times that the resistance becomes greater than the thrust of the screws. Under such circumstances the best of management will not insure a long flight, and at the best the speed accelerates very slowly. In one flight of 39 1/4 seconds the average speed over the ground was only 33 ft per second, a velocity only about 3 ft per second greater than that at startup. The wind averaged 12 ft per second. In a flight against a wind averaging 17 ft per second, the average speed over the ground was 42 ft per second, an average relative velocity of 59 ft per second and an indicated maximum velocity of 70 ft per second. We think the machine when in full flight will maintain an average relative speed of at least 45 miles an hour. This is rather more than we care for at present.

Our starting apparatus is approaching completion and then we will be ready to start in calms and practice circling.

Yours truly
Wilbur Wright."

Source: http://www.loc.gov/resource/mwright.06007/?sp=55, (Library of Congress,
Page 55 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904)

 

[jack action]

It might be possible to go faster in a stronger headwind if you consider the angle of attack. If there is no wind, the plane needed probably a more angled position to create the necessary lift at the expense of creating greater drag and probably less thrust as well (since the propeller was angled along with the plane).

Concerning the first post, it's not clear that the necessary conditions weren't met for the 600-ft flight. This one flight worked but all other failed. Maybe there was a proper temporary headwind, at the right time, at the place.

Although the «it is only after the speed reaches 27 or 28 miles that the resistance falls below the thrust» statement doesn't make sense to me as well, for the same reason you mentioned: How could the plane accelerate from zero to the take off velocity if Thrust was less than Drag at speeds below 27 - 28 mph?

There must be some misuse of words. I would think this velocity might refer to the lift-to-drag ratio more than to the drag-thrust relation.

Some references:
Wikipedia: Lift-to-drag ratio
How much thrust does an aircraft need to fly?

 

[rcgldr]

I"m wondering if the reference to thrust less than drag refers to the model in flight as opposed to the model on the track, where the lift is zero and the drag is less until the pilot pitches the aircraft upwards to begin flight once sufficient speed is acheived. I'm also wondering if all of the quoted speeds are below the speed of best lift to drag ratio, so that the lift to drag ratio would improve as speed increased. There's also the issue of a fixed pitch prop, and it's efficiency is also affected by the relative air speed.

 

[simplex1]

It might be possible to go faster in a stronger headwind if you consider the angle of attack. If there is no wind, the plane needed probably a more angled position to create the necessary lift at the expense of creating greater drag and probably less thrust as well (since the propeller was angled along with the plane).

Regarding the second letter from Aug. 28, 1904 and your remarks:

Using equations (1), (2) and the relation Ro * P = T * V (Propeller efficiency

* Engine power = Thrust * Ground Speed), where Ro = Ro(Vp+Vw) and T = T(Vp+Vw), four relations can be written:

T1 = Kd1 * (33+12)^2
T2 = Kd2 * (42+17)^2
Ro1 * P = T1 * Vp1, where Vp1 = 33 ft/s
Ro2 * P = T2 * Vp2, where Vp2 = 42 ft/s

which means that:
Ro1/Ro2 = 0.457 * (Kd1/Kd2)

The best that can be achieved to keep Ro2 as close as possible to Ro1 and Kd2 to Kd1 (starting from the premise that a 5 ft/sec increase in the wind speed can not modify Ro and Kd too much) is obtained when Ro1/Ro2 = sqrt(0.457) = Kd2/Kd1 which means that:
Ro2 = 1.48*Ro1
Kd2 = 0.676*Kd1

Mathematically it is possible but I do not know if such a small increase in the wind speed (just 5 ft/sec) could determine the efficiency of the propellers to go up 1.48 times while the drag constant reduced to 0.676 of what it was before.

 

[Andre]

Drag can be greater than trust at low speeds due to the nature of the drag being the sum of parasite drag and induced drag. See fig 2-5. halfway down. http://www.americanflyers.net/aviationlibrary/instrument_flying_handbook/chapter_2.htm Essentially when drag exceeds trust at low speed you find yourself in the stall situation,

 

[DaveC426913]

Drag can be greater than trust at low speeds

No, no. Trust definitely overcame drag in this case, or he never would have gotten in the plane!

 

[russ_watters]

Drag can be greater than thrust at low speeds due to the nature of the drag being the sum of parasite drag and induced drag. See fig 2-5. halfway down. http://www.americanflyers.net/aviationlibrary/instrument_flying_handbook/chapter_2.htm

I think you nailed it, but let me expand a bit on how that applies to the story of Wright Flyer told above:

The Flyer accelerates along the track at zero angle of attack, then pitches-up (rotation) and takes off. Since it is at zero angle of attack before rotation, there isn't much drag and it can accelerate...at least until friction with the ground (plus the relatively low drag) starts to limit it. If the plane is moving too slowly when it rotates, it may still lift-off, but it needs such a high angle of attack to generate the necessary lift that it has too much drag, slows down and falls back to earth. Above that critical speed, the angle of attack needed to take off is lower, the drag is lower, and it can both climb and continue to accelerate.

Essentially when drag exceeds thrust at low speed you find yourself in the stall situation...

If I remember correctly, you're a pilot, but that isn't quite right. You might find yourself in a stall or you might just be slowing down. A stall doesn't have anything directly to do with thrust, it's all about how air is flowing over the wing. As long as flow doesn't separate it isn't a stall, even if you are crashing.

 

[jack action]

The best that can be achieved to keep Ro2 as close as possible to Ro1 and Kd2 to Kd1 (starting from the premise that a 5 ft/sec increase in the wind speed can not modify Ro and Kd too much)

I'm not sure about those assumptions.

The lift coefficient ratio will be obtained from Kl1(33+12)² = Kl2(42+17)², where Kl2/Kl1 = 0.58. Based on a typical airfoil, assuming the angle of attack at low speed was 15°:

Then the high speed angle of attack was 6°. From that the ratio Kd2/Kd1 could be as low as 0.40. But we only need 0.58 from Kd1(33+12)² = Kd2(42+17)².

 

[simplex1]

jack action:

Then the high speed angle of attack was 6°. From that the ratio Kd2/Kd1 could be as low as 0.40.

If Kd2 = 0.4*Kd1 then Ro1/Ro2 = 0.457 * (Kd1/Kd2) = 0.457 / 0.4 = 1.14 (see my previous post) which means that Ro2 = 0.875 * Ro1. In other words the efficiency of propellers has to drop also despite the fact that, in the second case when the plane flies faster (42 ft/sec ground speed), the angle between the thrust vector and horizontal decreases from 15 to just 6 degrees. Logically the propellers should be more, not less, efficient when the attack angle of the wings is just 6 degree as compared with the case when the angle was 15 degrees (Thrust * cos (6) > Thrust * cos (15) ).

 

[jack action]

What I said is «could be as low as 0.40». I used data from a more modern airfoil (Clark Y), as I don't have the ones from the Wright airplane, to show that getting such a value for Kd2/Kd1 was possible and realistic. You seemed to doubt that a Kd2/Kd1 = 0.676 was possible.

As for the propeller efficiency, if we keep it the same for both cases, then Kd2/Kd1 = 0.457 and that is greater than 0.40, so it is a possible value. If the propeller efficiency increases, then Kd2/Kd1 increases as well, which makes it even more possible.

Also, from the graph below (source), you can see that going from 23 to 29 mph (33 to 42 ft/s), the increase in efficiency is about 1.29. So such an increase seems to be a possibility as well.

 

[simplex1]

If we take Ro1 = Ro2 = Ro then the useful power available in both situations is P' = Ro * P. Using this P' the plane flies at 33 ft/sec when the wind is 12 ft/sec and faster at 42 ft/sec when the wind is 17 ft/sec. Therefore, the machine is more efficient in the second case as long as it travels a longer distance in the same time.

The main question is: If Flyer II was capable to fly at 33 ft/sec against a 12 ft/sec wind, staying firmly in the air, and then, in the second case, to run faster at 42 ft/sec against a stronger 17 ft/sec wind, for what reason was this plane unable to accelerate from 33 ft/sec to above 42 ft/sec in a 12 ft/sec headwind?!

From the letter of W. Wright (Aug. 28, 1904) it appears that his plane could not accelerate above 33 ft/sec flying against a 12 ft/sec wind and this behavior does not make sense.

 

[russ_watters]

...this behavior does not make sense.

I don't mean to be rude, but you didn't respond to any of the above posts directly, so I can't be sure you read them: did you? The answer to the question was given and explained in detail above, more than once. I'll say it again, more concisely:

Higher airspeed means lower angle of attack, which means less drag.

 

[simplex1]

I have read everything.

russ_watters:

Higher airspeed means lower angle of attack, which means less drag.

The angle of attack is not a constant of the plane. The pilot can modify it. In the beginning (after the plane takes off and manages to float at a constant height) the angle of attack can be quite big but after that the pilot slowly lowers it reducing the drag. In consequence, the plane accelerates gaining lift due to the increase in speed and in the same time losing some lift because Cl decreases as the attack angle gets smaller.

In general:
Lift(t) = Kl(alfa(t)) * v(t)^2

where alfa(t) = the angle of attack as a function of time

In order to maintain the lift constant, as the speed of the plane grows, the following condition should be met:
Kl(alfa(t)) = k / v(t)^2
where k = a constant.

Only if Kl(alfa(t)) goes down faster than k / v(t)^2 the plane falls. It appears to me that you, russ_watters, assume Kl decreases more rapidly than k / v(t)^2 as the pilot, once flying at 33 ft/sec in a 12 ft/sec headwind, tries to accelerate the plane by lowering the attack angle.

 

[jack action]

The main question is: If Flyer II was capable to fly at 33 ft/sec against a 12 ft/sec wind, staying firmly in the air, and then, in the second case, to run faster at 42 ft/sec against a stronger 17 ft/sec wind, for what reason was this plane unable to accelerate from 33 ft/sec to above 42 ft/sec in a 12 ft/sec headwind?!

Because the headwind was stronger. The extra headwind adds «free» energy to lift the plane.

You are right, at 33 ft/s with a 12 ft/s headwind, the plane cannot accelerate further. But the sudden gust of wind that increase the headwind to 17 ft/s increases the lift force of the plane. If the pilot don't correct the angle of attack, the plane will take altitude. If he decreases the angle of attack, the plane will stay at the same altitude. But doing so, he also reduces the drag, thus the plane accelerate since it still have the same thrust (and maybe even a little more because of the angle correction of the propeller).

Does this makes sense? I'm just thinking out loud.

 

[russ_watters]

The angle of attack is not a constant of the plane. The pilot can modify it. In the beginning (after the plane takes off and manages to float at a constant height) the angle of attack can be quite big but after that the pilot slowly lowers it reducing the drag. In consequence, the plane accelerates gaining lift due to the increase in speed and in the same time losing some lift because Cl decreases as the attack angle gets smaller.

Correct, though that glosses over why the pilot modifies the angle of attack: he's trying to keep the plane aloft. So you should conclude that at a minimum, at all times the angle of attack is high enough to provide enough lift to remain aloft.

In general:
Lift(t) = Kl(alfa(t)) * v(t)^2

where alfa(t) = the angle of attack as a function of time

In order to maintain the lift constant, as the speed of the plane grows, the following condition should be met:
Kl(alfa(t)) = k / v(t)^2
where k = a constant.

Only if Kl(alfa(t)) goes down faster than k / v(t)^2 the plane falls. It appears to me that you, russ_watters, assume Kl decreases more rapidly than k / v(t)^2 as the pilot, once flying at 33 ft/sec in a 12 ft/sec headwind, tries to accelerate the plane by lowering the attack angle.

What is k / v(t)^2? Is that derived from an equation equating drag and thrust? I really think this is easier to conceptualize than trying to build equations to describe it. This was in Andre's link (graph at right):

As you can see, the power required to stay aloft at 75 (whatever unit of speed) is higher than the power required at 125.

 

[simplex1]

russ_watters:

What is k / v(t)^2? Is that derived from an equation equating drag and thrust?

$k=mg$

$m*g = Kl * v^2$ (Weight = Lift, for level flight)

In consequence, $Kl = m*g / v^2$ which means that if Kl does not have this dependency, mg will be smaller or greater than $Kl * v^2$ and so the plane will rise or descend. If the pilot is capable to adjust in such a way the angle of attack to always keep $Kl = m*g/v^2$ then the plane is always in level flight no matter how much the airspeed, $v$, is. The question is why should the pilot artificially force Kl to decrease so dramatically with speed?

 

[jack action]

The question is why should the pilot artificially force Kl to decrease so dramatically with speed?

Because the headwind increases ... which increases $v$ ...

If $v$ increases, then $Kl$ must be decreased to keep the equality $m*g = Kl * v^2$ true.

Doing so, you also reduce the drag; Then the plane can accelerate further.

You never really responded to that argument.

 

[simplex1]

russ_watter:

As you can see, the power required to stay aloft at 75 (whatever unit of speed) is higher than the power required at 125.

Source: http://www.americanflyers.net/aviationlibrary/instrument_flying_handbook/chapter_2.htm
If the 1904 plane had been able to fly in both reversed and normal command regions then, as can be seen in the general theoretical diagram (Figure 2-6.), for a given power P there are only two solutions, points 1 and 2. Assuming the total airspeed in 1 is 33+12 = 45 ft/sec the second point can have the airspeed 42+17 = 59 ft/sec and it appears that W. Wright described a real situation or at least physically possible, when, in his Aug. 28, 1904 letter, he wrote that his plane had flown faster in a 17 ft/sec headwind than against a 12 ft/sec wind.
However, in the same letter W. Wright also says:

"Dayton, Ohio, August 28, 1904.
... Since the first of August we have made twenty five starts with the #2 Flyer... . We find that the greatest speed over the ground is attained in the flights against the stronger breezes."

which means that, in general, he remarked that the airspeed (Vp+Vw), recorded by a device on the plane, increased as the wind intensified. However, as the diagram shows, only two solutions exist: (y=P, x=45 ft/sec) and (y=P, x=59 ft/sec) and nothing else.

 

[simplex1]

jack action:

If v increases, then Kl must be decreased to keep the equality m∗g=Kl∗v^2 true.

It appears that it would be much more logical to try to keep Kl constant and let Kl∗v^2 grow larger than m*g which means that the plane will gain altitude.
You can keep m∗g = Kl∗v^2 with a lift killer wing, like the ones used in Formula 1, but this kind of approach only increases the drag.

I have started from the assumption that Kl and Kd are constant for the range of air speeds [45, 59 ft/sec] of interest. People on the forum came with the idea that both these quantities are in fact variables:
Kl = Kl(alfa) = (for example) a*alfa^b (also it can be any formula you propose)
Kd = Kd(alfa) = (for example) c*alfa^d
where a, b, c, d are some known parameters which have to be given as numerical values.

From the two relations above we can find:
Kd = Kd(Kl)
and quickly calculate the amount of drag induced by a Kl = m∗g / v^2 and see if drag really goes down when the airspeed increases.

 

[russ_watters]

russ_watters:
$k=mg$

Ok, fair enough. So, back to the previous:

Only if Kl(alfa(t)) goes down faster than k / v(t)^2 the plane falls. It appears to me that you, russ_watters, assume Kl decreases more rapidly than k / v(t)^2 as the pilot, once flying at 33 ft/sec in a 12 ft/sec headwind, tries to accelerate the plane by lowering the attack angle.

That seems like a pretty odd way to arrange the lift equation, but yes, that is correct.

The question is why should the pilot artificially force Kl to decrease so dramatically with speed?

Because he wants to maintain steady flight.

which means that, in general, he remarked that the airspeed (Vp+Vw), recorded by a device on the plane, increased as the wind intensified. However, as the diagram shows, only two solutions exist: (y=P, x=45 ft/sec) and (y=P, x=59 ft/sec) and nothing else.

Two solutions for a power that equals drag. Clearly, if the plane is accelerating, the power doesn't equal the drag.

Can I ask what the point of all this is? It seems really obscure.

 

[simplex1]

russ_watters:

Because he wants to maintain level flight.

Wilbur Wright wanted to fly as much as possible. His intention was not to maintain level flight but to stay in the air by any means.
In fact all this story regarding the angle of attack is pure theory because Flyer II had a sinuous trajectory being up and down all the time, flying like a bat (this is the claim).
Climbing would have helped because, a bit later after reaching a certain height, Hmax, the plane would have used his potential energy mg(Hmax - Hmin) to accelerate to a high enough velocity to stay stably in the air at Hmin, even if the wind had dropped to zero.

russ_watters:

Two solutions for a power that equals drag. Clearly, if the plane is accelerating, the power doesn't equal the drag.

But the plane was not accelerating in the two cases (wind speed 12 and 17 ft/sec) W. Wright talked about in his Aug. 28, 1904 letter.

 

[russ_watters]

russ_watters:
Wilbur Wright wanted to fly as much as possible. His intention was not to maintain level flight but to stay in the air by any means.

I actually edited while you were typing to say steady flight, which would be more complete. Presumably, whatever he's doing when the gust of wind hits him he wants to keep doing. If he is in level flight he wants to stay in level flight. If he's in a slight climb he wants to stay in a slight climb, etc. This is no different from if you are walking down the street on a windy day, fighting the wind to stay on a straight path.

In fact all this story regarding the angle of attack is pure theory because Flyer II had a sinuous trajectory being up and down all the time, flying like a bat (this is the claim).

I'm really not sure what you are trying to say by "pure theory", but yes, the Flyer II was difficult to control because of its configuration. Planes with canards instead of stabilizers in back are inherrently unstable and require constant control input to keep steady.

You seem to not like "all this story". Why? As I said, this is all pretty obscure.

 

[simplex1]

russ_watters:

You seem to not like "all this story". Why? As I said, this is all pretty obscure.

Are you talking about the story of the powered flights the Wright brothers said they had performed between Dec. 17, 1903 and Aug. 7, 1908?

Anyway, because I do not want to open a different topic, I announce that I am searching for a plane (it can be unmanned) with a weight of around 335 kg and powered by a 12 HP engine. The only condition is to fly at 45 km/h or more. Is somebody on the forum aware about the existence of such a airplane?

 

[russ_watters]

russ_watters:
Are you talking about the story of the powered flights the Wright brothers said they had performed between Dec. 17, 1903 and Aug. 7, 1908?

Whatever "all this story" you were referring to.

Anyway, because I do not want to open a different topic, I announce that I am searching for a plane (it can be unmanned) with a weight of around 335 kg and powered by a 12 HP engine. The only condition is to fly at 45 km/h or more. Is somebody on the forum aware about the existence of such a airplane?

Clearly, you are referring to the Wright Flyer. What is going on here?

 

[jack action]

It appears that it would be much more logical to try to keep Kl constant and let Kl∗v^2 grow larger than m*g which means that the plane will gain altitude.

It is not a matter of logic. You asked: «How is it possible?» We answered: «If the plane keeps its altitude, it is.»

I have started from the assumption that Kl and Kd are constant for the range of air speeds [45, 59 ft/sec] of interest.

This assumption is wrong (or at least badly worded) because both of these variables are independent of speed (within reasonable limits). They only depend on the shape and position of the object within the air stream. The pilot controls both shape and position of the plane to steer it.

People on the forum came with the idea that both these quantities are in fact variables:
Kl = Kl(alfa) = (for example) a*alfa^b (also it can be any formula you propose)
Kd = Kd(alfa) = (for example) c*alfa^d
where a, b, c, d are some known parameters which have to be given as numerical values.

From the two relations above we can find:
Kd = Kd(Kl)
and quickly calculate the amount of drag induced by a Kl = m∗g / v^2 and see if drag really goes down when the airspeed increases.

You are again mixing positioning with speed. Your equations are solely dependent of alfa and you are concluding «[we will] see if drag really goes down when the airspeed increases.» You'll never be able to prove that with your equations, as they are clearly independent of speed.

What they are, is dependent of the pilot 's intention. If the airspeed changes and the pilot does nothing, the plane will instantaneously either go up and slow down (airspeed increases) or go down and speed up (airspeed decreases).*

But IF the pilot decides to change the shape and/or position of its plane in a precise way, the plane will stay at the same altitude. What he actually did was in fact reducing the lift coefficient. By doing so, he also has decreased the drag coefficient, which means the plane will accelerate under the same thrust. And, within reasonable limits of airfoil designs, drag and lift coefficients are proportional to one another: When one goes up, so does the other, and vice-versa. It is a fact for all airfoils.

So when you ask «How is it possible?» The answer is: «It is possible if the pilot decides to change the angle of attack of its plane.»

The problem now is that you seem to refuse that the Wright brothers have different angles of attack in the different passes, even though nothing seems to be mentioned to support whether it is or it is not.

All we can say right now is, if the angle of attack was different under different headwind, the speed can be faster in the larger headwind.

-----------------------------------
* When you say:

His intention was not to maintain level flight but to stay in the air by any means.

Choosing to gain altitude might not be an option. Even though the lift force increases and the plane wants to go up, the drag force also increases, slowing down the plane, which in turn reduces the lift force. There is a point where the plane will not be able to climb up anymore, no matter the strength of the headwind.

 

[jim hardy]

335 kg and powered by a 12 HP engine.

looks to me like you guys are all thinking steady state.

all math aside, let's apply some psychology and transient analysis to this question.

a craft so heavy and underpowered is going to feel like it's barely flying, much like a small boat trying to climb up over its bow wave onto plane where lift/drag is less.. If you ever water skied you've felt this "climbing your bow wave" when you first get up.

A gust of headwind cannot immediately decelerate the craft because of its substantial inertia so might push it "over the hump" so to speak.

Then there's the fact Wrights were sensitive about falls. That's one of the reasons they chose Kitty Hawk, for its soft sand to lessen impact..

So it's very plausible to me that a gust might well lift the plane causing Wilbur to push nose down with no other thought than to make his impending fall a shorter one. Surely the deceleration would feel a lot like prelude to stall.
Nose down would move center of lift aft on the wing thus reducing drag in this new stable attitude., one which he simply lacked the horsepower to reach unaided by outside forces.

(sorry, the images are in reverse order)

http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_2.htm

The Wrights knew plenty about AOA. Did their flyer have such a gage? I suppose a piece of yarn trailed from one of the support wires might have worked, if he wasn't too busy to watch it.

just curious, old jim

 

[simplex1]

1) Regarding Flyer I (1903) it looks like gravity supplied an important part of the thrust the plane needed to stay in the air and the vertical component of the wind also aided Wilbur's machine to float. If the well known photo with Orville Wright just taking off and his brother looking at the plane, was really taken on Dec. 17, 1903 (the image was first published in Sep. 1908), then it is quite clear from the detail you can admire (see the attached picture) that Flyer I performed a power assisted glide at best.

The first flight on December 17, 1903. Flyer I taking off and just about to go down along a slope..

2) "They carried the machine up on the Hill", John T. Daniels, eye witness
The fact that Flyer I 1903 just glided, aided partly by the engine, was confirmed apparently unwillingly by John T. Daniels, an eye witness, in a letter addressed to a friend:

"Manteo NC, June 30 —- 1933,

Dear friend,

I Don’t know very much to write about the flight. I was there, and it was on Dec the 17, — 1903 about 10 o’clock. They carried the machine up on the Hill and Put her on the track, and started the engine … and he went about 100 feet or more, and then Mr. Wilbur taken the machine up on the Hill and Put her on the track and he went off across the Beach about a half a mile …

John T. Daniels, Manteo NC, Box 1W"
Source: http://wrightstories.com/eyewitness-account-of-first-flight-by-john-daniels

3) The 2003 flying replica of Flyer I 1903 couldn't fly more than 115 feet (35 m)
The 2003 accurate replica of the Wright brothers' 1903 plane was not able to do more than short flights. None of its takeoffs came close to the claimed 59 seconds flight performed on December 17, 1903. What the 2003 experiment really showed was that Flyer I could have been theoretically able to take off and fly chaotically for 100 - 115 feet, no more. The 1903 machine was uncontrollable and not able to perform a sustained flight.


"On November 20, 2003, Dr. Kevin Kochersberger piloted the 1903 Wright Experience Replica Flyer. With 15-18 mph winds he flew a distance of nearly 100 feet."


"December 3, 2003 test flight of the Wright Experience 1903 Wright Flyer Replica. Dr. Kevin Kochersberger was at the controls and piloted the Flyer for a distance of 115 feet. Slight cross wind after initial rotation which is compensated with slight wing warp."

The instability of Flyer I had been already predicted by prof. Fred Culick who tested in the wind tunnel another replica, different from the one that flew in 2003:

""They built it and then drew as they went along," said Fred Culick, professor of aerodynamics at the California Institute of Technology and chief engineer on Cherne's team. …" Cherne's group, working mainly on weekends in a warehouse donated by a rocket company in El Segundo, finished what they considered an exact replica. Then in 1998 they tested it at NASA's Ames Research Center near Sunnyvale, Calif. Three weeks of wind-tunnel tests of their Wright Flyer replica "clearly showed how unstable it was and how it can't be flown safely," said Culick."
Source: http://community.seattletimes.nwsource.com/archive/?date=20031008&slug=wright08

4) "the brothers only “glided” off Kill Devil Hill that day. Their first real flight came on May 6, 1908", Alpheus W. Drinkwater, telegraph operator
"Wilbur and Orville Wright are credited with making their first powered flight in a heavier-than-air machine on Dec. 17, 1903. But Alpheus W. Drinkwater, 76 years old, who sent the telegraph message ushering in the air age, said the brothers only “glided” off Kill Devil Hill that day.
Their first real flight came on May 6, 1908, he said." Source: New York Times, Dec. 17, 1951.

The declaration of Alpheus W. Drinkwater corroborates well with the article "The Wright brothers in US and in France - The last tests of the Wrights' in US described by themselves", L'Aérophile, June 1908, pag. 222-223" ( http://gallica.bnf.fr/ark:/12148/bpt6k6550620m/f232.image.r=wright 6 mai.langEN ) where the two brothers talked, amongst others things, about a 337 m flight, against a 4-6 m/s headwind, that took place on May 6, 1908. As a remark, this is the first claimed powered flight mentioned by the Wright brothers after they stopped flying (also a pure claim) in October 1905.

 

[russ_watters]

Ok, that's enough. It's clear that you aren't looking to learn aerodynamics but rather are looking for fuel for conspiracy theory. We don't do conspiracy theory here. Thread locked.