# Aerospace The 1904 plane of the Wright Brothers a bit strange

1. Mar 15, 2015

### 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)

Last edited: Mar 15, 2015
2. Mar 15, 2015

### Staff: Mentor

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.

3. Mar 15, 2015

### simplex1

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.

4. Mar 16, 2015

### 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

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)

5. Mar 19, 2015

### 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?

6. Mar 19, 2015

### 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.

7. Mar 20, 2015

### simplex1

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.

8. Mar 20, 2015

### Andre

9. Mar 20, 2015

### DaveC426913

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

10. Mar 20, 2015

### Staff: Mentor

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.
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.

11. Mar 20, 2015

### jack action

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.

12. Mar 21, 2015

### simplex1

jack action:
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) ).

13. Mar 21, 2015

### 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.

14. Mar 22, 2015

### 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.

Last edited: Mar 22, 2015
15. Mar 22, 2015

### Staff: Mentor

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.

16. Mar 22, 2015

### simplex1

I have read everything.

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.

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.

17. Mar 22, 2015

### jack action

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.

18. Mar 22, 2015

### Staff: Mentor

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.
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.

19. Mar 22, 2015

### simplex1

russ_watters:
$k=mg$

$m*g = Kl * v^2$ (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?

20. Mar 22, 2015

### jack action

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.

21. Mar 22, 2015

### simplex1

russ_watter:

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.

22. Mar 22, 2015

### simplex1

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.
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.

Last edited: Mar 22, 2015
23. Mar 22, 2015

### Staff: Mentor

Ok, fair enough. So, back to the previous:
That seems like a pretty odd way to arrange the lift equation, but yes, that is correct.
Because he wants to maintain steady flight.
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.

Last edited: Mar 22, 2015
24. Mar 22, 2015

### simplex1

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.
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:
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.

Last edited: Mar 23, 2015
25. Mar 23, 2015

### Staff: Mentor

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.
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.