Something strange regarding the 1904 plane (Flyer II) built by the Wright Brothers

[selenius]

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 headwind 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 (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 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."

For the original hand written letter see:

http://www.loc.gov/resource/mwright.06007/?sp=52

(Library of Congress,
Page 52 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904)

 

[GregLocock]

I suspect he means that the aircraft's drag is too great to fly (ie take off) at very low speed due to the angle of attack required.

Cheers

Greg Locock


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[Sparweb]

It's been a while since I read the WB's book (How We Invented the Airplane) but I do recall being impressed by their aeronautical knowledge and the amount of research and testing they conducted before the Flyer made the history books.

Several thoughts come to mind about the letter.

I believe the WB's spent a lot of time on propellor design, so they may have also understood the "advance ratio" conundrum. One can design a fixed-pitch prop to have optimal thrust at one speed, but when flying faster or slower, that thrust will drop off. The static thrust can be much smaller than the thrust when travelling forward at the design speed. In later years, pilots referred to these different styles as "climb props" and "cruise props" to distinguish between propellors cut with an advance ratio suitable for low or high speeds, respectively.

The WB's were also flying aircraft very sensitive to the ground effect. It sounds to me like they were floating a few feet over their track but unable to climb away. This would account for their reference to lifting off at 23 MPH but below 27-28 MPH where they seem to gain full control of the aircraft. The early Flyers had such low W/S ratios that I'm not surprised they were also very touchy in gusts.

Getting stuck in ground effect with very low power can also lead to a problem that I'll call "pilot-induced-drag". By this I mean that the pilot, now fed up with wallowing in the weeds, pulls up on the elevator slightly to try to gain some altitude. However, by doing so, the aircraft may pitch up and actually increase its induced drag instead of its lift, slowing down in the process. This is a dangerous place to be, and the Wright Brothers faced it frequently.


STF

 

[selenius]

Wilbur Wright: 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.

As long as the resistance is less than the thrust the plane gets a negative speed if the initial speed relative to the ground is zero (which was the case of Flyer II at the time the letter was written) and this phenomenon is independent of the wind speed or the dynamic thrust of the propellers.
T(Vp+Vw) - Kd * (Vp + Vw)^2 = m * dVp/dt (Thrust - Drag = ma)
if T(Vp+Vw) < Kd * (Vp + Vw)^2 -> Vp < 0 where Vp_initial = 0

 

[GregLocock]

yes, and your constants aren't constant. You need a lift/drag/angle of attack model. Aero 101

Cheers

Greg Locock


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[selenius]

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)

 

[Compositepro]

The resistance to which the letters refer includes aerodynamic drag and rolling resistance. The aerodynamic drag is zero at zero speed and increases with speed. The rolling resistance decreases with speed because of lift.

At a certain wind speed the aircraft can take off and will actually be flying backwards with respect to the ground.

The letter in your first post is saying that (at zero wind speed) the aircraft can only get to 24 mph before it runs out of ramp, so it cannot take-off.

Another factor could be that until the aircraft takes off the wing angle of attack is not adjustable to get minimum drag. After take off, the pitch of the aircraft can be optimized for reduced drag.

 

[selenius]

I) The second letter, from Aug. 28, 1908 refers to Flyer II in horizontal flight, after the take off. Whatever complications, phenomena may appear at lift off they are no longer present in horizontal flight. That gain in the plane ground speed from 33 ft/sec to 42 ft/sec (see my previous post) can not be explained by a 5 ft/sec increase in the headwind velocity (from 12 to 17 ft/sec).

II) Compositepro:

The resistance to which the letters refer includes aerodynamic drag and rolling resistance.

Yes, rolling friction should be included. I added it:

1) T(Vp+Vw) - Kd * (Vp + Vw)^2 - miu * N = m * dVp/dt (Thrust - Drag - Rolling friction = ma)
2.1) m * g = N + Kl * (Vp + Vw)^2 (Weight = Normal reaction of the track + Lift) if y = 0
2.2) Kl * (Vp + Vw)^2 - m * g = m * dVvp/dt if y > 0

where
miu = rolling friction coefficient
Vv = the vertical speed of the plane when y > 0
Vp = the horizontal speed of the plane
y = the vertical coordinate of the plane

However, at 23 mph (see the first letter, from Aug. 8, 1904) N should have been zero. A sudden increase with 5 mph of the headwind speed, at the moment when Vp + Vw = 23 mph (N = 0) will simply increase Drag that will go from Kd * (23 mph)^2 to Kd * (28 mph)^2 which automatically implies that Vp will decrease and finally there will be no net gain in Vp + Vw.

The horizontal headwind can not communicate power to the plane so it can not help the airplane fly better.

 

[selenius]

Honestly, what Wilbur Wright wrote, in both his Aug. 8 and 28, 1904 letters, about the way his Flyer II machine worked, perfectly fits the behavior of a glider flying along an incline in a headwind blowing parallel to the slope.
In such a case, the headwind helps the glider stay in the air due to the airflow vertical component (which did not exist near Dayton, along that flat pasture).

A glider sliding on a track installed along a slope can find itself in the impossibility to take off and fly if the headwind speed in not high enough. Also, once in the air above the incline and flying, if the wind speed drops, the glider can fell to the ground because the vertical component of the wind decreases and the glider gets less lift.

I am getting more and more convinced that the things written by Wilbur Wright in his two letter addressed to Octave Chanute (Aug. 1904) are not descriptions of actual flights but how he envisioned Flyer II should behave, based on his real experiments with gliders at Kitty Hawk which took place the previous years.

 

[IRstuff]

You make a potentially erroneous assumption, that the drag coefficient is constant. In order to keep the angle of attack (AOA) high for low wind, the tail elevator must be cranked down, potentially substantially increasing the drag. This is in addition to the increased wing drag from the high AOA on the wing. What I see is a 70% increase in drag due to aircraft AOA and elevator drag.

TTFN
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[selenius]

IRstuff:

In order to keep the angle of attack (AOA) high for low wind, the tail elevator must be cranked down, potentially substantially increasing the drag.

Flyer II (1904) had no tail elevator and it was not stable in pitch. The Wright Brothers constantly maneuvered the front elevator up and down to keep the plane fly as close to the horizontal as possible, at least this is what they said and how a slightly improved version of the 1904 plane behaved after Aug. 8, 1908 (four years later) in France when the two american inventors started to give public demonstrations.

IRstuff:

What I see is a 70% increase in drag due to aircraft AOA and elevator drag.

Using eq. (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)

if Kd1 = 1.72 Kd2, as you say,
then Ro2(42+17) = 1.27 * Ro1(33+12)
which is an implausible large gain (27%) in the efficiency of the propellers for just an increase in the headwind speed of only 5 ft/sec. Also assuming that Kd becomes 1.71 times smaller when the wind gets 5 ft/sec stronger is quite unlikely. Such a small change in the headwind velocity can not lead to huge changes in Kd and Ro.

 

[GregLocock]

A couple of things

also Aero 101 assumes that a propeller driven aircraft is a constant power device, and you can take an efficiency of 80% for the prop.

Cheers

Greg Locock


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[selenius]

GregLocock:

Aero 101 assumes that a propeller driven aircraft is a constant power device, and you can take an efficiency of 80% for the prop

If Ro1 = Ro2 (see my previous message) than:

Kd2(42+17) = 0.457 * Kd1(33+12)

which means that a 5 ft/sec increase in the headwind velocity would have lowered the drag coefficient at 45.7% of what it was when the plane flew in a 12 ft/sec headwind with a ground speed averaging 33 ft/sec.

This idea that a small increase in the headwind speed improved the drag coefficient in flight, making it significantly smaller, implies that for some unknown reasons the angle of attack (the position of the airplane in the air) was quite unfavorable when it flew against a 12 ft/sec wind and after that, when the wind intensified at 17 ft/sec, Flyer II became much more aerodynamic by somehow changing its average angle of attack. Kd can not vary so much for such a such small range of velocities. Kd is a constant. If it had been so sensible to Vp+Vw the experts in aerodynamics would have defined it as Kd = Kd(Vp+Vw).

 

[Sparweb]

...implausible...
...unknown reasons...
...somehow changing its average angle of attack ...
...Kd is a constant ...

Respectfully, we are trying to explain it to you. Induced drag is dominant at low airspeeds and very susceptible to slight changes in headwind, angle of attack. Take some time to review some basic textbook aerodynamics and you will begin to understand.

Heh Heh...
Of course, there could be other reasons you want the WB's to be wrong. Vive Henri Farman! -Le vrai inventeur de l'avion!


STF

 

[GregLocock]

Well here's my take on it using the 'x' curves, which were from a fabric covered model.

If we assume it is flying at the nominated level. CL=0.62 CD=.105, alpha is about 4 degrees

If we slow by 20% then CL must be .62/.8^2=.97. From the curve we see CD is 0.22, and alpha is around 10 degrees.

So the drag force has increased from .105*1^2 to .22*.8^2=.141, so the drag horsepower has increased to .141*.8/.105=107% of the value at the nominal flying speed.

I really think you are trying to prove an unremarkable observation (that planes have a minimum takeoff speed) by using an argument that ignores the fundamentals of wings, an unwise approach when dealing with aircraft.

Cheers

Greg Locock


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[selenius]

GregLocock:

If we slow by 20% then CL must be .62/.8^2=.97.

What law are you using for CL?
According to your law if the plane is slowed down to 50%, of what that reference speed might be, then:
CL(50%) = .62/0.5^2 = 2.48
also
CL(80%) = .62/0.2^2 = 15.5
CL(100%) = .62/0^2 = infinity
As you see, the slower the plane flies or slides on the track the more lift it gets which can not be a correct assumption.

 

[selenius]

A remark, instead of "the more lift it gets which can not be a correct assumption." read: "the larger CL gets".
The lift for a certain speed V = p * V0, where p = a percentage and V0 = the speed at which CL = 0.62, will be:
Lift(p*V0) = k * 0.62/p^2 * (p*V0)^2 = k * 0.62 * V0^2 = Lift(V0).
In consequence, Lift = constant and independent of Vp + Vw. This does not agree with experimental facts.

 

[selenius]

SparWeb:

"It's been a while since I read the WB's book (How We Invented the Airplane) but I do recall being impressed by their aeronautical knowledge and the amount of research and testing they conducted before the Flyer made the history books."

Flyer I (1903) had a propeller placed underneath that revolved horizontally, according to an article signed Wilbur Wright and published in Feb. 1904!!

"One of the propellers was set to revolve vertically and intended to give a forward motion, while the other underneath the machine and revolving horizontally, was to assist in sustaining it in the air. … After the motor device was completed, two flights were made by my brother and two by myself on December 17th last."
Source, "The Experiments of a Flying Man", author Wilbur Wright, The Independent, Feb. 04, 1904, pag. 246, internet address:

http://www.loc.gov/resource/mwright.05001173/?sp=4

 

[GregLocock]

"What law are you using for CL?"

In order to fly the lift must equal the gravity force on the plane. Therefore if the plane is flying straight and level, the lift is constant.

The lift is proportional to CL*v^2, so if the plane is /flying/ at 80% of the original speed, CL must have increased to 1/.8^2 times the original figure.

This is utterly fundamental stuff.

Cheers

Greg Locock


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[selenius]

GregLocock:

"What law are you using for CL?"
In order to fly the lift must equal the gravity force on the plane. Therefore if the plane is flying straight and level, the lift is constant.

The Lift Equation for flying straight and level:
m * g = k * CL * (Vp + Vw)^2
In consequence:
CL = m * g / (k * (Vp + Vw)^2) (a)
and if Vp + Vw -> 0 then CL -> infinity which is not correct. Everybody knows that a plane, below a certain speed, can not stay in the air, no matter what the pilot does and how it tries to change the attack angle. Also we all know that a plane can not mentain a straight course if Vp + Vw increases more and more. The airplane will rise according to the equation:
k * CL * (Vp + Vw)^2 - m * g = m * dVvp/dt (b) (I have already explained it in a previous message)
I agree that CL = CL(Vp + Vw) but it does not have the dependency (a) and for small variations of the wind speed (5 ft/sec) it is constant.

Another contradiction:
Your assumption (a) plugged into (b) implies that a plane can never rise if its initial vertical speed is zero (Vvp=0) which is again incorrect.
By doing the math you get:
m * dVvp/dt = 0 which means that Vvp(t) = constant. If Vvp _initial = 0 (level flight) you will get Vvp(t) = 0 no matter how much Vp + Vw increases which comes into conflict with experimental facts.