Trial calculation of lift
As described in "Mechanism of lift generation", since the shape of the wing is roughly plate-like, "Principle 1" becomes dominant and the influence of "Principle 2" is considerably reduced.
Therefore, if my personal opinion is correct, as shown below, it should be possible to obtain approximate values of lift and drag by analysis based on "Principle 1".
Explanation of "Principle 1" and "Principle 2” is linked from the last of this page.
1. Calculation of lift in "Principle 1"
As shown in the graph below, the magnitude of the pressure drop that occurs when the air is stretched behind the object and the pressure rise that occurs when the air is compressed in front of the object, which is generated in "Principle 1", proportionately depends on the velocity of the object.
However, the magnitude shown in the graph is for a plane perpendicular to the flow.
・ Speed 1224km / h is the propagation speed of sound on the spot.
I may be the only one who considers the concept of sound when considering the mechanism of lift generation of the wing.
Given that sound is a longitudinal wave of compression and expansion, I couldn't move forward without considering the sound.
<Way of thinking>
* Air has great compressibility, viscosity, and mass.
Considering an object moving at a speed of 100 km / h in an atmosphere of 1 atm,
Air is stretched behind the object, and on the surface of the surface perpendicular to the direction of movement
100/1224 = 0.082 atm (kgf / ㎠) drop,
Air is compressed in front of the object, and on the surface of the surface perpendicular to the moving direction
100/1224 = 0.082 atm (kgf / ㎠) rise.
On a surface that is not perpendicular to the direction of movement, the force corresponding to the projected area is stretched, and the force exerted per unit area decreases from P to Psinθ.
Similarly, the velocity of the moving surface in the direction perpendicular to the plane decreases from V to Vsinθ.
In the calculation described later, the force per unit area is calculated when the moving speed of the plane in the direction perpendicular to the plane is reduced to Vsinθ.
<Let's calculate the lift and drag of a light aircraft based on the above theory>
Imagine a high-wing Cessna aircraft type with a simple shape.
According to Principle 1, the wing area is calculated to be 4.9 x 2 and the chord length is 1.45 m.
The figure below shows the outline of the wing shape.
<Case 1>
The wing cross section is constant as shown in the figure below, and is a rectangular wing.
LH is the horizon, point T is the highest point of the wing, U is the tip, S is the last point, which is also the lowest point in this attitude.
This airplane is now gliding on a runway at an altitude of 10m at a speed of 80km / h, and the angle of attack for a moment when the nose is lifted is 5 °.
The lift generated by the wings in Fig. 13 is considered to be equivalent to the lift generated by each surface of 9.8 m in length with the triangular STU surrounded by the lines ST, TU, and US in Fig. 14 as the cross section.
This is an important point in my opinion.
Points S, T, and U are important significance points.
1) Lift calculation on surface ST
The area of the surface ST is 1.12 x 9.8 = 11㎡, and the speed of the surface ST in the direction perpendicular to the surface (direction perpendicular to the surface) is
80km / h × sin10.5 ° = 14.6km / h
Therefore, the force acting at right angles to this surface is
(Vacuum degree at moving speed in the direction perpendicular to the plane) x (Force per unit area at the time of vacuum on the spot) x Area.
(Velocity in the direction perpendicular to the plane / speed of sound on the spot) x (atmospheric pressure on the spot / 1013hPa) x area.
In other words
(14.6km / h / 1224km / h) × (1012hPa / 1013hPa × 1kgf / ㎠ × 10000㎠ / ㎡) × (11㎡) = 1311kgf
1012hPa is the atmospheric pressure on the spot (elevation 10m).
Lift acting on surface ST (vertical force) is 1311kgf x cos10.5 ° = 1289kgf
The drag of the surface ST is 1311kgf x sin10.5 ° = 239kgf.
2) Surface TU
Area is 0.35 x 9.8 = 3.43㎡, elevation angle is -10 °, straight speed is 80km / h x sin10 ° = 13.9km / h
Therefore, the force acting at right angles to this surface
(13.9km / h / 1224km / h) × (1012hPa / 1013hPa × 1kgf / ㎠ × 10000㎠ / ㎡) × (3.43㎡) = 389kgf
Therefore, the lift (vertical force) is 389 kgf x cos10 ° = 383 kgf (downward).
Also, the drag is 489kgf x sin10 ° = 68kgf
3) Similarly, for surface US
Area is 1.45 x 9.8 = 14.2㎡, angle of attack is + 5 °, plane speed is 80km / h x sin5 ° = 7.0km / h
Therefore, the force acting at right angles to this surface
(7.0km / h / 1224km / h) × (1012hPa / 1013hPa × 1kgf / ㎠ × 10000㎠ / ㎡) × (14.2㎡) = 811kgf
Therefore, the lift is 811kgf x cos5 ° = 808kgf.
The drag is 811kgf x sin5 ° = 71kgf.
Therefore, in the whole wing
Lift: 1289-383 + 808 = 1714kgf, Drag (wing): 239 + 68 + 71 = 378kgf
Also, since the aircraft and tail wings other than the main wing are in the high-speed flow generated by their own propellers, the relationship between the flight speed and the main wing as described above should be clearly different.
I was wondering what to do with this treatment, but I interpreted this calculation method as being able to neglect the drag of the cabin in the high-speed current generated by my propeller.
Obviously, if the projected area of the landing gear outside the high-speed flow by the propeller is 0.2 m2, the drag force will be that much.
80/1224 × 1012/1013 × 10000 × 0.2 = 131kgf (drag in the case of flat plate)
Although it depends on the streamlined condition of the stays and tires, the drag force other than the wings is 131 x 0.35 because about 35% of the drag force of the flat plate for the projected area is the actual drag force from the numerical value derived from the author's experiment. = 46kgf
It can be calculated as.
Therefore, the total drag is 378 + 46 = 424kgf.
In other words, the lift when this airplane is gliding at 80km / h and the nose is raised to an angle of attack of 5 ° is 1714kgf.
If the airplane weighs a little over 1 ton now, it may be the moment when takeoff starts.
The calculation result of Case 1 is shown in the table below.
<Case 2>
Next, this small airplane is flying over 500m at a speed of 150km / h, and the angle of attack of the wings at that time is 3 °.
The atmospheric pressure above 500m is 955hPa, the temperature is 12 ℃, and the speed of sound is 1219km / h.
The positional relationship of each significant point is as shown in the figure below.
The rear end of the wing is S, the apex of the wing is T, the front end of the wing is U, and the lowest point of the wing is V.
The lift generated by the wing in the above figure is considered to be equivalent to the lift generated by a wing with a length of 9.8 m with a square STUV as the cross section.
Therefore, the lift in each aspect is calculated.
1) Lift calculation on surface ST
Since the area is 1.07 x 9.8 = 10.5㎡ and the angle of attack of the surface ST is 8 °, the speed in the direction perpendicular to the surface ST (on the surface) at a flight speed of 150km / h is
150km / h x sin8 ° = 20.9km / h
The atmospheric pressure above 500m is 955hPa, the speed of sound is 1219km / h, and the force acting on the surface ST is
20.9km / h / 1219km / h x 955hPa / 1013hPa x 1kgf / ㎠ x 10000㎠ / ㎡ x 10.5㎡ = 1697kgf
In the vertical direction, 1697 x cos8 ° = 1680kgf,
Drag is 1697kgf x sin8 ° = 1697 x 0.139 = 236kgf.
2) Similarly for the surface TU
Area is 0.4 x 9.8 = 3.92㎡, angle of attack is minus 11.3 °
The straight speed is 150km / h x sin11.3 = 29.4km / h
The force acting on the surface TU is 890 kgf (downward), the vertical direction is -873 kgf (downward), and the drag is 174 kgf.
3) About surface UV
Area is 0.95 x 9.8 = 9.31㎡, angle attack is 5.3°
The straight speed is 150km / h x sin5.3° = 13.8km / h
The force acting on the surface UV is 994 kgf, the vertical direction is 990 kgf , and the drag is 91 kgf.
4) About surface VS
Area is 0.51 x 9.8 = 5.0㎡, angle of attack is minus 2 °
The straight speed is 5.25km / h
The force received by the surface VS is 203 kgf (downward), -203 kgf (downward) in the vertical direction, and the drag force is 7.0 kgf.
Therefore, in terms of overall lift,
1680-873 + 990-203 = 1594kgf
The total drag of the wings is 236 + 174 + 91 + 7 = 508kgf.
The drag force other than the wings has a projected area of about 0.2 m2.
150km / h / 1219km / h x 955hPa / 1013hPa x 1kgf / ㎠ x 10000㎠ / ㎡ x 0.2㎡ = 232kgf
Taking 35% of that, 81kgf
Therefore, the drag of the entire airplane is 508 + 81 = 589kgf.
The lift when this light aircraft is flying at 150km / h in the attitude shown in the conditions is 1594kgf.
The results of Case 2 are shown in the table below.
I calculated under the following conditions.
<Case 3>
Altitude 2500m, barometric pressure 747hPa, speed of sound 1196km / h, flight speed 180km / h, angle of attack 3 °.
<Case 4>
Altitude 3000m, atmospheric pressure 701hPa, speed of sound 1181km / h, flight speed 180km / h, angle of attack 3 °.
The above calculation results are shown in the table below.
In case 4, the lift was 1451 kgf. If the weight of this airplane is a little over 1 ton, it seems that the ceiling is about 4000m.
In order to further increase lift, it is necessary to increase the speed or wingspan, or to increase the angle of attack by turning the attitude upward, but in each case the drag increases, so the propulsive force corresponding to it is required.
Actually, it is necessary to accurately determine the turbulence of the airflow caused by the propeller and the effect of the cabin and outriggers on the underside of the wing through experiments.
It is also possible that the viscosity of the air and the treatment of the boundary layer affect the calculation formula, but this is not my knowledge.
About the drag of the cabin.
Since the airplane itself is exposed to the high-speed flow caused by the propeller, the relationship between flight speed and wings should definitely be different.
Therefore, it was judged that the drag force of the main body of the aircraft can be ignored with this calculation method, but is this okay? I don't know.
* Actually, the wing tip of the Cessna has a slight roundness, but in the calculation, it has a shape with sharp corners as if it was cut off. Therefore, I think the calculation result is slightly larger.
* In my opinion, the air flow around the wing is guided by the pressure difference generated there, so even if the wraparound at the wing tip is reduced on a vertical surface such as a winglet, the lift will not increase.