Angle of Attack (Aoa) and the Lift Coefficient (Cl)

View previous topic View next topic Go down

Angle of Attack (Aoa) and the Lift Coefficient (Cl)

Post  Admin on Tue Dec 01, 2009 6:48 am

The angle at which the wings meet the flight path — more properly termed the geometric angle of attack — is near 16° at minimum controllable airspeed and around 2 to 5° when cruising at low altitudes; less at higher speeds, greater at higher altitudes. We will cover the close relationship between CL, angle of attack (aoa or alpha) and airspeed in the aerofoils and wings module.

The diagram shows a typical CL vs angle of attack curve for a light aircraft not equipped with flaps or high-lift devices. From it you can read the CL value for each aoa, for example at 10° the ratio for conversion of dynamic pressure to lift is 0.9.

Note that CL still has a positive value even when the aoa is –1°. This is because of the higher camber in the upper half of the wing; some highly cambered wings may still have a positive CL value when the aoa is as low as –4°. A light non-aerobatic aircraft pilot would not normally utilise negative aoa because it involves operating the aircraft in a high-speed descent, but we will discuss this further in the 'Flight at excessive speed' module.

Also note that the lift coefficient increases in direct relationship to the increase in angle of attack, until near 16° aoa where CL reaches a maximum of about 1.3 and then decreases rapidly as aoa passes 16°. A rule of thumb for light aircraft with simple wings is that each 1° aoa change — starting from –2° and continuing to about 14° — equates to a 0.1 CL change.

Also, it is not just the wings that produce lift. Parts of a well-designed fuselage — the aircraft body — can also produce lift and the vertical component of the thrust vector can supplement lift when that vector is angled upwards.

We can calculate CL for the Jabiru cruising at an altitude of 6500 feet and an airspeed of 97 knots (50 m/s). The wing area is very close to 8 m²:

• lift = weight = 4000 N
• r = 1.0 kg/m³ (the approximate density of air at 6500 feet altitude)
• V² = 50 × 50= 2500 m/s
• S = 8 m²

Lift = CL × ½rV² × S

Dynamic pressure = ½ ×1.0 × 2500 = 1250 N/m²

So, 4000 = CL × 1250 × 8, — thus — CL = 0.4.

All of the information here is accredited to http://www.auf.asn.au/groundschool/index.html#lift
avatar
Admin
Admin

Posts : 102
Info Rating : 1
Join date : 2009-11-30
Location : Melbourne, Australia

View user profile http://aviationtheory.forumotion.net

Back to top Go down

View previous topic View next topic Back to top

- Similar topics

 
Permissions in this forum:
You cannot reply to topics in this forum