Originally posted by Crumpp
Is not a finite wing. It's an area. Any shape can have an equal area. Finite defines the shape of the wing. Not just a random area.
Well, you have a logical problem here, if wing span is infinite, then also area is infinite and if wing area is finite then also wing span is finite.
Originally posted by Crumpp
If your not including the effects of induced drag along with the AR, it's not 3D.
Here you have another logical problem, the formula:
Cl = L / (A * .5 * r * V^2)
Is simply for Cl in 3D because it assumes finite wing. There is no need to know induced drag nor wing geometry for calculating required Cl at given conditions.
Please follow the
link justin_g gave and see how it is used to calculate 3D lift coefficient:
"
We can use this relationship to rearrange the lift equation and apply it in a new way. Since lift must equal weight for the plane to fly level, lift becomes a known value that we can use to solve for the lift coefficient. The lift equation then becomes:
We can again use the values provided above for the 747 to solve this equation.
Lo and behold, the lift coefficient is 0.52, exactly the value provided in the original data. This application of the lift equation may seem backwards, but engineers often use it during the preliminary design process of a new aircraft. A cruise speed as well as an approximate weight and size are typically specified when the design effort begins. These values can be used in the above equation to solve for the lift coefficient necessary to maintain cruise flight at those conditions. Once that coefficient in known, designers can determine what wing shape and airfoil section will best provide that lift coefficient while minimizing drag. Other factors obviously come into play since an aircraft does not spend its entire flight at steady and level conditions, but this technique is a common first step in the initial sizing and design of a new flying vehicle."
Note that induced drag and AR (ie wing geometry) are not needed to calculate required Cl.
Originally posted by Crumpp
Right. Fortunately people can read the thread. If they want to muddle through it.
Well, fortunately the original thread is
here but to save time of readers, here are the relevant parts from Wood's book (Karl D. Wood (1935):"Technical Aerodynamics"). Wood described a system to estimate e factor of an airframe by splitting e factor to two parts:
ew = efficiency of the wing
ef = efficiency of the fuselage
The e factor being:
e = ew * ef
For the wing (ew) Wood gave following chart:
The formula seen in the chart is for rectangular wing and for wing only. The problem was that it was claimed that this formula can be used alone (without ef) to determine the e factor of an airframe which is wrong way to use it. After some discussion that was sorted out, except in the case of the Mr. Crumpp who apparently still believes that there is some kind of generalized formula to calculate e factor from AR alone.
Originally posted by Crumpp
Not unless he is using instrumentation and a windtunnel.
Yet another logical problem; required Cl can be calculated simply knowing the lift, speed, density and wing area without instrumentation or wind tunnel. See the 747 example linked above.
gripen
