Throwing some light on rates of turn

[Holtzauge]

Originally Posted by Crumpp

We are not going to do another 100 pager because you lack formal education in aerodynamics.

EAS is the most common expression for velocity in all aircraft performance calculation. It is the preferred expression because it is so simple to use.

It is too easy to convert to TAS any performance derived with EAS and you don't have worry about density effects in the mechanics of the calculation. Just convert at the end.

It also a great approximation of Indicated Airspeed and very easy to convert to that with a PEC chart and a universal compressibility.






http://www.google.com/url?sa=t&rct=j...o9yenlVuG8g5Zw







If you are trying to quickly gauge relative performance you don't have to convert back to TAS. The specific numbers for rate and radius will change in proportion to density ratio which is a universal application.

The problem is that your turn time of 21 s at 20,000ft is physically impossible. No amount of posturing and posting irrelevant book quotes underlined in red will change that fact:

You have claimed R=900 ft turn radius and turn time T=21 s at 20,000 ft:

Since I'm a metrics guy I will convert R to SI units, i.e. 274.3 m

This gives a turn speed of 82.08 m/s (2*pi*R/T)

So from this we calculate the turn acceleration: a=v**2/R=24.56 m/s**2

So load factor is n= sqrt(a**2+g**2)/g=2.696

Let's calculate the Cl this would require:

n*m*g=0.5*ra*v**2*Cl*S

Spitfire data:

W=6000lb=2724 Kg
S=242 sqft=22.36 m**2
ra=0.65 (Approx at 6.1 Km alt)

Solving for Cl:

CL=(2.696*2724*9.81)/(0.5*0.65*82.08**2*22.36)=1.47

Now NACA claims Clmax for the Spitfire at 1.2 which is a bit low but according to RAE it is 1.36 tops. Your claim leads to a Cl of 1.47 which is clearly unrealistic and like you fails the sanity check.

BTW: I found a RAE report, R&M 2349, Notes on the turning performance of the Spitfire as affected by altitude and flaps.

On page 4 there is a figure 4 which gives the following results for the Spitfire at 20,000 ft: R=1045 ft and T=31.5 s

With my C++ simulations I get R=337 m (1106 ft) and T=31.65 s.

You claim 21 s turn time and 900 ft radius of turn. I get 31.65 s and 1106 ft while Morgan & Morris in R&M get 1045 ft and 31.5 s.

So on the one hand we have C++ simulation data and the data from the RAE report R&M 2349 which seems to tally and on the other hand we have your overbearing attitude and simplistic calculations leading to an off the chart Clmax. What could be the right number I wonder , 21 or 31 s?

Finally, I think the only thing we actually agree on is the other parties lack of formal aerodynamic training. We have been down this road before and as I've told you before I have an Mcs in aeronautical engineering from the Royal Institute of Technology in Stockholm from 1986 and more than 10 years in the business working in the defense industry for Ericsson and SAAB on the Viggen and Gripen fighter systems.

Tell me, What aeronautical companies have you worked with and the Msc in aeronautics from Embry-Riddle you claim to have, which year did you graduate and was that before or after your stint in US Special Forces?