It turned out that it was not really difficult to show that the assumptions in the report are perfectly valid. In the little octave/Matlab program below the formulae for the velocities in potential flow around a cylinder are implemented. The components in radial (uR = (1 - (R/r)^2)*Uo*cos(thta); and circumferential (uthta=-(1 + (R/r)^2)*Uo*sin(thta)) direction are calculated and then transformed to a body fixed cartesian coordinate system uX (flow parallel to the rotor disk) and uY (flow perpedicular to the rotor disk). Since we are interested in the velocities far away from the cylinder the flow can be assumed inviscid and the solution of the potential flow equations gives a very good approximation to the real flow speeds. The height of the rotor head for a Cavalon is about four times the fuselage radius (r=4*R) and a point on the windward side 60° from the negative x axis was calculated. Here the distance to the fuselage is 4.0/sin(60°) = 4.662*R. For a lateral gust of 25mph = 40km/h = 11.11 m/s the flow velocity at this point through the rotor disk is 0.45 m/s. At a flight speed of 60mph the constant induced velocity through the rotor is 1.18m/s thus the upward flow speed ist almost 40% of the induced velocity, which, of course, *is* significant. This actually is a lower bound since, due to the boundary layer, the distance of the flow cylinder to the rotor head is smaller. Assuming a thickness of the boundary layer of 5% the distance to the point of interest now is 4.4*R and the flow perpendicular to the rotor is 0.5 m/s

The important result line looks like this

Uo= 11.11 m/s -> uX= 11.40 uY= 0.45

The octave/Matlab program is this one:

clc

clear all

R = 0.50;

rN= 4.0*R;

psX = 60;

fprintf('\nrN = %3.2f*R\n',rN/R);

thta2= psX*pi/(180);

r = rN/sin(thta2);

fprintf('r = %3.2f*R\n',r/R);

thta = pi-thta2;

Uo=1.0;

uR = (1 - (R/r)^2)*Uo*cos(thta);

uthta=-(1 + (R/r)^2)*Uo*sin(thta);

uX = uR*cos(thta) - uthta*sin(thta);

uY = uR*sin(thta) + uthta*cos(thta);

fprintf('uthta %10.3f\n' , (uthta));

fprintf('uR %10.3f\n' , (uR));

fprintf('uRes %10.3f\n',sqrt(uR^2 + uthta^2))

fprintf('uX %10.3f\n' , (uX));

fprintf('uY %10.3f\n' , (uY));

uRes=sqrt(uX^2 + uY^2);

fprintf('uRes %10.3f\n',uRes)

Uo= 40/(3.6);

fprintf('Uo= %6.2f m/s -> uX= %8.2f uY= %8.2f\n',Uo,uX*Uo,uY*Uo);

fprintf('uRes %10.3f\n',uRes*Uo)