every vehicle has a "wall." this is when road load forces (namely aerodynamic drag) become equal to the accelerative force exerted by your rear tire on the road surface.
if you know the torque curve of the bike via a dyno ("at the wheels"), then it's just a matter of dividing the torque values by the moment arm (radius of the rear tire)...this will give you your accelerative force.
if you know the aerodynamic properties of the bike (Cd and A), you can then determine the aerodynamic drag at various speeds.
then, it's just a matter of finding the point at which the aero drag overcomes your accelerative force.
*the below taken from "Fundamentals of Vehicle Dynamics" by Gillespie. This was the class text for my Automotive Engineering class in college.
Da=(1/2)*rho*V^2*Cd*A
where:
Da=aerodynamic drag
rho=air density
V=velocity
Cd=aerodynamic drag coefficient
A=frontal area of the vehicle
rho=0.00236*(Pr/29.92)*(519/(460+Tr))
where:
Pr=atmospheric pressure in inches of mercury
Tr=air temperature in degrees Fahrenheit
*at standard conditions (59 degrees F and 29.92 inches of Hg), rho is simply 0.076 lb/ft^3...divide by gravity (32.2 ft/sec^2) for mass density and get 0.00236 lb*sec^2/ft^4
PS - if someone would be so kind as to post the Cd and A for the XX, i'll run the calcs and post graphs and whatnot.