Good video! I like this guy and have seen other videos of his.
I do have an important correction. In my 50 years of investigating crashes, I have yet to see a dry level traveled public road surface with a static friction coefficient of 1.0 - 1.1. Many tests all around the world have been done over many years and the average level dry pavement friction coefficient is .75. An expert rider threshold emergency braking can attain close to, and sometimes greater than 1.0. This is not the same as cornering friction. Not talking about race track surfaces, or superelevation here - just your garden variety public dry level road surface.
f=F/W The coefficient of friction (f) is simply a ratio of the force (F) (needed to slide locked tires) divided by the weight (W) of the vehicle. If 3750 lbs of force is necessary to slide a 5000 lb vehicle, the friction value would be .75. Adding a "g" for gravity (32.2'/sec/sec) to this ratio turns the value into acceleration or deceleration since gravity is a force. Therefor, .75g is .75 X 32.2 = 24.15'/sec/sec, which is acceleration in the positive and deceleration in the negative.
The friction value of the road surface is not necessarily the same as g-force or lateral acceleration.
As a rider, when in hoon mode on a good clean road surface, I consider that .75 value as my friction threshold. I also am aware of the friction circle, which means any braking (especially trail) while in the curve decreases the available lateral road friction by the square root of the sum of the squares of the horizontal and lateral forces. Example: You are at a fair 30° lean, therefor your lateral acceleration is .70g (tan 30°). You trail brake slightly, developing .3g linear deceleration. Your available friction=sqrt/(.7 squared + .3 squared), or .76. You may be in trouble here!
Note that just a 5° reduction in lean (25°) reduces this to .47g. This is a .23g difference and with the same .3g trail braking you are at .56g and well within a safe margin.
As you reach the threshold of friction in a curve, remember that every extra angle of lean costs increasing more reduction in your available friction.
HTH