In this context, it means a ripple in the air caused by a wake vortex from the wing of another plane.
The .3g means that it exerted an acceleration force of 3/10ths the force of gravity (32 feet/sec acceleration).
Can someone do a quick calculation and determine the lateral acceleration generated generated by a 20 mph right turn around the corner? I think it would help people understand the absurdity of this claim.
Can someone do a quick calculation and determine the lateral acceleration generated generated by a 20 mph right turn around the corner? I think it would help people understand the absurdity of this claim.It depends on the radius of the curve. The signs on sharp curves are generally posted to keep you under 0.25g. So, the 0.3g figure is not much higher than following the speeds posted on the signs exactly. Most people when driving will feel comfortable taking a corner at 0.4g to 0.5g. Most SUV's can generate 0.6g to 0.8g of cornering force. Automobiles generally can attain 0.7g to 0.9g, with the best sports cars capable of 1.0g or a little more on street tires.
Let's see...for a 40 ft turning radius:
(20 m/hr)*(5280 ft/sec)*(1 hr/3600 sec) = 30 ft/sec
a = v*v/r = (30 ft/sec)*(30 ft/sec)/(40 ft) = 23 ft/sec^2,
or about .7 g.
All figures approximate to 20 sig. figs.