flyingcheesehead
Touchdown! Greaser!
A friend pointed me towards a book today that I think contains an error. The book is "The Illustrated Guide to Aerodynamics" by Hubert Smith.
Check this page on Google Books: http://books.google.com/books?id=C8...page&q=ground roll+weight of aircraft&f=false
The last paragraph above "Jet Aircraft Performance" reads:
Emphasis mine - That does not seem to be correct to me.
First, the friction part of the equation: f = µsFn where µs is the static coefficient of friction, and Fn is the normal force, which will vary with weight. So it is true that there will be more friction with the heavier airplane, and all else being equal the bolded statement above would be true. So, say we have a 3000 pound airplane and a 2500 pound airplane of the same type on the same runway - According to Wikipedia, µs of rubber on concrete is 1.0, so f would equal 3000 lbf and 2500 lbf respectively.
However, the purpose of the landing roll is to dissipate kinetic energy. KE = (1/2)mv^2. Now, if we assume that velocity is equal, then the author of the book would appear to be correct as the difference in mass creates a linear relationship in both kinetic energy and friction, and they would in fact "cancel out."
However, the lighter airplane will be able to maintain the same angle of attack at a slower airspeed, so landing at the same angle of attack (be it critical for a full-stall landing or a degree or two shy of critical), v would vary with the square root of the proportion of the weights. So, if we pick Vo as the velocity that the ligher airplane can land at, the heavier airplane will land at Vo(sqrt(3000/2500)), or 1.095Vo. So, the kinetic energy of the lighter airplane would be 1250Vo^2 and the KE of the heavier airplane would be 1500(1.095Vo)^2 = 1798.5Vo^2.
Since the kinetic energy of the heavier airplane is 43.88% greater than that of the ligher airplane, and its braking force available is only 20% greater, it seems to me that the landing roll would in fact be longer with the heavier airplane.
So, who's right - Me, or the author of the Illustrated Guide to Aerodynamics?
Check this page on Google Books: http://books.google.com/books?id=C8...page&q=ground roll+weight of aircraft&f=false
The last paragraph above "Jet Aircraft Performance" reads:
Emphasis mine - That does not seem to be correct to me.
First, the friction part of the equation: f = µsFn where µs is the static coefficient of friction, and Fn is the normal force, which will vary with weight. So it is true that there will be more friction with the heavier airplane, and all else being equal the bolded statement above would be true. So, say we have a 3000 pound airplane and a 2500 pound airplane of the same type on the same runway - According to Wikipedia, µs of rubber on concrete is 1.0, so f would equal 3000 lbf and 2500 lbf respectively.
However, the purpose of the landing roll is to dissipate kinetic energy. KE = (1/2)mv^2. Now, if we assume that velocity is equal, then the author of the book would appear to be correct as the difference in mass creates a linear relationship in both kinetic energy and friction, and they would in fact "cancel out."
However, the lighter airplane will be able to maintain the same angle of attack at a slower airspeed, so landing at the same angle of attack (be it critical for a full-stall landing or a degree or two shy of critical), v would vary with the square root of the proportion of the weights. So, if we pick Vo as the velocity that the ligher airplane can land at, the heavier airplane will land at Vo(sqrt(3000/2500)), or 1.095Vo. So, the kinetic energy of the lighter airplane would be 1250Vo^2 and the KE of the heavier airplane would be 1500(1.095Vo)^2 = 1798.5Vo^2.
Since the kinetic energy of the heavier airplane is 43.88% greater than that of the ligher airplane, and its braking force available is only 20% greater, it seems to me that the landing roll would in fact be longer with the heavier airplane.
So, who's right - Me, or the author of the Illustrated Guide to Aerodynamics?
