izax1:
I believe they are all the same. I have heard all three terms used interchangeably. Typically, they are used in fatigue analysis or determination. People in our lab typically perform sinusoidal vibration to obtain the first three most prominent resonant frequencies (fn) and transmissibilities (Q). These are typically the frequencies with the highest transmissibilities. Then people in the lab perform random vibration tests but always calculate the "Zero Positive Crossings" first.
SINUSOIDAL TESTING
Get fn(1-3) and Q(1-3). The number of fatigue cycles accumulated during a resonant dwell of a sinusoidal test is (resonant frequency x dwell time).
RANDOM TESTING
The number of fatigue cycles in a random test is harder to calculate. The number of fatigue cycles must relate to the frequencies that do the most damage, however, all of the exciting frequencies are present at once.
S. O. Rice,"Mathematical Analysis of Random Noise",pp. 353-461, offers a great explanation of zero crossings both + and -. It is defined in terms of per unit time or a rate, so the units are Hz. (Integration based equation.)
D. Stienberg,"Vibration Analysis For Electronic Equipment," further defines a positive zero crossing. The number of POSITIVE ZERO CROSSINGS (N0+), is the number of times, on average that the displacement trace crosses the zero axis with a positive slope.
For lightly damped structures with several resonant peaks, D. Stienberg offers a relatively simple theoretical equation that our lab uses. It is based on the assumption that each resonant peak acts as a single degree of freedom. On lightly damped structures we have used it with some success.
N0+=(1/(2 x pi))x (SQRT(numerator/denominator)) Hz
numerator
a/b + c/d + e/f)
a
pi/2)x(PSD @ fn1)x Q1
b:[(2 x pi) x (fn1)]^2
c
pi/2)x(PSD @ fn2)x Q2
d:[(2 x pi) x (fn2)]^2
e
pi/2)x(PSD @ fn3)x Q3
f:[(2 x pi) x (fn3)]^2
denominator
g/h + i/j + k/l)
g
pi/2)x(PSD @ fn1)x Q1
h:[(2 x pi) x (fn1)]^4
i
pi/2)x(PSD @ fn2)x Q2
j:[(2 x pi) x (fn2)]^4
k
pi/2)x(PSD @ fn3)x Q3
l:[(2 x pi) x (fn3)]^4
D. Stienberg EXAMPLE
IF:
fn1:138 Q: 5.0
fn2:279 Q:17.0
fn3:534 Q:4.0
THEN:
N0+ = 196 Hz
Further note that there are more peaks than Zero Positive Crossings. Why? The displacement will sometimes reverse before crossing the zero axis.
Finally, The +zero question was posed to T. Paez during a "Random Vibrations" course(1998) involving the text you have mentioned above. As I recall, he agreed with the explanation.
Hope this helps!
Kaiserman
