Vibration Profile Question

[stockdam]

Hi...sorry for cross posting but I was advised that this section may be also help.

Can anyone help to explain a profile in Mil-STd-810.

The profile is shown in Fig 514.5C-9.

It's a vibration profile for propellor aircraft.

L0 is the g2/hz value for the first prop frequency. The profile rolls off at 6db/octave. How do I work out the g2/hz for the other frequencies (harmonics)?

f1 is twice f0 so that's an octave. Does this mean the g2/hz at f1 is L0/2 or is it L0/4 - is the g2/hz rolling off at 6db/octave or is the g rolling off at this rate and therefore g2/hz is a factor of 4 per octave lower?

 

[stockdam]

I'm going to reply to my own post as I don't see how to edit it.

Since I was talking about PSD then I'm almost certain that a 6dB drop is a quarter of the PSD.

 

[CESSNA1]

STOCKDAM: There are two equations you need the first is for the ocrave band and the second for Decibels. the octave band is

(f2/f1)=2^n
then
octaves = n=(10^(f2/f2))*log(2)

and
Db = 20*log(w2/w1)

Where "log" denotes to the base 10

The slope of -6 is in Db/octave

Divide Decibels (Db) by Octave and equate it to -6 then solve for the unknown amplitude

Since this is a log-log scale the slopes that look linear are not, they are curved, so you cannot use straight trigonometry

The plot shows that the band is between 15 and 2000 Hz.

Regards
Dave

 

[stockdam]

Hmmm

I thought n=(log(f2/f1))/log(2)


and since we are talking PSD (power) then

dB=10*log(w2/w1)

????

 

[CESSNA1]

STOCKDAM: Congratulations you are correct. I apologize. I was doing it from memeory but by brain let me down

Sorry
Regards
Dave