I am putting together some simple ride comfort models that use road surface PSDs from ISO 8606 (road surface profiles) as inputs to a quarter car suspension model to produce body acceleration spectra. These are then weighted according to ISO 2631 (human exposure to whole body vibration).
The road surface data is vertical displacement PSDs referencing spatial frequency (wave number; inverse of wavelength). The suspension model is simply a gain of body displacement per unit of ground displacement over a range of temporal frequencies. The input to the weighting process needs to be an RMS acceleration over a range of temporal frequencies. So, the problem is to convert the road surface profile equation from a PSD against spatial frequency to an RMS against temporal frequency.
There may be a number of ways of achieving this but the one I seem to be having most success with so far is:
- Convert road surface displacement PSD to acceleration PSD (given in ISO 8608)
- Convert acceleration PSD from spatial frequency to temporal frequency (some clues in ISO 8608)
- Convert acceleration PSD to acceleration RMS (method readily available on the net)
- Weight acceleration RMS using filter described in ISO 2631.
The specific issue I want to ask about is one of units when completing steps 2 and 3.
In step 1, the road surface profile is created using the following equation: $$G_d (n)=G_d (n_0 )\cdot(n/n_0)^{-w}$$ Where:
$G_d(n)=displacement\,PSD,\,m^2/cycle/m\,(=m^3)$
$G_d(n_0 )=displacement\,PSD\,at\,reference\,spatial\,frequency,\,m^3$
$n=spatial\,frequency,\,cycle/m$
$n_0=reference\,spatial\,frequency,\,cycle/m$
$w=fitting\,exponent$
This is a straight line on a log-log plot with slope -w. The ISO standard shows that the acceleration PSD is:
$$G_a(n)=(2πn)^4\cdot G_d(n)$$
Where:
$G_a(n)=acceleration\,PSD,\,1/m$
This gives a straight line on a log-log plot of slope w.
To convert spatial frequency to temporal frequency I need to select an appropriate vehicle velocity and use the relationship:
$$f=n\cdot v$$
Where:
$f=frequency,\,cycle/s$
$v=velocity,\,m/s$
At this point I make the assumption that my PSD will have the same values at each input of temporal frequency as it did when given the equivalent spatial frequency multiplied by speed. To clarify, if the acceleration PSD equation returned a value of 2.7 for a spatial frequency input of 1.62 (and using a reference output of 6.55E-02 and fitting exponent of 2), then it should also return a value of 2.7 when given a temporal frequency of 36.1 at a speed of 22.22 m/s.
This allows me to plot the acceleration PSD against two individual independent variables – spatial and temporal frequency. An example of this is in the following image, where the speed was chosen to be 22.22 m/s (80 km/h).
If this thinking is correct, then the acceleration PSD equation becomes:
$$G_a(n)=(2π f/v)^4\cdot G_d(n_0)\cdot ((f/v)⁄(n_0))^{-w}$$
The complication I’m having is in understanding what the units of the PSD are. When the RMS is calculated in step 3 the process involves finding the area under the PSD within given frequency bands. After this calculation the units are those of the dependent variable multiplied by those of the independent variable. The result is then square rooted to give the RMS level in units of, in this case, acceleration in $m/s^2$.
To make this work, the units of the PSD would need to be:
$$(m/s^2)^2/(cycle/s)$$
When multiplied by the independent variable of temporal frequency, the units would then be:
$$(m/s^2)^2/(cycle/s)\cdot cycle/s=(m/s^2)^2$$
Then once square rooted the units would be correct for RMS acceleration. My conundrum is that the units for the acceleration PSD don’t seem to have been converted and remain as 1/m. After going through the process of calculating the RMS the units would end up as:
$$1/m\cdot cycle/s=1/ms$$
Square rooting would then give units of:
$$(ms)^{-0.5}$$
This obviously isn’t correct so where have I gone wrong?
Thanks.
