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From: DaZ on 20 Feb 2010 18:29
I am looking at sinusoidal motion (shaker tables) and am trying to
calculate peak displacement given RMS acceleration. I can do so using:

a_peak = 4 * pi^2 * f^2 * x_peak
a_rms = 0.707 * a_peak

as: x_peak = a_rms / (0.707 * pi^2 * f^2 * 4)

But isn't there a way to do this without involving frequency? I
thought I would be able to use 1/4 cycle at the RMS acceleration to
find the peak. I even verified that the RMS value of a sinewave is
still 0.707 even over only 90 degrees.

Here is what I was trying:

x_peak = 1/2 * a_rms * (1/120)^2 * 1.414

This doesn't seem to be correct, however. Can anyone help me figure
out why?

Thanks.

From: Cwatters on 21 Feb 2010 11:55

"DaZ" <davidd31415(a)gmail.com> wrote in message
news:df824845-1f18-4150-aefa-58dab84dd992(a)v1g2000yqk.googlegroups.com...
>I am looking at sinusoidal motion (shaker tables) and am trying to
> calculate peak displacement given RMS acceleration. I can do so using:
>
> a_peak = 4 * pi^2 * f^2 * x_peak
> a_rms = 0.707 * a_peak
>
> as: x_peak = a_rms / (0.707 * pi^2 * f^2 * 4)
>
> But isn't there a way to do this without involving frequency? I
> thought I would be able to use 1/4 cycle at the RMS acceleration to
> find the peak. I even verified that the RMS value of a sinewave is
> still 0.707 even over only 90 degrees.
>
> Here is what I was trying:
>
> x_peak = 1/2 * a_rms * (1/120)^2 * 1.414
>
> This doesn't seem to be correct, however. Can anyone help me figure
> out why?
>
> Thanks.
>

You can't. Consider a system with fixed displacement but variable frequency.
It seem reasonable to expect the average acceleration to increase as the
frequency of oscillation increases.

I might be wrong as it's been years but...

x = x_peak.sin(wt)
v = x_peak.w.cos(wt)
a = -x_peak.w^2.sin(wt)

a_rms = x_peak.w^2/sqrt(2)

so
x_peak = a_rms.sqrt(2)/w^2

and w=2.p.f







From: DaZ on 21 Feb 2010 15:15
On Feb 21, 11:55 am, "Cwatters"
<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
> "DaZ" <davidd31...(a)gmail.com> wrote in message
>
> news:df824845-1f18-4150-aefa-58dab84dd992(a)v1g2000yqk.googlegroups.com...
>
>
>
>
>
> >I am looking at sinusoidal motion (shaker tables) and am trying to
> > calculate peak displacement given RMS acceleration. I can do so using:
>
> > a_peak = 4 * pi^2 * f^2 * x_peak
> > a_rms = 0.707 * a_peak
>
> > as: x_peak = a_rms / (0.707 * pi^2 * f^2 * 4)
>
> > But isn't there a way to do this without involving frequency? I
> > thought I would be able to use 1/4 cycle at the RMS acceleration to
> > find the peak. I even verified that the RMS value of a sinewave is
> > still 0.707 even over only 90 degrees.
>
> > Here is what I was trying:
>
> > x_peak = 1/2 * a_rms * (1/120)^2 * 1.414
>
> > This doesn't seem to be correct, however. Can anyone help me figure
> > out why?
>
> > Thanks.
>
> You can't. Consider a system with fixed displacement but variable frequency.
> It seem reasonable to expect the average acceleration to increase as the
> frequency of oscillation increases.
>
> I might be wrong as it's been years but...
>
> x = x_peak.sin(wt)
> v = x_peak.w.cos(wt)
> a = -x_peak.w^2.sin(wt)
>
> a_rms = x_peak.w^2/sqrt(2)
>
> so
> x_peak = a_rms.sqrt(2)/w^2
>
> and w=2.p.f- Hide quoted text -
>
> - Show quoted text -

I was thinking that the time factor would take care of any average
acceleration changes due to frequency. In the example above I
multipled by 1/120 sec in order to find the displacement during a
quarter of a 30 Hz sinewave.

From: Cwatters on 22 Feb 2010 06:37

"DaZ" <davidd31415(a)gmail.com> wrote in message
news:911b98f9-d59b-47f5-b9e5-aecdb589c87c(a)l26g2000yqd.googlegroups.com...

>I was thinking that the time factor would take care of any average
>acceleration changes due to frequency. In the example above I
>multipled by 1/120 sec in order to find the displacement during a
>quarter of a 30 Hz sinewave.

x_peak = a_rms.sqrt(2)/w^2

isn't an equation for x it's x_peak

Putting aside equations for the moment... It should be obvious that if
something is oscillating faster and faster the average acceleration must
also be getting faster or the peak displacement smaller?

Try running up and down on a tennis court. Your body limits how fast you can
accelerate. You can do more widths than lengths in 1 min. The frequency with
which you return to the start is higher when running shorter widths.


From: DaZ on 22 Feb 2010 19:19
On Feb 22, 6:37 am, "Cwatters"
<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:

>
> Putting aside equations for the moment... It should be obvious that if
> something is oscillating faster and faster the average acceleration must
> also be getting faster or the peak displacement smaller?
>

Ok. If I hold a sinusoidal acceleration constant, say 3 Gpk, RMS
acceleration will be constant as well- 2.1 Grms.

Assuming displacement starts at x=0, it should always peak after 1/4
of a cycle, regardless of frequency, right?

If I consider RMS the 'average' acceleration here, then after 1/4 of a
cycle I would expect x to be the peak displacement; or at least the
peak RMS displacement.

x_rms = 1/2 * a_rms * (1/f)^2

would at least behave in the way you describe- decreasing as
acceleration increases.

I realize my thinking is wrong here, it just hasn't completely clicked
yet. I've used RMS for multiplying voltage and current together to
find RMS power, from which I can also find peak power. I still don't
completely see why this can't be done for position.

Thanks.
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