Finding peak displacement from RMS sinusoidal acceleration without using frequency...?
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From: Cwatters on 23 Feb 2010 04:07 "DaZ" <davidd31415(a)gmail.com> wrote in message news:206d271c-b1b8-4cc6-8077-29bf7b362b21(a)q29g2000yqn.googlegroups.com... >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? Yes >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. You loose me when you start talking about "peak RMS" The answer is to write the equations for displacement and then differentiate it to give the equation for velocity and acceleration. Then note that _frequency_ comes outside the trig function not _phase_
From: PD on 23 Feb 2010 09:48 On Feb 20, 5:29 pm, DaZ <davidd31...(a)gmail.com> wrote: > 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. RMS is a way of averaging out an oscillation. It basically says, what is the value of a *constant* function that would have the same area under it as the area under the sinusoidal curve? Thus, there is no such thing as an RMS "peak". If you want to know peak values, then you should work with the sinusoidal function itself rather than RMS values.
From: DaZ on 23 Feb 2010 13:16 On Feb 23, 9:48 am, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 20, 5:29 pm, DaZ <davidd31...(a)gmail.com> wrote: > > > > > > > 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. > > RMS is a way of averaging out an oscillation. It basically says, what > is the value of a *constant* function that would have the same area > under it as the area under the sinusoidal curve? Thus, there is no > such thing as an RMS "peak". If you want to know peak values, then you > should work with the sinusoidal function itself rather than RMS values.- Hide quoted text - > > - Show quoted text - I shouldn't have used the term 'RMS peak.' With electricity I have found peak from RMS. For instance, RMS power is RMS voltage * RMS current. Peak power is 1.414 * RMS power. Apparentley these concepts do not apply to kinematics of motion.
From: PD on 23 Feb 2010 13:31 On Feb 23, 12:16 pm, DaZ <davidd31...(a)gmail.com> wrote: > On Feb 23, 9:48 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Feb 20, 5:29 pm, DaZ <davidd31...(a)gmail.com> wrote: > > > > 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. > > > RMS is a way of averaging out an oscillation. It basically says, what > > is the value of a *constant* function that would have the same area > > under it as the area under the sinusoidal curve? Thus, there is no > > such thing as an RMS "peak". If you want to know peak values, then you > > should work with the sinusoidal function itself rather than RMS values.- Hide quoted text - > > > - Show quoted text - > > I shouldn't have used the term 'RMS peak.' With electricity I have > found peak from RMS. For instance, RMS power is RMS voltage * RMS > current. Peak power is 1.414 * RMS power. > > Apparentley these concepts do not apply to kinematics of motion. Actually, they do. RMS amplitude is amplitude/sqrt(2), and RMS acceleration is peak_acceleration/sqrt(2). Just remember that the real acceleration changes all the time in an oscillation, and it passes the (constant) RMS value of acceleration 4 times in any given cycle. But at 1/4 cycle, the real acceleration of the system is peak_acceleration, not the RMS acceleration.
From: DaZ on 23 Feb 2010 22:01
On Feb 23, 1:31 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 23, 12:16 pm, DaZ <davidd31...(a)gmail.com> wrote: > > > > > > > On Feb 23, 9:48 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Feb 20, 5:29 pm, DaZ <davidd31...(a)gmail.com> wrote: > > > > > 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. > > > > RMS is a way of averaging out an oscillation. It basically says, what > > > is the value of a *constant* function that would have the same area > > > under it as the area under the sinusoidal curve? Thus, there is no > > > such thing as an RMS "peak". If you want to know peak values, then you > > > should work with the sinusoidal function itself rather than RMS values.- Hide quoted text - > > > > - Show quoted text - > > > I shouldn't have used the term 'RMS peak.' With electricity I have > > found peak from RMS. For instance, RMS power is RMS voltage * RMS > > current. Peak power is 1.414 * RMS power. > > > Apparentley these concepts do not apply to kinematics of motion. > > Actually, they do. RMS amplitude is amplitude/sqrt(2), and RMS > acceleration is peak_acceleration/sqrt(2). Just remember that the real > acceleration changes all the time in an oscillation, and it passes the > (constant) RMS value of acceleration 4 times in any given cycle. But > at 1/4 cycle, the real acceleration of the system is > peak_acceleration, not the RMS acceleration.- Hide quoted text - > > - Show quoted text - But with electricity I can find the peak power using RMS votlage and RMS acceleration... With sinusoidal motion I can not find peak displacement using RMS acceleration and RMS position it seems. I'm not sure why I'm not bridging the gap here- I'm going to have to give this a rest and then try looking at it again at a later date I think. |