Accelerometer Frequently Asked Questions

Measuring "vibration level" requires the use of a "vibration level meter." Vibration level is calculated by correcting the vibration signal for human vibration perception (frequency correction followed by level conversion <finding the effective value>), determining how many times greater the value is than ^10⁻⁵ m/ ^s², and then displaying the logarithm of that value. The electrical configuration is the same as that of a sound level meter, so please refer to "VR-6100 Vibration Level Meter" for details. A level converted without frequency correction is called "vibration acceleration level."

Since frequency correction is not possible in this case, we will explain a method of "converting to an equivalent vibration acceleration level." However, this method does not involve the level conversion process defined by JIS or other standards. Therefore, please treat it only as a reference value.

<Calculation example>

For example, if the voltage value is 1.39 mV (PP), then when converted to 0-P,

P-P = 1.39 mV ÷ 2 = 0.695 mV （V_0-P）

When you convert voltage to vibration acceleration B m/ ^s²​

B = 0.695 ÷ A （ m/s²）

A = Sensitivity of NP-3120 mV / (m/ ^s²) If we set this to 1.00 mV,

B = 0．695／1 m/s² = 0.695 （ m/s² _0−P）

Converting this to the effective value C,

C = B／√2 = 0.695／√2 = 0.491 （ m/s² rms）

To convert this to vibration acceleration level D, take the logarithm of the ratio with ^10⁻⁵ m/ ^s².​

D = 20 Log （0．491／10⁻⁵） = 20 Log （0.491） + 100= 93.8 dB

This is the result. (Since time waveforms consist of various frequencies, frequency correction is not possible, and therefore they cannot be represented by vibration levels.)

That's correct. Please calculate using a sensitivity of A that is 10 times higher (= 10.0 mV) than in the above <Calculation Example>.