About Vibration Isolation Calculator (Transmissibility)
The vibration isolation calculator sizes machine vibration isolators using single-degree-of-freedom theory. Enter the isolator static deflection (or the spring stiffness and supported mass) to get the natural frequency fn, then the machine speed in RPM or the disturbing frequency in Hz to get the frequency ratio r = f/fn, the transmissibility T, and the isolation efficiency (1 - T) x 100%.
Transmissibility includes viscous damping through the ratio zeta, so elastomeric mounts (zeta around 0.05) and damped spring mounts can be compared realistically. Isolation only occurs above r = sqrt(2); between r = 0.8 and 1.4 the tool flags a resonance condition, and below sqrt(2) it reports the amplification instead. A reverse mode returns the required static deflection and natural frequency for a target isolation percentage at a given disturbing frequency, which is how isolator catalogues are normally entered.
How It Works
- Choose how the natural frequency is defined: from the isolator static deflection delta in mm (fn = 15.76 / sqrt(delta_mm)), or from the total spring stiffness k in N/mm and the supported mass m in kg (fn = (1/2pi) sqrt(1000 k / m)).
- Enter the disturbing frequency directly in Hz or as machine speed N in RPM (f = N / 60), and the viscous damping ratio zeta (default 0.05 for elastomeric isolators).
- The calculator evaluates the frequency ratio r = f / fn and the damped transmissibility T = sqrt(1 + (2 zeta r)^2) / sqrt((1 - r^2)^2 + (2 zeta r)^2), then reports the isolation efficiency (1 - T) x 100% and classifies the operating point as isolation, amplification, or resonance.
- In reverse mode, enter a target isolation percentage: the tool inverts the undamped relation T = 1/(r^2 - 1) to get the required frequency ratio, natural frequency, and static deflection.
Worked Example
A fan running at 1500 RPM disturbs its base at f = 1500/60 = 25 Hz. Rubber-in-shear mounts deflect 25 mm under the fan weight, so the natural frequency is fn = 15.76 / sqrt(25) = 3.15 Hz and the frequency ratio is r = 25 / 3.15 = 7.93. With negligible damping the transmissibility is T = 1/(r^2 - 1) = 1/61.9 = 0.016, so only 1.6% of the disturbing force reaches the structure and the isolation efficiency is about 98.4%. Since r is far above sqrt(2) and well outside the 0.8-1.4 resonance band, the selection is a comfortable one.
Formulas
- Natural frequency from static deflection
fn = (1/2pi) * sqrt(g / delta) = 15.76 / sqrt(delta_mm) = 3.13 / sqrt(delta_in)- Natural frequency from stiffness and mass
fn = (1/2pi) * sqrt(k * 1000 / m)- Frequency ratio
r = f / fn, f = N / 60- Damped transmissibility
T = sqrt(1 + (2*zeta*r)^2) / sqrt((1 - r^2)^2 + (2*zeta*r)^2)- Isolation efficiency and reverse selection
eta = (1 - T) * 100%; r = sqrt(1/T + 1), fn = f / r, delta_mm = (15.76 / fn)^2
Standards & References
- Classical SDOF forced-vibration theory (Den Hartog, Mechanical Vibrations)
- ASHRAE Handbook -- HVAC Applications, ch. Sound and Vibration Control
- ISO 2017-1 resilient mounting of machines
Frequently Asked Questions
What is transmissibility and what value should I aim for?
Transmissibility T is the ratio of the force (or motion) transmitted to the supporting structure to the disturbing force generated by the machine. T = 0.1 means 10% is transmitted, i.e. 90% isolation. Typical HVAC and machine-mount practice targets a frequency ratio of at least 3, giving roughly 85-90% isolation, with critical installations aiming for 95% or more.
Why does isolation only start above a frequency ratio of sqrt(2)?
At r = sqrt(2) the transmissibility equals exactly 1 for any damping ratio, so the mounts neither help nor hurt. Below sqrt(2) the transmitted force is larger than the disturbing force (amplification), with the worst case at resonance r = 1. Only above sqrt(2) does T fall below 1, and it keeps decreasing as the ratio grows.
How does static deflection relate to the natural frequency?
For a linear isolator, the deflection under the supported weight fixes the natural frequency regardless of the actual mass: fn = 15.76 / sqrt(delta_mm), or 3.13 / sqrt(delta_in). This is why isolator catalogues are organised by rated deflection. A 25 mm deflection mount has fn of about 3.15 Hz, while a 5 mm pad is around 7 Hz and only suits high-speed machines.
Does adding damping improve vibration isolation?
It depends on the operating region. Near resonance, damping is essential because it caps the transmissibility peak, which is unbounded for an undamped system. In the isolation region above r = sqrt(2), however, damping slightly increases the transmitted force, so highly damped mounts isolate a little worse at high frequency ratios. Elastomeric mounts around zeta = 0.05 are a common compromise.
What happens if the machine runs near the isolator natural frequency?
Operating in the band 0.8 < r < 1.4 means the disturbing frequency is close to resonance, where the mounts amplify vibration severely; the calculator flags this condition. The fix is to soften the isolators (more deflection, lower fn) so that r rises above 3, or occasionally to stiffen them so the machine passes quickly through resonance during run-up only.
How do I use the reverse mode to pick an isolator from a catalogue?
Enter the disturbing frequency (or machine RPM) and the isolation percentage you need. The tool inverts the undamped transmissibility relation to give the required frequency ratio, the maximum natural frequency, and the minimum static deflection. You then choose a catalogue mount whose rated deflection under your actual equipment weight meets or exceeds that value.