Calculate spring rate k, solid length, spring index C=D/d, and natural frequency for helical coil and leaf springs in real time. Force-deflection and strain energy charts update instantly.
$$k = \frac{Gd^4}{8D^3 N_a}$$
コイルばね定数 [N/mm]：$ 横弾性係数 [MPa]、$ 線径 [mm]、$ コイル中径、$ 有効巻数
$$C = D/d, \quad K_W = \frac{4C-1}{4C-4} + \frac{0.615}{C}$$
ばね指数とワール応力修正係数（曲率効果）
$$f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
固有振動数 [Hz]：\(m\) 取付質量 [kg]。共振回避の設計に使用
The primary equation for the stiffness (spring constant) of a helical compression or tension spring is derived from the torsion of the wire. It relates the material's shear modulus, the wire geometry, and the coil configuration.
$$ k = \frac{G \cdot d^4}{8 \cdot D^3 \cdot N_a}$$$k$: Spring constant [N/mm]
$G$: Shear modulus of the material [GPa]. (~78.5 GPa for steel, ~26 GPa for aluminum).
$d$: Wire diameter [mm]
$D$: Mean coil diameter [mm] (center-to-center of the wire)
$N_a$: Number of active coils (coils that deflect under load)
The natural frequency of a spring-mass system determines its dynamic behavior. A spring with an attached mass will vibrate at this inherent frequency when set in motion.
$$ f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$$f_n$: Natural frequency [Hz]
$k$: Spring constant [N/m] (Note: consistent units with mass)
$m$: Mass attached to the spring [kg]
This shows that a stiffer spring or a lighter mass leads to higher, faster vibrations.
Automotive Suspension: The spring constant is carefully tuned to balance ride comfort (softer spring) and handling stability (stiffer spring). Engineers use calculations like these to select wire diameter and coil count before prototyping. A miscalculated spring can lead to a car that bottoms out over bumps or feels overly harsh.
Precision Valvetrains: In internal combustion engines, valve springs must have a high enough natural frequency to avoid resonating at engine operating speeds. If the spring's frequency is too low, it can't close the valve fast enough at high RPM, causing catastrophic engine failure.
Consumer Electronics (Click Buttons & Keyboards): The tactile "click" in a mechanical keyboard or a push-button is often created by a small helical or leaf spring. The spring constant defines the required actuation force, which is critical for user experience and product feel.
Industrial Vibration Isolation: Large machinery is often mounted on springs to isolate its vibrations from the building floor. Here, engineers aim for a very low natural frequency (using soft springs and heavy mass) so the system doesn't amplify the machine's operating vibration, protecting the structure.
First, there is the misconception that "the spring constant hardly changes even if you change the material." While it's true the formula includes the shear modulus $G$, for example, between steel ($G \approx 79,000$ MPa) and titanium alloy ($G \approx 44,000$ MPa), the spring constant for the same shape becomes about 44% smaller. A common mistake is switching to titanium for weight reduction only to find the spring becomes softer than expected. When you switch materials in the tool, always check if you can adjust the shape parameters to maintain the target $k$ value.
Next, there's a tendency to simplistically think that "the number of active coils $N_a$ is just the total coils minus the ends." In reality, the range considered "active" changes slightly depending on the end shape (closed and ground, plain ends). Especially for designs requiring high precision, you should refer to calculation methods defined in catalogs or standards. Treat the tool's calculated value as a theoretical guideline and ultimately rely on measured values from prototypes.
Finally, overlooking the maximum load. The tool calculates "the theoretical limit this spring can withstand," but in actual design, it's standard practice to apply a large safety factor. For instance, even a spring that can theoretically handle 100N, when used in applications with repeated loading, needs a safety factor of 1.5 to 2.0, meaning the actual operating load should be kept below 50N to 67N. Using a spring near its maximum load will quickly lead to permanent set (failure to return to original shape) and loss of function.
A compression spring for automotive suspension: wire diameter 4.5 mm, mean coil diameter 28 mm, 8 active coils, free length 85 mm. Using Wahl's formula with shear modulus G=81 GPa for music wire: K = (G × d⁴) / (8 × D × n) = (81000 × 4.5⁴) / (8 × 28 × 8) ≈ 18.4 N/mm. Solid length ≈ 40.5 mm (9 coils × 4.5 mm). Spring index = 28/4.5 ≈ 6.2. Natural frequency ≈ 15.3 Hz with 0.5 kg mass.