Formula Used
Choose the relation that matches your available data:
- k = √(I/m) where I is mass moment of inertia and m is mass.
- k = √(I/A) where I is area second moment and A is cross‑sectional area.
- k = √(Σ mᵢ rᵢ² / Σ mᵢ) for discrete point masses at distances rᵢ.
How to Use This Calculator
- Select the method that matches your known quantities.
- Enter values and pick units; the tool converts everything to SI internally.
- Click Compute to see k and intermediate steps.
- Use the export buttons to download a CSV table or a PDF summary.
Example Data Table (Point Masses)
Sample dataset about a reference axis. Use it to validate your setup.
| # | r (m) | m (kg) | m·r² (kg·m²) |
|---|---|---|---|
| 1 | 0.100 | 2.000 | 0.020000 |
| 2 | 0.200 | 1.500 | 0.060000 |
| 3 | 0.350 | 0.800 | 0.098000 |
| 4 | 0.500 | 0.600 | 0.150000 |
| Totals | 4.900 | 0.328000 | |
| Example k (m) | 0.258725 | ||
Why Radius of Gyration Matters
Radius of gyration, k, compresses a distributed mass or area into one meaningful length. It answers: “At what distance from an axis could everything be concentrated to produce the same moment of inertia?” This makes k a practical bridge between geometry, dynamics, and stability checks.
Two Definitions Used in Practice
In mechanical dynamics, the mass form uses k = √(I/m) with I in kg·m² and m in kg. In structural calculations, the area form uses k = √(I/A) with I in m⁴ and A in m². Both return a length.
Mass-Inertia Workflow (k = √(I/m))
When you know a rigid body’s inertia about an axis, the calculator converts your inputs to SI and evaluates the square‑root ratio. Example: I = 0.085 kg·m² and m = 2.4 kg gives k = √(0.085/2.4) ≈ 0.1882 m. This is common for rotors, flywheels, and instrument mounts.
Area-Inertia Workflow (k = √(I/A))
For beams and columns, k describes how cross‑sectional area is distributed about a bending axis. If you have a section property table, enter I and A directly. This form is frequently used with slenderness ratios and buckling, where smaller k indicates a section that is more “compact” about the chosen axis.
Quick Reference Values for Common Shapes
These are centroidal, area‑based results: solid circle k = r/2 (so r = 50 mm gives k = 25 mm). Rectangle about its centroidal x‑axis: k = h/√12, and about y‑axis: k = b/√12. Uniform rod about its centroid: k = L/√12.
Hollow Sections and “Where the Material Lives”
Moving material away from the axis increases k. For a hollow circle with outer radius rₒ and inner radius rᵢ, k = 0.5·√(rₒ² + rᵢ²). With rₒ = 50 mm and rᵢ = 40 mm, k ≈ 0.5·√(4100) ≈ 32.02 mm—larger than the solid case.
Units, Scaling, and Typical Magnitudes
Because k is a length, scaling the geometry by a factor s scales k by s as well. A millimeter‑scale mechanism often yields k in mm, while machine rotors and frames frequently land in centimeters. The export keeps both the output unit and the SI‑meter value for auditability.
Quality Checks Before You Export
Confirm the axis: switching from a centroidal axis to an edge axis can dramatically change I and k. For point masses, ensure each distance r is measured from the same reference line. Finally, sanity‑check the result against your geometry—k should be on the same order as the object’s characteristic size.
FAQs
1) Is radius of gyration the same as a physical radius?
No. k is an equivalent distance derived from inertia. A thin ring has k ≈ r, while a solid disk has k = r/2 about the same axis.
2) Which method should I use: mass, area, or point masses?
Use mass‑inertia for rotating bodies with known I and m. Use area‑inertia for cross‑sections in bending or buckling. Use point masses when your system is a set of discrete components.
3) What if my inertia is about a different axis?
You must use I about the same axis you want k for. If needed, convert using the parallel‑axis theorem before entering the value, then compute normally.
4) Can k be larger than the outer radius?
For a bounded solid section about its centroidal axis, k is typically less than the maximum radius. However, for offset axes or composite point masses placed far away, the effective k can exceed a local radius reference.
5) Why does the point‑mass method require absolute distances?
Because the formula uses r². Distances must be measured from the same reference axis, not between points. Mixing reference lines silently inflates Σ(m r²) and produces a misleading k.
6) How are unit conversions handled?
Inputs are converted to SI internally (meters, kilograms, and matching inertia units). The calculator then converts the final k into your selected output unit, while still showing the SI meter value for verification.
7) Where does k appear in buckling calculations?
In Euler‑type slenderness checks, engineers use r = √(I/A) (often called radius of gyration) and compute ratios like L/r. A larger k generally reduces slenderness for a given length.