Spring calculation - How we calculated

Bee

2 months ago

Federberechnung – Die Physik dahinter

For those who want to know the details

This page shows you:

→ How the forces act on the vehicle
→ Why narrow frames are critical
→ What the calculation formula does (and what it doesn't)
→ The physical principles

Teil 1: So funktioniert’s – Interaktive Visualisierung

Understand the forces acting on the vehicle

Move the sliders and observe how the forces change:

Track width (distance between springs) 0.85 m

Typical for trucks: 0.8-1.0 m | Typical for wide frames: 1.6-2.0 m

Cabin weight 2000 kg

Center of gravity height above spring plane 1.0 m

Lateral force (cornering)

10000 N

Tilting moment

10000 Nm

Additional load per spring (curve)

11765 N

Required spring rate

192 N/mm

⚠️ Watch: If you halve the track width, the additional load from cornering doubles! That's why narrow truck frames need much stronger springs.

The comparison: Narrow vs. wide frame

Truck ladder frame (0.85 m)

Additional load curve per spring

11765 N

Required spring rate

192 N/mm

→ High stress!

Wide frame (1.8 m)

Additional load curve per spring

5556 N

Required spring rate

131 N/mm

→ Only 68% of the load

💡 What is happening here physically?

When cornering, lateral acceleration creates a force at the center of gravity. This force generates a Tilting moment, because the center of gravity lies above the plane of the spring.

This tipping moment must be supported by the springs. Each narrower the distance between the springs (Track width), the greater the force that each individual spring has to absorb.

Formula: Additional load = tipping moment / track width

→ Half the track width = double the additional load!

Teil 2: Die technischen Grundlagen – Vollständige Herleitung

2.1 Static load distribution (lever law)

Initial situation:

Weight force F_G = m × g = m × 10 N

(Simplification: g = 10 m/s² instead of 9.81 m/s² for easier calculation)

Lever law (moment equilibrium):

F_{in front} × L_{Spring fixed bearing} = F_G × L_{Main fixed bearing}

It follows:

F_{in front} = F_G × (L_{Main fixed bearing} / L_{Spring fixed bearing})

Per page:

F_Page = F_{in front} / 2

Per spring (with n springs per side):

F_{static, spring} = F_Page / n

📘 Physical background

The lever principle applies because the system is in static equilibrium. The sum of all moments about the pivot point (fixed support) must be zero.

∑M = 0

→ Weight force × lever arm = support force × lever arm

2.2 Dynamic loads: Cornering (lateral dynamics)

Lateral acceleration:

a_across = 0.5 × g = 5 m/s²

Shear force (centrifugal force):

F_across = m × a_across = m × 0.5 × g

Tilting moment (about spring plane):

M_tip = F_across × h_focus

Additional load per side:

F_{Additional information, curve, page} = M_tip / Track gauge

Additional load per spring:

F_{Additional, curve, spring} = F_{Additional information, curve, page} / n

🔍 Where does 0.5 g come from?

Lateral acceleration during cornering:

a = v² / r

Example calculation:

Speed: 50 km/h = 13.9 m/s
Curve radius: 40 m (tight curve)
→ a = 13.9² / 40 = 4.8 m/s² ≈ 0.5 g

This is a realistic value For off-road driving situations. Higher values (e.g., 0.8g for road vehicles) are deliberately not used, as expedition vehicles drive slower off-road.

2.3 Dynamic loads: Braking (longitudinal dynamics)

Braking delay:

a_brake = 0.5 × g = 5 m/s²

Braking force:

F_brake = m × a_brake = m × 0.5 × g

Nodding moment (around fixed bearing):

M_Nick = F_brake × h_focus

Additional load on all front springs:

F_{Additional brakes, all} = M_Nick / L_{Spring fixed bearing}

Additional load per side:

F_{Additional brakes, side} = F_{Additional brakes, all} / 2

Additional load per spring:

F_{Additional parts, brakes, spring} = F_{Additional brakes, side} / n

2.4 Overlay and safety factor

Combined load per spring:

F_combined = F_static + F_{Additional information, curve} + F_{Additional brakes}

Design force (with safety factor):

F_Design = F_combined × 1.8

🔍 Why a safety factor of 1.8?

The factor 1.8 covers:

Dynamic impacts: Potholes, bumps (~20%)
Asymmetrical loading: Diagonal entanglement (~15%)
Tolerances: Manufacturing tolerances, assembly errors (~10%)
Uncertainties: Center of gravity estimate, weight (~15%)
Aging: Material fatigue over lifetime (~20%)

Total: ~1.8 (conservative estimate)

2.5 Suspension travel and spring rate

Required suspension travel (from frame twisting):

s_erf = tan(α) × L_{Spring fixed bearing}

s_erf = tan(4°) × L × 1000 mm ≈ 0.07 × L × 1000

Total suspension travel (with reserve):

s_{in total} = s_erf / 0,8

Spring rate:

c = F_Design / s_erf

📘 Derivation of frame twist angle

Geometric considerations:

At maximum articulation (e.g., diagonally driving over an obstacle), the frame rotates by a certain angle α.

Example:

Axle articulation: 250 mm
Wheelbase: 3500 mm
→ α = arctan(250/3500) = 4.1°

The approach of 4° is therefore realistic and on the safe side.

🔍 Why only 80% spring utilization?

The spring should never compress to its full block dimension:

Hard impacts: Destroy spring and bearing
Loss of damping: No more suspension = direct power transmission
Extreme situations: 20% Reserve for overload situations

Factor 0.8 = 80% usage It is a compromise between maximum spring utilization and sufficient reserve.

2.6 What the calculation does NOT take into account

⚠️ Limits of preliminary dimensioning

The following will not be taken into account:

effect	impact	Why neglected?
Asymmetrical entanglement	Local overload of individual springs	Covered by a safety factor of 1.8
Dynamic impacts (potholes)	Short-term peak loads up to 2-3g	Covered by a safety factor of 1.8
damping	Reduces vibrations	Separate component (shock absorber)
Fatigue strength	Material fatigue after millions of cycles	By selecting the right spring (high-quality compression springs)
Temperature effects	Change in spring characteristic curve at -30°C / +60°C	Vernachlässigbar bei Stahlfedern (<5%)
Friction in bearings	Hysteresis, energy loss	This is compensated for by a reserve (0.8 factor).
Resonance frequencies	Oscillation at certain speeds	Expedition vehicles drive slowly

✅ Conclusion: When is preliminary dimensioning sufficient?

This calculation is suitable for:

Expedition vehicles with ladder frames
Speeds up to 80 km/h
Off-road use with moderate articulation
Pre-dimensioning and spring selection

Not suitable for:

Extreme off-road competitions (trial, rock crawling)
High-speed desert race
Standard-compliant static analyses
TÜV approval (separate inspection required)

2.7 Summary of the formulas

Complete calculation process:

1. F_G = m × 10
2. F_{in front} = F_G × (L_SP-FL / L_F-FL)
3. F_Page = F_{in front} / 2
4. F_{stat, spring} = F_Page / n
5. F_across = m × 0.5 × 10
6. M_tip = F_across × h_SP
7. F_{Curve, side} = M_tip / B_track
8. F_{Curve, spring} = F_{Curve, side} / n
9. F_brake = m × 0.5 × 10
10. M_Nick = F_brake × h_SP
11. F_{brakes, all} = M_Nick / L_F-FL
12. F_{Brakes, side} = F_{brakes, all} / 2
13. F_{Brakes, spring} = F_{Brakes, side} / n
14. F_combination = F_stat + F_curve + F_brakes
15. F_Note = F_combination × 1.8
16. s_erf = 0.07 × L_F-FL × 1000
17. s_total = s_erf / 0,8
18. c = F_Note / s_erf

Legend:
L_SP-FL = Distance between center of gravity and fixed bearing | L_F-FL = Distance between spring and fixed bearing | h_SP = Height of center of gravity | B_track = Track width | n = Number of springs per side

📚 Further reading (if available):

Heißing/Ersoy: „Fahrwerkhandbuch“ – Grundlagen der Fahrzeugdynamik
DIN EN 1789: Requirements for ambulance bodies (similar problem)
ISO 2631: Assessment of the effects of mechanical vibrations
Manufacturer specifications from chassis manufacturers (MAN, Mercedes, etc.)

🔧 Ready to do some math?

Now that you understand the physics, you can calculate your own springs:
→ To the spring calculator