Results
Example data table
| Scenario | Mass (kg) | Speed (km/h) | Stop distance (m) | mu | Grade (%) | Decel (m/s²) | Braking force (N) |
|---|---|---|---|---|---|---|---|
| Compact car emergency stop | 1500 | 100 | 45 | 0.85 | 0 | 8.57 | 12,900 |
| Loaded truck stop | 12000 | 80 | 70 | 0.65 | 0 | 3.53 | 42,300 |
| Preset deceleration target | 1800 | 60 | — | 0.75 | -5 | 6.00 | 11,900 |
| Torque-based estimate | 1400 | 50 | — | 0.80 | 0 | 5.53 | 7,740 |
| Traction-limited braking | 2000 | 90 | — | 0.70 | 0 | 6.86 | 13,700 |
| Wet road braking example | 1500 | 80 | 70 | 0.55 | 0 | 3.53 | 5,290 |
Values are rounded and assume no extra drag unless specified.
Formula used
- a = v² / (2d) from kinematics (speed + stopping distance).
- F_required = m·a − F_roll − F_aero − m·g·sin(θ) (braking force at tires).
- F_roll = Crr·m·g·cos(θ) rolling resistance.
- F_aero = ½·ρ·Cd·A·v² aerodynamic drag.
- F_from_torque = (T·wheels·η)/r from brake torque and radius.
- F_max = μ·m·g·cos(θ) traction limit (friction cap).
How to use this calculator
- Select the method that matches your known data.
- Enter vehicle mass and initial speed with units.
- Fill the method fields (distance, decel, torque, or mu).
- Optionally add grade, rolling resistance, and drag.
- Click Calculate to see force and deceleration.
- Use CSV or PDF buttons to export results.
Braking force guide with practical data
1) Mass and speed set the kinetic energy
Braking demand rises fast with speed. Kinetic energy is ½·m·v², so doubling speed quadruples energy the brakes must remove. For example, a 1500 kg car at 80 km/h (22.22 m/s) carries about 370 kJ, while at 120 km/h (33.33 m/s) it carries about 833 kJ.
2) From stopping distance to deceleration
If you know stopping distance, use v² = 2·a·s. A vehicle at 60 km/h (16.67 m/s) stopping in 25 m needs a ≈ 5.56 m/s². Braking force is then F = m·a, so a 1800 kg SUV would need roughly 10,000 N at the wheels before losses.
3) Deceleration and g units for comfort and limits
Many road tests report deceleration in g. Since 1 g ≈ 9.81 m/s², a 0.7 g stop corresponds to about 6.87 m/s². For a 2000 kg van, that is about 13.7 kN of net braking force. Passenger comfort in practice typically feels mild near 0.2–0.3 g, firm near 0.5 g.
4) Tire-road grip caps the usable force
The traction limit is Fmax = μ·m·g·cos(θ). Typical μ values: dry asphalt 0.8–1.0, wet asphalt 0.4–0.6, packed snow 0.2–0.3, ice 0.05–0.15. If calculated force exceeds Fmax, wheels will lock or ABS will intervene.
5) Using brake torque and wheel radius
When torque data is available, estimate wheel force by F = (T·η·Nwheels)/r. With 1200 N·m per wheel, radius 0.30 m, η 0.90, four wheels produce about 14.4 kN. Larger tires increase radius and reduce force for the same torque.
6) Grades, rolling resistance, and drag matter
Downhill grade adds a component m·g·sin(θ) that opposes braking. A 5% descent adds about 0.49 m/s² of “extra” acceleration to overcome. Rolling resistance (Crr ≈ 0.008–0.015) and aerodynamic drag (½·ρ·Cd·A·v²) reduce net braking force, especially above 90 km/h.
FAQs
1) What is braking force in this calculator?
It is the net force slowing the vehicle along the road. The tool can compute it from mass and deceleration, from stopping distance, from brake torque and wheel radius, or from the traction limit using friction coefficient.
2) Why does the calculator show a traction limit?
Tires can only transmit up to μ·m·g (adjusted for slope). If your required force is higher, wheels will lock or ABS will reduce brake pressure. The limit helps you judge whether a stop is physically possible.
3) Which method should I use?
Use “Stopping distance” when you have measured distance. Use “Deceleration” when you have m/s² or g data. Use “Brake torque” for component sizing. Use “Friction limit” to estimate maximum braking on a surface.
4) Do aerodynamic drag and rolling resistance reduce braking distance?
Yes. Both act opposite motion, so they help slow the vehicle. Drag grows with speed squared and becomes noticeable at highway speeds. Rolling resistance is smaller but steady. Include them when you want a more realistic net force estimate.
5) How do slopes affect braking force?
Uphill slopes assist braking and downhill slopes resist it. A 5% downhill adds about 0.49 m/s² of acceleration you must overcome. The calculator applies grade using sine and cosine components, affecting both required force and traction limit.
6) What units should I enter for best results?
Any supported units work, but be consistent. For speed, km/h is common. For distance, meters are easiest. For mass, kg is standard. If you switch units, the calculator converts internally before computing force in newtons.