Advertisement

G-force calculator

Compare acceleration with standard gravity using a measured velocity change, or calculate the gravitational field of a spherical mass.

Calculator

Signed acceleration
9.80665
Acceleration in standard g units
1.00000000
Acceleration scale, not a human tolerance table
Standard g unitsAcceleration (m/s²)
0.000000000.00000000
0.500000004.90332500
1.000000009.80665000
2.0000000019.61330000
3.0000000029.41995000
5.0000000049.03325000
10.0000000098.06650000

Advertisement

How this calculator works

Standard gravity provides a fixed acceleration scale: divide metres per second squared by 9.80665 to obtain a value in g units. Velocity mode calculates average acceleration along one signed axis; it cannot recover peak acceleration from endpoints alone. A negative value indicates direction, such as braking with forward velocity positive. Gravity mode instead estimates the field outside a spherical source. Neither scalar model predicts what an accelerometer or passenger feels. Proper acceleration depends on support forces: a freely falling accelerometer reads near zero while a supported stationary one near Earth reads about 1 g.

Formula

a_average = (v_final - v_initial)/t; g_units = a/9.80665; t = (v_final-v_initial)/(g_units x 9.80665). For a spherical gravitational source, a = GM/r².

Worked example

Speed increasing from 0 to 19.6133 m/s in 2 s gives average acceleration 9.80665 m/s² or 1 g. Braking from 19.6133 m/s to rest in 1 s gives -19.6133 m/s² or -2 g along the chosen forward axis.

Advertisement

Frequently asked questions

Is a negative result a negative load on my body?

It specifies direction along the chosen axis. Felt load also depends on gravity, support direction and rotation, which this average-acceleration calculation does not combine.

Why does free fall feel weightless?

Gravity still changes velocity, but an unsupported accelerometer falls with its case and registers little proper acceleration.

Can I use average g to estimate an impact peak?

No. An impact pulse can have a much larger peak than the average over the entered interval. Time-resolved measurements are needed.

Sources

Advertisement