Follow Us

Vehicle Recovery Calculation


Angle/ slope:

• 15 deg: X 25 % vehicle weight

• 30 deg: X 50 % vehicle weight

• 45 deg: X 75% vehicle weight

Stuck depth:

• Wheel rim depth: 1 X vehicles weight

• Brake disc/drum depth: 2 x vehicle weight

• Axle/chassis depth: 3 x vehicle weight

Extracting a vehicle weighing 2500 kg stuck in a muddy gully up to brake disc depth with the gully exit point of +- 15 degrees would require the following force for extraction:

Vehicle weight 2500 kg x 25 % (15 degree slope) = 3125 kg. 3125 kg x 2 (disc height stuck depth) = 6250 kg of force required. Your vehicle is equipped with a 9500 lb winch. (9500 lb = 4320 kg). This is the maximum pull at the last turn on the drum. A single line winch recovery is now insufficient, hence a double line winch recovery is now required, increasing the winch power to 8640 kg, sufficient for the 6250 kg requirement.

Having calculated the above, we can now see why we should rather purchase recovery equipment rated above our vehicle weight requirements. An example of this would be if the same vehicle is stuck in a mud hole with a 45 degree exit bank and stuck up to chassis height, the recovery force required would be 13125 kg. Triple line winching would now be essential with some additional assistance from the engine, sand tracks digging etc. a second snatch block would also be necessary.

The same calculation would be used for a snatch/kinetic recovery.

Recovery training should be done before attempting either of these two potentially dangerous recovery methods.

(The figures used are rule of thumb estimations for training purposes and were not scientifically proven).