# Weight effect on efficiency due to braking

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Simulation and calculations

Weight reduction effect on fuel usage due to braking can be calculated using the following methods and equations.

The effect of weight on fuel efficiency is largely caused by braking which converts all of the inertial energy into heat in the brake pads. A very heavy car that accelerates up to 55mph and coasts down to 0 mph will use almost the same amount of fuel as a very light car doing the same acceleration and coast. The equations below only account for the effect of weight when braking from a specific speed.

Kinetic energy
KE = 0.5 * m * V²
KE is kinetic energy given in joules

Where:
m is the mass in kg
V is velocity im meters per second

Energy used to brake 50kg from 60 km/h to 0 km/h derived by user Saand is as follows
Velocity difference is 16.67 M/s

Energy = 0.5 * 50 * 16.67²7
Energy = 6944 j

Energy available in petrol
Translating the joules required into fuel volume is done using the following

Each gallon of fuel has 114500 BTU which is given by wikipedia
Each BTU is equivalent to 1055.056 joules
So joules per gallon is 120.8 million joules (120803895)

Accounting for losses in converting petrols chemical energy to momentum
The drive train is about 95% efficient

using 6944 j for the energy in the mass The energy the engine has to put out is 7310 j

The efficiency of a cars engine is around 18% The energy that the fuel has to provide to the engine is 40611 j

Petrol required
Using the energy available in a gallon of petrol found above and the energy required by the petrol to accelerate the weight found above we can find the petrol required to accelerate that mass

gallons = 40611/120803895 gallons = 0.000336 liters = 0.00127

Using this information
You probably want to know what effect this has on your cars efficiency if you remove some weight.

The numbers calculated assume decelerating by braking from 60 to 0 km/h and a mass of 50 kg. So for every 50 kg reduction the calculated volume of petrol will likely be saved each acceleration.

If you are likely to remove 100kg then use double the volumes calculated, if you more frequently braking from a different speed down to 0 kph you will have to run through the calculations again as the velocity is a squared relationship.

If you remove 100kg you will save 0.0025 per 60 to 0 kph deceleration by braking.

To equate this to a efficiency improvement you will need to estimate how frequently you brake from 60 to 0 during a standard 100 km distance. Assuming you brake from 60 to 0, 30 times every 100 km then every 100km you are saving 76ml after removing 100kg of weight from your car. That is 0.076 liter per 100 km

To equate this to an efficiency improvement divide the number above by the standard liter per 100 km efficiency of your car.

assuming 6 liter per 100km this gives 1.27% improvement

Note: The calculations above assume there is no coasting, it only accounts for braking.