# Comparing accelerated vs constant motion; inspired by real life example. Is reducing mass the way to save energy in vehicle use?

Energy consumption in jerky vs steady motion.

An average European car travels 30 km per day in urban areas mostly. This requires starting and stopping due to traffic lights and congestion and results in jerky motion.

Let’s assume a car accelerates constantly to 20 m/s and slows down to a stop in 60 seconds. It stands still for 12 seconds and repeats such cycle for an hour. This way it covers 600 m per cycle which lasts 72 seconds. The whole trip takes 3600 s and is 30 000 m long.

Let’s compare this to a case where the car covers the same distance in a steady motion. First, it accelerates for 20 seconds to reach its average velocity, continues for 3560 seconds and finally decelerates to a halt in the last 20 seconds. This way it covers 80 metres in the acceleration phase. Next 29840 metres take 3560 seconds to cover at an average velocity of 8,4 [ 8,382022] m/s. The last 80 metres are covered in the deceleration phase; 30 000 metres in total.

The car is lightweight; its mass equals 1000 kg. Its frontal area is $$2$$×$$1,5 = 3 m^2$$ and the aerodynamic coefficient is 0,3. The average rolling coefficient is 0,01. Let’s assume the drive-chain is 100% efficient and there is no AC nor heating. The brakes use no recuperation unfortunately, for ease of calculation.

This took me quite a while to find the right formula to calculate this. I hope it is convincing:

$$E_{el} = E + E_l$$

Where: $$E_el$$ – is total energy consumed by the motor, $$E$$ – energy consumption of the motion, $$E_l$$ – drive train and idling losses [ $$E_l = 0$$ in our case ]

$$E = E_{rr} + E_{air} + E_{gr} + E_k$$

Where: $$E_{rr}$$ – energy lost to rolling resistance, $$E_{air}$$ – energy lost to air resistance, $$E_{gr}$$ – enrgy lost to climb grades = 0 as assumed, $$E_k$$ – energy used to increase kinetic energy.

$$E = ½ mv^2$$ - during acceleration only

$$E_{rr} = mgC_{rr}(L_a + L_d)$$ and $$E_{rr} = m g C_{rr} L_{avg}$$

Where $$m$$ is mass, $$g$$ – gravity, $$C_{rr}$$ – rolling resistance coefficient, $$L$$ - distance covered, $$a$$ – acceleration, $$d$$ – deceleration, $$avg$$ – average velocity for the steady case.

$$E_{air} = AC_d(V_{avg}^2L_a + V_{avg}^2L_d)$$ and $$E_{air} = AC_d V_{avg}^2L_{avg}$$

Where $$A$$ - frontal area, $$C_d$$ - air friction coefficient [ $$\rho = 1$$, negligible ]

There are basically 3 qestions:

1] How much energy is consumed during the way described in ether of the two cases: jerky motion vs steady motion.

2] How does the total kinetic energy of jerky motion compare to the rolling resistance and air drag.

3] How much energy can be saved by reducing mass of the vehicle?

It took qute a long time to put all these calculations together and I am still doubtful whether they are correct. I am going to be very grateful to anyone willing to verify this example and present a solution.

**The energy of deceleration phase was completely omitted and treated as a loss of energy. I hope this is correct according to the Laws of conservation of energy.

What do you say?

• This seems to be about as useful as those Government Fuel Consumption figures... No relation with reality. Apr 16 at 7:40
• Yes of course reducing vehicle mass will reduce energy thrown away braking, when speeds cannot be steady. Also if you're driving routes with large hills, and the speed limit on the way down forces you to throw away the potential energy you worked to get on the way up. Modern electric or hybrid vehicles can recover a lot of this energy though, but in that case you could say a lighter vehicle will allow a smaller battery. Apr 16 at 16:08
• A car with regenerative braking is half as sensitive to increases in mass as one with conventional brakes, as a rule of thumb. (1) well that's what your calculation is for (2) you can't compare KE with forces (3) again your model should tell you. Apr 17 at 0:15
• Your Eair equations are incorrect and your way of using average speeds introduces a further error . Also a 20% error in rho is not a great place to start. Apr 17 at 1:30
• Could you please pinpoint the error in the Eair equation; Greg Locock; I agree that rho = 1,2 could be taken into account but that's easy to recalculate later I guess Apr 18 at 7:45