Thank you to Andres for starting this thread. It has prompted me to investigate the topic somewhat rigorously, looking at the physics involved. What does Mr. Newton have to say?
I've put together some simple calculations of an example to demonstrate the principles, without going into calculus or numerical models. Hopefully this gets to the gist of what's correct. It should be quite easy to reproduce the spreadsheet used in Section 1 and get the same results.
A simple fuel efficiency model is constructed based on a Camry and a known BSFC plot. This is used to investigate how to drive most efficiently. The most efficient method is to accelerate at high throttle % up to a slow speed.
This post will consider optimum driving behaviour for minimum fuel consumption on flat roads with corners, but no wind.
1. Steady State
6. Relative Importance
General Principle: Two factors are in play: 1. Minimise the work required to be performed on the vehicle by the engine. 2. Produce the work using the least mass of fuel possible.
Driving can be split into four regimes: Steady state, deceleration, cornering, and acceleration. These regimes can then be combined into actual road use.
1. Steady State
Objective: Drive a given distance at a constant speed, using the least fuel.
Air and rolling resistance require the vehicle to be pushed with a force F to cover a distance D.
The energy required E is force * distance, E = F * D.
To minimise the energy required to move any given distance, minimise the force. This means, travel as slowly as possible! Not much fun.
Thankfully, our engines are inefficient at producing very small amounts of power.
We can decrease the fuel required for any given distance by performing *more* physical work, but at a higher efficiency.
Note that this has nothing to do with the commonsense argument, "travelling faster takes more power, but for less time".
The power required to travel at higher speeds increases as the third power of speed, however the time decreases only linearly. Therefore decreased time cannot be enough to make travelling faster more efficient.
To look at it another way, the total energy required is E = F * D, but the resistance force increases with the square of speed. Therefore the energy to travel any distance rises with the square of speed.
As an example, let's double the speed. The power required is 8 times as high, but the time is only halved. Therefore, 4 times as much energy is required.
A calculation was constructed in MS Excel to model the fuel use of a 1992 Toyota Camry equipped with the engine shown in the post by user "RSCRS" on page 1. The choice of engine is simply because the BSFC map is available but it is reasonable because the engine appears to have similar power to a Camry of the period, with similar engine speeds. The model used the following properties:
Rho (air) = 1.225kg/m^3, Cd = 0.31, frontal area = 2.275m^2, weight = 1500kg, tyre pressure = 2.41 bar (~35 psi), fuel density = 0.7kg/L.
Transmission losses were ignored.
Air resistance was calculated as per the standard formula, and rolling resistance as per the "Rolling Resistance Cars" section on EngineeringToolbox.com.
Force and power were calculated for the air drag, rolling resistance and combination of the two, at speeds from V = 1m/s to 59m/s, which is the maximum speed of the car in the model.
Energy output / distance was also calculated (J/m).
Assuming ideal gearing, minimum BSFC figures were obtained for speeds matching the lines of constant power on the engine BSFC diagram.
BSFC in between these lines was linearly interpolated as a function of speed.
BSFC and J/m for each speed were combined to calculate fuel efficiency in g/m, kg/100km and L/100km (fuel density = 0.7kg/L).
The BSFC map of the engine, initially presented on page 1 of the thread:
For all power outputs available on the diagram, fuel efficiency (L/100km) improved with reduced power output. (Lowest line: 5kW, 350g/kW-h, 18m/s, 4.15L/100km)
Extrapolating down to 4kW with some guesstimating showed a worsening of fuel efficiency. (~4kW, 400g/kW-h, 16m/s, 4.2L/100km)
Initial Conclusion: Assuming perfect gearing, optimum speed is around 18m/s, however this region is right at the limit of the BSFC chart, so accuracy is poor.
Due to the difficulty in reading the map with optimum gearing, and the difficult nature of the low-end results as the best occurred at 800rpm, higher engine speeds were chosen to examine some effects.
RPM ------------------5kW (18m/s)--------------------------------------10kW (24m/s)---------------------Note (speed increase 33%)
------------BSFC(g/kW-h)---Efficiency (L/100km)--------BSFC(g/kW-h)---Efficiency (L/100km)
1000-------375---------------4.45----------------------------305----------------5.16-----------------------Efficiency decreases by 16%.
1500-------445---------------5.28----------------------------325----------------5.50-----------------------Efficiency decreases by 4%.
2000-------500---------------5.93----------------------------348----------------5.89-----------------------Efficiency increases by 0.6%!
2500-------550---------------6.52----------------------------405----------------6.86-----------------------Efficiency decreases by 5%.
Producing the same power (and hence road speed) with higher engine speed is less efficient. As expected.
There is an interesting engine speed around 2000rpm where fuel consumption is actually less at 24m/s than at 18m/s! However, note, this is still not the best overall fuel consumption available.
None of the engine speeds showed a better efficiency than the already discovered 5kW at 800rpm.
The optimum speed for the 1992 Toyota Camry with the engine considered is ~18m/s at 4.15L/100km.
The increase in efficiency at speeds higher than near-zero is solely due to running the engine in a more efficient condition, despite the greater total work involved in travelling faster.
Any vehicle at a given mass (passengers, cargo, fuel) will have an optimum cruising speed for minimum fuel consumption. This speed will likely increase slightly as the weight of the fuel decreases, due to lower rolling resistance. (Aero effects due to change in ride height ignored.)
Why is this so? What is timelike in the model? Not moving the air or pushing against the wheels, this relates only to distance and speed.
The pumping and turning in the engine, which cannot be reduced below a minimum level. Running the valvetrain, moving the cylinders up and down, and pumping the water and oil.
The power for these processes is a function of engine speed, and hits a minimum at the idle speed of the engine. The total energy for these processes is linear with time, for any given engine speed.
In reality most cars have only a small number of fixed gears to chose from. This will mean that for any given road speed, only a few set RPM can be selected. This means the BSFC will most likely not be optimum but must be chosen from a set of sub-optimal options.
Objective: Use the least fuel.
Fundamental Principle: The vehicle requires no energy input from the engine.
Analysis: This is quite straightforward, none is required.
If the engine accessories (eg power steering) are not required, switch it off.
If engine accessories are required, place engine and transmission into least fuel use state. Eg, clutch in for a manual, or neutral for an automatic.
Objective: Use the least fuel around the corner.
Scope: The corner is assumed to take the same distance regardless of the speed.
Fundamental Principle: Losing speed loses energy which must be put back in again.
Analysis: There are two aspects to consider:
a. The fuel used to drive around the corner itself. The slower the speed below the optimum, the more fuel is used simply to drive around the corner.
b. The fuel used to get back to the optimum steady state speed. The lower the minimum speed, the higher the fuel required.
c - not considered). Roll, yaw and tyre scrub will all increase drag forces on the car. Consequently the optimum speed around a corner may be slightly lower than the straight line optimum.
There is a fine point here, that if making a sharper, slower turn at the apex can allow an increased speed in other parts of the corner, it may be possible that a lower corner minimum speed may be optimal.
Taking an example of a normal car and a corner slightly below the optimum steady state speed:
Typical car sensitivity to lowered minimum speed:
Toyota Camry from above:
E = 1/2*m*v^2, mass (kg) and velocity (m/s).
E (15m/s) = 0.5*1500 * 15^2 = 1500 * 225 = 168,750 J
E (16m/s) = 0.5*1500 * 16^2 = 1500 * 256 = 192,000 J
Difference = 23,250 J
Corner 50m long. Compare 1. 50m at 16m/s with 2. 40m at 18m/s and 10m at 15m/s.
1. 50m * 0.029385g/m = 1.46925g
2. 40m * 0.029064g/m + 10m* 0.029365g/m =1.191925g.
Saves 0.28g of fuel driving around the corner, but requires additional (at best BSFC 240g/kW-h) 23,250J * 0.0000667 g/J = 1.05g of extra fuel to gain back the lost 1m/s of lost speed!
Slowing down slightly more to go faster around most of the corner is clearly not more efficient in this case.
An extremely light vehicle in a very long corner, with an engine exhibiting a very sharp low-power efficiency drop may give an opposite result.
Result: Sacrificing minimum speed for higher speed elsewhere in the corner is less efficient.
Conclusion: Drive around the corner to maximise the minimum speed.
Further discussion: What if we can massively increase the minimum speed by increasing the distance?
Eg, travel an extra 200m at no loss of speed, compared to a tight corner at 1/2 speed?
200m at 18m/s: 0.029064g/m * 200m = 5.8128g of fuel. (Increased fuel use due to yaw and scrub not considered.)
18m/s compared to 9m/s: Energy difference = 0.5*1500*(18^2 - 9^2) = 750*(324-81) = 182,250J, at 0.0000667g/J = 12.15g of fuel.
Less than 1/2 the fuel is used by running wide. Clearly, if a wide runoff area is available, it is better to not lose any speed at all!
There may be some optimum level of minor speed loss to reduce the corner distance sufficiently.
4.1 Acceleration - neglecting air and rolling resistance, and transmission losses
Objective: Increase from one speed to another using the least fuel.
Principle: This can be considered as an energy problem. The vehicle has energy A (joules), and we must increase it to energy B (joules). To do this we must add C joules of energy to the vehicle.
4.1.1. - Single gear.
To add energy to the vehicle we have an engine whose speed is dependent on road speed, so at any road speed the engine speed is determined.
At any given engine speed, the engine can be commanded to produce varying levels of power, which have different efficiencies (fuel mass per energy produced, typically g/kw-h).
It is clear that to obtain the required kinetic energy with the least fuel use, the engine must be run at the most efficient power (best BSFC) at every given engine speed.
This is the NOT the maximalen line in the BSFC drawing. At engine speeds below peak BSFC, higher power gives better BSFC. Best example is at 1800rpm.
There must be some other line "maxiBSFC-Linen" which must be used.
4.1.2. - Multiple gears.
Assuming gear changing is instantaneous and requires no energy, gears must be changed to keep the engine running at the most efficient combination of power and engine speed.
IE, around the "maxiBSFC-Linen" line at the best BSFC. In the example above this is around 2000-3000rpm and 85% to 100% maximum power.
In reality gear changes take some time, so in a vehicle which accelerates quickly (car, motorbike) it may be optimum to ignore some shifts.
4.1.3 - CVT.
Run at the Be (min) point on the BSFC graph.
4.2 Acceleration - including air and rolling resistance
Objective: Accelerate from speed A to speed B using the least fuel.
Principle: Accelerating more quickly shortens the distance, which decreases work required on the air and tyres.
Total energy = change in kinetic energy of vehicle*efficiency + drag work + rolling resistance work.
Accelerating more quickly than optimal in 4.1, ie at a lower efficiency, decreases drag and rolling resistance work.
Analysing this requires an acceleration computational model, or calculus, and has not been performed.
Whether accelerating more quickly than the optimum in 4.1 minimises fuel use depends on how badly the BSFC drops off at higher throttle %.
If it only drops off slightly then higher (even full) throttle % may provide the most efficient acceleration.
If so, the difference is likely to be slight, as the kinetic energy of the car vastly outweighs the wind and rolling resistance on short distances (see figures above).
Note, this is *not* to minimise the time spent at a "less efficient speed".
Those lower speeds were only less efficient because we were running at a very low power output to maintain the same speed.
Accelerating more quickly simply minimises the distance, and hence the energy devoted to air and tyre drag.
NB: In real life with a straight afterwards, accelerating more quickly than the "maxiBSFC-Linen" simply increases the length of the straight at steady speed, so any benefit may be reduced.
4.3 Acceleration - including transmission losses
Analysis: BSFC maps of engines show the output power of the engine crank shaft. They do not account for transmission losses due to spinning the input shaft and gears more quickly.
Higher engine speeds have an efficiency penalty not reflected on the BSFC diagram.
Result: Shifting to a higher gear must occur earlier than optimal according to any analysis based purely on the BSFC. How much earlier will depend on the transmission design and engine speed regimes in play.
4.4 - Acceleration Conclusion: Accelerating from speed A to B, using the least energy, accounting for air and rolling resistance, and transmission losses:
Accelerate at power above the "maxiBSFC-Linen" line, and shift gears earlier than suggested by a pure BSFC analysis.
5. Combining the Four Regimes
5.1 Steady to Braking to Cornering
Begin coasting to reach corner entry speed at exactly the right time.
5.2. Cornering to Accelerating to Steady
Accelerate at the "maxiBSFC-Linen" line, as tyre grip allows, up to the optimum speed.
5.3 "Short" Straights
On a short straight the extra energy required to reach the optimum speed may vastly outweigh the advantage of travelling at it.
For example, a 100m straight at 14m/s and 18m/s:
Fuel use at 14m/s = 0.029254g/m * 100m = 2.9254g.
Fuel use at 18m/s = 0.029064g/m * 100m = 2.9064g. A saving of 0.019g.
However, the difference in fuel required to reach the speeds is: 243,000J - 147,000J = 96,000J of energy, at max BSFC of 240g/kW-h or 0.0000667g/J, 6.4g!
This difference is so huge that travelling at 18m/s instead of 14m/s only pays off after 30+km! This is hard to believe, but apparently is correct with this particular engine. (?!)
5.4 Highways or backlanes?
Which is more efficient, to travel faster non-stop, or travel at the optimum speed (18m/s), stopping and starting all the time?
Let's consider a 100km journey at 100km/h (28m/s) or our optimum speed, 18m/s.
28m/s in our model can be performed at 4.25kg/100km.
18m/s in our model can be performed at 2.9kg/100km, so the optimum driver has 1.35kg to spare.
Accelerating from 0 to 18m/s requires 243,000J, at best BSFC = 240g/kW-h or 0.0000667g/J this takes 16.2081g of fuel.
So, the slow driver can stop and start 82 times before he exceeds the faster driver!
What's going on here? This does not match with experience, which is that meandering country lanes take more fuel than the freeway, both in L/100km and total volume.
a. The backlanes are longer in distance.
b. Nobody ever sticks to 60km/h in the backlanes. More likely, the driver is thrashing it to the next corner, braking hard, then thrashing it again to 100km/h if he can.
c. Even on the straights, the driver is likely not keeping a set speed, and is constantly decelerating and accelerating slightly.
d. The backlane is full of bumps and holes, which bounce the suspension around more and bleed off energy in the dampers.
e. The driver is not accelerating in the most efficient way. He may be at around half throttle, which is a good bit less efficient.
6. Relative Importance
6.0 The steady speed is very important no matter how long the straight is. If the straight is short it must not be too high simply due to the kinetic energy concerns, and if the straight is long it must be chosen optimally to drive at the most efficient speed.
6.1 The shorter the straight, the more important it is that the acceleration is performed efficiently.
6.2 The shorter the straight, the less important it is that the steady speed is driven at the most efficient gearing and engine torque.
6.3 The longer the straight, the more important it is that the steady speed is driven at the most efficient gearing and engine torque.
The above analysis explains two apparently contradictory aspects of driving.
First, the claim that acceleration should be performed at high power for best efficiency, and second, the conventional wisdom and personnel experience of everyone that being "lead-footed" results in high consumption, and that driving gently results in low consumption.
What is happening here?
The effects depend on the driving regime. In city areas with short straights, the "enthusiastic" driver is almost certainly accelerating far past the optimum speed, and not coasting enough.
Further, he is also likely using higher engine speeds than required when he could shift up
In contrast the conservative and gentle driver is likely reaching much lower speeds on short straights, using lower engine speeds, and if he is trying for efficiency, coasting more.
Out in the country or on the freeway, the "lead-footed" driver is simply driving faster. This is less efficient.
To accelerate most efficiently, do so at high throttle and low RPM.
On short straights, the chosen speed should be much lower than the optimum speed.
The optimum speed generally is quite low (for the car considered, ~65km/h).