Today we are seeing considerable changes in our transportation systems. Electrical cars! Impressive that humans in the end understood that electrical energy is much more efficient than oil in transportation.
But what is transportation efficiency?
Do we really understand what transportation efficiency is? Car manufacturers have been hiding the truth for decades, and we can assume that they have promoted the combustion engine relentlessly with the help of Mr. Oil, without beeing to concerned about the effects.
Successfully they have created an extreme and global pollution machine. Almost no other alternatives have been explored.
But is this the best technology for transportation?
Lets find out …..
The human as a reference
Let’s first start with some simple mathematical expressions to establish the most efficient machine in the world – the human.
Theoretically, your jogging speed should average around 9-11 km per hour, if you are an average human in good health.
We assume that in this speed you will burn around 800-900 kilocalories (ca. 3600 kilojoules). This averages to around 1 kilowatt-hour. Remember that this is a theoretical simplification to create a very basic conversion between the kilocalorie to the kilowatt-hour for an average human with 60-90 kilogram of mass.
This simplification leads us to the efficient human machine that can transport itself 10 kilometers in one hour using one kilowatt of energy. This is not far from the truth! Again – we can use this simplification to create an energy-efficiency ratio where the human is its reference-value.
We call this ratio the human-machine transportation efficiency ratio (HMTER), and assign this the value of 1 kilowatt-hour per 10 kilometer. We will assume that this factor will be a factor of scale, and assume same values for women with lower weight as for men with higher weight since their muscular systems are adapted to their physics. Anyway we will neglect political and gender-based discussions, and massively simplify to state the fact that the HMTER is expressed as the following:
HMTER = 0,1 kWh/km
The transportation weight coefficient (TWC)
Let’s establish one more mathematical identity.
We all know that when you transport something, you need to transport it with something, and that this most likely is a car. Then we have two things that need to be moved, the thing transported and the car itself.
Let’s assume that you are transporting yourself, 80 kilogram of mass, in a car with a mass of 1500 kilogram.
Then we can swiftly create an coefficient called the transportation weight coefficient (TWC), by taking the transported weight (TW = 80kg) and divide it by the mode of transport weight (MTW = 1500kg). Since we are using the human as a reference we assume that when MTW is zero you will transport yourself (you will not use your car but walk), and the coefficient will be equal to one.
Transportation weight coefficient= transported weight / (transported weight + mode of transport weight)
TWC = TW/(TW+MTW)
In our example the TWC will be equal to 80/(80+1500) = 0,0506
This is an dimensionless universal coefficient. It will be one when the MTW is zero, and it will be close to zero when the MTW is getting close to infinity.
It was impossible to find such coefficient anywhere, in other transportation efficiency calculations, so we can assume that this is its first use.
The motor efficiency factor (MEF)
Until now we know the simplified efficiency of humans, and we know that 0,1 kWh per km is very efficient. We also know that modes of transportation are related to mass; the more people we can fit into our car, the more efficient the transportation is. Such ideas are fairly obvious.
The one thing that is not entirely obvious is the efficiency of a petrol combustion engine compared to the electrical engine.
Motor efficiency factor (MEF) is the amount of energy contributing in creating forward movement or force
We will not elaborate on the truths with regards to motor efficiency, but rather state a well-established facts:
The MEF is for the different motors equals:
- Combustion engine efficiency = 0,2 (20%)
- Electrical engine efficiency = 0,9 (90%)
These numbers will give us a good indicator of how efficient the transportation is performing.
Because of the very easy conversion measurements of resistance, voltage and currents in electrical motors they can be used as a reference for measuring other motor efficiency factors. See the notes about combustion engines below.
Have in mind that the combustion engine efficiency have almost no variations. The fuel consumption specified as liter per kilometer (or gallon per mile), is directly related to friction forces in the wheels, motor and gears. The weight of the car and its air friction (drag) is also important parameters, when the car manufacturers specify the fuel consumption of their cars. But the combustion engine efficiency should be fairly static at 0,2. This indicates that only 20% of the energy is converted to forward forces.
Now, we can summarize the information into one simple equation, which will give us an extremely good indicator to evaluate which car manufacturer are the best in ecological mathematics.
The transportation efficiency (TE)
Let’s do this simple and provide an example.
Again we assume that you are going to transport yourself in an electrical car. We assume your mass is 80 kilogram and the cars mass is 1500 kilogram. It is important to understand that most civilized countries have maximum speed limits, and that this introduces relative speed. We also need to assume something about the human reference, and thereby stating that the car has 25kWh of motor power, and that it can drive 120 kilometers on this power.
This gives us these parameters:
- Transported weight (TW) = 80 kg
- Mode of transport weight (MTW) = 1500 kg
- Engine Power (EP) = 80kW
- Energy Available (EA) = 25kWh
- Range (R) = 120 km
- Motor efficiency factor (EEEF) = 0,9 (90%)
Maybe we would need to know the HMTER of the car?
We know that the engine have a maximum of 80kW of power, we know the range of 120 km and we know the energy contained in the battery of 25 kWh. It is fairly simple to understand that full throttle will empty the battery fast, but we also know that at a maximum range of 120 km, the car need to use exactly 25 kWh when driving 120 km/hour. This is not exact science but car-manufacturer science.
This gives us the following:
HMTER of 25/120 * 1,1 = 0,2292 kWh/km
Since this value is submitted to 10% reduction of motor energy efficiency, and we multiply with 1,1 for correction.
But we do understand that this is related to the weight of the car, and is probably relative to the amount of people transported. So this number is not an identity, and can contain a lot of uncertainties. We do understand that with an overloaded car, it will probably affect the range.
When we look at the tests provided by US Environmental Protection Agency (EPA) and New European Driving Cycle (NEDC), they are not stating anything about weight. This is strange, and their drive cycle tests create high differences in the car specifications.
This makes the HMTER value a not too good indicator, because we can probably assume that overloading the car for a holiday drive, the specifications will no longer have any meaning.
We can use the HMTER, to provide a suggestive comparison of the car efficiency against the extremely efficient human, but we cannot include the value in transportation efficiency calculations.
Can we use the speed as a factor in the efficiency of the car?
I would assume that if we are getting to where we want to go faster, we are more efficient, because efficiency is relative to time. The answer to this is bluntly no!
Efficiency is relative to time if we had unlimited speed limits. But we can assume a maximum speed limit of 100 km/hour. This makes the car more efficient than the human, because it can go 100 km/hour instead of 10 km/hour, exactly 10 times more efficient.
But because of the HMTER, the car is energy-inefficient compared the human reference, and we need to downgrade the time-efficiency to get an overall efficiency measurement. It is exactly 10/2.292 = 4,363 times more efficient than the human if both factors are included.
But this is not really a good reference to include in the discussion either, because of speed limits, rush hours and other time-dependent factors we can include into the discussion. It is better to leave such relative factors behind and rather try to provide a clean unit of measurement.
Let’s find out how the transportation efficiency can be measured.
We now know from our example that the TWC will be equal to 80/(80+1500) = 0,0506.
If we multiply this with the MEF the TWC will be lowered because of a not 100% optimal engine. This is intuitive and sound mathematics.
We will now get the Compensated TWC (CTWC) in our example:
CTWC = 0,0506 * 0,9 = 0,04554 = Transportation Efficiency
This is probably the best transportation efficiency calculation provided. It considers only mass and energy efficiency. This means that the CTWC calculated above can be used as transportation efficiency.
Conclusion
Transportation efficiency = Transportation Weight coefficient * Motor efficiency factor
TE = TWC * MEF
= TW/(TW+MTW) * MEF
This will lead to a dimensionless universal number, which will be close to the motor efficiency factor (MEF) with higher efficiency and closer to zero with lower efficiency. The TE curve passes the 0,5 mark where the transported weight (TW) is equal to the mode of transport weight (MTW).
The great thing about this measurement is that it is only relative to the mass contained in the transportation and the energy efficiency in the motor. All other parameters are subject to discussions and will be relative to other parameters.
By looking at the below curve you can see the effect of a 20% inefficient motor, compared to the efficient electrical motor at 90%.
If you try to find other explanations about transportation efficiency you will be shocked. It is almost impossible to understand that people use liter/km (gallon/mile) as a transportation efficiency measurement.
If you look at national transportation efficiency calculations they often specify this as energy usage per passenger travel length (kW/passenger-km or kW/passenger-mile). I would assume that this measurement is not taking into account the weight of either the passenger or the mode of transportation, and should be considered a non scientific and stupefying political measurement.
A small list of land based modes of transportation and it’s respective TE:
This list is based on transportation of one passenger (as cargo), and will change when you add TW. It should not be necessary to comment on the different brands, because the science is its own judge. It is for you to expand the list with your favorite brands. Remember that your choice is a vital part of our future. The car manufacturer is not!
Notes about compensations
We can with high certainty establish the above Transportation Efficiency as a universal measurement. But we have to make some notes about the compensations for different modes of transportation.
Forces like drag friction and Newton’s second law are the only forces contributing to fuel consumption!
An airplane will give us a good scientific value to add a compensation factor for airplanes.
For an air plane this is the total drag of the airplane. The force needed to maintain speed and lift of the airplane are the main source of fuel consumption.
Lets try to find out something about this force!
The lift/drag ratio of a Boing 737-700 is 13.82, and we assume that this ratio is at cruise speed.
The empty weight is 37.648, and full weight is 70.080, because of the fuel. Lets just average this for simplifications to 54000.
The number of passengers are 128-149, and with an average passenger-weight of 70 kg, the average cargo weight is approximately 9700.
For an airplane the average Transport efficiency would then be (9700/(9700+54000))*0,3 = 0,04568.
This is not too bad. But lets find out the drag. This should be around 54000/13.82 = 3907 kilogram (around 3,9 metric ton).
If we divide the drag per passenger just for the fun of it, we get around 26 kilo per passenger in a full airplane.
Lets be more specific, and define the drag as Fd = 1/2pv2CdA. Then we provide the car as an example.
To just massively simplify and state that averag CdA of cars are 0,3 * 0.790 m2* = 0,237 (average drag coefficient * average drag area).If we then consider the air density and assume an average speed of 18 m/s and an air density at sea level, 1,225 kg/m3. Then we have a 1/2pv2N as 0,5*1,225*324 = 198,45.
Total drag force is then averaged to 47 Newton (approximately 4,8 kilo) on a car.
If we assume that only you are driving, we can get a drag of 4,7 kilo for one passenger. With more passengers this number will be better (5 passengers will give us nearly 1 kilo of drag per passenger). If we include rolling frictions from gears and wheels the number will be more difficult to obtain, but we neglect this for the sake of simplicity.
Compared to the airplane, the drag-number looks very nice. ( 1 kilo compared to 26 kilos). This makes the forces to maintain speed extremely high on airplanes compared to cars.
Lets consider normal fuel consumption on a car (cruising speed). This can be around 0,05 liters/km, and the airplane around 26 times higher around 1,3 liters per km.
This is very close to the truth.
By this statement, we can establish a very simple compensation factor called total friction force per kilo of transported weight (drag is a form of friction). This number cannot be used directly, but will tell us something about the efficiency of the mode of transportation. As we can see the car outperforms the airplane 26 times, and will give us strong indications that the Transportation Efficiency TE must be multiplied by a factor called the mode of transportation compensation coefficient (MTCC).
If we divide the transportation efficiency of the airplane with 26 we will get a TE of 0,001757. This is not too much to be happy about, and we know that this is even lower due to lower passenger counts and higher total drag during takeoff and landing.
It is difficult to create a proper compensation factor, but this discussion have been trying to give some indication about where to start. The biggest problem is to find a satisfying scientific reference value.
The pure physics of drag, friction and Newton’s second law will impact transportation efficiency
With regards to solar-powered units and their compensation factor, it is necesaary to understand something vital. Such transportation uses available energy directly from the sun for work. This is more efficient that humans and their HMTER because humans need to convert energy to use it, and will indicate an above 100% compensation factor. This compensation will increase TE above unity for such modes of transportation. The most interesting part, is that in such cases mass is taken out of the calculations. Why, is a very interesting question regarding this effect!
If you can transport 10 tons of mass with solar power, it might not be interesting to calculate TE?
Normal physics is shown in the figure below will help us to find compensation factors for all modes of transportation. We might update this article later with more information concerning this issue.
Notes about combustion engines
Let’s do this simple and provide a quick example.
A liter of gasoline contains 34,2 mega-joules of energy. This is about 9,5 kWh. If we assume that a car contains 50 liters of gasoline, we that this car have 475 kWh of energy.
By multiplying this with the combustion engine’s MEF we get around 475*0,2 = 95kWh of pure energy.
This is around 95/25 = 3,8 times the energy of the electrical car from the example above. This makes mathematical sense, and the car can drive 120*3,8 = 456 kilometers, about four times the length of the electrical car.
Have in mind that on a full gas-tank, we are wasting 380 kWh of energy by using a combustion engine. This would give us 380/25 = 15,2 full tanks on the electrical car in our example if we could find better ways of exploiting this energy.
Pure stupidity is when you know what is right, and politicians are applauding the opposite
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