How to manipulate numbers and get any result you want

This post hast little to do with physics, let alone quantum mechanics; I’m just writing it because I saw reports in the media about a study done by three German professors that had the incredible conclusion that electric vehicles emit more CO$_2$ than diesel vehicles. I’ll not focus in debunking this “study”, as it has already been thoroughly debunked in the newspaper articles that I’ve linked, but rather I’ll explain how the calculation is done, and how one would go about manipulating it to get the result you want. They are simple mistakes, that would have been caught by even cursory peer-review, so maybe the lesson here is that non-peer-reviewed “studies” like this one are better ignored altogether.

So, we are interested in the total amount of CO$_2$ that a vehicle emits over its lifetime. It is the emissions caused by producing it in the first place, $P$, plus the amount of CO$_2$ it emits per km $\eta$ times the distance $L$ it travels over its lifetime: $P + \eta L$. To get a number that is easier to relate with, we divide everything by $L$ and get the effective emissions per km $\frac{P}{L}+\eta$. We want to compare a diesel vehicle with an electric vehicle, so we want to know whether
\[\frac{P_E}{L}+\eta_E\quad\text{or}\quad\frac{P_D}{L}+\eta_D\] is bigger.

Assume that $P_E > P_D$, because of the extra emissions needed to produced the battery of the electric vehicle, and that $\eta_E < \eta_D$, as it is much more efficient to extract energy from oil in a big power plant than in an internal combustion engine 1.

Now, what could you do to make the electric vehicles look bad? Well, since their production causes more emissions, you want to emphasise that in the equation, and since they emit less CO$_2$ when running, you want to downplay that. How? We have three variables, so we have three ways of manipulating the numbers: we can multiply $P_E$ and $P_D$ by some large number $n_P$ (e.g. by assuming that the factories producing the cars are powered purely by oil shale), we can divide $\eta_E$ and $\eta_D$ by some large number $n_\eta$ (e.g. by assuming the cars are ran always at maximal efficiency), and we can divide $L$ by some large number $n_L$ (assuming that car is scrapped after a few kilometres).

What is the effect of doing that? If the real numbers say that electric vehicles are better, that is, that
\[\frac{P_E}{L}+\eta_E < \frac{P_D}{L}+\eta_D,\]which is equivalent to
\[ \frac{P_E-P_D}{L(\eta_D-\eta_E)} < 1,\]then the manipulations of the previous paragraph imply in multiplying the left hand side of this inequality by $n_Pn_Ln_\eta$; if we want to flip it we just need to make $n_P,n_L,$ and $n_\eta$ large enough so that \[ n_Pn_Ln_\eta\frac{P_E-P_D}{L(\eta_D-\eta_E)} > 1.\]

And what the authors of the study did? All of the above. Most interestingly, they used the NEDC driving cycle to calculate $\eta_D$ and $\eta_E$, a ridiculously efficient driving cycle that has been discarded in favour of the less unrealistic WLTC. They did this because WLTC numbers hadn’t yet been released for the Tesla Model 3, the electric car they used for the comparison. They claim that this is not a problem, because NEDC favours city driving, where electric cars excel, so if anything this incorrect assumption would be tilting the scales in favour of electric cars. As we have seen, though, this is not the case: pretending that the cars are more efficient than they are tilts the scales in favour of the diesels.

Another mistake the authors made is to assume that cars only last 10 years or 150.000 km before going to the junkyard, which is about half of the actual number. Again this tilts the scales if favour of the diesels, as the production of electric cars causes more emissions. The reason they made this mistake is because they assumed that the battery of an electric car would only last this much, which is false for two reasons: first because a Tesla battery retains more than 90% of its capacity after 250,000 km, hardly junkyard material, and second because batteries that have in fact degraded too much to be useful in a car, say retaining only 70% of their capacity, do not go to the junkyard, but instead are reused for applications where the energy/weight ratio doesn’t matter, like grid storage.

The third mistake the authors made is exaggerating the emissions caused by production, using a discredited study that claimed that producing the lithium-ion battery causes 145 kg CO$_2$/kWh. The peer-reviewed number I could find is 97 kg CO$_2$/kWh for production in China. Even that seems too high, though, as Tesla’s batteries are produced in the Gigafactory 1 in Nevada, which has a cleaner energy mix, and should eventually be powered fully by rooftop solar. One thing that might look like a mistake but isn’t is that the authors don’t consider the emissions caused by producing the components that are common to both electrics and diesels2: wheels, body, seats, etc. Of course, ignoring that means that the number you get is not effective emissions per km, but it doesn’t change which car is the best, as that depends only on the difference $P_E-P_D$.

With the theory done, let’s get to the numbers. The authors use $P_E = 10,875,000$ gCO$_2$, $\eta_E = 83$ gCO$_2$/km, $P_D = 0$ gCO$_2$, $\eta_D = 144$ gCO$_2$/km, and $L=150,000$ km, which results in the effective emissions
\[ E = 155\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km},\]their absurd conclusion that electric vehicles emit more CO$_2$. Now what I find amazing is that this conclusion requires all three mistakes working together; correct any of the three and it flips.

First we correct $\eta_E$ and $\eta_D$ using the WLTC numbers (which are still too optimistic, but are the best I’ve got), which are already available for both the Model 3 (16 kWh/100 km) and the Mercedes C 220 d (5.1 l/100 km)3, resulting in $\eta_E = 88$ gCO$_2$/km and $\eta_D = 163$ gCO$_2$/km, and the effective emissions
\[ E = 160\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 163\text{ gCO}_2/\text{km}.\] Next we keep the wrong $\eta_E$ and $\eta_D$ and just correct $L$, setting it to $250,000$ km, resulting in the effective emissions
\[ E = 126\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km}.\] Next we keep the wrong $\eta_E, \eta_D$, and $L$, correcting only the emissions caused by the production of the battery. Putting 97 kg CO$_2$/kWh results in $P_E = 7,275,000$ gCO$_2$ and effective emissions
\[ E = 131\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km}.\] To finalize, let’s calculate the true numbers, correcting all three mistakes at once and also taking into account the emissions caused by producing the parts common in both vehicles. I couldn’t find a good number for that, just some estimates that put it around 20 tons of CO$_2$. Using this results in $P_E = 27,275,000$ gCO$_2$, $\eta_E = 88$ gCO$_2$/km, $P_D = 20,000,000$ gCO$_2$, $\eta_D = 163$ gCO$_2$/km, and $L=250,000$ km, and effective emissions \[ E = 197\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 243\text{ gCO}_2/\text{km}.\]

It doesn’t look very impressive, though. Only 19% less emissions? Is all the trouble worth it? The point is that none of the emissions of electric vehicles are necessary: as the grid cleans up both their production and operation will be CO$_2$-free. Diesels, though, will always burn diesel, so at best they will cause only the tailpipe emissions4, and the ultimate numbers will be \[ E = 0\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 135\text{ gCO}_2/\text{km}.\] There is no need to wait, though: electric vehicles are better for the environment than diesels. Not in the future, not depending on magical technologies, not in Norway, but right here, and right now. And this is only about CO$_2$ emissions; electric vehicles also have the undeniable benefit of not poisoning the atmosphere in densely populated cities.

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2 Responses to How to manipulate numbers and get any result you want

  1. Michael Wagner says:

    Hello Mateus,

    Explained very nicely – a joy to read!

    Thank you. Danke. Obrigado.

  2. Mateus Araújo says:

    I’m glad to hear that, thanks for your comment!

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