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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10 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!

  3. Thomas Cohoe says:

    Is there some reason to think that long-chain hydrocarbons cannot be produced by reducing CO2 using the grid as energy source. My ideal world is what I call the photon, electron, proton order, where solar photons produce electrons for energy transport, which produce hydrogen for consumption. But hydrogen is difficult, so reducing CO2 to methane or longer chain hydrocarbons, if that technology can be developed, would be better. Then, tailpipe emissions of CO2 would be part of a cycle that would result in no net emissions of CO2.

    Of course, electric is probably better in the end for other reasons, but CO2 emitting vehicles need not be as bad as we think.

  4. Mateus Araújo says:

    The technology to do that is well-known. One produces hydrogen from electricity via electrolysis, and then coverts the hydrogen to methane via methanation.

    The problem is that the pathway electricity -> hydrogen -> methane -> kinetic energy is terribly inefficient, as compared to the pathway electricity -> charged battery -> kinetic energy. The efficiency for electricity to methane is around 50%, and from methane to kinetic energy around 30%, so 15% in total. For comparison, charging a battery has around 90% efficiency, and from the battery to kinetic energy again around 90%, so in total 80%.

    This means that you would need more than 5 times as much energy to run a car on methane than you would need to run it directly on electricity. It is a fatal problem. It makes much more sense to build clean power plants in order to shutdown dirty power plants, instead of using them to produce methane. Even if all the dirty power plants were already shut down, it would mean that the methane car would be at least 5 times more expensive to run.

  5. Thomas Cohoe says:

    Current methanation technology is indeed inefficient. This just means that there is a long way to go, which is why I mentioned developing the technology. Here is a report by the Karlsruhe Institute of Technology on a new methanation technology, already demonstrated at 76% and projected to reach 80% or better. Really, if the electrical conversion is direct rather than heat and pressure based, I don’t see any reason why the conversion could not reach an efficiency similar to battery charging. Look at the efficiency of biological conversion in general. We have been putting a lot of money into battery technology for a long, long time. Methanation is in its infancy. The ideal would be to eliminate conventional batteries altogether and go to some kind of fuel cell conversion straight back to electricity. This takes out the diesel, of course, because there is no arguing with the laws of thermodynamics that make heat based conversion so inefficient. but the big advantage of methanation and longer chain hydrocarbon production is that it uses up CO2. The cycle when it is used as fuel would make no net difference, but there is no need to stop at using hydrocarbons as fuel. As a feedstock for the chemical industry, the excess CO2 could end up in our clothes, our houses, as road surfacing, and in many other things that would actually consume CO2.

  6. Mateus Araújo says:

    This hoped-for 80% doesn’t take into account the energy needed to compress the gas (to around 300 bar) which is necessary to make it fit in a car-sized tank. And then we would need to take into account the efficiency of the fuel cell (if one for methane is even possible), that in the best case scenario would be around 50%. All in all I would be very surprised if a methane car achieved more than 30% efficiency from electricity to kinetic energy.

    More importantly, what is the point of pursuing this magical technology that might one day become almost as efficient as an electric car? Electric cars already exist, and we are in a climate emergency. There’s no time to waste in waiting for new technologies, we must do all we can now to reduce emissions.

  7. TG says:

    Your statement “The peer-reviewed number I could find is 97 kg CO2/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.” is wrong; the largest part of CO2 emissions in producing Li-ion batteries doesn’t come from mixing the components and assembly operations in Gigafactory, but from the very energy-intensive synthesis steps involved in producing the very special graphitic carbon (up to 2800 oC) and LiMn(x)Ni(y)Co(z)O2 cathode material, typically at 800-900 oC. I’m not even going to mention the energy required to produce the metals (stainless steel, copper, aluminum, as well as polyethylene separator, carbonate esters and the ultra-pure, ultra-dry and ultra-nasty LiPF6 salt in the liquid electrolyte. ) NONE of these components are manufactured at Gigafactory, it’s just an assembly operation.

  8. Mateus Araújo says:

    [citation needed]

  9. TG says:

    If the request for a citation is directed at my post, it’s a bit like asking for the proof that the grain for the flour used to bake your bread is not being grown and milled, resp., in the back of a grocery store… ;-) Most finished industrial products are assembled from often hundreds or thousands of components sourced from as many external suppliers.

  10. Mateus Araújo says:

    Yes, it was directed at your post. I know the Gigafactory is an assembler, I wanted the citation for the assertion that “the largest part of CO2 emissions in producing Li-ion batteries doesn’t come from mixing the components and assembly operations in Gigafactory”.

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