Talk:Energy consumption of cars

A brief review and comments

Hi, Paul. As requested, I am responding to your request that I review this article. In particular, I have reviewed the introduction and the numerical examples. Here are some comments:

If a reader simply plugs your stated parameter values into your equation

$P_{\mathrm {drag} }+P_{\mathrm {rol} }=4\left({\frac {1}{2}}\,v^{3}\,A\,\rho +rvm_{\mathrm {c} }g\right)=86.4\;\mathrm {kW} ,$

The answer the reader gets is not 86.4 kW. Instead, the reader gets an answer of 3,503,720 which is meaningless because the dimensional units are wrong. The problem lies in the fact that g is 9.8 m/s² and your stated v is 110 km/h.

I suggest that you state the value of v to be 30.6 m/s (110 km/h or 70 miles/h) and then rewrite your equation as

$P_{\mathrm {drag} }+P_{\mathrm {rol} }=4\left({\frac {1}{2}}\,v^{3}\,A\,\rho +rvm_{\mathrm {c} }g\right)=86,492\;\mathrm {W} =86.5\;\mathrm {kW}$

Your next two equations also have the same problem ... that v needs to be expressed as m/s and the equations also need similar rewriting:

$P_{\mathrm {tot} }=4\left[{\frac {1}{2}}v^{3}{\Big (}{\frac {m_{\mathrm {c} }}{d}}+A\rho {\Big )}+rvm_{\mathrm {c} }g\right]=43\;\mathrm {kW} .$

$P_{\mathrm {tot} }=1/(0.38)\left[{\frac {1}{2}}v^{3}{\big (}{\frac {1}{2}}{\frac {m_{\mathrm {c} }}{d}}+A\rho {\big )}+rvm_{\mathrm {c} }g\right]=18.3\;\mathrm {kW} ,$

Yes, I know that an experienced physicist or engineer would make the necessary conversion just as I did ... but other readers will just plug your stated values of v into the equations and would get incorrect and meaningless answers.

I hope the above comments are helpful.

As for the remainder of the article, where you derive the basic equations, I am quite sure that you have correctly derived them and therefore I did not check your derivations in detail. Milton Beychok 05:05, 5 January 2010 (UTC)

Thank you Milton. I could have sworn that I mentioned the speed in meter per second, but apparently I didn't. I will adapt the text somewhat. However, I get slightly different values than you, even though I took g at the round value of 10 m/s². I get 86.395 where you get 86.492. This must be a matter of rounding. I used a little Matlab program and Matlab works internally with full 8 byte precision. I could put my program on a Code subpage, but since Matlab is commercial I'm hesitant. How did you calculate the number? For the time being I will stick to my Matlab numbers.

Another thing: Is the message clear that the electric car does not run free of charge? As I said before, I get annoyed by politicians and their entourage who act as if electric cars will be the ultimate solution to the energy problem. Whenever they talk about electric cars they pretend that electric energy is either very cheap or completely free, while at the same time EU politicians have forbidden (electric) incandescent lamps of 100 watt and more. In a few years all incandescent lamps will be forbidden in the EU, because "they cost too much energy". It seems to me that policy makers don't understand that electric power is electric power, independent of its use. (To avoid misunderstanding, I'm not an energy guzzler, my wife and I have one car only, a Toyota Prius).--Paul Wormer 08:58, 5 January 2010 (UTC)

I did my calculations using my Hewlitt Packard scientific calculator (HP 35s) which uses reverse Polish notation that I love greatly. Your message about the cost of electrical power could be made more clear.

As for us, I drive a 1997 Saturn and my wife drives a 1999 Toyota Echo ... both quite small cars since both of us are uncomfortable driving large cars. Milton Beychok 10:55, 5 January 2010 (UTC)

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Milt, do you have any suggestions how my message about the cost of electrical power could be made clearer? (Without sounding like a proselytizer, of course).--Paul Wormer 13:12, 5 January 2010 (UTC)