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)