The article discusses the forces acting on a falling object in respect to the drag forces and reynolds numbers. The article suggests for varying Reynolds numbers, the force of drag equation is written as such:

$F[d] = kv^n$

So the forces acting on the object will be described by the function:

$−mg − kv^n = ma$

When the solution of that DE will derive from the reynolds number.

k is the air resistance coefficiant, and its units are N/sec.

How do the units make sense for reynolds numbers greater than 1?

  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Jul 18, 2018 at 9:04
  • $\begingroup$ If units bother you, you also have the option of non-dimensionalizing the equation. $\endgroup$
    – Deep
    Commented Jul 18, 2018 at 12:54

1 Answer 1


The units of k changes accordingly. When n changes, the whole relation between velocity and drag force changes, it would be logical that the proportionality constant changes. Knowing this, it could be useful to write k as $k_n$, but in the article it is probably assumed.


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