I know the velocities (v) of an object after it fell for respective distances (s). From that alone, is it possible to calculate the drag force at a specific moment in time? If not what else would I need to find? So far I have looked at the linear air drag hypothesis and also the following equation, however the falling object I am dealing with is not a sphere. $$F_D=\frac{1}{2}C_D\rho_{air}Av^2$$

Thanks in advance.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jan 6, 2019 at 6:13

1 Answer 1


Preliminary information on the problem can be found in every fluid dynamics textbook or on line. See for example wikipedia.

However, a key point to take into account when using the expression $$ F_D=\frac{1}{2}C_D \rho_{air} A v^2 $$ is that $C_D$ is constant only over limited interal of velocity. For example, in an answer to a related question you may see a plot of $C_D$ for a sphere, as a function of the Reynolds number, which is proportional to the flow velocity $v$.

So, while from the final velocity of a falling object it is possible to extract the value of $F_D$ and then of $C_D$, strictly speaking, that value of $C_D$ will be valid only for the corresponding Reynolds number. It is clear that at very low Reynolds numers ($Re \ll1$) a Stokes-like formula where drag force is proportional to $v$ should be more appropriate. In general, one has to measure $F_D$ at different velocities, in order to assess the value of $C_D$.

  • $\begingroup$ Could I use this $$W_{0,s}=\int_0^sF_{drag}\left(ds\right)=\frac{AC_dρ}{2}v^2\left(s\right)$$ to find $$C_d$$? $\endgroup$
    – Marcell
    Jan 6, 2019 at 8:38
  • $\begingroup$ @Marcell What is s? displacement? You can use that formula only if you are sure that $C_d$, over the interval of velocities of interest is almost a constant, or, with reference to the formula for work, if $C_d$ does not varies too much over the integration path. This is something you cannot assess theoretically. You need data. As a first approximation, you may refer to the case of a sphere, using the figure at the link physics.stackexchange.com/questions/73108/… it will be a new example of spherical cow approximation. $\endgroup$
    – GiorgioP
    Jan 6, 2019 at 8:57
  • $\begingroup$ thank you so much for your help, it seems I am nearing a dead end here then. Could I differentiate the work done by air resistance with respect to distance to find the force of drag? $$F_{drag}=\frac{d}{ds}\left(mgh-\frac{1}{s}mv_{observed}^2\right)$$ $\endgroup$
    – Marcell
    Jan 6, 2019 at 9:14
  • $\begingroup$ @Marcell If the object has got its final velocity of falling you have not to bother about the exact form of the drag force as function of velocity. You have just to consider that in that situation the total force on the body is zero. So $F_D=m g-F_{Arch}$, where $F_{Arch}$ is the Archimede's force. In air, if m is not too small, you can neglect $ F_{Arch}$ with respect to $m g$. $\endgroup$
    – GiorgioP
    Jan 6, 2019 at 11:19

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