Talk:Drag coefficient

Correction
The previous version stated that the drag coefficient, for a sphere, increased from 0.1 in laminar flow to 0.47 in turbulent flow. This was wrong since in laminar flow, Stokes equation applies, i.e. Cd = 24/Re, which gives high values at low velocities and Reynolds numbers (Re). —Preceding unsigned comment added by Cbriens (talk • contribs) 06:21, 31 January 2010 (UTC)

Hi there, im not firm with editing wikipedia articles. I just found a misstake in the graph of cp vs cf, it must be L/D not D/L, i guess (?). — Preceding unsigned comment added by 178.7.136.166 (talk) 17:02, 1 June 2013 (UTC)
 * Why removed the diagram ? Dear editors,Please read discussion page for Drag coefficient CD and leave no blanks in wikipedia article after your works, or no discussions.will inherit by lazy editors, thimk twice of yours trim and be kind to post a notification to the open questions.Myself I will have no reason but leave Wikipedist — Preceding unsigned comment added by 188.25.59.135 (talk) 02:00, 30 September 2023 (UTC)

Unclear combination of metric and imperial
The article changes from imperial to metric measurements half way through - could be clarified - Cd measurements are given for the cars in both, meaning that they differ by a factor of 10.7. —Preceding unsigned comment added by User: (talk • contribs) This statement is absolutely ridiculous. As an Aero student, I can assure you that the Cd value is entirely unitless, and the reference area determines the units of the answer (as does, if one is observing the entire drag equation, velocity and density).


 * I am afraid that this statement is not ridiculous at all. In the metric world, Drag D = Cd*½*ρ*V^2*A. D in Newton, dynamic pressure in N/m^2, and area in m^2. In the USA, D = Cd*ρ*V^2*A, not sure what the preferred units are. The factor ½ has disappeared into the Cd, which must also account for air density being in obscure units etc. In the footpound world, physical equations usually contain some sort of constant, required because of the lack of any logic at all in defining the units. How strong was the horse of the horsepower definition? What size foot was the proto for the length definition? Koyovis (talk) 01:19, 7 April 2013 (UTC)

Cd can be obtained through a method called dimensional analysis. This method allows comparison of scaled conditions that are similar enough to represent the actual scale. It also reduces the amount of parameters that need to be considered to fully understand a given condition.

Inaccuracy
The article states "The boundary layer will remain attached longer if it is laminar than if it is turbulent." It is actually the other way around. See http://imaging.me.jhu.edu/PDF/327lec-drag.pdf.

Drag coefficient of an Sphere worse than of an Hemisphere?
The Cd for an hemisphere is stated in the article chart as 0.42, while an Sphere is attributed 0.47, thus the Sphere having more drag under the same air flow than the Hemisphere. It sounds strange that an Sphere, even when not fully streamlined, but looks as more Streamlined than an Hemisphere, may have a worse, higher drag, higher Cd, and from a non-experimental, intuitive point of view, one will tend saying that the Cd of an Hemisphere is worse, higher than of an Sphere, and that the figures in chart should be interchanged, Sphere Cd= 0.42, Hemisphere Cd= 0.47. How rigt or wrong is this rationale?. To what extent Cd figures of an Sphere and an Hemisphere may result in different actual drags according to speed and Reynolds number? For common people, the involved fluid almost always would be Air, that anyway can change in density because of temperature and height of place, humidity content... Thanks. Salut +--Jgrosay (talk) 22:46, 16 December 2014 (UTC)

shown direction of relative movement between Fluid and object
As in picture "Drag coefficients in fluids ..." compared to "Flow around a plate ...":

I believe it would be quite advantageous for many readers to keep this direction uniform. 176.7.136.147 (talk) 08:22, 4 February 2024 (UTC)