|WikiProject Physics / Fluid Dynamics||(Rated Start-class, Mid-importance)|
- 1 Why this isn't on the drag coefficient page
- 2 Can people clarify
- 3 velocity squared
- 4 deleted section
- 5 Low velocity
- 6 units table = useless
- 7 I don't understand
- 8 Okay what in the world?!??!?!?
- 9 Land Speed
- 10 Merge "Drag (physics) derivations" into "Drag equation"
- 11 Deleted sentence
- 12 Example possible ?
- 13 "Derivation"
- 14 External links modified
Why this isn't on the drag coefficient page
Nothing on this page is really something that should not be included on the drag coefficient page as far as i can see... (i have a doctorate in theology and i've based my thesis on this! ) byo 01:01, 26 March 2007 (UTC)
- Drag equation is a good well bounded topic, and this article is about complete. Drag coefficient is where all the details come it. That page could be greatly expanded in the future, if anyone was interested. Drag Coefficient is however an advertising term, and as you see, that article is overrun by car pictures. Meggar 05:48, 9 April 2007 (UTC)
Can people clarify
- Is the drag coefficient a property of the object, or of the object and the fluid in combination? (i.e. if I switch fluids, does the coefficient change?)
- What is the "reference area"? Is it something like "the projection of the object onto a plane perpendicular to the direction of motion"?
- Cd is a function of the Reynolds number, which depends on the viscosity and density of the fluid, among other things. Reference area is now described (I won't go so far as to say defined) in the article. -- Heron
--Ryguasu 00:23 Jan 27, 2003 (UTC)
- Of particular importance is the V² dependence on velocity, meaning that fluid drag increases with the square of velocity. A car cruising on a highway at 50mph may require only 10 horsepower to overcome air drag, but that same car at 100mph requires 40hp. In other words, doubling the car's velocity quadruples the power needed to overcome air resistance. Contrast this with other types of friction that generally do not vary at all with velocity.
- I don't get it. In the above example, since the drag force quadruples with a doubling of velocity, wouldn't the power required increase by a factor of eight?
- You are absolutely correct, and I am sorry for the error. I didn't translate properly from my head into the text :) Jcobb 20:55, Apr 3, 2004 (UTC)
Could someone look this up or at least explain this? Looking at drag/horsepower calculators on the internet, there seems to be a split between camps which claim that the power required goes up by a factor of 4 or 8. 18.104.22.168 12:54, 2005 September 13.
- See if the last sentence is any better now. Meggar 03:37, 14 September 2005 (UTC)
- I am also confused. If ones velocity goes up by a factor of two, v^2 goes up by a factor of 4 NOT 8. I'm confused as to why some think that it would go up by a factor of 8... Fresheneesz 06:01, 30 May 2006 (UTC)
- Ok i'm not confused anymore. I fixed up the page and added a section on power. Fresheneesz 06:43, 30 May 2006 (UTC)
While a dimensional analysis can be done with any equation, it doesn’t necessarily lead to a result pertinent to the article. As I understand the policy page the entry remains original research while without a supporting reference demonstrating a direct relation. There are other reasons I believe that it shouldn’t be included; original research ought to be the easiest. Meggar 05:35, 16 February 2006 (UTC)
- Its more than a dimensional analysis, its a quick derivation of the drag equation (to within a constant). The derivation of an equation is certainly relevant to an article devoted to that equation. The only thing better would be a full derivation. A reference is "Dimensional Analysis" by H.E. Huntley, page 103. What other problems are there? PAR 03:19, 17 February 2006 (UTC)
The second of the "drag" discussions omits the low-velocity case discussed in the first, where drag is linear with velocity. (Marty, 8 May 2006)
- As speed becomes low and drag is more a linear function of speed then Cd will change and the result will be hold correct. Of course this is a dynamic formula, so in a situation where viscous drag predominates it would not be concidered an appropriate tool. Meggar 04:20, 9 May 2006 (UTC)
units table = useless
Any units for any of those parameters can be used. Why have an incomplete units table to take up more space? Fresheneesz 06:05, 30 May 2006 (UTC)
|SI||fps gravitational||fps absolute|
|density||kilograms per cubic meter||slugs per cubic foot||pounds per cubic foot|
|velocity||meters per second||feet per second||feet per second|
|area||square meters||square feet||square feet|
I don't understand
I don't get it. Can someone please explain to me why a derivation of the drag equation does not belong in an article on the drag equation? PAR 01:32, 1 June 2006 (UTC)
- I will try to explain, but this might take some work for both of us. It is not derived from anything. It is an exact representation of the physical model mentioned in the article. That is all there is to it. Seems too simple; but Rayleigh was known for taking bafflingly complex problems and making simple solutions that, as in this case, are as good as ever a hundred years later. Pause a while and admire the coincidence of both elegance and utility in this small equation. Meggar 05:24, 1 June 2006 (UTC)
- But the physical model (as noted in the article) is unrealistic. The dimensional analysis method is not; it shows that a realistic model (whatever it may be) will give the same results as long as the given parameters are sufficient to describe the model. I do admire the coincidence of both elegance and utility in this small equation, but then the dimensional analysis method is known for taking bafflingly complex problems and making simple solutions that do not rely on unrealistic models which happen to yield correct results. This is not to put down Rayleigh - I will bet that his intuition was dimensional in nature and he cooked up this simple model as a way to convey the essential aspects of the problem and its solution. Dimensional analysis is a more elegant and general approach to the problem. I know that the application of dimensional analysis is not as simple and straightforward and may take some time to appreciate, but its not like the reader is being forced to understand it in order to read the rest of the article. If the learning curve is too steep, then they can skip it, if not, they can learn and appreciate it. PAR 15:09, 1 June 2006 (UTC)
Okay what in the world?!??!?!?
I've been snooping around the internet today (as alaways) and I see that on every page in the world there is a different drag equation!
Can someone explain this? Is there more than one drag equation out there, because I need an official set-down equation.
And I actually think the equation on this article is COMPLETELY wrong, because on NASA's site, it has a different equation (like despite the fact it is arranged differently, the whole negitive 1/2 thing isn't right)... Hyperkraz
- I looked at NASA's "modern" version an it is the same except for a sign change, and NASA uses r instead of ρ for density. The sign change is not a big problem, just remember that the force is opposite to the direction of the velocity. The Wiki article defines the plus direction as the direction of the velocity, the NASA article defines the plus direction as the direction of the force. PAR 03:44, 4 October 2006 (UTC)
- Agree. There is only one equation but for arrangement and symbols. The 1/2 is correct. The negitive sign is optional. It makes sense if we look at velocity as the body moving through the fluid, not as the fluid is moving over the body. We should leave it off to be more general, drag only acts one way or we would have to call it something else. Meggar 02:22, 20 November 2006 (UTC)
- I agree with Meggar in that the equation should be generalized to the positive form, leaving the user of the equation to determine whether or not the negative sign should be used, based upon the assumed directions of the specific problem. I would like to point out that the equations on the drag coefficient/equation pages should be the same, otherwise wiki users who aren't familiar with the particular equation will become easily confused, as the user who added this discussion section was. I am rather adamant about this correlation, so I will post a comment on the coefficient discussion page to see if this inconsistency is corrected. If there is no change within a week, I will change the one on this page since it is the one that uses an assumption in direction. —Preceding unsigned comment added by Rlboyce (talk • contribs) 14:53, 24 April 2008 (UTC)
Merge "Drag (physics) derivations" into "Drag equation"
I propose to merge Drag (physics) derivations into here, since they contain in large parts the same information. This article (i.e. "Drag equation") seems to be more dynamic and developing, so I suggest to keep this one and to make "Drag (physics) derivations" a redirect to here. -- Crowsnest (talk) 21:55, 30 July 2008 (UTC)
- There were no objections, so I performed the merger. -- Crowsnest (talk) 11:16, 10 October 2008 (UTC)
I deleted this sentence from the intro:
- "The reference for a wing would be the plane area rather than the frontal area."
What is the plane area? area of the plane? (what plane?) Area of the airplane? (what part of the surface of the airplane?). If you want to reinsert the sentence, please revise it to make it clear. Thank you. Paolo.dL (talk) 14:35, 1 September 2008 (UTC)
The conventional definition of "wing area" for Drag, Lift and Performance Calcs is usual Planform Area (the area if you drew it on a piece of paper looking from above). But other disciplines/industries/professions use reference areas more relevant to themselves. The key thing is you make sure you use the consistent one for the domain you are working in. 22.214.171.124 (talk) 13:36, 31 January 2014 (UTC)tom
Example possible ?
A Boeing 707 had 4 engines with 19.000 pound drag force each. In Newton that gets around 4 * 4.44 * 19000 = 337400 N or 337.4 kN for all four engines together. Can the acceleration be calculated from this (airspeed= 0 , altitude = 0 m above sea level) supposed a full thrust when pilot releases the breakes ? Acceleration in m/s2, and speed at any given time, and how much runway that remains at any given time (f.i. 3300 m, at release of the breakes) all 337.4 kN used all the time at runway ? I would be very glad if someone can help me with this example. If possible. 126.96.36.199 (talk) 23:56, 30 April 2013 (UTC)
- Yes, the initial acceleration can be calculated very easily. The acceleration in m/s2 is simply the total thrust (in N) from all engines divided by the mass of the airplane (in kg). It is only the initial acceleration that can be calculated easily in this way because as speed increases the drag on the airplane and the rolling friction force on the wheels become increasingly significant and those two forces are unknown. Airplane manufacturers have to measure actual take-off distances in order to publish information about the required take-off distance. Dolphin (t) 00:03, 1 May 2013 (UTC)
- Thanks very much. At max gross weight the B707 weight 143335 kg. (Source is Microsoft Flight Simulator 2004 and a downloaded B707) "Ideal acceleration" (without concideration of air and rolling friction) should then be 337400 N / 143335 kg = 2.35 m/s2 !? I've experimented a bit with this issue and have come to the conclution that the average acceleration up to 180 knots is around 1.6 m/s2 in this simulation. As You state friction makes the difference 188.8.131.52 (talk) 02:36, 3 May 2013 (UTC)
I strongly think the current "derivation" is completely bogus, since it is not directly grounded on physical laws and starting with the definition of the drag coefficient (the line ), immediately reveals itself to be (quasi-)circular logic, since one had no idea that such a coefficient would have such a dependence without the drag equation itself.
If it is merely an empirical result such as that with a coefficient of friction, then we should leave it at that. If, as I suspect, more fundamental physical laws can be used to derive it, then that should be given as the derivation. One cannot "derive" an equation in physics without it being either a fundamental physical law, or using other already-known equations (including assumptions).--Jasper Deng (talk) 21:26, 19 September 2014 (UTC)
- Yes, "derivation" is maybe too strong a word, and should be reserved for arguments from more basic principles. The argument is not bogus, though. It's a dimensional analysis argument, and *assuming* you have collected the relevant physical parameters, the conclusions are sound. PAR (talk) 03:36, 8 November 2015 (UTC)
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