How is the total air resistance force exerted upon a body (aircraft or car) related to its speed relative to air at far away? A simple Newton's force momentum analysis suggests that the force $F \propto v^2$ where $v$ is the speed of the body. I am sure it is not exactly speed squared but how close is it for a car?
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1$\begingroup$ In the aircraft case, are you assuming a fixed angle-of-attack (such as would be the case if we mounted the aircraft in a wind tunnel and varied the wind speed), or are you assuming the angle-of-attack is varied as needed to maintain level flight with lift equal to weight? It will make all the difference as to the correct answer; the one answer given so far appears to be assuming the latter case. Of course you can see why an aircraft considered according to the latter assumption, would be governed by very different equations than a car where the weight is carried by the wheels. $\endgroup$– quiet flyerFeb 24, 2020 at 22:34
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$\begingroup$ @quietflyer: I did not consider the distinction. It would be very informative if you would write an answer elucidating the two scenarios in details. $\endgroup$– HansFeb 25, 2020 at 18:16
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1$\begingroup$ Peter Kampf's answer assumed the latter case-- hence the statement "... induced drag ... decreases with speed squared." and that would be the normal assumption when talking about an aircraft in the context of actual flight. Just understand that the same conclusions wouldn't be valid for talking about what happens when you mount an aircraft at a fixed angle-of-attack in a wind tunnel and vary the wind speed. No component of drag will decrease as speed increases in the fixed-in-wind-tunnel case, but lift won't stay equal to weight either. $\endgroup$– quiet flyerFeb 26, 2020 at 14:18
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2 Answers
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Subsonic aircraft drag is composed of two major components:
- Pressure drag. Most of this is induced drag, but also some is from viscous effects which result in less pressure on the rear sections of aircraft.
- Friction drag from boundary layer effects.
At high speed, friction can be the dominant drag contributor and here we have a less than quadratic dependency because with increasing speed the influence of friction is diminished. This is expressed by the Reynolds number and for first order approximations the friction drag coefficient changes in proportion to $Re^{-0.2}$. Since Re increases linearly with speed, the friction drag force is approximately proportional to $v^{1.8}$ at high speed.
At low speed induced drag is dominant and this decreases with speed squared. Of course, there is still a contribution from separation (which is a major problem at low speed), so the proportionality is smaller. Still, at low speed aircraft drag drops over speed.
With cars things are simpler. Here we have a rather blunt body with lots of separation at the backwards-facing surface. The suction on that separated flow is the dominant contributor to a car's aerodynamic drag and is much less influenced by Reynolds number effects. However, a small decrease in the friction drag coefficient over speed on the surfaces with attached flow remains, so the proportionality is a whisker below 2. For engineering, using $v^2$ is good enough, however.
answered Feb 22, 2020 at 15:50
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$\begingroup$ Thank you. With regard to cars, could you please give a reference to the statement "a small decrease in the friction drag coefficient over speed on the surfaces with attached flow remains, so the proportionality is a whisker below $2$"? $\endgroup$– HansFeb 22, 2020 at 16:42
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$\begingroup$ @Hans: Some of the total drag is of course friction, and here things are like on aircraft with a bit less than $v^2$ proportionality. Only that pressure drag from separation is much bigger on cars as a proportion of total drag, so the influence from friction is smaller. $\endgroup$ Feb 23, 2020 at 13:45
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$\begingroup$ +1 and accepted. It would make it easier to read if you write out the explicit formulae taking into account your ponts 1. and 2. It took me a double take to realize by "the friction drag changes in proportion to Re$^{−0.2}$" you meant to say the frictional drag coefficient $c_f$ rather than the friction drag itself is proportional to Re$^{−0.2}$. More explicitly $D_f\propto c_fv^2\propto Re^{-0.2}v^2\propto v^{1.8}$ . $\endgroup$– HansFeb 25, 2020 at 18:17
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$\begingroup$ @Hans: Yes, you are correct. That should be the coefficient. Sorry, corrected! $\endgroup$ Feb 26, 2020 at 11:59
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1$\begingroup$ @Hans There is no simple formula. Instead, drag is a bit like accounting where all contributors need to be added up without counting one twice. Next, you have an influence from surface imperfections down to the roughness of the paint job or details of the body contour like air intakes or bumps. $\endgroup$ Feb 27, 2020 at 9:18
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Keep in mind that a car always presents the same "angle-of-attack" to the airflow (ignoring crosswinds, and ignoring changes in the car's pitch attitude due to effects related to the stiffness of the suspension springs), and so does an aircraft rigidly mounted in a wind tunnel. Thus the drag force can be expected to vary according to velocity squared in such cases, except for complications due to changes in Reynolds number, etc.
However, an aircraft in actual flight is constrained to have lift equal to weight (assuming the simple case of straight-line horizontal flight with no upward or downward tilt in the thrust vector), and thus must fly a higher angle-of-attack at low airspeed than at high airspeed. Therefore the lift coefficient and drag coefficient vary dramatically throughout the flight envelope, and so total drag cannot be expected to vary according to airspeed squared or anything close to that.
Here is a link to an excellent on-line resource explaining how to calculate the drag created by an aircraft in flight -- https://aerotoolbox.net/drag-polar/ . Scroll down to the interactive graph entitled "Variation in Aircraft Drag with Velocity" near the bottom of the webpage, specifically the third-to-the-last figure, with Velocity (kts) on the horizontal axis and Drag force (Newtons) on the vertical axis. You'll see that the drag force is nowhere near proportional to the square of the velocity for this part of the flight envelope, due to the fact that the induced drag component is decreasing as the airspeed is decreasing. At some vastly higher airspeed, where the induced drag component is trivial, this particular aircraft will experience a drag force that is closer to proportional to the square of the velocity (at least for subsonic airspeeds) than is shown in the part of the flight envelope illustrated here.
However, if the aircraft were intended to fly that fast most of the time, it would be a waste to design it to have such a large wing, due to the profile drag and skin friction drag created by such a large structure. With a smaller wing, the aircraft would fly at a higher angle-of-attack at any given airspeed, and induced drag would no longer be trivial at some given higher airspeed where it was trivial with the larger wing. In fact, the minimum total drag (and thus the minimum thrust requirement for horizontal flight) occurs at the airspeed where the induced drag contributes half the total drag, so there is no point in a designing an aircraft with an overly large wing in an effort to minimize induced drag at the cost of increased profile drag and skin friction drag. In short, in the part of the flight envelope where a given aircraft typically operates, induced drag is usually not trivial, and therefore total drag is nowhere near proportional to airspeed squared.
answered Feb 26, 2020 at 16:29
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$\begingroup$ Yes as also alluded to by another comment under another answer, this answer doesn't consider changing flap settings, etc. Even when you take those things into account, you still get a drag versus airspeed curve shaped more or less like the one illustrated in the graph referenced by the link given in this answer. $\endgroup$ Feb 26, 2020 at 17:27
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$\begingroup$ "In short, in the part of the flight envelope where a given aircraft typically operates, induced drag is usually not trivial, and therefore total drag is nowhere near proportional to airspeed squared."-- and likewise is nowhere near proportional to airspeed ^ 1.8. $\endgroup$ Feb 26, 2020 at 20:49
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