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# Air resistance

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can somebody give some advice on modelling air resistance for a fairly streamlined body - such as an airplane? ---------------------------- http://www.digitalsentience.pwp.blueyonder.co.uk

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For objects moving at relatively large speeds (speeds of baseballs, airplanes, cars, etc) the air resistance can be modeled with the following equation: F = (1/2)*D*rho*A*v^2, where D is the drag coefficient, rho is the density of the object falling, A is the cross sectional area parallel to the velocity vector, and v is velocity. Im sure if you do a quick search, you can find far more complex and accurate modelings. But all things aside, the air resistance on your airplane will pretty much be proportional to the square of the speed it is traveling.

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Short answer: at low speed, friction proportional to v, at high speed, friction proportional to v². For most real-world uses, v is correct.

Cédric

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In response to Cedric-
Sorry for being picky, but for his application of an airplane, the friction is most definitely going to be proportional to the square of the speed. Quoting from Serway''s "Physics for Scientists and Engineers" - "For large objects moving at high speeds through air, such as airplanes, sky divers, and baseballs, the resistive force is approximately proportional to the square of the speed". When we did a lab in high school physics with coffee filters, they had a square force acting on them, even though the terminal velocity was less than 5 m/s. I think "high speed" depends upon the makeup of an object, because its clear that high speed for an airplane is far different than high speed for a coffee filter.
Brendan

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I agree that using the square of the speed is better even at low speed. Not only is is mathematically correct it also corresponds better to experience: in anything where air resistance is the main limiting factor (driving a car, cycling) if air resistance were linear you would expect the speed to vary linearly with power, and it doesn''t.

This is true for airplanes as much as anything. The main difference with them is as well as being streamlined they use some of the drag to lift and turn the plane, so the drag will vary as the plane is flown.

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wasnt sure if it would be like this:

F=av+bv²

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From my understanding, it should strictly be just the squared term, however, a linear term may arise in some situations. In general though, the v^2 term should be much larger than the v term such that it is very well approximated by the v^2 term alone, if there is any linear component at all. Best of luck.
Brendan

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You should never have both a squared component and a linear component. The linear approximation is really valid for "creeping" or "Stokes" flows. These have the characteristic that the Reynold''s number is 1 or less, where Reynold''s number is a function of velocity and viscosity, and the size of the object. The larger the object, the slower it has to be moving before the linear drag approximation is good.

Even without knowing about Reynold''s number, the linear approximation makes mathematical sense only for slow velocities because for values of v around 1.0, v2 is approximately equal to v, and therefore v can be used instead. You can try this easily using Excel or a small program. For v ranging from 0.9 to 1.1, the percentage error in v compared to v2 is at most around 10%. That isn''t a very large range of values, but the linear drag model isn''t actually a very good approximation outside that range. The range of applicable values can be shifted a bit by using different coefficients.

The linear drag model makes for some very convenient closed form solutions, but is always an approximation. The squared version is usually more accurate in most real circumstances, even for slow speeds. If your object is not traveling slow in molasses, the squared version is more accurate.

Here''s a few sets of lecture notes that I found related to this subject:

http://www.scri.fsu.edu/~jac/Physics/drag/drag1b.html

http://www.sfu.ca/~boal/101lecs/101lec8.pdf

http://www.sfu.ca/~boal/211lecs/211lec3.pdf

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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thanks guys, i now have more confidence that my model is somewhat correct.

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