What factors determine whether or not wind resistance will have an important effect on the trajectory of a projectile?


As a general rule of thumb, you can often assess the importance of air resistance by considering the mass of the air swept out by an object compared to the mass of the object itself. Though if you are being a little more accurate, you should estimate the ratio of the drag force to a relevant force for your problem.

Air Drag

To start, let's first try to figure out what the form of air resistance is. By understanding the form it takes, we'll have a better intuition about its importance. By dimensional analysis, we expect the form of the drag force to be

$$ F \sim \rho_{\text{air}} A v^2 $$

which agrees well with the canonical form $$ F = \frac 12 C_D \rho_{\text{air}} A v^2 $$

here $\frac 12 C_D$ taking the place of the unspecified dimensionless constant. Being a dimensionless constant, we should expect $C_D/2$ to be of order ~1. Typical values for $C_D/2$ are closer to ~0.1. The details of the value of $C_D$ depends on the shape and speed of the object, in particular the Reynold's number, but typical values are around 0.5 for smooth spheres, or 0.3 for most cars or so, wikipedia lists several values, and nasa has values for a baseball between 0.5 and 0.2 depending on speed. But we will not be too worried about the particular coefficient in front since we are going for a rough rule of thumb at this point.

We can additionally understand this form for the air drag with a simple model. In a very simple model, we could treat all of the air molecules at rest, wherein an object moving through the air will experience a drag force because it will continuously come into contact with the air molecules. Imagine that the air has density $\rho_{\text{air}}$ and each molecule has mass $m_{\text{air molecule}}$, as our object moves through the air, with velocity $v$, every time it collides with an air molecule it will have to transfer some momentum $\sim 2 m_{\text{air molecule}} v$ to each molecule it collides with. Each air molecule occupies on average a volume $m_{\text{air molecule}}/\rho_{\text{air}}$, and if our object moves with velocity $v$ and cross section $A$, in a small amount of time it will trace out a volume $ v A \Delta t$, so in total, the momentum we must transfer is

$$ \Delta p = \left( \text{ momentum transferred per molecule } \right) \times \\ \left( \text{ molecules per volume } \right) \times \left( \text{ volume traced in a small time } \right) $$ $$ \Delta p = \left( 2 m_{\text{air molecule}} v \right) \left( \frac{ \rho_{\text{air}} }{ m_{\text{air molecule}} } \right) \left( A v \Delta t \right) \sim \rho_{\text{air}} A v^2 \Delta t $$ which gives us a force of $$ F \sim \frac{\Delta p}{\Delta t} \sim \rho_{\text{air}} A v^2 $$

which agrees in form to both the dimensional analysis and the known formula.

To be clear, I'll point out as a small aside that these arguments only apply to air drag in the limit of large Reynold's number (which is totally applicable to everyday objects in air), where we can treat the air as turbulent. Stoke's Drag ( $ F \propto v $) occurs in the low Reynold's number limit. For more on Reynold's number, see this answer .


In physics, we cannot really say whether any dimensionful quantity is by itself big or small. By changing units one could always change the form of the number. 1 meter sounds like a regularly sized number, but it is $10^9$ nm, or $10^{-16}$ light years. But, dimensionless numbers we can really say something about. So, let's try to determine whether air resistance is important by taking its ratio with another force. If that ratio is small, we are probably safe in ignoring the impact of air resistance. As our other force, let's take gravity, since we most often are interested in the effects of air resistance when it comes to ballistic problems. We are interested in the ratio

$$ \epsilon = \frac{ \rho_{\text{air}} A v^2 }{ m g } $$

This ratio is a dimensionless number, and so should reveal whether air resistance is important or not in a real, quantifiable sense. If you were doing another sort of problem, you should instead determine the ratio of the air drag force to whatever force you thought was relevant.

But, if we are interested in a decent rule of thumb, we can simplify further. In a ballistic problem, the scale for the velocity of the object can be taken to be $ v \sim \sqrt{ g L } $, where $L$ is something like the range of the projectile, on purely dimensional grounds, at which point, our ratio becomes

$$ \epsilon = \frac{ \rho_{\text{ air }} A g L }{ m g } = \frac{ \rho_{\text{air}} A L }{ m } = \frac{ \text{ mass of air swept out }} { \text{ mass of object }} $$

wherein we recover the rule of thumb I gave at the top.

Another interesting way to rework the rule of thumb (though less accurate) would be to express the mass of the object in terms of it's own density.

$$ \frac{ \rho_{\text{air}} A L }{ m } = \frac{ \rho_{\text{air}} A L }{\rho A R } = \frac{ \rho_{\text{air}} }{ \rho } \frac{ L }{ R} \sim \frac{ L }{ 100 R } $$

Where in the last step, I've taken the typical densities of solid objects to be $\sim 1 \text{ g/cm}^2 $, a factor of 1000 less than air, and tried to be slightly more accurate by putting back in another factor of 0.1 for the $C_D/2$ we dropped. This last rule of thumb suggests that air resistance is important when an object moves 100 times its own size through air, but is the most inaccurate form we have, and assumes the object under consideration has a density near to water, or said another way, that it nearly either floats or sinks, but this form is easy to do in your head. This form is reminiscent of Newton's approximation for the impact depth.


Let's demonstrate this rule of thumb with a simple example. I'll try to simulate throwing a baseball at different speeds, both with and without air resistance taken into account and compare the effect it has on the range the ball travels. For my baseball, I'll use

$$ m = 145 \text{ g} \quad A = \frac{\pi}{4} \left( 9 \text{ in} \right)^2 \quad \rho_{\text{air}} = 1.2 \text{ kg/m}^3 \quad g= 9.8 \text{ m/s}^2 \quad \theta_0 = 20^{\circ} \quad y_0 = 2 \text{ m} \quad C_D = 0.3 $$

Here I show the results of the simulation. The darker lines are with air resistance taken into account, and the lighter lines are without. The colors correspond to different initial speeds.

Simulated Baseballs

And here is a table summarizing the distances travelled with ($L'$) and without ($L$) air resistance, the percent error you would have gotten in the range if you ignored air resistance $\Delta L/L$, and our estimates for the relative importance of air resistance from above.

$$ \begin{array}{c|lllll} v_0 (\text{mph}) & L (\text{ft}) & L' (\text{ft}) & \Delta L/L & \frac{\frac 12 C_d \rho A v_0^2 }{ m g } & \frac{\rho A L}{m} \\ \hline 5 & 4.96 & 4.77 & 0.03 & 0.03 & 0.51 \\ 10 & 11.2 & 10.3 & 0.08 & 0.10 & 1.2 \\ 15 & 18.8 & 16.5 & 0.12 & 0.23 & 1.9 \\ 20 & 28.1 & 23.3 & 0.17 & 0.42 & 2.9 \\ 25 & 38.9 & 30.2 & 0.22 & 0.65 & 4.0 \end{array} $$

So, in particular, notice that when the ratio of the mass of air swept out to the mass of the object is small, air resistance isn't that important. And furthermore, if you actually calculate the ratio of the air resistance force to the other relevant force, this gives you a more accurate measurement of whether air resistance is important, and in particular, if the value is small, it gives you an quantitative estimate of the expected deviation of your answer.

Further Reading

  1. Introductory physics: The new Scholasticism. Hogg and Mahajan arXiv:physics/0412107
  2. Air Resistance. David Hogg arXiv:physics/0609156
  3. Real World Ballistics: A dropped bucket. David Hogg arXiv:0709.0107
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    $\begingroup$ If I'm considering the mass of the air swept out, doesn't that imply a linear relationship with the velocity and not a quadratic one? $\endgroup$ – Jerry Schirmer Sep 4 '14 at 17:35
  • $\begingroup$ @JerrySchirmer not sure what you're referring to. The fact that drag is $\propto v^2$ has one $v$ from sweeping and another $v$ from momentum imparted. The air swept out is $\propto L$, and as a simple scaling argument for a projectile $v \sim \sqrt{ gL }$. These are all consistent with one another. Drag $\propto v$ is only for low Reynold's number, in which the simple model is flawed. $\endgroup$ – alemi Sep 4 '14 at 17:39
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    $\begingroup$ Great answer. One comment: When you first define $\epsilon$ you leave out the dimensionless factor $C_D/2$. Any chance you could mention typical order of magnitudes of $C_D$ at that point? Otherwise I don't know how small $\epsilon$ really needs to be in order for the ratio of the forces to be small. $\endgroup$ – BMS Sep 4 '14 at 17:56
  • $\begingroup$ @BMS tried to clarify. being dimensionless we expect $C_D \sim 1$ or so. Typical values are closer to ~0.1. Included links to wiki and nasa for some values. $\endgroup$ – alemi Sep 4 '14 at 18:06

The drag force [1] is proportional to the square of the velocity, the density of the surrounding medium and the product of the cross section of the projectile and the drag coefficient [2] of your projectile. You need to compare this force to other forces involved to determine if you can omit it in your calculations.




The forces which a fluid through which a solid body is moving can be separated on the basis of their direction: the force along the body's direction of motion is a drag. The force perpendicular to the direction of motion (assuming for simplicity the body to be rotationally symmetric with respect to the velocity axis) is called lift.

The drag comprises various terms, called respectively skin friction, form drag, interference drag, induced drag and wave drag. The total drag on an object depends on which of these components dominates. For streamlined objects, total drag decreases at first with increasing air speed, so that the total drag has a minimum at some intermediate speed, then increases as airspeed increases further. For non-streamlined objects, the law discussed elsewhere ($\vec F\propto -\vec v$) is roughly correct.

For compact and dense bodies like a bullet, neglect of the lift is easily justifiable, leaving only the resistance term. For less dense, and elongated bodies the lift can become the dominant (though by no means the only force). The ratio of the two can be expressed in terms of a coefficient called $L/D$, which is the major term influencing the efficiency of streamlined bodies like wings, fuselage, whole airplanes, and so on.

Wikipedia has good discussions of all of these terms.


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