We all know that a jetliner can generate enough blast behind it to toss around cars, light aircraft, and even the occasional school bus, and also can generate enough suction to inhale (and promptly destroy) unwary ground crew.

However, if some foolish ground person drove a car (let us assume it's a typical passenger car of exactly 1 ton weight) into the intake hazard area of a GE90-115 engine during a full power runup, would that engine generate enough suction to pick the car up and attempt to ingest it? Or would they have to get so close to the intake that they'd notice that they'd better put the pedal to the metal and get out of Dodge before they went for one last, most unfortunate ride?

  • $\begingroup$ 1 ton is actually a pretty light car. For example, a SMART car has an 1800lb curb weight (the FourTwo is a little lighter at 1650). I'm guessing though that the answer is "no" because the air is pulled in from a much larger area and lower velocity (and relative pressure) than the outlet. $\endgroup$
    – Ron Beyer
    Commented Sep 15, 2016 at 2:22
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    $\begingroup$ @RonBeyer -- it wouldn't surprise me if it was a "no" -- but it'd be interesting to see a solid set of calculations showing so, nonetheless. $\endgroup$ Commented Sep 15, 2016 at 2:24
  • $\begingroup$ I think this is bordering on a physics.se question, aside from it being a jet engine, you are asking what relative suction pressure it would take to move a specific mass, and then if the GE is capable of generating that pressure differential at some distance from the inlet. $\endgroup$
    – Ron Beyer
    Commented Sep 15, 2016 at 2:27
  • $\begingroup$ @RonBeyer -- if you feel it should be migrated, flag it for migration -- I'm certainly not opposed. $\endgroup$ Commented Sep 15, 2016 at 2:41
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    $\begingroup$ That container looks like an AKE can. Even the all-aluminum AKE's don't weigh much more than 200lbs empty. I think the tare wt listed on the ones we use with canvas doors is 82kg $\endgroup$
    – TomMcW
    Commented Sep 16, 2016 at 18:31

1 Answer 1


This question would probably be better answered by an automobile aerodynamicist, so be wary of my assumptions. Hopefully, I can at least outline how to get a rough estimate.

Assuming that a "typical passenger car" is less streamlined laterally than longitudinally, let's consider a car crossing orthogonally to the intake airstream of the engine. Assume also that the intake air only generates a force on the car pointed towards the center of the inlet, parallel to the ground, and low enough to not tip the car over (this may not be a good assumption, as crosswinds can generate other aerodynamic forces on a car, especially if it's moving forward or backward).

The problem then basically reduces to overcoming the friction between the tires and the surface on which they rest; let's assume it's concrete. We can start right at the inlet, because if the engine can't move the car there, it can't move the car anywhere in the intake hazard area.

Here's a helpful paper (if I'm interpreting it correctly) that experimentally determined some dimensional sidewind coefficients $K$ for typical vehicles under a somewhat analogous situation. It's a bit odd to use a dimensional coefficient (and thanks to Koyovis for pointing that out; always check your units!), but with some unit conversion we can make it work. Let's take $K=.003(239.6)=.7188\;\mathrm{kg/m^3}$ as a typical value. The side force on our car is thus \begin{equation} F=KAV^2\,\text{,} \end{equation} where $A$ is the area facing the flow and $V$ is the air velocity. Let's say a good value for the area of a side of a car is $5\;\mathrm{m^2}$ based on this calculation from these numbers.

The mass flow rate $\dot{m}$ of a GE90 at full thrust is $1350\;\mathrm{kg/s}$ and the intake radius $R$ is $1.562\;\mathrm{m}$. Taking a standard air density of $1.225\;\mathrm{kg/m^3}$, \begin{equation} V=\frac{\dot{m}}{\rho\pi R^2}=144\;\mathrm{\frac{m}{s}} \end{equation} or about 320 mph.

You've specified a mass $M$ of $907.185\;\mathrm{kg}$, so we'll use that. The static coefficient of friction $\mu$ between dry concrete and rubber is around $.75$. So the force we need to overcome is \begin{equation} f=\mu Mg=6675\;\mathrm{N\,.} \end{equation} Meanwhile, assuming the whole side area of the car is exposed to the maximum air velocity, \begin{equation} F=KAV^2=74525\;\mathrm{N\,.} \end{equation}

So, under our assumptions, the engine has more than enough suction to ingest our car at the inlet. However, I think a more realistic mass $M$ is about $1270\;\mathrm{kg}$, which yields \begin{equation} f=\mu Mg=9344\;\mathrm{N\,,} \end{equation} which is still plenty below what the engine produces.

So we've determined that at the inlet the engine should be able to suck up a car. But what about the entire hazard zone? We calculated the inlet velocity right at the lip of the nacelle, but the area over which the engine draws air is much, much larger even inches away from the intake, seeing as the hazard zones extend behind it. Thus, at a distance away equal to a typical car's width, the velocity will be quite a bit lower. Let's model the increased suction area as a half-circle with a radius $R_0$ extending from the point on the ground directly beneath the hub of the fan.

First, let's calculate the actual airflow velocity needed to move the 1270 kg car: \begin{equation} V_0=\sqrt{\frac{f}{AK}}=51\;\mathrm{\frac{m}{s}}\,. \end{equation}

Now, we can calculate $R_0$ based on that velocity: \begin{equation} R_0=\sqrt{\frac{2\dot{m}}{\pi\rho V_0}}=3.7\;\mathrm{m}\, \end{equation}

or about 12 feet. The intake hazard zone for the much smaller CFM56 has a greater radius than this. So if we assume that all the air going into the engine comes from the hazard area and neglect the presumably increased hazard area for the GE90 (quite conservative), our car will still be okay unless it is right in the inlet. So, "would they have to get so close to the intake that they'd notice that they'd better put the pedal to the metal and get out of Dodge before they went for one last, most unfortunate ride?" I say yes.

However, because vehicles in jet blast tend to get tipped over and lifted from the ground, a simple statics problem is probably not the best model, but I don't have enough background in automobile aerodynamics to tackle the full dynamics problem.

  • $\begingroup$ The value of K=0.03 does not seem right. The article you mentioned uses feet and miles etc. In SI units I would expect the value to be between 0.5 and 1 based on frontal area from sideways direction of the car (flat plate perpendicular to wind has $C_W$ value of 2). The C-value is in relation to a reference area, for long slender bodies like aeroplanes the wetted area is taken and symbol is $C_D$, for short blunt bodies the frontal area is taken and symbol is $C_W$. $\endgroup$
    – Koyovis
    Commented Jul 14, 2017 at 4:01
  • $\begingroup$ Wikipedia lists some values for front side direction of the car which is more streamlined (using $C_D$, confusingly). For toppling over, you could use moment arms, CoG between wheels. $\endgroup$
    – Koyovis
    Commented Jul 14, 2017 at 4:02
  • $\begingroup$ Oh, excellent catch in noting that $K$ is not dimensionless. I've updated my answer accordingly; many thanks. $\endgroup$ Commented Jul 14, 2017 at 12:04
  • $\begingroup$ Are you sure OP meant 1 ton as 907 kg? Couldn't it also be 1000 kg? (As an aside, this is one big reason why I dislike it when people use units that have different definitions in different locations.) $\endgroup$
    – user
    Commented Jul 14, 2017 at 16:50
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    $\begingroup$ @Koyovis: I don't see where the authors took density into account, so it looks to me like $K$ has units of density based on their formula $K=F/(AV^2)$. Please correct me if you see something else in the paper. $\endgroup$ Commented Jul 14, 2017 at 21:49

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