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As of Feburary 2017 the Tesla Model S has the fastest 0-60mph time of any production car. With the fastest 0-60mph time recorded as 2.28 seconds.

This got me thinking, is there any aircraft that whilst on the ground can achieve a similar 0-60mph time?

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    $\begingroup$ 0-160mph in 2 seconds for a carrier launch. $\endgroup$
    – Simon
    Feb 12, 2017 at 11:51
  • $\begingroup$ I definitely had carrier launches in mind, but that doesn't really count as the aircraft isn't producing all of the acceleration. $\endgroup$
    – Darth Vader
    Feb 12, 2017 at 12:00
  • $\begingroup$ Then no. 0-150 yes, but any reasonable performance car will beat an aircraft 0-60. After that the story changes. $\endgroup$
    – Simon
    Feb 12, 2017 at 12:08
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    $\begingroup$ A glider in a winch launch isn't much different from a carrier launch and faster at 60 mph than any Tesla. $\endgroup$
    – Peter Kämpf
    Nov 18, 2017 at 21:07
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    $\begingroup$ @user3528438: If you're going to allow those, why not the self-guided projectiles launched from a rail gun? 0-Mach 7 in a few dozen feet :-) $\endgroup$
    – jamesqf
    Nov 19, 2017 at 19:32

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Airplane performance really isn't gauged that way, partly because quick acceleration 'drag race' data is not a pertinent aspect of their flight envelope. That being said, there are a few examples, one being from the TV show Top Gear where a Bugatti Veyron races a Eurofighter Typhoon - and gets smoked by the jet.

I don't have data on jet fighters as far as linear acceleration; a quick calculation could reveal some basic data.

Take the case of an F-22. Assuming a 55,000 lbf takeoff weight, lineup, 70,000lbf thrust at full power, then brakes off and rolling, we expect to see an acceleration around:

$$(70/55)*32.2 ft/s^2 = 41 ft/s^2$$

This also assumes no friction from the wheels or air resistance during the takeoff roll.

By basic kinematics, the jet would reach 60mph or 88ft/sec in 88/41 = 2.15 seconds.

So could an F-22 or another high performance jet fighter beat an Model S in a drag race? Well it's theoretically possible and and as the above example shows, it has happened and it didn't end well for the auto.

As for airliners, the car is going to smoke them until the car reaches Vr for the jet or thereabouts.

And there is no way in hell a Model S is going to do 0-1000 mph.

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  • $\begingroup$ It would be interesting to find out how much of a difference taking into account frictional forces would make. $\endgroup$
    – Darth Vader
    Feb 12, 2017 at 12:41
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    $\begingroup$ Well, if the friction is 0, the aircraft will have the highest acceleration, whereas the car will just sit there spinning its wheels and not moving at all. $\endgroup$
    – Jörg W Mittag
    Feb 12, 2017 at 14:35
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    $\begingroup$ @JörgWMittag if you go that far, Aircraft will get no acceleration either, because there is not thrust without friction/viscosity... To get the acceleration for both, just make the integral of power (~constant), this give speed as a square root of time. More power/weight = win $\endgroup$
    – MrBrushy
    Feb 20, 2017 at 13:35
  • $\begingroup$ Ground takeoff isn't ideal because the jet need time to ramp up power. If some kind of "wheel blocker" is used, like ones on ski-jump carriers, the jet would perform better. $\endgroup$
    – user3528438
    Nov 17, 2017 at 22:22
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    $\begingroup$ 55/70 is less than 1. How can 32.2 become 41 after multiplication by a factor < 1? $\endgroup$
    – Koyovis
    Nov 18, 2017 at 3:00
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That would be the aircraft with the highest thrust-to-weight ratio. This site lists the fighters of the world with a bit of a puzzling caveat for TO weight:

TWR or T/W ratio = (Max Thrust of Engine / (Empty Weight + (3.505 Tonnes of Fuel & Weapons, or only Internal Fuel)))

  • 1.30 - Su-35BM
  • 1.29 - F-15K
  • 1.26 - Su-27S
  • 1.25 - Eurofighter
  • 1.24 - Mig-35
  • 1.23 - Su-27SK & J-11A
  • 1.19 - Rafale C
  • 1.19 - Mig-29M/M2
  • 1.19 - F-15C
  • 1.18 - F-22 (T/W = 1.37 with Round nozzles)

Let's take the F-22 with the round nozzles, whatever those may be: thrust-to-weight of 1.37. Weight W = m * g, so

$$ a = \frac{T}{W} \cdot g = 1.37 * 9.81 = 13.44 ~\text{m/s}^2$$

60 mph = 26.82 m/s, and with V = a * t we get for t = V/a = 2.0 sec. The F-22 with the round nozzles reaches 60 mph in 2.0 seconds.

Or with TWR the thrust-to-weight ratio, and all SI units: $$ t = \frac{V}{g \cdot TWR}$$

Edit

The above is of course for frictionless circumstances, as @Manu H pointed out in a comment. Also, @Zeus pointed out that the "F-22 with the round nozzles" never became anything more than a gleam in some engineer's eye. So what do we get if we take the aircraft listed as highest T/W - the Su-35 - and make an attempt at a ROM for friction effects?

Let's take an SU-35 at TWR of 1.3. Resisting friction is caused by:

  • Rolling friction of the tyres. This Wikipedia site lists rolling resistance coefficients, let's take 0.006 (about the minimal car tyre value), so tyre drag = 0.006 the weight of the aircraft.
  • Aerodynamic friction. Let's assume that this is caused by lift-less drag $C_{D_0}$ only. This Wiki site gives subsonic drag coefficient of 0.021 for a Phantom, let's take that. This Wiki site gives further data on the Su35, such as max thrust and wing area. In order to match the TWR of 1.3 with the stated afterburner thrust of 284 kN, we'll take a mass of 22,269 kg.

The aerodynamic drag is a quadratic function of the velocity. At 60 mph = 26.82 m/s, the aerodynamic drag = $ C_{D_0} \cdot \frac{1}{2}\rho \cdot V^2 \cdot S$ = 0.021 * 0.5 * 1.225 * 62 * 26.82$^2$ = 574 N = 0.2% of thrust. That is at the end speed, the average weighted value is a third of that = 0.07% of thrust

Tyre drag = 0.006 * m * g = 1,311 N = 0.5% of thrust

So if we account for drag, we need to use about 99.4% of thrust. Time to reach 60 mph now becomes

$$ t = \frac{V}{g \cdot 0.994 \cdot TWR} = \frac {26.82}{9.81 \cdot 0.994 \cdot 1.3} = 2.1 ~\text{sec}$$

What a way to spend a Sunday...

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  • $\begingroup$ you should mention this is only theorical calulation not taking into account several factors such as drag. $\endgroup$
    – Manu H
    Nov 18, 2017 at 16:33
  • $\begingroup$ Or limited airflow at low speed $\endgroup$
    – Antzi
    Nov 19, 2017 at 3:11
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    $\begingroup$ If we are not limited to 'production' aircraft, then P-42 (a stripped-down Su-27 that set the world climb record) had TWR (in static conditions) in excess of 2. (It couldn't stand on the brakes at full power and had to be tied). This will easily win over Tesla, but then it would be fairer to compare it to dragsters. If we only account for production models, then we can't use "F-22 with the round nozzles"... $\endgroup$
    – Zeus
    Nov 19, 2017 at 3:35
  • $\begingroup$ @Zeus Oh right-o, I was wondering what that was. $\endgroup$
    – Koyovis
    Nov 19, 2017 at 3:46
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The fastest acceleration of an aircraft from a stationary position is found on catapult launched naval aircraft. For one like the FA-18, it goes 0-165mph in two seconds, which exceeds the Tesla figure by a wide margin.

Of course, that's with some assistance from the catapult.

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The fastest self-accelerating airplane ever built as far as I know was the Sukhoi π-42, a modified Cy-27. Its FAI recorded takeoff weight was 14,100 kg, and its thrust was recorded as 2x13,600 kg, giving nominal T:W at takeoff of 1.93, giving 0-60 mph in 1.41s but this is kinda theoretical: most fighter engines produce less than their rated thrust while the aircraft is static, and substantially more at higher indicated speed. Based on scaling the performance of the Su-27 in DCS (18,000 kg takeoff weight, 2 x 12500 kg thrust), these nominal numbers give the π-42 a likely 0-60 mph time of 1.8 s. Based on the same testing and assuming the same SFC, The 3000m climb woukd have required about 500 kg of fuel, and in practice there's probably some margin on that number, so a min fuel π-42 might be able to squeeze under 13,600 kg and over 2.00 nominal T:W for a 1.75 second 0-60.

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