A Tesla Roadster hovers on 28.3 kilograms of nitrogen a second, and 25.6 seconds is the ceiling

A Tesla Roadster has a kerb mass of 1,235 kg, and the bottles to hover it weigh another 376. Moog’s largest nitrogen cold gas thruster runs 57 seconds of specific impulse, so 1,611 kg throws away 1,611 ÷ 57 = 28.3 kilograms of nitrogen a second. Fifty bottles buy 5.03 seconds. No tank of 300-bar nitrogen beats 25.6.

Most of the inputs are quoted and three are mine. Tesla’s service manual gives the masses, Moog’s two datasheets the thrusters, NIST the gas, Luxfer the bottle, the BIPM both constants, NASA the class. I chose the bottle count, the tank temperature and one carry-across ratio that arrives in the last section. Nobody here has run a thruster or weighed a car, so the arithmetic is the whole argument.

Can a Tesla Roadster fly on cold gas thrusters?

No, and the number that ends the argument is 28.3 kilograms of nitrogen a second. Four quoted figures set it: the car’s mass, the thruster’s specific impulse, standard gravity, and the nitrogen a bottle holds. Tesla’s Roadster Service Manual gives, in the first row of its Technical Data Weights table, under “Curb weight (no occupants or luggage)”, 2,723 lb (1,235 kg). That is the original Roadster, the only car the manual covers, and no kerb mass is published for its replacement.

The thruster is Moog’s model 058-118, from the Cold Gas Thrusters datasheet, Form 500-1171 0121, whose Performance Characteristics table gives thrust as “3.6 N (0.8 lbf) @ 15.7 bar (230 psia) GN2” and, in the row immediately below it, “Specific Impulse (Nom) 57 sec”. Standard gravity comes from the 3rd CGPM’s 1901 declaration on the unit of mass, item 3, which fixes it at “980.665 cm/s2” — 9.80665 m/s² in modern units.

Three inputs are mine: fifty 300-bar composite bottles, a tank temperature of 293.15 K, and the carry-across ratio at the end. None is a measurement. We don’t own a thrust stand, and the gas and the bottle arrive in the third section with their own links.

Start with F = mg, because hover thrust is just weight

Hovering means thrust exactly equals weight, so the problem is one multiplication and one division. The bare car weighs 1,235 × 9.80665 = 12,111.21 N. Exhaust velocity comes out of the specific impulse through the same constant: ve = Isp × g0 = 57 × 9.80665 = 558.979 m/s. Thrust is mass flow times exhaust velocity, so the flow you need is weight over exhaust velocity.

Leave the symbols in and something falls out. Mass flow = m × g ÷ (Isp × g0), and at the Earth’s surface g is taken as g0 by convention, which is the convention the 1901 declaration fixed, so the two cancel and mass flow = m ÷ Isp.

The gravitational constant cancels.

To hover you throw away your own mass divided by your specific impulse, every second, and the seconds in “specific impulse” are those seconds. Off Earth it stops cancelling and the local g comes back into the sum.

The cancellation isn’t mine. An undated back-of-envelope at ScopeViews, “Can SpaceX Really Make a Tesla Fly?”, runs the identical substitution and prints “m = 2000 Kg / 70s = 28 Kg/s”. Different car mass, different specific impulse, same 28 — a coincidence of two roundings, not a confirmation. Where that page and this one part company is the tank, in section six.

NASA says what cold gas buys you, in chapter 4.0 of its State-of-the-Art of Small Spacecraft Technology:

Gaseous nitrogen is a commonly used propellant for cold gas systems… The low specific impulse of these systems limits their ability to provide large orbital correction maneuvers… Cold gas thrusters are often attractive and suitable for small buses due to their relatively low cost and complexity.

Small manoeuvres, small buses. Hold on to that.

The mass flow rate to hover a Tesla Roadster is 28 kilograms a second

28.265 kilograms a second, for a car with fifty bottles of nitrogen bolted to it. Fifty is a number I chose, picked to sit clear of the limit derived two paragraphs down. The bottle is Luxfer’s LCX-15Y carbon composite cylinder, part L87A: 300 bar, 9 litres of water volume, 4.8 kg.

The ladder runs from bars to kilograms a second, and every rung is printed. 300 bar is 30,000,000 Pa. At that pressure and 293.15 K, the NIST Chemistry WebBook isotherm for nitrogen at 300 bar and 20 °C returns 302.44 kg/m³. Nine litres is 0.009 m³, so 0.009 × 302.44 = 2.72196 kg of nitrogen per bottle. Fifty bottles hold 50 × 2.72196 = 136.098 kg of gas and weigh 50 × 4.8 = 240.0 kg empty. Add the car: 1,235 + 136.098 + 240.0 = 1,611.098 kg on the ground. Multiply by standard gravity, 1,611.098 × 9.80665 = 15,799.47 N, then divide by exhaust velocity, 15,799.47 ÷ 558.979 = 28.265 kg/s.

The schoolbook figure misleads, and it misleads on the second rung. PV = nRT at 300 bar and 293.15 K, with NIST’s molar gas constant of 8.314462618 J/(mol·K) and the WebBook’s molecular weight for nitrogen of 28.0134, gives 30,000,000 × 0.0280134 ÷ (8.314462618 × 293.15) = 344.80 kg/m³ and 3.1032 kg in the bottle. 344.80 ÷ 302.44 = 1.1401 is the compressibility factor you skipped, so the real figure sits (344.80 − 302.44) ÷ 344.80 = 12.29% below the schoolbook one.

Further down the ladder sits a licensing problem, not a physics one. The same Weights table gives a gross vehicle weight of 3,273 lb (1,485 kg) “with two 220 lbs (100kg) occupants and 110 lbs (50 kgs) of luggage”, so headroom above kerb mass is 1,485 − 1,235 = 250 kg. Each bottle brings 2.72196 + 4.8 = 7.52196 kg, and 250 ÷ 7.52196 = 33.2, so thirty-three bottles is the most the car is rated to carry. Fifty puts it 1,611 − 1,485 = 126 kg over that before anybody sits in it.

Counted a second way: 4,389 thrusters

Divide the weight by one thruster: 15,799.47 ÷ 3.6 = 4,388.74, so 4,389 Moog 058-118s once you round up to whole valves. Each flows 3.6 ÷ 558.979 = 0.0064403 kg/s, and 4,388.74 × 0.0064403 = 28.2648 kg/s against 15,799.47 ÷ 558.979 = 28.2649 from the weight-over-exhaust-velocity route. Two roads, five figures, one digit of daylight.

You would be bolting four thousand three hundred and eighty-nine of them to a car.

NASA doesn’t survey anything bigger. Table 4-5, “Cold Gas Propulsion”, in the In-Space Propulsion chapter of the State-of-the-Art of Small Spacecraft Technology, lists seven nitrogen devices and the largest makes 3.6 N. Table 4-1 near the top of the same chapter caps the whole category at “10 μN – 3.6 N”, so one Moog part number sets the ceiling of an entire propulsion class in NASA’s own survey. Table 4-5 sits below the whole chapter’s technology descriptions and four earlier tables, and when you get there the chapter has labelled its hydrazine table “Table 4-1” as well — two tables, one number, and no Table 4-2 on the page at all. Small manoeuvres and small buses, restated as a table.

Counting devices surfaces two figures the first route hides. 4,389 thrusters at 23 grams each is 100.95 kg of valve bodies that were never in the 1,611 kg: put them in and the mass at ignition becomes 1,712.05 kg, the flow 1,712.05 ÷ 57 = 30.04 kg/s, and the hover time falls from 5.03 seconds to 4.72. I haven’t iterated the count, which doesn’t converge quickly — at 1,712.05 kg you need 1,712.05 × 9.80665 ÷ 3.6 = 4,664 valves. Each draws 30 W, so 4,389 × 30 = 131,670 W, and 131.67 ÷ 215 = 61.2% of the 215 kW the service manual gives the traction motor goes on holding valves open.

How long will 50 nitrogen bottles keep a car off the ground?

5.03 seconds. The car gets lighter as it burns, so the honest form is the rocket equation with delta-v = g × T. Dry mass is 1,611.098 − 136.098 = 1,475.0 kg, so T = 57 × ln(1,611.098 ÷ 1,475.0) = 57 × 0.0882579 = 5.0307 s. Check it through velocity: 558.979 × 0.0882579 = 49.3343 m/s of delta-v, and 49.3343 ÷ 9.80665 = 5.0307 s.

More bottles help, less than you’d expect.

Mass at ignition, mass flow and hover time against bottle count, at 2.72196 kg of nitrogen and 4.8 kg of shell each:

Doubling the gas buys 8.397 − 5.031 = 3.37 seconds and halving it costs 5.031 − 2.793 = 2.24, sub-linear both ways, because every kilogram of nitrogen arrives with 4.8 ÷ 2.72196 = 1.7634 kg of bottle that never leaves. The 33-bottle row is the last the car’s rating allows, and it is the row I’d build if somebody insisted: 3.561 seconds, street legal, and still a car that cannot cross a car park.

The bottle ceiling: 25.6 seconds, whatever tank you bolt on

No arrangement of 300-bar nitrogen bottles holds anything off the ground for longer than 25.6 seconds, and the proof never mentions the car. Call it the bottle ceiling: maximum hover endurance on gas stored in pressure vessels is T_max = Isp × ln(1 + m_gas ÷ m_bottle), which turns on the propellant’s specific impulse and the vessel’s gas-to-shell ratio and nothing else. Add bottles without limit and the car underneath stops mattering, because the mass ratio tends to one full bottle over one empty one.

m_gas / m_bottle = 2.72196 / 4.8      = 0.567075
mass ratio       = 1 + 0.567075       = 1.567075
T_max = 57 x ln(1.567075)             = 57 x 0.4492108  = 25.605 s
check = 558.979 x 0.4492108 = 251.10 m/s; 251.10 / 9.80665 = 25.605 s

251.10 m/s is 251.10 × 3.6 = 904.0 km/h, which is what that gas could do pointed sideways in a vacuum instead of downward in air.

Now the ceiling earns its keep. ScopeViews reaches the opposite verdict from the same algebra — “yes, SpaceX CGT tech really could make a Tesla Roadster fly, allowing around 15-20 seconds of hover time and a slow drift across a car park” — on 500 kg of air in a one-cubic-metre vessel at 10,000 psi. Run the rocket equation on that page’s own three numbers. A 2,000 kg car carrying 500 kg of air starts at 2,500 ÷ 70 = 35.71 kg/s, because the air has to lift itself, and the 2,000 ÷ 70 = 28.57 behind the published 28 is the flow at the instant the tank runs dry. With a vessel of zero mass the burn lasts 70 × ln(2,500 ÷ 2,000) = 15.62 s, so fifteen seconds is the figure for a tank that weighs nothing and twenty is unreachable at any tank mass. Put Luxfer’s shell ratio on it — my carry-across, not theirs, from a 300-bar cylinder to a 690-bar one — and 500 × 1.7634 = 881.70 kg of vessel gives 70 × ln(3,381.70 ÷ 2,881.70) = 11.20 s. A purpose-built vessel at 690 bar should beat a breathing-air cylinder, so 11.20 s is the pessimistic end and I wouldn’t defend the second digit.

Now derate the specific impulse, the input doing most of the work. Halve it to 28.5 s and fifty bottles give 28.5 × 0.0882579 = 2.52 seconds; quarter it to 14.25 s and they give 1.26. Multiply by four and you land on 228 seconds, which is not a round number I picked but the published nominal for Moog’s MONARC-22-6 hydrazine thruster. Hold the mass ratio at the fifty-bottle figure and 228 × 0.0882579 = 20.12 seconds, or 20.74 at Moog’s best monopropellant row of 235 s. Four times the specific impulse buys fifteen extra seconds and 136 kg of hydrazine to carry. The decision doesn’t flip either way, because the binding constraint was never the propellant — it is 1,235 kg of car that cannot be thrown out of a nozzle.

Why a vacuum nozzle loses most of its thrust at sea level

Because thrust is F = mdot × ve + (pe − pa) × Ae, and the second term is zero only in vacuum. Ambient pressure pushes back on the exit plane, and the 10th CGPM’s 1954 resolution on the standard atmosphere fixes how hard: “1 standard atmosphere = 1 013 250 dynes per square centimetre”, or 101,325 N/m².

Bound the penalty from printed numbers. Moog gives the 058-118’s envelope as Ø6.6 mm, so its exit area cannot exceed π × 0.0033² = 34.212 mm², and at one standard atmosphere 101,325 × 0.000034212 = 3.4665 N of back pressure sits against a thruster rated at 3.6 N. That is 96.3% of it taken straight back by the air.

The real figure is lower, and the next number is inferred where the others were read. NIST’s isotherm at Moog’s own 15.7 bar inlet and 20 °C gives Cp = 1.0665 and Cv = 0.74688 J/(g·K), so nitrogen’s ratio of specific heats at that state is 1.0665 ÷ 0.74688 = 1.4279. (At the 300 bar storage state the same table gives 1.3681 ÷ 0.79575 = 1.719, which is why the state has to be named, since the gas alone does not fix the ratio.) A choked-flow check at 15.7 bar and 293.15 K puts the throat at 1.755 mm² and caps the expansion ratio inside a 6.6 mm body at 34.212 ÷ 1.755 = 19.5:1. Below roughly 19:1 the 058-118 can’t be a high-area-ratio vacuum design, so the penalty lands nearer half the thrust than all of it: at 10:1 the exit is 17.55 mm², the back pressure 101,325 × 0.00001755 = 1.778 N, and 1.778 ÷ 3.6 = 49.4%.

A quoted peak rate is not a sustained resource flow

The same error that makes a car look flyable makes a throughput claim look affordable: a peak figure gets quoted and the flow behind it never gets costed. Moog’s own datasheet has it. The 058-118’s Life (Cycles) row reads > 10,000 and the response rows above it give < 4 msec to open and < 4 msec to close, so at the sheet’s own bounds a full cycle is 8 ms, the valve modulates at 125 Hz, and the 10,000-cycle floor is about eighty seconds of duty. Eighty, from a floor multiplied by a ceiling, so the true life is longer and the true cycle shorter. The quirk is sharper than that one figure: the four other models on the same page are rated 500,000–2,000,000, 20,000, 1,000,000 and > 100,000 cycles, so the thruster NASA puts at the top of the class is the worst valve on its own datasheet.

Systems claims take the identical shape. A pipeline quoted at a peak events-per-second implies bytes, IOPS and egress for every hour it holds that rate, and the replay arithmetic behind a six-hour streaming backlog is the same division in another unit. Before quoting a software build I ask for the sustained figure and the hours it has to hold, because a peak is a number somebody measured once.

Moog publishes a specific impulse with no ambient pressure to read it against

The weakest input is the one Moog doesn’t print. The Cold Gas Thrusters datasheet gives no nozzle expansion ratio for any model and no ambient reference condition for 3.6 N and 57 sec: it names the inlet pressure and says nothing about what the nozzle exhausts into. Moog’s Monopropellant Thrusters datasheet, Form 500-934 0424, same house style and the same Performance Characteristics heading, prints “Nozzle Expansion” for every one of its seven models, from 40:1 to 135:1. A thrust figure whose ambient condition isn’t stated is not a specification, and I wouldn’t design an atmospheric system from that sheet.

Costing the gap takes four lines. Nitrogen’s ideal exhaust velocity at 293.15 K is sqrt(2γ ÷ (γ − 1) × Rs × T0) = 762.03 m/s, at γ = 1.4279 and Rs = 8.314462618 ÷ 0.0280134 = 296.803 J/(kg·K), so 762.03 ÷ 9.80665 = 77.71 s is the ceiling in vacuum. Expanded optimally from 15.7 bar to one standard atmosphere it falls to 570.31 m/s, 58.16 s. Moog’s 57 is 57 ÷ 77.71 = 73.35% of the vacuum ceiling, and the same 73.35% carried across to sea level — my assumption, not Moog’s — gives 58.16 × 0.7335 = 42.66 s, then 1,611.098 ÷ 42.66 = 37.77 kg/s and 42.66 × 0.0882579 = 3.77 seconds of hover. I’d want that ratio from Moog before quoting it to anybody.

Two smaller admissions. Until I opened NASA’s Table 4-5 I would have told you nitrogen cold gas was good for about 70 seconds, and the 70 in that table belongs to the Benchmark Starling and the Moog 58E163A, neither of which makes a newton and a half. And Luxfer’s page carries no note against its weight column, so I can’t tell you whether 4.8 kg includes the valve; its own air capacity works out at 2,434 ÷ 9 = 270.44 litres per litre against an ideal 300, a (300 − 270.44) ÷ 300 = 9.85% discount where nitrogen gave 12.29%. Different gas, no stated reference temperature, and I’m not averaging them.

One last division, because it answers the engineering question and not the physics one. 1,235 × 9.80665 ÷ 445 = 27.2, so twenty-eight MONARC-445s lift the Roadster’s kerb mass where 4,389 cold gas valves were needed, a factor of 4,389 ÷ 28 = 157 on device count bought by changing nothing but the propellant. On paper. That thruster’s published 50:1 nozzle has a 14.8 cm exit, and at sea level 101,325 × π × 0.074² = 1,743.1 N of air pushes back on it, 3.92 times its own rated thrust, so the nozzle separates and the 445 N means nothing down here. And the size of the thruster was never the lever anyway. Flow per newton is 1 ÷ (Isp × g0), so at 57 seconds it is 1 ÷ 558.979 = 0.00179 kg/s whether you bolt on one valve or 4,389. A bigger nozzle sells you thrust. Only specific impulse sells you seconds.