Can the Roadster SpaceX jump, fly, drive ? The full calculation.

Motion. The car is a variable-mass point. Position \((x,y)\), velocity \((v_x,v_y)\), mass \(m\):

$$\frac{dx}{dt}=v_x,\quad \frac{dy}{dt}=v_y,\quad \frac{dv_x}{dt}=\frac{F_x}{m},\quad \frac{dv_y}{dt}=\frac{F_y}{m}-g,\quad \frac{dm}{dt}=-\dot m$$

4th-order Runge-Kutta integration, 2 ms step during thrust. On flat ground, liftoff only happens if \(F_y \ge m g\). On a ramp of slope \(p\): \(\theta=\arctan(p)\), \(v_{x0}=v_0\cos\theta\), \(v_{y0}=v_0\sin\theta\).

Tank. The stored mass comes from the equation of state with compressibility factor \(Z\) (air, 293 K): \(m_0 = \dfrac{P_0 V}{Z(P_0)\,R\,T_0}\) with \(R=287.05\) J/(kg·K). Venting in under a second is adiabatic: the entropy of the remaining gas stays constant, which gives the tank pressure and temperature as a function of the remaining mass fraction \(f=m/m_0\). These curves \(P(f)\), \(T(f)\) are tabulated from a reference equation of state (Lemmon et al. for air) and interpolated.

Thrust. For a choked fixed-throat nozzle, thrust is proportional to the supply pressure: \(F=C_F A_t P_a\). With a regulator set to \(P_{reg}\): \(P_a=\min(P_{tank},P_{reg})\), hence

$$F(t)=F_0\cdot\min\!\left(1,\frac{P_{tank}(t)}{P_{reg}}\right)$$

A variable-throat nozzle holds \(F_0\) as long as \(P_{tank}\) is above a floor (60 bar). Only the lift nozzles fire during the jump; mass flow is \(\dot m = F_y/V_e\), with an exhaust velocity that follows the gas cooling: \(V_e = I_{sp}\,g\sqrt{T_{tank}/T_0}\).

Landing. On the way down, braking starts at height \(h_f=\dfrac{v_y^2-v_t^2}{2a}\) with \(a=F_y/m-g\) and \(v_t=3\) m/s the targeted touchdown speed (the simulator applies a 1/0.6 margin because thrust drops during braking). On the ground, the suspension (0.15 m of travel) absorbs \(\bar F = \dfrac{m v_t^2}{2 s}\).

Clearance. The car's length (4.4 m) is taken into account: the front and the rear, at the floor height, must clear each obstacle along the whole trajectory, not just at the apex.

Energy balance. Two checks run on every calculation: the work \(\int F\cdot v\,dt\) corrected for the ejected-mass term must equal the car's change in mechanical energy; and the car cannot receive more energy than the jet contains, \(\tfrac12 m_{gaz}V_e^2\).

Stored mass. \(m=\dfrac{PV}{Z(P,T)\,R\,T}\). The factor \(Z\) comes from a reference equation of state (Lemmon, Jacobsen, Penoncello and Friend, 2000, for air; Span et al. for nitrogen). At 690 bar and 293 K: \(Z_{air}=1.544\), \(Z_{N_2}=1.603\). The Redlich-Kwong equation used in the previous version of this page gave 1.463, a 5.6% difference on mass.

Real pressure for 250 kg in 600 L. Density \(\rho=250/0.6=417\) kg/m³; inverting the equation of state gives \(P=437\) bar (6,330 psi).

Adiabatic venting. In under a second, the gas does not exchange heat with the wall. The gas remaining in the tank undergoes isentropic expansion: \(s(\rho,T)=s_0\). For an ideal gas this gives \(T/T_0=(\rho/\rho_0)^{\gamma-1}\) and \(P/P_0=(\rho/\rho_0)^{\gamma}\) with \(\gamma=7/5\). The chart values come from the real equation of state, which departs from the ideal law by 10 to 15% at these pressures.

Recoverable energy (exergy). The maximum work the gas can deliver as it returns to ambient: \(E_x=m\left[(h-h_0)-T_0(s-s_0)\right]\). Cross-checked with the ideal-gas isothermal formula, \(E_x\approx mRT_0\ln(P/P_0)=48.6\) kWh, within 3.6% of the real-gas calculation.

Burst energy. Isentropic expansion of the gas down to 1 atm: \(E_b=m\,(u-u_e)\) with \(u\) the internal energy. The TNT equivalent uses 4.184 MJ/kg. A pneumatic burst releases about 40% of this energy as a shock wave (section 13).

Tank mass. For carbon-wound composite tanks (type IV), mass is proportional to the product of pressure and volume: \(m_{res}\approx \dfrac{PV}{100\text{ to }115\text{ kJ/kg}}\), figures taken from automotive 700 bar hydrogen tanks.

A useful law. With \(PV=Z\,m_{gaz}RT\) and \(m_{res}=PV/k\): \(\dfrac{m_{res}}{m_{gaz}}=\dfrac{Z\,R\,T_0}{k}=0.785\,Z\). At 690 bar (\(Z=1.54\)) the tank weighs 1.21 times the gas; at 440 bar (\(Z=1.25\)) 0.98 times. Impulse per kilogram of complete system goes from 0.25 to 0.28 kN·s/kg (effective exhaust velocity 549 m/s).

For a gas with a heat-capacity ratio \(\gamma=1.4\) and a choked throat (upstream pressure at least 1.9 times downstream pressure), mass flow depends only on supply pressure and temperature:

$$\dot m = \frac{A_t P_a}{\sqrt{R T_a}}\;\sqrt{\gamma}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}} = 0.6847\,\frac{A_t P_a}{\sqrt{R T_a}}$$

Thrust is given by \(F=C_F A_t P_a\). For ideal expansion down to ambient pressure \(P_e=P_{amb}\):

$$C_F=\sqrt{\frac{2\gamma^2}{\gamma-1}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}\left[1-\left(\frac{P_e}{P_a}\right)^{\frac{\gamma-1}{\gamma}}\right]}$$

Exit Mach: \(\dfrac{P_a}{P_e}=\left(1+\dfrac{\gamma-1}{2}M_e^2\right)^{\frac{\gamma}{\gamma-1}}\); exit temperature: \(T_e=\dfrac{T_a}{1+\frac{\gamma-1}{2}M_e^2}\). Air liquefies around 80 K at 1 bar, which caps \(M_e\) at about 3.65. Effective exhaust velocity with the pressure term: \(V_e=\dfrac{F}{\dot m}=u_e+\dfrac{(P_e-P_{amb})A_e}{\dot m}\).

At every instant, flow \(\dot m = F/V_e\) and \(F = F_0\,P_{tank}/P_0\) (fixed, choked throat). For an ideal gas, the tank density \(\rho\) follows \(\dot\rho = -\dot m/V\) and pressure \(P=P_0(\rho/\rho_0)^{\gamma}\). Combining them, the isentropic venting of a tank through a choked throat gives a closed-form law:

$$\frac{P(t)}{P_0}=\left(1+\frac{\gamma-1}{2}\,\frac{t}{\tau}\right)^{-\frac{2\gamma}{\gamma-1}}=\left(1+\frac{t}{5\tau}\right)^{-7}\qquad \tau=\frac{V}{A_t\,\Gamma\,\sqrt{R T_0}}$$

with \(\Gamma=0.6847\). For the Thrust, \(\tau\approx0.94\) s; for the Rocket, \(\tau\approx2.8\) s. The table values use the real equation of state rather than the ideal law, which shortens the times slightly.

Total impulse is \(\int_0^\infty F\,dt\). The "average thrust over 5 s" is this impulse divided by 5 s. Simple check: 106 kg ejected at 549 m/s gives 58 kN·s total; a fixed nozzle only exploits 37 of it because the end of venting happens at low pressure and low temperature.

Holding 0.59 s with a fixed nozzle would require 117 kN at the start. Maximum Thrust mass to stay under 0.9 s with real venting: 1,905 kg.

Tractive force available at the wheel, limited by grip, power and torque:

$$F_{roue}=\min\Big(\mu\,(mg+D),\;\frac{\eta P_{max}}{v},\;F_{cap}\Big)$$

with \(\mu=1.26\) (grip coefficient deduced from Tesla's "1.26 g on the tires" claim), \(D=46.1\) kN of suction downforce, \(\eta=0.92\), \(P_{max}=1\) MW. \(F_{cap}=28.5\) kN is obtained by inverting the announced 1.6 s. Equation of motion: \(m\,\dot v = F_{roue}+F_{tuyères}(t)\), integrated up to \(v=26.82\) m/s. Nozzle thrust follows section 5.

Torque cross-check: \(F=\dfrac{C\cdot i}{r}=\dfrac{1\,302\times 9}{0.35}=33.5\) kN for a gear ratio of 9 and a 0.35 m wheel radius; the order of magnitude matches the inverted 28.5 kN.

The simulator shows that with a fixed nozzle behind a 380 bar regulator, the pressure left after the climb is no longer enough to brake a 2,475 kg car: vertical thrust drops below weight. Landing like in the demo requires either a variable-throat nozzle, a pressure reserve dedicated to braking, or a lighter car. A full jump (climb + braking) consumes 100 to 130 kg of gas depending on mass, a third to half a tank.

Height reached after thrust: \(H=y_1+\dfrac{v_{y1}^2}{2g}\) where \(y_1\), \(v_{y1}\) are the height and vertical velocity at the end of thrust (integrated, section 1). For a constant thrust \(F_v\) over a duration \(T\) and constant mass: \(v_{y1}=(F_v/m-g)T\), \(y_1=\tfrac12(F_v/m-g)T^2\).

Free fall: \(v_{imp}=\sqrt{2gH}\). Braking at thrust \(F_v\): deceleration \(a=F_v/m-g\), trigger height \(h_f=\dfrac{v_{imp}^2-v_t^2}{2a}\), duration \(\dfrac{v_{imp}-v_t}{a}\), gas \(\dfrac{m(v_{imp}-v_t)}{V_e}\). Suspension: \(\bar F=\dfrac{m v_t^2}{2s}\) with \(s=0.15\) m.

Chassis flex on landing, treating the tanks (700 kg) as a load at mid-wheelbase \(L_e=2.7\) m resting on two supports: \(M=\dfrac{m_{res}\,a\,L_e}{4}=15\) kN·m at 3.2 g, 23 kN·m at 5 g. Minimum section modulus at 300 MPa with a 1.5 safety factor: 74 to 116 cm³. A carbon shell absorbs it; this is not the hard point.

Inertia of a homogeneous box of mass \(m_c\): \(I_{roulis}=\dfrac{m_c(W^2+H^2)}{12}\), \(I_{tangage}=\dfrac{m_c(L^2+H^2)}{12}\), \(I_{lacet}=\dfrac{m_c(L^2+W^2)}{12}\), to which the tanks are added (0.35 m-radius cylinder for roll, point mass at 0.8 m for pitch and yaw). Dimensions 4.4 × 1.9 × 1.1 m, body 1,700 kg, tanks 700 kg.

To turn by an angle \(\theta\) over a duration \(\Delta t\) at constant speed: \(\omega=\theta/\Delta t\), angular momentum \(L=I\omega\), impulse of a nozzle with arm \(r\): \(J=L/r\), once to start, once to stop. Gas: \(2J/V_e\). Attitude error: \(\delta\theta=\theta\cdot\delta J/J\); valve delay: \(\delta\theta=\omega\,\delta t\).

Thrust = weight: \(\dot m\,V_e=mg\) with \(V_e=I_{sp}g\), so \(\dot m = m/I_{sp}\) and

$$m(t)=m_0\,e^{-t/I_{sp}}\qquad t_{max}=I_{sp}\ln\frac{m_0}{m_0-m_{gaz}}$$

Gas fraction needed for a duration \(T\): \(1-e^{-T/I_{sp}}\). Rotor lift (momentum theory): induced power \(P=\dfrac{1}{\mathrm{FM}}\sqrt{\dfrac{(mg)^3}{2\rho A}}\) with a figure of merit \(\mathrm{FM}=0.7\) and \(\rho=1.2\) kg/m³.

Vacuum: \(\Delta P = D/A_{eff}\). Air entry speed under the skirt: \(u=\sqrt{2\Delta P/\rho}=107\) m/s. Flow the fans can extract: \(Q=\eta_f P_{fan}/\Delta P\). Equivalent clearance: \(e=\dfrac{Q}{C_d\,u\,L_{per}}\) with \(C_d=0.6\). Maximum lateral acceleration: \(a_y=\mu\,\dfrac{mg+D}{m}\). Driving on a ceiling requires \(D>mg\).

At 370 km/h, the tire tread undergoes 3,000 g of centrifugal acceleration (2,800 rpm): special-category tires are needed, and a 1.2 m² wing there produces 11 kN of downforce for 78 kW of drag.

Consumption per kilometer at constant speed \(v\):

$$e(v)=\frac{\tfrac12\rho\,C_dA\,v^2 + C_{rr}\,m\,g}{\eta}+\frac{P_{aux}+P_{fan}}{v}$$

with \(\rho=1.2\) kg/m³, \(C_dA=0.55\) m² (assumption, Tesla has not published the coefficient), \(C_{rr}=0.010\), \(\eta=0.88\), \(P_{aux}=400\) W. Range: \(E_{utile}/e(v)\) with 95% of nominal capacity usable. Top speed: \(P_{roues}=\dfrac{\tfrac12\rho C_dA v^3 + C_{rr}mgv}{0.9}\). Dynamic pressure at 370 km/h: \(q=\tfrac12\rho v^2=6.3\) kPa. Tread centrifugal acceleration: \(\omega^2 r\) with \(\omega=v/r\).

Recharging at 500 kW from 15 to 80% of 150 kWh takes 11.7 minutes, consistent with the announced "under 15 minutes". The 1 MW claim is tenable provided there is a 150 kWh pack and low-resistance cells.

Number of cells: \(n=E/97\) Wh (4680 cell). Architecture: \(n_s=243\) in series for 900 V (3.7 V per cell), \(n_p=n/n_s\) in parallel. Pack resistance: \(R=\dfrac{n_s R_{cell}}{n_p}\). Heat: \(Q=I^2R\) with \(I=P/U\). Adiabatic heating: \(\Delta T=\dfrac{Q\,t}{m_{cell}\,c_p}\) with 355 g and 900 J/(kg·K) per cell. Heat at partial power scales as the square of power: \(Q(P)=Q_{1MW}(P/1\,\mathrm{MW})^2\).

Shock wave. TNT equivalent \(W=0.4\,E_b/4.184\) MJ (a pneumatic burst releases about 40% of the isentropic energy as a pressure wave). Distances via the Hopkinson scaling law: \(R=Z\,W^{1/3}\), with \(Z\approx3.5\) m/kg^{1/3} for 5 psi and 18 for 1 psi (Kingery and Bulmash charts).

Noise. Mechanical jet power \(P_j=\tfrac12 F V_e\). Acoustic power \(P_a=\eta_a P_j\) with \(\eta_a=0.2\) to 1% for a cold supersonic jet. Power level \(L_w=10\log_{10}(P_a/10^{-12})\); level at distance \(r\): \(L_p=L_w-20\log_{10}r-8\).

Blast. On-axis speed of a free turbulent jet far from the nozzle: \(u(x)\approx 7\,\dfrac{\sqrt{F/\rho}}{x}\), where \(F\) is the nozzle's thrust.

Recharge energy: \(E=E_x/\eta_c\) where \(E_x=50.4\) kWh is the gas exergy (section 3). Time: \(E/P_{comp}\). Instantaneous propulsive efficiency of a nozzle on a mover at speed \(v\): \(\eta_p=\dfrac{2v/V_e}{1+(v/V_e)^2}\); integrated over the 0-60 (6.5 m, 0.45 s, 520 kJ of work for 65 kg of gas) it is 5.3%. Equivalent electrical cost: \(W/0.9\).

• Storage between 400 and 500 bar, not 690: tank mass per kilogram of gas dictates the lowest pressure compatible with the volume (lead 6).
• Regulated, constant thrust on the Rocket version (regulator or variable throat), otherwise the 0.59 s figure isn't reproducible (section 6).
• Locked gas reserve for landing, 40 to 75 kg, invisible to the driver (section 7).
• Flight time presented as a "jump": the physical ceiling of 6 to 8 s won't move without changing fluid (section 9, lead 4).
• Station recharge via a cascade of pre-compressed banks, 3 to 12 minutes, with pre-chilled gas to offset the +117 K of fill heating; never via an onboard compressor (section 14).
• 1 MW as a peak, 10 to 30 s, 150 kWh pack (lead 8).
• Fans controlled by driving mode, switched off on the road to hold the 300 miles (sections 10 and 11).
• Thrusters locked out off-track by geofencing: noise and blast incompatible with public roads (section 13).

These predictions follow from this page's equations under the assumptions of section 15. They will be verifiable once the spec sheets are published.