The physics of the Roadster wall-flip simulation

Report on sim/wall_flip_sim.py, written 2026-10-17. Every number on this page comes from the code, from a run of it, or from PROJECT.html. Code links go to the file on GitHub, by line. Other pages: videos, 3D replay, development journey.

Part 1: the theory

0. Setting: clocks, axes, symbols

Clock. All times are seconds on the VideoFromSpace clock (eqOfmtTJD1A). The PCMag clip runs 61.35 s later: PCMag time = sim time + 61.35 s. The header of data/hud_wall_flip.csv still says 61.0 s. That value is old. The gauge readings corrected it to 61.35 s on 2026-10-16.

World axes. \(x\) is horizontal, along the wall's plane. \(x = 0\) is the foot of the curve. The car starts at negative \(x\), climbs toward positive \(x\), and the overhang carries it back over negative \(x\). \(y\) points to the side, and \(z\) points up. The flight model keeps the car in the plane \(y = 0\) (see Known limits).

Body axes. \(x_b\) points forward, \(y_b\) left and \(z_b\) up, from the car's centre of gravity (CG). The rotation matrix \(R\) takes a body vector into world axes: \(\mathbf{v}_{\text{world}} = R\,\mathbf{v}_b\).

Altitude. The HUD ALTITUDE is the CG height above its height when parked. The sim stores the CG at \(z = h + h_{cg}\), with \(h_{cg} = 0.45\) m.

1. Rebuilding the wall from the HUD

No source gives the wall dimensions. During the first, slow climb (0:05.5–0:27 in VideoFromSpace), the HUD shows PITCH \(\theta\) and ALTITUDE \(h\). These two gauges are enough to rebuild the shape of the CG path.

Step 1: the slope relation. Let the car move a short distance \(ds\) along the track at pitch \(\theta\). The CG then rises \(dh = \sin\theta\,ds\) and moves \(dx = \cos\theta\,ds\) sideways. Divide the two to remove \(ds\):

Here \(\bar\theta_k\) is the mid-point pitch of step \(k\). The code steps \(\theta\) from 0° to 170° in 0.1° steps.

Step 2: the entry curve. Near \(\theta = 0\), \(\tan\theta \to 0\). A 0.1 ft read error in \(h\) then gives a large error in \(x\). So the code drops readings below 20° and uses a circle there. A circle of radius \(R\) that starts flat on the road has \(h = R(1-\cos\theta)\). The least-squares radius minimises \(\sum_i \left[h_i - R(1-\cos\theta_i)\right]^2\). Set the derivative with respect to \(R\) to zero:

The sum uses the readings between 20° and 80° (27.6° to 76.4°). The result is \(R = 8.656\) m.

Step 3: the blend. Above the circle, a shape-preserving cubic (PCHIP) passes through the readings \(h(\theta)\). Between 20° and 40°, a linear weight \(w = (\theta - 20°)/20°\) moves from the circle to the data: \(h = (1-w)\,R(1-\cos\theta) + w\,h_{\text{PCHIP}}(\theta)\).

Step 4: the ceiling. From the first 170° reading, the pitch gauge stays at 170°. The code makes the rest a straight line at 170°. It must reach the release altitude \(z_{rel}\), which is the lag-corrected HUD altitude at 52.25 s (§10). Its length is

\(L_{lip}\) is the extra ceiling past the release point: 1.5 m in the default hud variant, 0 m in the short variant. Last, the code adds 40 m of flat road in front of the foot and integrates the arc length \(s = \int\sqrt{dx^2 + dz^2}\).

2. Motion on the wall

On the wall, the model does not predict the motion. It reads the motion from the HUD of the flip run and then computes the forces that this motion needs. This is an inverse (kinematic) model.

Altitude to track position. On the wall, \(dh/ds = \sin\theta > 0\) for \(0 \lt \theta \lt 180°\). So \(h(s)\) increases strictly and has an inverse \(s(h)\). Each lag-corrected HUD altitude gives one track position.

Start position from the G gauge. The car starts from rest at 45.65 s, when G steps from 1.0 to 1.1. The altitude stays below 0.1 ft until 46.3 s. On a flat road, the accelerometer measures the vector difference of the car's acceleration \(a\) and gravity. Its size is \(\sqrt{a^2 + g_0^2}\). So

With constant acceleration for 0.65 s (45.65 to 46.3 s), the car travels \(\tfrac12 \cdot 0.46\,g_0 \cdot 0.65^2 = 0.95\) m. The code puts the start that far before the point where the altitude first reads 0.1 ft.

Smoothing. The readings \(s_i\) are noisy, and the loads need the second derivative. A smoothing spline \(f(t)\) minimises

Speed is \(v = f'(t)\) and acceleration along the track is \(a = f''(t)\). The code samples them every 0.01 s from 44.0 s to the release at 52.25 s. Before 45.65 s it sets \(v = a = 0\).

3. Loads on the wall: tyres, suction, G, power

Path frame. At pitch \(\theta\), the unit tangent along the track and the unit normal away from the surface (the car's "up") are

Acceleration on a curved path. Differentiate the velocity \(\mathbf v = v\,\hat{\mathbf t}\). Because \(d\hat{\mathbf t}/ds = \kappa\,\hat{\mathbf n}\) and \(ds/dt = v\):

The second term is the centripetal acceleration. On the concave wall it points away from the surface, into the car's "up" direction.

Forces. Five forces act on the car: tyre traction \(F_t\,\hat{\mathbf t}\); the surface's push \(N\,\hat{\mathbf n}\) with \(N \ge 0\); suction \(-F_s\,\hat{\mathbf n}\), which pulls the car onto the surface; gravity \(m\mathbf g\); and air drag \(-D\,\hat{\mathbf t}\) with \(D = \tfrac12\rho\,(C_dA)_x\,v^2\). Gravity along the two directions is \(m\mathbf g\cdot\hat{\mathbf t} = -mg_0\sin\theta\) and \(m\mathbf g\cdot\hat{\mathbf n} = -mg_0\cos\theta\). Newton's second law along each direction gives

Friction limit. The tyres can transmit at most \(\mu N\): \(|F_t| \le \mu N\). Put in \(N\) and solve for the suction:

On the ceiling, \(\theta = 170°\) and \(\cos\theta = -0.985\). The last term then becomes \(+0.985\,mg_0\): suction must first cancel almost all of the car's weight, and then supply the normal load for traction.

Accelerometer. An accelerometer measures the specific force \(\mathbf a - \mathbf g\), not \(\mathbf a\). At rest it reads \((0, 0, g_0)\), that is 1 G. The model's G gauge is

Power. The power at the wheels is \(P_w = F_t\,v\).

4. Free flight: the rigid body

After release at 52.25 s, the car is a free rigid body with six degrees of freedom (DOF): three of position and three of attitude. The initial state is the last wall state: position \(\mathbf p\), velocity \(v\,\hat{\mathbf t}\), attitude from the path frame \(R = [\hat{\mathbf t},\ \hat{\mathbf n}\times\hat{\mathbf t},\ \hat{\mathbf n}]\), and \(\boldsymbol\omega = 0\).

Translation (Newton). With all non-gravity forces collected in \(\mathbf F_{ng}\) (world axes):

The mass falls as gas leaves: \(m = m_{dry} + m_{gas}(t)\). For a body that ejects mass, the thrust force \(\mathbf F_{thrust}\) already contains the momentum carried away by the gas. So \(m\dot{\mathbf v} = \mathbf F\) holds with the current mass and no extra \(\dot m\,\mathbf v\) term.

Rotation (Euler). The angular momentum in body axes is \(I\boldsymbol\omega\). Its rate in an inertial frame, written in rotating body axes, is \(I\dot{\boldsymbol\omega} + \boldsymbol\omega\times I\boldsymbol\omega\). That rate equals the applied moment \(\mathbf M\):

Attitude (quaternions). A unit quaternion \(q = (w, x, y, z)\) stores the attitude without the singularities of Euler angles. With \(\boldsymbol\omega\) in body axes, its rate is

where \(\otimes\) is the quaternion product. The rotation matrix follows from \(q\):

Integration. The step is \(\Delta t = 1\) ms, semi-implicit Euler: first \(\mathbf v\) and \(\boldsymbol\omega\), then \(\mathbf p\) and \(q\) with the new rates. After each step the code divides \(q\) by its norm, so it stays a unit quaternion. Over 5.75 s that is 5,750 steps.

5. Forces in flight: nozzles, drag, wheels

Nozzle wrench. Nozzle \(i\) sits at body position \(\mathbf r_i\) and pushes along the unit body direction \(\mathbf d_i\) with thrust \(f_i \ge 0\). Its force is \(f_i\mathbf d_i\) and its moment about the CG is \(\mathbf r_i \times f_i\mathbf d_i\). All ten nozzles together give a six-component wrench, linear in the thrusts:

Drag. The code uses a separate drag area on each body axis \(j\):

Wheels. Each of the four wheels is a contact point \(\mathbf r_w = (\pm\tfrac{WB}2,\ \pm\tfrac{TR}2,\ -(h_{cg}+\delta_0))\) in body axes. \(\delta_0 = mg_0/(4k) = 72.0\) mm is the static spring deflection. When the world height \(z_w\) of a wheel point is below 0, the ground pushes up with a spring-damper force, and friction acts on the sliding velocity \(\mathbf v_t\):

At low sliding speed, friction acts as a 5,000 N·s/m damper. At higher speed, it is Coulomb friction of size \(\mu N\). The damper part removes the jump at \(\mathbf v_t = 0\), which a 1 ms integrator cannot handle. Each wheel force also gives a moment \(R^{\mathsf T}\bigl((R\mathbf r_w)\times\mathbf F\bigr)\) in body axes.

6. The gas: Abel-Noble tank and blowdown

Why not the ideal gas law. The ideal gas law \(Pv = RT\) treats molecules as points. At 690 bar, nitrogen molecules fill a real part of the volume, and the gas is stiffer than the ideal law says. The Abel-Noble equation removes a co-volume \(b\) per kilogram from the space the molecules can move in:

The compressibility factor is \(Z = Pv/(RT) = 1 + bP/(RT)\). With \(b = 8.0\times10^{-4}\) m³/kg, at 690 bar and 293 K, \(Z = 1.635\). The code comment says the co-volume was fitted to \(Z = 1.65\) for N₂ at 690 bar. The value this \(b\) gives at 293 K is 1.635, which is 0.9% lower.

Start state. 250 kg in 0.600 m³ gives \(v_0 = 2.4\times10^{-3}\) m³/kg. Then \(P_0 = RT/(v_0 - b) = 296.8\times293/(1.6\times10^{-3}) = 543.5\) bar. The ideal law would give 362 bar.

Blowdown. Follow the gas that stays in the tank. Its specific volume grows as gas leaves. The first law for one kilogram is \(du = \delta q - P\,dv\). For an Abel-Noble gas, internal energy depends on \(T\) only, because \((\partial u/\partial v)_T = T(\partial P/\partial T)_v - P = TR/(v-b) - P = 0\). So \(du = c_v\,dT\). With no heat flow:

This uses \(R/c_v = \gamma - 1\). The tank wall gives heat back to the cold gas, so the real process lies between adiabatic (\(n = \gamma = 1.4\)) and isothermal (\(n = 1\)). The code uses the polytropic form with \(n = 1.25\):

Each step removes \(\dot m\,\Delta t\) of gas (§7) and then updates \(T\) and \(P\).

7. The nozzle: thrust and mass flow

Choked flow. At a pressure ratio of hundreds to one, the flow at the nozzle throat reaches the speed of sound. The mass flow then depends only on the tank state and the throat area \(A^*\). For an ideal gas with isentropic flow from the tank to the throat:

For \(\gamma = 1.4\), \(\Gamma = 0.6847\). The code writes the same quantity as \(c^* = \sqrt{\gamma RT}/\Gamma'\) with \(\Gamma' = \sqrt{\gamma}\,\Gamma = 0.8102\). At 293 K, \(c^* = 430.7\) m/s.

Thrust. The thrust coefficient \(C_f\) collects the exit speed and the exit pressure terms: \(F = C_f\,P\,A^*\). Divide the two relations to remove \(A^*\) and \(P\):

Pressure limit. With a fixed throat, \(F \propto P\). So each nozzle's maximum thrust falls with the tank pressure: \(F_{\max,i}(P) = F_{\max,i}^{(0)}\,P/P_0\). The valve can only reduce the thrust below that limit.

Throat sizes (derived). From \(A^* = F_{\max}^{(0)}/(C_f P_0)\): rear 230 mm² (17.1 mm diameter), corner lift 402 mm² (22.6 mm), front 172 mm² (14.8 mm). At full pressure one lift nozzle uses 50.8 kg/s.

8. The flight controller

The demo car was autonomous, so the model flies it with a controller, not a human. The controller runs at 100 Hz (every 10th 1 ms step). It has three references and two feedback laws.

Reference 1: height. \(z_r(t)\) is a smoothing spline (\(\lambda = 10^{-3}\)) through the lag-corrected HUD altitude plus \(h_{cg}\), from 52.20 s on.

Reference 2: horizontal braking. The car leaves the lip with horizontal speed \(v_{x0}\). The controller brakes it to zero by \(T_b = 55.0\) s with a minimum-jerk profile. Let \(u = (t - T_{rel})/(T_b - T_{rel})\), clipped to [0, 1], and

\(S\) goes from 0 to 1 with zero slope and zero curvature at both ends, so the braking starts and ends without a jerk. Because \(\int_0^1 (1 - S)\,du = \tfrac12\), the braking distance is \(v_{x0}(T_b - T_{rel})/2\).

Reference 3: attitude. The car must half-roll from upside down to upright, between \(t_0 = 52.35\) s and \(t_1 = 53.10\) s, and lose its 10° release pitch by 53.40 s:

\(R_{up} = \mathrm{diag}(-1, -1, 1)\) is upright with the nose toward \(-x\). \(\theta_0 = -10°\) is the release pitch as a rotation about \(y\).

Feedback law 1: position. The wanted acceleration is the reference acceleration plus proportional-derivative (PD) corrections:

A PD law \(\ddot e + k_d\dot e + k_p e = 0\) has natural frequency \(\omega_n = \sqrt{k_p}\) and damping ratio \(\zeta = k_d/(2\omega_n)\). Horizontal: \(\omega_n = 2\) rad/s, \(\zeta = 1.0\). Vertical: \(\omega_n = 3\) rad/s, \(\zeta = 0.7\). Before the burn, \(a_z = -g_0\) asks for zero vertical non-gravity force: the car falls free. The jets still fire for the roll and the braking, so a small vertical thrust can remain. The required non-gravity force in body axes is

Feedback law 2: attitude. The attitude error is the rotation from the reference to the actual attitude, \(R_e = R_r^{\mathsf T}R\). For a rotation by angle \(\phi\) about unit axis \(\mathbf k\), \(\tfrac12(R_e - R_e^{\mathsf T}) = \sin\phi\,[\mathbf k]_\times\). So the vector

is a small-angle rotation error (\(^\vee\) takes the three entries of a skew matrix). The reference rate comes from \([\boldsymbol\omega_r]_\times = R_r^{\mathsf T}\dot R_r\), by a central difference over ±0.01 s. The wanted moment is

That is \(\omega_n = 6\) rad/s and \(\zeta = 0.7\) on each axis. The last term cancels the gyroscopic term of Euler's equation (§4).

9. The valve allocator

The controller asks for a wrench \(\mathbf w = (\mathbf F_b, \mathbf M_b)\): six numbers. There are ten thrusts, each between 0 and its pressure limit. The allocator finds the thrusts that come closest:

\(W = \mathrm{diag}(10^{-3}, 10^{-3}, 10^{-3}, 3\times10^{-3}, 3\times10^{-3}, 3\times10^{-3})\) per N and per N·m: one N·m of moment error counts 3 times as much as one N of force error. \(\varepsilon = 10^{-5}\) is a small penalty that picks the lowest total thrust when several answers fit equally. SciPy's lsq_linear solves this bounded-variable least-squares problem.

Valves. A valve does not open at once. The actual thrust follows the command with a first-order lag:

and is clipped to \([0, F_{\max,i}(P)]\).

Landing cut. When the wheels carry more than half the weight for more than 0.05 s, all commands go to zero and stay there.

10. The HUD lag: a free-fall fit

The first full run failed its own check (sensitivity). One cause: the HUD altitude is late. The G gauge drops from 1.0 to 0.4 between 52.2 and 52.3 s, but the altitude keeps rising until 52.6–52.7 s. A car in free fall after release can rise only for \(v_{z0}/g_0\), about 0.14 s.

Model. Before release, the CG climbs the straight 170° ceiling at a constant vertical speed \(v_{z0}\). After release it falls freely. The HUD shows this height \(\ell\) seconds late. With \(\tau = t - \ell - T_{rel}\):

A parabola alone has only two free parameters, so it cannot separate \(\ell\) from \(z_0\) and \(v_{z0}\). The kink at \(\tau = 0\), where the straight climb changes to the parabola, fixes \(\ell\). The fit uses the 18 HUD readings from 52.2 to 53.95 s, before the burn.

Result. \(\ell = 0.337\) s, \(v_{z0} = 1.329\) m/s, \(z_0 = 20.768\) m, rms residual 1.258 ft. Check: the sim's release state is 8.139 m/s along a 170° ceiling, so \(v_{z0} = 8.139\sin10° = 1.413\) m/s, 6% above the fit. The code uses \(\ell = 0.30\) s. The second stunt's HUD showed about 0.2 s. The record gives no other reason for 0.30 s in place of 0.34 s.

11. Fit metrics and the self-check

The report compares the sim with the HUD on the HUD's own sample times:

• Altitude RMSE in flight: \(\sqrt{\tfrac1N\sum_i\bigl(h_{sim}(t_i - \ell) - h_{HUD}(t_i)\bigr)^2}\) over the HUD samples with \(T_{rel} + \ell \lt t_i \le 57.2\) s.
• G RMSE: the same form, without lag, on the wall (46.0 to 52.2 s) and in flight. The G gauge is taken as not lagged.
• Apex, touchdown, landing tilt, gas, clearance: from the flight record. Touchdown is the first time the wheels carry more than half the weight. Clearance is the smallest distance from a body corner to the wall surface, from 0.3 s after release.

The run stops with an error unless all seven checks pass: apex within 1.5 ft of the HUD; touchdown within 0.4 s; altitude RMSE under 2.5 ft; landing tilt under 10°; gas used between 0 and 250 kg; wall clearance over 0.1 m; no body corner more than 0.05 m below the ground.

sim/wall_flip_sim.py:340–373 report; 399–408 self-check; 314–329 clearance

Part 2: the numbers

Sources: PROJECT.html §3–§8 and the source table in the README. "Derived" values were calculated for this report by out/physics/make_charts.py or by the sim run.

The car

The wall (ramp)

The nozzles

Ten nozzles: 4 rear main, 4 corner lift, 2 front for braking and yaw (event, Top Gear, TechCrunch). The force per nozzle is not published. Body axes: x forward, y left, z up, from the CG.

Code: sim/wall_flip_sim.py:58–66, \(C_f\) line 36, valve lag line 46.

The tank and the gas

Timing and the controller

Part 3: results against the HUD

Run of 2026-10-17, default hud overhang. The self-check passes.

A check on the braking. §8 gives a braking distance of \(v_{x0}(T_b - T_{rel})/2 = -8.015 \times 2.75 / 2 = -11.02\) m. The release is at x = −10.45 m, so the car should stop at −21.47 m. The sim lands at x = −21.47 m.

Until 2026-10-18, PROJECT.html §8 listed the first passing run: 2.04 ft, 3.90 g, 187.8 kg, 0.59 m, touchdown 56.19 s. Those values are from before the half-roll window changed from 52.45/53.35 s to 52.35/53.10 s on 2026-10-16 (§9 of that page). python check.py found them; §8 now shows the current run. This page uses the current code.

The wall

Reading Fig. 2. The speed is about 6.8 m/s in the entry curve and falls to 4.6 m/s on the overhang (49.0–50.5 s). Up to 51.0 s the minimum suction peaks at 3,779 kgf, against 4,700 kgf available: 80% of it. The median from 48.5 to 51.0 s is 1,989 kgf. Wheel power up to 51.0 s peaks at 207 kW, against more than 1,000 kW published. So suction, not motor power, limits the climb.

Why the suction curve is a saw. The swings come from the acceleration \(a = f''(t)\), not from the wall. Example: the HUD altitude jumps from 48.1 to 50.6 ft between 50.2 and 50.3 s. After the 0.30 s lag shift, \(a\) rises to 6.4 m/s² at 49.8 s and falls to −3.2 m/s² at 50.0 s. The suction need follows: 3,779 kgf, then 1,579 kgf.

The ceiling values are noise. On a 170° ceiling, one 0.1 ft altitude step is 0.176 m of track. The sim's peaks there, 5,003 kgf of suction at 51.40 s (speed 10.14 m/s, acceleration 10.62 m/s²), 294 kW and 10.46 m/s, are in this band. They are not physical results. PROJECT.html §4.1 gives the typical overhang need as about 3,300 kgf and the peak power as about 190 kW; those values were sampled every 0.5 s.

Altitude and G

Reading Fig. 3. On the wall, the sim leads the raw HUD by the lag, by design. In flight the sim follows the shifted HUD. The largest gap is in the fall: from 53.1 to 53.4 s (lag-corrected) the sim is 5.3 to 6.3 ft below the shifted HUD.

Reading Fig. 4. On the wall, the sim's G swings by about ±0.5 g around the HUD. The cause is the same acceleration noise as in Fig. 2. In free fall the sim reads 0.0 to 1.4 g; the HUD reads 0.2 to 0.6 g. The sim spikes at 52.6 s and 53.1 s are the lift pairs that start and stop the roll. Both curves rise at about 53.95 s. The sim peaks at once, 3.65 g at 54.04 s; the HUD peaks at 3.3 g at 54.4 s. The spike of 3.78 g at 56.19 s is the wheel impact at touchdown.

Tank and nozzles

Reading Fig. 5. The run uses 76.7% of the gas (191.8 of 250 kg), but the pressure falls by 89% (543.5 to 58.6 bar). At touchdown each nozzle can give only 11.3% of its full-tank thrust.

Reading Fig. 6. Before the burn, the front pair brakes the forward speed and the lift pairs make the roll: first the right pair (FR, RR), then the left pair (FL, RL) to stop it. The outer rear nozzles fire one at a time (rear0 at y = +0.45 m, rear3 at y = −0.45 m). One-sided, they give a yaw moment and some forward push. From 53.95 s the four lift nozzles run at or near the pressure limit until about 55.0 s, then below it for the hover. The burn is limited by pressure, which is why the lift nozzles had to be 35 kN each (sensitivity).

Sensitivity findings

Each row changes one input and reruns the flight (current code, same wall phase). Pass/fail uses the self-check limits: touchdown within 0.4 s of 56.30 s, altitude RMSE under 2.5 ft.

• Tank heat flow decides the landing. With n = 1.4 the gas cools most, the pressure and thrust fall too early, and the car lands 0.92 s early. Note that n = 1.1 uses less gas than n = 1.25 but ends at a higher pressure: warmer gas has a higher \(c^*\), so each newton costs less mass (\(\dot m = F/(C_f c^*)\)). The change log recorded 55.75 / 56.19 / 56.27 s for the three exponents before the roll-window change; the table shows the current values.
• Nozzle sizing. With the first-guess sizes the lift nozzles cannot stop the fall: the car hits the ground at 54.78 s with a 9.81 g wheel impact. The 35 kN lift size is the derived value that gives the 3.3 g burn and the hover.
• HUD lag. Without a lag model, the controller chased a gauge 0.3 s late and fought the free fall at the top. The first full run touched down at 55.17 s with a peak of 11.0 g (DEVELOPMENT-JOURNEY.html §5.3).
• Tolerance 2.0 → 2.5 ft. The altitude RMSE limit was 2.0 ft at first. At n = 1.25 the run gave 2.04 ft. The limit was raised to 2.5 ft because the pure free-fall fit to the same HUD data already leaves 1.26 ft: the gauge's 0.1 ft steps and the simple lag model set that floor. The reason is in the code at line 402. With the current code, all three exponents give more than 2.0 ft. A reader should know that the limit moved after a run failed it.

Known limits

Part 4: the flip

4.0 Timeline and symbols

Symbols added in this part:

4.1 Before: riding the ceiling

The force balance upside down. On the straight ceiling, the curvature is \(\kappa = 0\). The normal equation of §3 then has no centripetal term:

The surface can only push, so \(N \ge 0\). The suction must first carry 98.5% of the weight, 2,314 kgf, before it presses the tyres on at all. The tyres must then carry the along-track load \(F_t = m a + m g_0\sin\theta + D\).

Worked case: the least suction at constant speed. Take \(a = 0\) at the release speed \(v = 8.139\) m/s. The drag is \(D = \tfrac12\rho\,(C_dA)_x v^2 = 27.8\) N, so \(F_t = 4{,}030\) N. With \(\mu = 1.1\):

That is 57% of the 4,700 kgf that the HUD showed as the peak downforce. The ceiling values of the sim's \(F_{s,\min}\) in Fig. 2 are larger, but they come from gauge noise in \(a\) (Part 3).

No spin at the start. On the curved overhang the car pitches at \(\dot\theta = v\kappa\); its largest value from 50.0 s to 51.31 s is 38.2°/s in the sim. On the straight ceiling \(\kappa = 0\), so \(\dot\theta = 0\). A car that lets go of a straight ceiling therefore starts its flight with \(\boldsymbol\omega = 0\), which is the sim's initial condition (§4). Any turn after the release must come from a moment applied in the air.

The accelerometer cannot see the suction. While the car is attached at \(a = 0\), the non-gravity force is \((N - F_s)\,\hat{\mathbf n} + (F_t - D)\,\hat{\mathbf t} = m g_0\cos\theta\,\hat{\mathbf n} + m g_0\sin\theta\,\hat{\mathbf t}\), whose size is \(m g_0\). The G gauge reads 1.0 for any suction above the minimum. It drops only when the car leaves the surface and the contact force goes to zero. So the G step from 1.0 to 0.4 between 52.2 and 52.3 s dates the separation, not the moment the suction was switched off. The HUD reads 1.0–1.1 g from 51.3 to 52.2 s, in agreement.

4.2 Letting go: separation from the ceiling

The condition. When the suction falls below \(0.985\,mg_0 + |F_t|/\mu\), the tyres slip; when it falls below \(0.985\,mg_0\), the push \(N\) reaches zero and the car leaves. The sim models this as an instant switch at \(T_{rel} = 52.25\) s.

The gap. After release, gravity has a part \(g_0|\cos\theta| = 0.985\,g_0\) along the normal away from the ceiling and a part \(g_0\sin\theta = 0.174\,g_0\) against the motion along it. With no thrust, the gap below the plane of the ceiling grows as

where \(h_0 = 0.120\) m is the distance from the body to the surface at release in the sim's box model. The free-fall part reaches 0.5 m after \(\tau = 0.322\) s. The sim's own gap, which includes the roll, is 0.170 m at 52.35 s, 0.312 m at 52.45 s and 0.485 m at 52.55 s.

How much ceiling is left. In the hud variant the wall ends 1.5 m past the nominal release point. The smoothing spline of §2 puts the sim's car 0.392 m past that point at 52.25 s, so only 1.108 m of ceiling is ahead of the CG. Solving \(1.108 = v\tau - \tfrac12(0.174\,g_0)\tau^2\) gives \(\tau = 0.138\) s: the CG passes the lip at 52.388 s, when the free-fall gap is 0.092 m. The rear bumper, 2.35 m behind the CG, passes it at 52.696 s. In the short variant the lip is at the release point: the CG passes it at 52.25 s and the rear bumper at 52.548 s. Fig. 8 shows the hud case.

A check the self-check skips. The wall-clearance check of §11 starts 0.3 s after the release, because the car starts on the surface. Over the skipped 0.3 s, the smallest body-to-ceiling gap in the sim is 0.120 m, at 52.25 s itself. The gap only grows after that, also during the start of the roll. No contact is hidden in the skipped window.

4.3 One model for the whole window

The phases before and after the release use one rigid body with one set of equations. Only the active forces change. Let \(c(t) = 1\) while the car touches the ceiling and \(c(t) = 0\) after it leaves. In world axes:

and in body axes:

\(B_F\) and \(B_M\) are the force and moment rows of the nozzle matrix \(B\) (§5). The contact obeys the complementarity conditions

that is, either the car touches and the surface pushes, or there is a gap and no push. The aerodynamic terms are

with \(\mathbf r_{cp}\) the offset of the centre of pressure from the CG and \(M_{damp}\) the roll damping (§4.8). \(\Delta\mathbf F_{jet}\), \(\Delta\mathbf M_{jet}\) are the forces that the jets induce on the body through the surrounding air: entrainment in free air and suction near a surface (§4.9). The table lists which terms the sim contains.

4.4 The half-roll as a rigid-body problem

The roll equation. Take the \(x_b\) row of Euler's equation:

The sim's peak \(|q| = 0.321\) rad/s and \(|r| = 0.279\) rad/s give a gyroscopic term of at most 15.3 N·m. That is 0.059% of the peak jet moment, so in practice \(I_{xx}\dot p = M_x\). With \(\varphi\) as defined in §4.0 and small pitch, \(\dot\varphi \approx p\).

The moment and the thrust it needs. The moment at the peak is \(I_{xx}\,\ddot\varphi_{\max} = 650 \times 32.25 = 20.96\) kN·m. The lift nozzles push along \(z_b\) at \(y = \pm d\), so a nozzle of thrust \(f\) gives a roll moment \(\pm d f\). Two nozzles on one side give the moment with \(f = M/(2d) = 12.33\) kN each, which is 35% of their 35 kN rating. The angular impulse to reach the peak rate is \(I_{xx}\dot\varphi_{\max} = 5{,}105\) N·m·s for each half of the roll.

The estimate that sets it. \(I_{xx} = 650\) kg·m² is an estimate (Part 2). The thrust need scales with it:

4.5 Jets that only push: the roll throws the car sideways

A wing can make a pure moment. A set of push-only jets cannot always do so. For the lift nozzles the cost is unavoidable.

The other jets do not escape this: the rear and front nozzles push in the \(x_b y_b\) plane, and their roll arm is at most 0.034 m per newton (the front pair), which gives at most 438 N·m in the sim, 1.7% of the peak jet moment.

Where the push goes. In world axes, the body \(z_b\) axis at roll angle \(\varphi\) (pitch neglected) is \((0,\ \sin\varphi,\ \cos\varphi)\). The sim confirms it: at \(\varphi = 90°\) (52.80 s) the world \(y\) part of \(z_b\) is +0.999. During the whole roll \(0 \lt \varphi \lt \pi\), so \(\sin\varphi \gt 0\): every pulse, the one that starts the spin and the one that stops it, pushes toward \(+y\).

The size. \(J_y/m = 1.72\) m/s. Without a sideways correction, the car would drift 0.64 m by the end of the roll and 2.11 m by the burn. The vertical parts cancel: the first pulse pushes down (body \(z_b\) points down while inverted) and the second pushes up. Their magnitudes add up to 5.11 m/s of velocity change, all spent to turn the car.

What the sim does with it. The sim's controller holds \(y = 0\) (§8). Fig. 11 (bottom) shows the result: the sim's \(v_y\) still reaches +0.987 m/s at 52.96 s, the controller then pulls it back to −0.521 m/s at 53.34 s, and the largest offset is 0.333 m. The open-loop line rises to 1.72 m/s and stays there.

Link to the video. In shot B and in the replay R1, the real car drifts to \(+y\) (screen left) (Known limits 1). The model gives a mechanism with that sign, if the real car rolled the same way as the sim (roof toward \(+y\) at mid-roll). The 10 fps roll frames are too small to show the roof side, so the roll direction of the real car is not measured. The triangulation that would measure the drift failed its gate.

4.6 What the sim's roll did

Reading Fig. 9. The reference meets the video: it passes 90° at 52.725 s and is upright at 53.10 s. The sim does not follow the reference closely. It passes 90° at 52.800 s, 0.075 s after the reference and 0.10 s after the end of the video's "on its side" interval. It passes upright at 53.00 s, overshoots to −34.3° at 53.18 s, and is still at −3.0° at 53.60 s. At 53.10 s, when the video shows the car upright, the sim is at −28.9°. Its peak roll rate is 9.02 rad/s against 7.854 rad/s, and its peak roll acceleration 39.8 rad/s² against 32.25 rad/s².

The cause. The attitude law of §8 has a natural frequency of 6 rad/s and damping 0.7, and it uses the reference rate \(\boldsymbol\omega_r\) but not the reference acceleration \(\dot{\boldsymbol\omega}_r\) (Known limits 7). The moment therefore rises only after an attitude error has built up, so the roll starts late, and the error then drives an overshoot. The valve lag of 20 ms adds to the delay. A feed-forward moment \(I\dot{\boldsymbol\omega}_r\) is the standard fix. It is not applied here, because this part does not change the sim. The landing tilt is still 0.0°.

Reading Fig. 10. The jet moment peaks at 26.0 kN·m and \(I_{xx}\dot p\) at 25.9 kN·m; the two curves lie on each other. This checks that the recorded attitude obeys the roll equation with no hidden term. The first pulse (FR + RR) peaks at 24.9 kN at 52.640 s (FR 13.4 kN, RR 11.5 kN, limit 33.0 kN), at \(\varphi = 157°\). The second pulse (FL + RL) peaks at 30.1 kN at 53.060 s (FL 15.5 kN, RL 14.6 kN), at \(\varphi = -21°\), that is, after the car has already rolled past upright. The lower panel shows that the jet moment is 2 to 3 orders of magnitude larger than every other term during the roll.

4.7 The G gauge during the roll

Fig. 11 (middle) shows a clear difference. During the roll, 52.35–53.10 s, the HUD shows 0.2–0.4 g; the sim shows 0.00–1.40 g. The roll pulses cause the sim's two humps.

The numbers. With \(G_{\max} = 0.4\), \(T = 0.75\) s: \(I_{xx} \le 351\) kg·m². If the roll took \(T = 0.85\) s (one more video frame), \(I_{xx} \le 451\) kg·m². For the minimum-jerk shape itself, the bound is 243 kg·m². The estimate is 650 kg·m², and a uniform box of the car's size and mass gives 1,042 kg·m². So at least one of these is true:

1. The G gauge does not show the unfiltered specific force. It may be filtered or averaged. The HUD altitude is already known to lag by 0.30 s (§10).
2. The roll inertia is at most about half of the estimate (351 / 650 = 54%).
3. Part of the roll comes from something other than the lift jets, for example a moment applied before the car leaves the ceiling.

An example of the third case: if one side's suction let go first, the still-attached wheel line would act as a hinge. Gravity's normal part about that line, \(0.985\,mg_0 \times d = 19.3\) kN·m, would then start the roll with \(8.22\) rad/s² (inertia about the hinge 2,348 kg·m²) and no jet force. The record has no evidence for or against this.

4.8 Aerodynamics along the sim's path

Dynamic pressure. The airspeed is 8.13 m/s at release and 14.10 m/s at the burn. So \(q = \tfrac12\rho V^2\) goes from 39.6 Pa to 119.2 Pa. For scale, the car's weight spread over its 9.40 m² plan area is 2,452 Pa. The air is weak during the flip.

Drag (in the sim). The largest drag force in the window is 929 N, 4.03% of the weight, at the burn, when the car falls fastest. During the roll the drag is at most 327 N.

Roll damping (not in the sim). A rolling car moves its floor, roof and sides through the air. Two simple models bound the moment.

1. Cross-flow. A strip of the plan at lateral position \(y\) moves at \(p\,y\) normal to itself and feels a pressure \(\tfrac12\rho\,C_{D}\,(py)^2\). The moment of all strips is

since \(\int_{-a}^{a} y^2|y|\,dy = a^4/2\). The drag coefficients come from the sim's own drag areas: \(C_{Dz} = 8.0/(L W) = 0.851\) and \(C_{Dy} = 2.0/(L H) = 0.370\). This gives \(M_{cf} = 1.257\,p|p|\) N·m.

2. Slender wing. In the forward flow \(V\), a strip at \(y\) sees an angle of attack \(py/V\). Slender-wing theory gives a lift slope \(C_{L\alpha} = \pi A/2\) with aspect ratio \(A = W/L = 0.426\), so \(C_{L\alpha} = 0.668\). The moment is

On the sim's path the peaks are 102 N·m and 119 N·m: 0.39% and 0.46% of the peak jet moment. If no controller corrected it, the sum of the two would change the roll angle by 2.51° over the roll. The two constants are equal by coincidence of the numbers, not by a physical rule.

Centre-of-pressure offset (not in the sim). The sim puts the aerodynamic force at the CG. Each 0.1 m of offset adds a moment of \(0.1\,|\mathbf F_{aero}|\): at most 93 N·m in the window, 0.36% of the jet peak, or 0.032 rad/s² in pitch.

Jet-induced lift loss in free air (not in the sim). A jet drags the nearby air with it, which lowers the pressure on the body around the nozzle. Kuhn, Bellavia, Corsiglia and Wardwell (NASA TM 102268, 1991, p. 5) give, for open air,

with \(S\) the plan area, \(A_j\) the total jet area, \(\Sigma\pi d\) the total jet perimeter, \(d_e\) the diameter of one jet with the total area, and NPR the nozzle pressure ratio. For one active pair (two jets), \(\Sigma\pi d/d_e = \sqrt2\,\pi\). The jet diameter is not known. The throat is 22.6 mm; ideal expansion from NPR 536 to 1 bar gives an exit Mach number of 5.01, an area ratio of 25.2 and an exit of 114 mm. The result is \(\Delta L_\infty/T = -0.10\%\) to \(-0.55\%\), at most 131 N on a 24 kN pair. The suckdown data in the 1985 survey were taken at NPR 1.45–2.4 (NASA TM 86825, Figs. 3 and 9); here the NPR is 435–536, 181 to 370 times larger. Take the value as an order of magnitude only.

4.9 The jets and the ceiling

Where the exhaust goes. The lift nozzles push along \(+z_b\), so their exhaust leaves along \(-z_b\). Upside down, \(-z_b\) points up, at the ceiling. The sim's first lift thrust above 0.5 kN comes at 52.400 s, when the CG is at \(x = -11.66\) m, already 0.2 m past the hud lip (\(x = -11.46\) m). The rear nozzles are 1.4 m behind the CG, so the rear right nozzle (RR) is still under the ceiling.

The geometry in the sim. In the hud variant, the axis of the RR exhaust meets the ceiling from 52.40 to 52.56 s, at 0.36–0.87 m from the nozzle, with up to 9.5 kN of thrust. In the short variant it meets the ceiling only from 52.40 to 52.42 s, with at most 1.2 kN. No other lift nozzle points at the ceiling while it fires.

The physics. A jet that hits a nearby surface spreads along it as a wall jet. The wall jet drags air out of the gap between the body and the surface, and the low pressure in the gap pulls the body toward the surface ("suckdown"). Here the surface is the ceiling, so the pull is upward, against the RR thrust, which pushes the car down and starts the roll. The pull therefore reduces both the push and the roll moment of RR. For a single jet in hover over a large ground board, Wyatt's correlation (Kuhn and Eshleman, NASA TM 86825, 1985, Fig. 2) is

where \(h\) is the height of the plan surface above the ground, \(D\) the effective diameter of the plan, and \(d\) the jet diameter. Its plotted data span \(h/(D - d) \approx 0.17\) to 1.5. For the car, \(D \approx 3.46\) m (circle of equal area) and \(d = 0.114\) m, so the range is \(h = 0.57\) to 5.02 m. At \(h = 0.57\) m, the correlation gives a loss of 70% of the jet thrust. The real gap during the event is 0.12 to 0.49 m, below the fitted range, where the formula gives more than 100%. The correlation cannot give a size here, for three reasons:

1. It is for a hovering body over a still surface. Here the car slides along the ceiling at 8 m/s and falls away from it.
2. It is for a large ground board. The ceiling is 3.3 m wide, less than the plan diameter of 3.46 m, and it ends at the lip, 1.22 m from the RR nozzle at 52.40 s.
3. The gap is below the fitted range, and the NPR (435–536) is 181 to 370 times the 1.45–2.4 of the survey's data.

A scale for the effect. At the bottom of its range the correlation gives 70% of the thrust, and at \(h = 0.49\) m it gives 100%. So take full cancellation of the RR thrust as the scale: while its axis meets the ceiling, the RR impulse is 769 N·s, or 0.327 m/s of speed. Its roll part is 654 N·m·s, 12.8% of the angular impulse of one half of the roll. This is the only aerodynamic term in the window that could matter at the 10% level, and it is the only one that depends on the overhang variant: the hud ceiling exposes it for 0.16 s, the short ceiling for 0.02 s.

4.10 After: into the burn, and what the model captures

Into the burn. By 53.40 s the release pitch of −10° is gone in the reference (the sim shows +0.87°). At 53.90 s the sim is upright within 2°. From release to the burn, the jets use 58.6 kg of gas. By thrust impulse, the front pair (braking and yaw) takes 48%, the lift nozzles (the roll and its correction) 42%, and the rear nozzles 10%. At 53.95 s the four lift nozzles open together for the burn of §8.

What the model captures.

• The inverted force balance on the ceiling and the reason the G gauge marks the separation (§4.1).
• The fall away from the ceiling and the time left under it, for both overhang variants (§4.2).
• The rigid-body roll: the moment, the thrust it needs, and its dependence on \(I_{xx}\) (§4.4).
• The coupling that push-only jets force between turning and translating, with a sideways push toward \(+y\) (§4.5).
• The size of every aerodynamic term on the sim's own path (§4.8).

What it misses, and where it is discussed.

• The sim's roll lags its reference by about 0.1 s and overshoots by 34.3° (§4.6). A feed-forward term would repair it; the sim is unchanged by request.
• The G gauge during the roll does not agree with a jet-only roll at the estimated inertia (§4.7).
• The release is an instant switch; the suction decay and any one-sided release are not modelled (§4.2, §4.7).
• The flight is planar (\(y = 0\) is enforced by the controller), so the sideways push is removed, not predicted (§4.5).
• The jet-ceiling interaction has a scale, not a value (§4.9).

Reproduce

Charts: matplotlib; colours checked for colour-blind separation on the dark card colour. Math: KaTeX 0.16.9 (needs internet).