The Variable Pitch Screw Launch system converts electricity into kinetic energy with electric motors before the spacecraft is launched. This kinetic energy is stored as rotational energy within the spinning screws and the flywheels inside these screws. During a launch, some of the flywheel's momentum is transferred through the screws to the launch train (which comprises the spacecraft, launch sled, and the adaptive nut). This claim asserts that it is possible to transfer this momentum directly - that is, without converting the flywheel's kinetic energy into electrical energy, and then back again to kinetic energy.

To begin, we'll calculate the amount of kinetic energy that is being transferred into the launch train as a function of its position along the variable pitch screw acceleration section. Note that some descriptions of the system include a feeder system that pre-accelerates the launch train to facilitate the adaptive nut's engagement with the screws; therefore, for this analysis, let's assume that the launch train is accelerated by the screws from an initial velocity, $v_1$ at position $x_1$ to a final velocity, $v_2$ at position $x_2$. The rate of kinetic energy gain as a function of time is given by

$$dE_K/dt = d/dt (½ m v^2) = mva$$

where $m$ is the mass of the launch train and $a$ is it s rate of acceleration. The kinetic energy gained per meter travelled is

$$dE_K/dx = (dE_K/dt) / (dx/dt) = (m v a) / v = m a$$

If we want to know how much energy is transferred by, for example, a one-meter segment of screw, we would integrate...

$$∫(dE_K/dx) dx = ∫(m a) dx → ΔE_K = m a (1 m)$$

From this, we can see that the kinetic energy transfer per meter of screw is constant. What changes as the vehicle accelerates is the duration of time over which this transfer occurs, which is given by $\Delta t=\Delta x/v$. If $v=11060 \space m/s$ at the end of horizontal acceleration, the time it takes to traverse one meter is 90.4 microseconds. However, the adaptive nut is longer than one meter. If we assume it is $l_{an}=100 \space m$ in length, then the flywheels will transfer their momentum over a time period of $100/11060 = 9 \space milliseconds$.

Next, we need to figure out how the rotational rate of the flywheels will change during the momentum transfer. Let's assume that each flywheel is made from steel ($\rho=7850 kg/m^3$) and is cylindrically shaped with an outer radius of $r_o=0.145m$, an inner radius of $r_i=0.1 m$, and a length of $l_{flywheel}=4.5m$. Let's assume that its initial rotational rate is $\omega_1$ and its final rotational rate is $\omega_2$, which is the same rotational rate as the screws, which are spinning at 1,060 radians per second. The moment of inertia for the flywheel is

$$I=½\pi\rho l_{flywheel}(r_2^4 - r_1^4)=0.5(3.1415)(7850)(4.5)(0.145^4-0.1^4)=18.98 \space kg \cdot m^2$$

and the final kinetic energy in the flywheels (that is, when they have decelerated to $\omega_2$) is

$$E_{K,\omega 2}=½I \omega_2^2 = 0.5(18.98)(1060^2) = 10.663 \space MJ$$

The kinetic energy absorbed by a screw, that is, the Power entering the screw, when flywheel braking occurs, is

$$P_{IntoScrew}=\tau\omega_{screw}$$

where $\tau$ is the braking torque and $\omega_{screw}$ is the rotational rate of the screw.

The total energy that flows into the screw during the flywheel breaking event is the time-integral of that power:

$$E_{toScrew} = \int_{t_1}^{t_2} \tau(t)\,\omega_s(t)\,dt$$

The screw’s angular velocity is, by design, constant during the transfer, so we can pull $\omega_s$ out of the integral:

$$E_{toScrew} = \omega_s \int_{t_1}^{t_2} \tau(t)\,dt$$

But the time-integral of torque is the angular impulse applied to the screw, which is equal to the angular impulse applied by the flywheel. And the angular impulse applied by the flywheel is its change in angular momentum, or its moment of inertia times its change in angular velocity. Therefore:

$$\displaystyle \int_{t_1}^{t_2} \tau(t)\,dt = I(\omega_1 - \omega_2)$$

Substituting, we get:

$$\boxed{E_{toScrew} = I\,\omega_s\,(\omega_1 - \omega_2)}$$

Assuming losses in the magnetic coupling are negligible, $E_{toScrew}$ is the same as the energy gained by the launch train per screw length of distance traveled, which is $m a l_{screw} / N_{screws}$. Note that since two screws are driving the launch train, we divide its kinetic energy by $N_{screws}$, which is two.

$$E_{toScrew}=m a l_{screw} / N_{screws}=38,940kg*80 m/s^2*5m / 2=7.788 MJ$$

Rearranging, we can solve for $\omega_1 - \omega_2$

$$\omega_1 - \omega_2 = \Delta\omega = {E_{toScrew} \over {I \omega_s}} = {7.788e6 \over {(18.98 kg⋅m^2)(1,060 rad/sec) }} = 387 rad/sec$$

If the flywheel's final rate of rotation is the same as the screw's rate of rotation, then the flywheel's initial rate of rotation is

$$\omega_2 = \omega_s + \Delta\omega = 1,060 + 387 = 1,447 rad/sec$$

The flywheel's maximum rate of rotation is 2,305 rad/s, assuming steel with a yield stress of 690MPa and an engineering factor of 1.5 (see flywheelMaxRateOfRotation in the Digital Twin).

Next, we need to calculate the flywheel's kinetic energy before braking occurs. The difference is the energy that must be added to spin the flywheel back up between launches.

$$K_{flywheel, \omega_1} = ½I \omega_1^2 = 0.5(18.98)(1,447^2) = 19.873 \space MJ$$

The energy added back between launches is $19.873 - 10.663 = 9.210 \space MJ$, and if we assume an electric motor efficiency of 80%, then $9.210/0.8 = 11.512 \space MJ$of energy must be supplied to the flywheel's motor between launches to spin the flywheel back up to $\omega_1$.

The difference between the energy added back and the energy that enters the screw is energy that is converted to heat:

$$E_{heat} = 9.210 - 7.788 = 1.422 \space MJ$$

Next, we need to determine whether this amount of energy would overheat the material at the braking interface. Let's assume that a 0.5 mm surface layer on each side of the braking interface absorbs the heat. Assume:

• Heat dissipated at the interface: $E_{heat} = 1.422e6 \space J$
• Steel properties: $\rho = 7850 kg/m^3, c_p \approx 500 \space J/(kg·K)$
• Flywheel outer radius and length: $r_o = 0.145 \space m, L = 4.5 \space m$
• Heated layer thickness: $\delta = 0.001 \space m$

Heated volume:

$$V = 2\pi r_o L\,\delta$$

Heated mass:

$$m_\text{layer} = \rho\,V = \rho\,(2\pi r_o L\,\delta)$$

Bulk-average temperature rise of the layer:

$$\Delta T = \frac{Q}{m_\text{layer} c_p} = \frac{Q}{\rho\,(2\pi r_o L\,\delta)\,c_p}$$

Numerical evaluation:

$$V = 2\pi (0.145)(4.5)(0.001) \approx 4.10\times 10^{-3}\ \text{m}^3$$ $$m_\text{layer} \approx 7850 \times 4.10\times 10^{-3} \approx 32.2\ \text{kg}$$ $$\Delta T \approx \frac{1.422\times 10^6}{(32.2)(500)} \approx 88\ ^\circ\text{C}$$

To summarize, the flywheel will initially be spinning about 37% faster than the screw and will brake against the inside of the screw to rapidly slow itself down. At the same time, the adaptive nut will be traveling alongside the screw, engaging with the screw flights. In this manner, the screws momentum will transfer from the flywheel into the screw and then from the screw into the launch train.

The next question we need to answer is whether designing a braking mechanism (referred to elsewhere as "electromagnetic clutches") to affect the braking action between the flywheel and the inside of the screw can be regarded as lying within charted engineering territory or if it ventures into completely uncharted territory.

The change in the rate of rotation is $\Delta\omega=387.1 \space rad/sec$ (about 53 revolutions per sec). Earlier, we estimated that in the worst case (at the end of the acceleration section), the braking would have to occur over a period of 9 milliseconds (Note: the value generated in the digital twin is actually 8.88 milliseconds, so we'll use that). The change in rate of rotation versus time is $d{\omega}/dt=\Delta\omega/t=387.1/0.00888=43,591 \space rad/s^2$.

The relative angular distance over which this deceleration will occur is $\theta=v^2/2a=(387.1^2)/(2*43,591)=1.72 \space rad$. To put this in perspective, the relative rim speed at the flywheel-screw interface is $387.1*0.145=56 \space m/s$ and the stopping distance is $1.72 \space radians*0.145 m= 0.25 \space m$, so the resulting acceleration at the flywheel's outer rim, using $v^2=2ax$, is $a=v^2/2x=56^2/(2*0.25)=6,321 \space m/s^2$, which is in the range of accelerations that, for example, pneumatic nail guns routinely achieve. In fact, piston acceleration in a Formula 1 engine can reach approximately $95,000\ m/s^2$.

Rapidly decelerating a flywheel creates a torque force on the screws that will travel through the screw flights to the grapplers. The force on the screw flights was modeled in a Fusion360 stress simulation, and these results can be seen in the peer-reviewed VPSL paper (see Figure 15).

The connection between the adaptive nut and the screw flights functions similarly to a magnetic worm gear, eliminating the need for the rapidly switching electromagnets of a traditional linear motor. The momentum is transferred from the screw to the adaptive nut through magnetic attraction, without any physical contact. The grapplers' pads engage with screw flights in a symmetric configuration, such as in opposing pairs or triangular arrangements, to ensure that the resulting forces are balanced. This minimizes net loads on the screw’s bearings and support structures, preserving mechanical stability during operation.