# Hyperloop: Like a Bullet?

Thunderf00t, a great chemist you may be, but you need to review your mechanical engineering.

Hyperloop for all its challenges – and particularly the danger of a vacuum failure – will not “shoot like a bullet” in the case of tube and vacuum failure.

In a gun, the projectile fits tightly in the barrel. It is propelled by massive pressure behind it. In the case of a 9mm pistol, the pressure is 2,400 atmospheres, 35,000 psi (2400 Bar). When the hyperloop tube fails behind a pod, there will be an onrush of pressure at the speed of sound to increase the pressure behind the pod to one atmosphere (1.0 Bar). Yes, it will give the pod a push, but not a big one.

In your ball bearing and glass tube demo, the ball fit relatively tightly in the tube, as a bullet would in a barrel. In Musk’s design, and likely Hyperloop One’s, the pod does not take up all the sectional area of the tube. In other words, it is not a tight fit. Your ball bearing was accelerated from a standstill until it broke through the glass at the end of the tube. The situation in an operating hyperloop is different. The pod is already moving at or near the speed of sound. The pressure wave approaching from behind it arrives at a low relative speed. When it reaches the pod, it is not constrained by the pod and will leak around. Within a short period, the pressure is equalized ahead and behind the pod. Air resistance increases and the pod will slow. We don’t have enough information to calculate perfectly the rate of deceleration, but we can try. The real problem then will be rescue.

### In depth

Let us do the physics. Assume the worst case scenario, that the pod is a tight fit in the tube and that the pod is at a standstill. In the SpaceX test track, the cross section is 2.5 m^{2}. The pressure ahead of the pod is .001 Bar (negligible in this back-of-the-envelope exercise). The pressure behind is 1.0 Bar. A pressure of 1.0 Bar acting over a surface of 2.5 m^{2} is 250 kN. We don’t have any idea what a pod will weigh, but let’s assume, *arguendo*, that it is 5000 kg. A force of 250 kN acting on a mass of 5000 kg in an otherwise frictionless environment is calculated thusly:

F = ma (1)

a = F/m (2)

a = 250000/5000 = 50 m/s^{2} or a bit over 5 times the acceleration of gravity.

This is the situation in your glass tube: (a) a tight fit and (b) acceleration from a dead stop. The initial acceleration would be quite a jolt for the passengers, but not likely to kill them. However, that’s not it would be like in a hyperloop tube.

In the case of a tube failure behind the pod, the pressure wave arrives at the speed of sound (there isn’t enough energy in the system to trigger a blast wave). If the pod is already moving faster than sound, it won’t reach the pod at all. If the pod were moving slower than sound, the pressure wave would arrive at *the difference in speeds*. Once it arrives, the excess pressure would leak around the pod, equalizing the pressure everywhere to one atmosphere. The propulsive force on the pod, if there were any in that section of tube, would be insufficient to maintain speed. Indeed, the aerodynamic drag would then be larger than whatever propulsive force was available. To calculate the maximum acceleration and deceleration requires that we know how long it takes the pressure to equalize. That we cannot do here. In any case, the deceleration is the unknown, not the acceleration, which we determined in the worst case was 5 gs.

Let’s look first at the aerodynamic drag on the pod going 1200 km/h in the low-pressure condition.

F_{d} = c_{d} ½ ρ^{2} v^{2} A (3)

where

F_{d} = drag force (N)

c_{d} = drag coefficient

ρ = density of fluid

v = flow velocity

A = characteristic frontal area of the body

Drag coefficient we know from automobile advertisements. A Toyota Prius has a *c _{d }*of 0.26. A human standing in the wind has a

*c*of 1.0 to 1.2. We’ll assume the pod is the same as a Prius. It’s a bad approximation;

_{d }*c*increases with velocity, reaching a peak at the speed of sound, but we’ll live with it for this investigation.

_{d}c_{d} = 0.26

A = 2.0 m^{2}.

The density of the fluid (depressurized air) ρ can be calculated using the universal gas law.

ρ = (MW)P/RT (4)

MW = 29 kg/kgmole = 29

P = .001 Bar = 101 Pa

T = 59°F = 288 K

ρ = (29)(101)/(8314.5)(288) (kg/kgmole)(N/m^{2})/[(joules/kgmole-K)(K)] = .001223** kg/m**^{3}

v = 340 m/s

Now we calculate the aerodynamic drag according to formula (3).

F_{d} = c_{d} ½ ρ^{2} v^{2} A = (0.26) ½ (.001223)^{2} (340)^{2} 2 = 0.004 N

Wow, almost no drag at all! That’s why the evacuated tube is so important to hyperloop. Now let’s see what the drag is when atmospheric pressure air is reintroduced to the tube.

P = 1.0 Bar = 100000 Pa

ρ = (29)(100000)/(8314.5)(288) (kg/kgmole)(N/m2)/[(joules/kgmole-K)(K)] = 1.211** kg/m**^{3}

F_{d} = c_{d} ½ ρ^{2} v^{2} A = (0.26) ½ (1.211)^{2} (340)^{2} 2 = 4408 N

The aerodynamic drag has increased a million-fold. What would that feel like to the passengers?

a = F/m (2)

a = 4408/5000 = .88 m/s^{2} or a about a tenth of a g. If we asumed instead that c_{d} had increased by a factor of five at Mach 1.0 (typical for a bullet), the passengers would feel a half-g deceleration as a result of aerodynamic drag. That’s hardly a calamity.

Aerodynamic drag is not the only force working on the pod. Levitation induces drag. The best estimate we have today for maglev is 1:200 lift/drag for the speeds we are discussing. The propulsion system needs to supply a constant force of 250 N to maintain the pod speed. If power to the tube failed at the same time the vacuum system did, the pod would decelerate at .05 g because of the power failure. These forces are additive. Still, a deceleration of 0.55 gs is not unusual for an automobile.

A vacuum failure behind the pod, simultaneous with a power failure, would have pressure wave approach the pod from behind, while it was already slowing down.

The worst-case scenario is a vacuum failure ahead of the pod combined with a power failure. It would combine the immediate drag associated with increased air pressure with the drag from levitation. Using formula (3), the velocity *v* is the sum of the speed of the pressure wave and that of the pod. If both are Mach 1.0, the drag goes up by a factor of four. That’s two gs. Add another twentieth of a g for the levitation failure and you still have two gs. Uncomfortable, but hardly life threatening.

Thunderf00t, there are plenty of valid arguments against the Hyperloop, but vacuum failure is not one of them. Getting passengers out of a stranded pod is.