# Power to run the hyperloop

### Executive Summary: Hyperloop wonâ€™t go at the speed of sound

Even with reduced pressure in the hyperloop tube it will take a lot of power to go 1220 km.h. Â To maintain any fast speed will require powered sections of tube more frequent than one each 70 km. Â Powered sections need to carefully engineered to make the ride comfortable.

## Nothingâ€™s free in life

Speed costs money.Â Thereâ€™s no getting around that.Â The Concorde went 2.0 Mach, but it was a gas hog.Â The MiG-25 does 3.2 Mach but warriors donâ€™t count gallons.Â Your bicycle, your car, high speed rail and the hyperloop are all bound by the following formula.

P = Â½ÏCAv^{3}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (1)

**P** is the power required to overcome aerodynamic drag.

In simple terms, the power goes up with the cube of the speed.Â Thatâ€™s the simple form.Â Itâ€™s worth it to hear a short physics lecture.Â After that, youâ€™ll be better equipped to judge claims about high speed rail and hyperloop.

**Ï** (rho) is the density of the fluid.Â You and I, your bike and car, and the California bullet train operate in air at sea-level pressure.Â Yes, the air gets thinner in Denver or Mexico City.Â With the exception of the electric train, most engines and animals lose more from lack of oxygen at altitude than they gain from less air resistance.

Hyperloop intends to to reduce the density of the air in the tube to 1/1000 of that at sea level.Â As the hyperloop leaves the station, life will very good.

**A** is the area of the front of the hyperloop capsule.Â The frontal area of a Tesla automobile is 2.3 m^{2}.Â The absolute maximum size of a hyperloop capsule in the first test track could be the same, although it will likely be smaller.

C is the coefficient of drag.Â It is also called Cd.Â The Tesla has a Cd value of 0.24.Â The best Cd on an experimental car was 0.19 on the Volkswagen XL-1.Â Donâ€™t use Cd values you look up for airplanes.Â They are related to wing area, not frontal area.

Unfortunately, as you approach the speed of sound, Cd increases.

Between about 0.9 Mach and 1.0, Cd jumps by a factor of five.Â Thatâ€™s one reason airliners generally donâ€™t go over 0.82 Mach.Â High powered rifle bullets travel well over 2.0 Mach, where the Cd gets low again.

Letâ€™s put some numbers to all of this.Â Weâ€™ll use the Tesla as an example.

### Force

To overcome air resistance at 70 mph (113 km/h) requires 77 pounds of force (343 N).Â This is 14 HP (10.4 kW).

To overcome air resistance at 125 mph (200 km/h) requires 91 HP (68 kW).Â To overcome air resistance at 187 mph (300 km/h) requires 306 HP (228 kW).Â The energy required to go half again as fast requires three times the power.Â Itâ€™s at speeds just above 200 mph is that high speed rail starts getting very expensive, pushing all that air out of the way.Â In the hyperloop tube, with the air mostly evacuated, pushing the air out of the way is close to free, until you approach the speed of sound, when the Cd starts increasing.

### Drag as a result of magnetic levitation

Alhborn *describes* the â€œpassive magnetic levitation system â€¦ makes the capsule levitate through motion so there is no energy involved.â€Â Perhaps he means that the system will not require a constant flow of electricity to the track, but the system will certainly require energy.

Post and Dyutov (*2000*) estimated magnetic L/D at 200:1 for typical Halbach track configurations and speeds.Â Using the Tesla automobile 4,785 lb (2,175 kg) as an example, the pull required to levitate the car will be 23.9 lb (105.9 N).Â The good news is that this force, and the subsequent energy required to levitate the car, doesnâ€™t increase with speed.

We can calculate how much pull and energy will be required to levitate a Hyperloop pod.Â Weâ€™ll start with the optimistic assumption that a Hyperloop capsule can be built, on a per passenger basis, as light as a B-787-9, which can seat 420 passengers.Â For every pound of structure, it can carry one pound of fuel, passengers, or cargo.Â A Hyperloop capsule carrying 28 passengers weighing 200 lb each would weigh 5,600 lb (2,545 kg).Â Fully loaded, a capsule would weigh 11,200 lb (5,090 kg).Â That would require a pull force of 28 lb (125 N).Â The pull force will be relatively constant over various speeds but the power required does go up with speed.

P = FvÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2)

If we choose to run the Hyperloop at .82 Mach, the same as an airliner, and before the Cd starts increasing, the average velocity will be 281 m/s (629 mph).Â That will require an average Â 35.1 kW (45.7 hp) along the route.Â A smooth ride in the capsule requires that the power be applied continuously.Â Yes itâ€™s possible to inject this energy to the system at intervals.Â Muskâ€™s white paper describes linear accelerators every 70 km, for a total of 1 percent of the tubeâ€™s length.Â His plan also includes a battery-powered fan on board the capsule that powers air-skis.Â This is different from the magnetic levitation strategy offered by HTT.

If one depends on magnetic levitation alone, the capsule will decelerate constantly in every section of tube not equipped with a linear motor ( the unpowered sections).Â This will be slightly unnerving to a passenger, but not overly so.Â When the capsule reaches a powered section, the sensation will change from deceleration (braking) to acceleration.Â In order to maintain a limit of 1.5 g felt by the passenger, the acceleration along the direction of travel must be limited to 1.12 g (as explained in *Donâ€™t Spill the Coffee*).Â The linear motors need to be supply a relatively limited amount of power at their beginning and end of each power segment, else the passenger will experience the sensation of jerk (the rapid transition between acceleration and braking).Â One can use equations from freshman physics to calculate the braking and the acceleration forces.

F = maÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (3)

Then rearranging terms.

a = F/mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (4)

We calculate deceleration on unpowered sections of track.

a = .05 m/s^{2Â }Â [.0051 g]Â Â (5)

This value is exactly the result we would expect given the L/D ratio of the levitation system.Â It will be barely perceptible to passengers.

At the start of the trip it will be advantageous to accelerate at the highest rate.Â The limit will be 1.12g.Â That acceleration cannot begin instantaneously.Â It is limited by the maximum jerk.

The maximum speed profile is achieved when the capsule accelerates from rest at the maximum jerk, up to its maximum allowed acceleration, which it then maintains until it approaches the maximum allowed speed. Â Acceleration is reduced at the maximum jerk until acceleration is reduced to zero as the capsule attains maximum speed. Â This process is reversed at the other end of the trip, as the capsule comes to a stop.Â In between the accelerations, the capsule should travel at its maximum speed.

### What is jerk?

Jerk is also described as jolt, surge, or lurch.Â It is the rate of change of acceleration.Â Mathematically, it is the derivative of acceleration, the second derivative of velocity, and third derivation of displacement (position).Â A human can tolerate large accelerations (g-loads) but when acceleration starts or stops suddenly, the effect is disconcerting, and may be injurious.

Jerk is well studied in two human transportation systems: trains and elevators.Â Train designers have rules of thumb for jerk limitations in the y-axis (turns).Â Elevator designers have similar rules for jerk in the z-axis (vertically).Â Jerk in the x-axis is not a frequent topic of study in the civilian world.Â Planes, trains, and automobiles do not accelerate fast enough to have problems with jerk.Â Subways may, but for cost reasons, both jerk and acceleration are kept down.Â Jerk was a big problem in the deployment of the first electromagnetic catapults on aircraft carriers.Â Steam, a compressible fluid, would round off the jagged edges of the acceleration curve.Â The first generation of linear acceleration systems produced loads that aircraft could not withstand. Â Naval aviators are trained to withstand more acceleration and jerk than hyperloop passengers.Â The details are probably classified anyway.

For the purposes of our analysis here, weâ€™ll use rules of thumb from the elevator industry.Â Using z-axis measures may not be entirely appropriate for x-axis applications, but itâ€™s a starting point.Â Hospitals limit jerk in their elevators to 0.7 m/s^{3}.Â Generally tolerable is 2.0 m/s^{3}.Â Generally unattractive is 6.0 m/s^{3}.

### Hyperloop wonâ€™t go at the speed of sound

Even with reduced pressure in the hyperloop tube it will take a lot of power to go 1220 km.h. Â To maintain any fast speed will require powered sections of tube more frequent than one each 70 km. Â Powered section need to carefully engineered to make the ride comfortable.

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