The Propulsion Cars in suspended light beam traffic systems

For every action, there is an equal and opposite criticism.

he propulsion car used in suspended light beam traffic systems is that part of the beam vehicle that normally is totally invisible, because it is entirely inside the beam. In a manner of speaking, suspended vehicles have their bogeys above the passenger cabin or cargo hold, instead of below. Construction-wise, this means that the cabin or cargo hold has to be of a sturdier construction, since its weight has to be carried by the roof. But apart from this, the difference is not so big, when compared to supported vehicles.		Together, the propulsion car and the carrige constitute the beam vehicle or beamcar. On this page we will look at: General What the propulsion car has to do Connecting the propulsion cars The traction motor Table of possible electrical traction motors How powerful motors? Regulating the speed Braking the car Methods of traction The specific design that pertain to FLYWAY® are shown on a separate page.

1. General

et us start by defining our terminology. The propulsion car is the controlling and driving unit for every vehicle. It is the "invisible" part of the beam vehicle that is inside the hollow beam. In figure 1:1 we have used a FlyWay® beam vehicle as an illustrative example. Figure 1:1

The carriage is the part of the vehicle that hangs underneath the beam and is transported along it by the propulsion car. This carriage could be a passenger cabin if it is meant to take people. Or it could be a customer-designed cargo hold, a flatcar (for carrying goods and motorcars), grapple hooks for container transport or something else.Together, the propulsion car and the carrige constitute the beam vehicle or beamcar. In between the carriage and the propulsion car there would have to be equipment that holds the carriage steady and serve as a cushion. In the case of FlyWay, this assembly also includes an elevator.

well-designed propulsion car for suspended beam vehicles needs to include many things, if it is to fulfill the SwedeTrack System requirements for service in a well managed and scalable beam network. It should have the following equipment and attributes (See figure 1:2): Small cross-sectional area relative to the beam, so that the air-resistance in the beam is kept at a minimum Propulsion by means of an electrical motor (linear or rotational), which feeds the energy back into the power conduits whenever it is slowing down on its own volition A mechanical emergency break, which clamps around the bottom flange of the beam or on the wheels. It would be controlled automatically by the beamcarīs computer, from the Network Control Center and by the passengers by means of an on-board emergency brake. Some kind of mechanism for switching. A transceiver for communication with the computers that control the network. Transponder to communicate with sensors in the beams. An addressable computer which controls all the actions of the vehicle, including the automatic locking of the doors. Properly designed so that it can negotiate sharp turns on the beam. Good traction so that it can handle climbs and downward slopes.

Figure 1:2: A cross-sectional view of the FLYWAY design.

Figure 1:3

2. What the propulsion car has to do

3. Connecting the propulsion cars

Figure 3:1

wedeTrack System is open to the possibility of physically interconnecting beam vehicles. This should not be confused with "platooning", which is something quite different. There could be 2 situations when the propulsion cars need to be coupled together: When a cabin or flatcar has 2 suspension points (A in the figure above). When two or more cars are joined together, forming a train (B in the figure) Long cars might, for stability reasons, need two suspension points, each controlled by a propulsion vehicle. The reasons why those propulsion cars should be joined together inside the beam are mainly two: If the two propulsion cars, for some reason, do not pull together (or breake equally hard), the resulting mechanical strain should be taken up by those propulsion vehicles, rather than being transferred to the cabin (or flatcar) below. This makes for better design, and also simplifies the interface between the propulsion car and the cabin (or flatcar) below.

In sloping beams the traction against the beam might be uneven between the two cars. Here, again, that strain should not be transferred to the cabin (or flatcar) below. Regarding the train situation, the same two reasons would apply. In addition, one could conceive of a situation where one car pulls some other cars that are not equipped with proper propulsion engines. The reason for that might be economics; such cars would be cheaper to manufacture.

While the connection point at A in the figure above is permanent, the connection at B should be automatically controlled by the cars' computers. There would have to be strain indicators mounted at those connection points, signalling whether a proper connection has been attained or the connection has been disengaged, as the case may be. When passageways for passengers are used between the cabins, as in the illustration, the same kind of safety measures would have to apply there. As a rule, however, when there are no passageways between cars, the cabins should not come within touching distance of one another. Another safety aspect comes into play if these cars are equipped with elevators to lower them to the ground. Then, obviously, all elevators have to operate in concert with one another. This would have to be closely monitored, with safety mechanisms to stop the operation of the elevators if they should get out of synchronization.

4. The Traction Motor

The demands on this motor are: It should be electrical It should be at least 85 % efficient at all times It should be able to accelerate at least 5 km/hour per second with maximum load (corresponding to about 1.4 meters/second2) It should be silent It should feed at least 80 % of the braking energy back into the powerlines at normal braking It should be dimensioned to handle the emergency braking required.

We have 3 primary candidates for the job: The Linear Induction Motor The 700-750 Volts DC Motor The Synchronous Motor, controlled by a VFD. (And the winner is.... ?)

4.1 The Linear Induction Motor

This motor has its appeal because it has no moving parts. Compared to ordinary, rotating electrical motors, it has these advantages:

it is cheaper to manufacture it is cheaper to maintain (less wear) there is less energy losses (because of lack of moving parts) the vehicle can accelerate and decelerate regardless of wheel to rail adhesion itīs more silent than a conventional rotating electric motor the beamcar can use the rods in the beam to calculate its position.

It is stated above that LIMs are cheap to manufacture. But they are (at present) not so cheap to buy. The problem with LIM prices is that they are made in too small a quantity, and we seldom see a real mass production environment for them. So, although they are fairly commonly used, they are not made in anything like the volumes that rotary motors are, and consequently there is not the competitive inducement to make them available inexpensively. A good pedagogic information about how a LIM functions can be found on the Mc Taggart Co. website, which describes the "Seraphim". In our very simplified illustration above, we use a car with 3 solenoids, labeled A, B and C.

Solenoids are electric wires formed into cylinders. When they contain an iron rod and voltage is applied to the wire, the ensuing current creates a electromagnetic field, and they become electromagnets. The beamcar travels along a giudeway where iron bars are placed at regular intervals (they are numbered in the figure for illustration purposes). These intervals should be slightly longer than the distances between the solenoids on the car. Thus, in the illustration, the rightmost magnet (C) happens to be closest to an iron rod (6). It is therefore energized (indicated by red), and thus attracted to the nearest iron bar (nr. 6). As they come opposite each other, the voltage is cut from that electromagnet, and applied to the next magnet (B), which will then have a correspondingly short distance to bar number 5. As that magnet comes opposite number 5, that voltage is cut from that magnet and applied to the next (C), which will then be pulled towards bar number 4, after which magnet A will once again be closest to a bar (number 7), and so on. The power of thrust would be proportional to the applied electrical power to each magnet, to the number of magnets and inversely proportional to the distance between the stationary bars. As can be seen, it is quite simple to regulate the carīs speed with this system.

4.2 The 700-750 Volts DC Motor

The Direct Current Motor is well tested and widely used for moving electrical locomotives. The subway cars in Stockholm use 700 Volts DC and in the USA 750 Volts DC is common. It is reliable, silent, cheapest of the 3 proposed motors, and it is simple to use, considering that we will be providing 750 Volts DC in the power rails mounted in the beams.

The DC-motor is, however, not efficient enough for our demands. It remains to be seen if it will be widely used in beam traffic systems.

4.3 The Synchronous AC-Motor, controlled by a VFD

5. Other Electrical Motors

lectrical motors are the most efficient at a certain speed of revolution, where their effectiveness can reach 95 %. This is a function of design for each individual motor, and depends on which parameter one uses to alter the speed. Varying the motor's speed away from the motorsī optimum thus usually means that some excess power is wasted as heat.

Maintaining constant motor speed and using transmission gears reduces effectiveness due to the transmission mechanics, and adds to the cost to such a degree that this solution has not been deemed wortwhile. The most effective way to control an electrically-powered vehicle where load and speed varies all the time is to vary the frequency. But this is also the most complicated method, demanding the most expensive control equipment.

An overview of how to regulate the number of revolutions per second
for various motor types:

Type of Motor and method	Comments
The Asynchronous Motor The equation for speed of revolution (= n rev/sec.) is: n = 2*f (1-s) / p, where f = stator frequency p = number of poles s = drag	Generally speaking, regulating the speed for these motors with a high degree of efficiency can only be accomplished by varying the frequency. This is also the preferred method.
Pole switching By switching between having the magnetic poles connected in series and in parallell, the machine would function as alternatingly having 2 and 4 poles (p in the equation).	This method is of limited use in this context.
Rotor feed Altering the drag (s in the equation) by feeding excess induced current in the rotor winding to a resistance.	The wasted energy could be used by, for instance, a DC-machine, but this is rather awkward and rarely used.
Rotor resistance Altering the drag (s) by use of serial resistance.	Poor efficiency, especially at low speed.
Primary voltage Altering the drag (s) by regulating the voltage.	Efficiency depends on how the voltage is regulated.
Variable frequency The input frequency ( f in the equation) is varied continuously or in steps.	The most optimal method. In priciple without losses. Not widely used, as the regulating equipment is expensive.
The Direct Current Motor The equation for speed of revolution (= n rev/sec.) is: n = [U / k*f] - [2pRa*T / (k*f) 2 ], where k = N * p / c U = Primary voltage Ra = Resistance in anchor winding N = Total number of conductors in the anchor winding p = Number of electro-magnetic poles c = Number of parallell anchor windings T= Torque f = Electro-magnetic flow	For a certain Torque, n can thus be influenced by altering U, f or Ra. All 3 methods can be used for shunt motors as well as for serial motors. Torque is reversely proportional to angular velocity; as load increases, the motor slows down, while the torque increases, until a new equilibrium is found.
Series resistance By increasing the resistance in series with the anchor winding, the Torque as function of the angular velocity is reduced.	This indirectly influences the angular velocity by reducing the power of the motor. It also reduces the efficiency.
Current regulation One tyristor per phase is used to regulate the current by chopping it up.	A common method, and far better than using resistors.
The Ward-Leonard system	This is the "classical" method of controlling the speed of DC motors.
The AC Commutator Motor The commutator winding produces a 2-component AC-voltage, one is proportional to the magnetic field, and one proportional to the alterations in this field, as experienced by the rotor winding.	This motor has, like a DC motor, a commutator winding in the rotor, which functions as a current alternator. The 1-phase kind of these motors are often used for traction in locomotives. It could, with proper design, be fed with either AC or DC voltages.
Moving the contactor brushes
Variable transformer	In the picture, 1 = 3-phase stator winding 2 = Commutator winding in the rotor 3 = the regulating transformer
Triac

hen a constant force is applied to an object, the speed increases linearly (at least as long as we donīt approach the speed of light!). It is then easy to assume that if a constant effect is applied to an object, the speed will increase linearly as well. But that is not the case! If you consider that the live energy of a moving object is proportional to the square of its speed, you will realize that it takes increasingly more power to maintain a certain acceleration, the faster the object moves. Power for Propulsion Let us first look at the propulsion and the equations involved. F = force (in Newton, N) m = mass (in kg) a = acceleration (in meters/second2) s = length of road (in meters) v = velocity (in m/sec.) W = work (i.e. energy/time, in = Joule or Nm) P = effect (in kW) The connections between these can be expressed with these formulas: F = m * a W = F * s P = W / t momentary power: P = dW/dt Expressing effective power P as a function of F, we get: P = dW/dt = d(F * s)/dt = dF/dt * s + ds/dt * F Assuming that the applied force F is constant over time, the first term becomes zero, and we can write this as: P= F* ds/dt and since ds/dt = v we could write: P = F * v Letīs add some figures to this. Letīs assume that the vehicle as a whole has a weight of 2 tons, out of which the propulsion car weights 500 kg. Letīs assume an electric motor that can supply 70 kW of propulsive power. Applying all this power at the moment of start would make the vehicle perform like a dragster. This is not good for neither the vehicle, the beams nor the travellers.

We will therefore set an acceleration limit of 4 m/s2, which is reasonably fast but still comfortable. To maintain this acceleration we need a force of: F = m * a = 2000 * 4 = 8000 N Ideally, we would like to accelerate like this until our top speed is attained. In the FLYWAY case, this would be 120 km/hour, or thereabouts. 120 km/h = 33.3 m/s. Is this possible? Well, at the end of 8 secondsī acceleration, our speed would be 8 * 4 = 32 m/s., which is almost there. Now, from second 7 to second 8, we need to accelerate from 28 to 32 m/s, and since P = F * v = m * a * v, we get: P = 2 000 * 4 * 28 = 8000 * 28 = 224 kW. This is considerably more than our motor of 70 kW can muster. So, what happens in reality with our acceleration? Well, it maintains a 4 m/s2 acceleration until we give it full throttle, then the acceleration will (of course) taper off. If we could maintain a constant acceleration of 4 m/s2 for 8 seconds, we would have (almost) reached our cruising speed in this short time, using only s = a * t2/2 = 4 * 64/2 = 128 meters of beam. How long stretch of beam would it take, then, to aquire our cruising speed of 120 km/hour with only 70 kW at our disposal? That takes some calculating to find out. P = F * v = 8000 * v = 70 000 W v = 70/8 = 8.75 m/s It thus takes only 8.75/4 = 2.1875 seconds (2.2 sec:s rounded off) before we can apply full throttle and this corresponds to v * t/2 = 8.75 * 2.1875/2 = 9.57 m. Letīs put that figure at 10 meters. From then on, the applied force is no longer constant over time, and we get from above: P = dW/dt = dF/dt * s + ds/dt * F At constant speed, a moving object has a kinetic energy, such that Wk = m * v2/2 which in our case comes to: Wk = 2000 * 33.32/2 = 1109 * 103 Nm and the energy W that has to be imparted on the vehicle is the difference between the "full throttle" (ft) energy and Wk. Thus, W = Wk - Wft = 1109 - 76.56 = 1032 * 103 Nm (rounded off).

From P = W / t we get: t = W/P = 1032 * 103/70 000 = 14.7 seconds, and total acceleration time comes to: 2.2 + 14.7 = 16.9 seconds, which is not unreasonable. How much beam would be required for this acceleration? Since dW/dt = F * v = m * a * v = constant during these last 14.7 seconds, we get: dW/dt = 70 000 = 2000 * a * v => a * v = 35. The final speed v = 33.3 m/s and at that moment in time the acceleration would thus be a = 33.3/35 = 1 m/s2 (rounded). The acceleration rate is not linear but we could approximate it as such, giving an average speed from time = 2.2 sec:s to time = 16.9 sec:s of 8.75 + (33.3 - 8.75)/2 = 21.0 m/s, which gives 21.0 * 14.7 = 309.1 meters. For a more correct figure, we can note that: m * v * a = dW/dt = 70 000 => v * a = 70 000/2000 = 35, a constant value and; v(t) = vo + v * t where t is any time in an interval, and vo is the speed at the start of that interval. The covered stretch of beam, s, can be found from ds = v(t) * dt where v(t) = vo + a * t v(t) = vo + 35 * t/v This involves solving a 2:nd degree equation and then integrating the result over the time concerned. 309.1 + 9.57 = 318.67 meters, which could safely be rounded down to 300 meters. This is because acceleration is highest in the beginning, and average speed is thus more than half of the final speed of 33.3 m/s. 300 meters is quite a stretch to go, before a beamcar reaches maximum cruising speed. Deceleration could be quicker, since we are not dependent on the motorīs capacity for that, other than that it has to be able to transfer breaking energy back into the power supply. But one can thus see that only trunklines can be expected to keep speeds over 90 km/hour. These trunklines must not have sidings going directly to stops, but only sidings leading to local beams having lower speed limits. Beams that have sidings leading to stops would probably have to keep permitted speeds well below those 90 km/hour.

Sloping beams Accelerating up a sloping beam will of course require a longer beam. FLYWAY is not intended for steeper slopes than 5 degrees, which is roughly 5%, meaning that for 100 meters of horizontal travelling, the beam will rise 5 meters. Thus, as an example, letīs say that we want an acceleration of 4 m/s2, the speed when we hit the slope is 20 m/s (corresponding to 72 km/h.) and the total weight of the vehicle is 1500 kg. If we assign: Ps = The extra effect required for maintaining the acceleration in the slope Pa = Effect required for maintaining the acceleration horizontally g = Earthīs gravitational force = 9.81 m/s2 (= almost 10 m/s2) then we would get: Ps = m * g * 0.05 * v Ps= 1500 * 10 * 0.05 * 20 = 15 kW Pa = 1500 * 4 * 20 = 120 kW and the total required effect comes to: Ptot = Ps + Pa = 15 + 120 = 135 kW.

Power for the elevators Vertical lifts are more straightforward to compute, at least in our examples. The lifts do not start and stop with "jerks", the accelerate/decelerata rather smootly. On top of this, there is a maximum acceleration/deceleration of about 3 m/s2, which probably will not be reached for the short duration of up and down lift. The FLYWAY beamcars will not have lifts for higher lifting heights than 5 meters, and the calculated time for the lifting/lowering is 5 seconds. So, letīs assume that the motion is linear. Then, using the acronyms from above: The work W = F * s Letīs assume a cabin with its cargo and/or passenger to weight 1500 kg. to lift this 5 m straight up during 5 seconds, the energy required would be: W = 1500 * 10 * 5 = 75 000 Joule and the required effect P = dW/dt = 75 000/5 = 15 kW.

Hierarchical Beams Figure 6:1 illustrates the principle with hierarchical beams that has to be adopted. On top is the trunkbeam, which can only be accessed or leaved via a local beam; it should not have direct access to a station. Because of the relatively small speed difference between these categories of beams, acceleration beams can be kept reasonably short. Because the local beams have a lower speed, their access beams to stations can also be relatively short. In some areas with many stops, a third category of station beams could be motivated. These beams could (temporarily) have queues, and therefore the speed on these beams have to be adopted to suit the circumstances. A 4:th beam (at the bottom) also a "station beam" would handle queues of cars that are leaving the station for the local beam.

The following factors influence the speed of the beamcar: The maximum allowable speed on the current beam section (info from local node) Any restrictions that apply for this car during this journey (info from central computer and/or locally stored information in the carīs computer) Whether it has to negotiate a shunt, with consequential changing of direction (info from beam sensor and from stored information regarding this trip) Whether it goes through a sharp turn (info from beam sensor) Whether itīs about to stop (info from beam sensor and from stored information regarding this trip) If it receives an alert from local node, central computer or other vehicle that immediately influenses its speed (complementary stored information regarding this trip) If the radar on the carriage notes an unexpected obstacle in its path If running in a timeslot; directives from the local node computer If the radar on the propulsion car detects a vehicle ahead (illustration at right).

Figure 7:1 The internal radar inside the beam functions as an electronic buffer. It also has a doppler function, enabling the car to calculate the speed of an encountered car by measuring their relative speeds and knowing its own speed. The beamcar then behaves as a human driver; it regulates its own speed so that it keeps a safety distance commensurate with the current speed, i.e. the higher the speed, the longer the required distance to the car up ahead. For trunkbeams, we calculate with speeds upwards of 140 km/hour (corresponding to 90 miles/hour). It has been calculated that the safety distance for that speed should be at least 76 meters. But the general condition for maximum speed on the beams is that the internal radar should be able to see far enough. Long, straight beams should not present any impediment to these 76 meters.

	As soon as the beam bends, the allowable maximum speed would be reduced to that commensurate with visibility. The beamcar could be told by sensors whenever this allowed speed is altered, or the inside of the outer wall of the bending beam segment could have reflectors, telling the carīs radar that the beam is bending, and how far away this is.

The beamcar has to be able to brake. There are situations such as: Passing a booking point and being ordered by the node to regulate speed Approaching a shunt Going through a curve Passing a downward-sloping beam Stopping at a station Handle an emergency. At such times, the motor power is reduced or, in the case of an VFD-controlled Synchronous AC-motor, the frequency of the supplied power is reduced. The ensuing mechanical torque is fed back to the power rail.

The propulsion motor should be designed to generate enough braking torque to suffice in most cases. This not only conserves energy (by feeding it back into the power rails) but also saves on the wear of the mechanical brakes. Under certain circumstances, however, the beamcar might be required to brake at maximum force to prevent a possible accident. This "maximum force" has to take due consideration to possible passengers; an empty car could brake even harder. It is stated above about the traction motor that "It should be dimensioned to handle the emergency braking required". This means that if the electrical power is cut altogether, the beamcar will breake at a rate of approximately 2g (= 20 meters/second2). To complement this, there is also a mechanical brake, functioning in principle like the illustration at right, and applied on all four wheels of the propulsion car.

Figure 8:1; Mechanical brake assembly When the beamcar has to emergency-brake, the slow-down speed is monitored. If it isnīt sufficiently quick, the mechanical brakes will be applied. The braking force would be applied at F in the above figure by means of, for instance, in electrical relay.

MagLev is a developing technology, but it is already being used, with good results, on real railway trains around the world. See, for instance, a good description on the Japanese Chuo Shinkansen website. Itīs an expensive investment as compared to wheels, but there are positive side-effects from MagLev as well.

One could for instance combine it with a LIM propulsion system and use this as a very exact positioning system. Since a positioning system would be needed (at least in Flyway), that would mean an additional saving in investments. Good presentations on MagLev can be found on the Internet. Check for instance out Meissner-shield Maglev, and an article in Popular Mechanics.