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B
eams have to cross each othersī path here and there. That situation is usually solved in either of 2 ways:
In order to avoid too steep slopes and too great height over the ground at the stop areas, so that the elevator won't reach all the way down, the beam traffic network should be limited to 2 levels on those parts that are above ground. This limitation need not apply, then, to bridges, tunnels, indoor facilities and garages for the beam cars. |
On this page will briefly look at:
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![]() Figure 1Figure 2 |
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n the case of two-way traffic, there is still no problem for cars wanting to turn right (assuming there is right-hand traffic on the beams) or going straight ahead in a 2-plane crossing. A left turn will require a more complex solution, such as a roundabout. The alternative in figure 2
requires quite a lot of space, but can in an urban environment be implemented by, for instance, putting one beam around a neighboring block (B in the figure). The B route will enable cars coming from the north (or up in the figure) to turn east.
Figure 2:1 |
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roundabout as shown figure 3 enables shunting in all directions, while keeping the intersection in one plane.Same plane croosings can be used even at high traffic throughputs, for instance by enabling cars turning right to use a separate roundabout, outside the first one, as is shown in figure 2:1. In the case of very high traffic intensity, one will have to use a 2-plane intersection, as is shown in figure 2:2. In this case, the routes marked B will have to go close to the ground, resulting in street-level traffic. Excellent place for a stop. Figure 2:2 |
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The optimal solution to the 2-way, 2-level crossing problem is to do as in figure 2:3, provided that there is space enough. The beamcar at A, coming from north, has the options of going straight ahead (1), to turn west by following path 2, or to turn east by following path 3. The advantage with this solution is that vehicles going straight ahead need not slow down, as they have to do in a roundabout.
Figure 2:3 |
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o let the beams cross each other in the same plane would be practical in many situations. 3 problems would arise out of this arrangement:
1) Problem: The wheels of the propulsion car would have to cross the slit of the intersecting beam.
This slit is broad enough to require supports for the wheels, so they won't get stuck or destroyed.
2) Problem: The risk for collisions will have to be eliminated. Possible solution: Both nodes closest after the crossing on their respective beams would have to operate as one node of the converging kind. Figure 7 illustrates how this would work. Nodes 1, 2 and 5 are normal divergence nodes, while 3 and 6 are normal convergence nodes. | ![]()
Suppose (to simplify matters) that nodes 2, 3, 5 and 6 are symmetrically positioned around the crossing. Then 5 and 2 (or, rather a point a bit south of node 2) would be booking points for node 6, and 2 and 5 (almost) would be booking points for node 3. Thus far, everything would function in accordance with the rules for "point synchronous networks as described elsewhere. |
ut apart from this, nodes 3 and 6 would have to function as one common convergence node, with booking points at 2 and 5. This whole planning would be less complicated if we were to assign common booking points for the whole intersection. These booking points would then preferably have to be moved to
to the next foregoing nodes, according to standard procedure. That is to say, node 1 and a point equally far removed
from the intersection, but on the other beam (i.e. point 4) would be booking points. |
As a tangible illustration, if the vehicle A is travelling straight ahead, then A would have to let B pass, since B has passed its booking point and A has not. This would apply regardless whether B is about to make a turn or travel straight on. If A however is about to make a right turn, then A, for the same reason, would have to wait for both B and C to pass (considering that B has already passed its booking point). | ![]() However, 2 is a divergence node, and is thus informed how B is going to shunt. Consequently, if node 2 reports that B is going to turn left, then A should not have to take B into consideration. Simple, when you think about it, isn't it? |
| Copyright Đ 2004, SwedeTrack System. | Last Updated: 2007-01-17 |
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