Here is a topic to know how we can calculate the diameters of the pulleys that allow the reduction of speed on the systems of launchers and retarders in the ski lift stations.
Essential elements of a ropeway, the stations allow the boarding and disembarkation of passenger but
also the training of the cable(s). There are several types of train stations, it depends first on all on the type of mono-cable ropeways, stations can
therefore be equipped with a system for disconnecting or engaging the clamps on the cable (if the ropeway is detachable).
The upper part of the clamp is in contact with the rubber tires in the stations. The latter are driven by a system of pulleys/belts that allow speed reduction, to slow or accelerate vehicles (seats or cabins).Â
 Calculations
By observing the stations of detachable gondola, we realize that their length varies between 15.5 and 24m approx. We will seek to find out what is the minimum length of a station for a cable speed of 4.0m/s and in station of 0.15m/s, this with the aim of freeing up as much ground space as possible. For this study, we focused more specifically on detachable gondolas, by fixing a maximum angle of the cabin with the vertical of 6.5° (measured on gondolas). The goal is to maintain comfort in the cabin, avoiding knocking down the passengers. If we change this angle, the final length will also change.
We will then apply the PFD (Fundamental principle of dynamics) to the cabin but in a non-Galilean frame of reference because the cabin decelerates (or accelerates). It is then necessary to ‘add the forces of ‘inertia. T is force exerted by the hanger on the cabin (the hanger being the rigid metal bar which connects the cabin to the clamp). In the diagram opposite, the cabin is modeled in red, the suspension in gray and the forces are spotted in orange.
For this study, let us assume that the ‘angle q between the cabin and the vertical is constant. The cabin therefore undergoes an acceleration of 1.12m/sÂ
However, the cabins go from 4.0m/s to 0.15m/s.
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The deceleration (or acceleration) time is calculated:
t = (vf - vi) /a
With vf = final speed; vi = initial speed; a = acceleration found above
We then find, t = 3.44s
Or distance d= 0.5 * a * t² + v0 * t
By replacing, we obtain: d=7.14m
Now, in general we found that the deceleration (or acceleration) is done on about 7.2m. Attention, in reality, this result is different according to the speeds of the apparatus (in line and in the contour), and the deceleration does not necessarily take place in a regular manner.
For a gondola going from 4.0m/s to 0.15m/s over a distance at the station of 7.2m, let's calculate the diameters of the pulleys. The deceleration length is known. The diameter of the tires is fixed at 0.36m and the’ spacing between the tires of 0.4m.
nb of tires = 7.2 / 0.4 = 18
Tangential speed wheel 1: 4.0 m/s first wheel (which has the same speed as the cable)
Tangential speed wheel 2: to be determined.
There is therefore a speed difference of 0.21 m/s between two wheels.
The diagram above makes it possible to understand the operation of the speed reduction system.
It is known that v1 = v2 because the belt speed is constant (no slippage). The goal is therefore to look for the diameter of pulley 2.
