World's steepest funicular railway opens in Switzerland, featuring a 110% gradient

midian182

Posts: 10,653   +142
Staff member

For those who don’t know, a funicular railway is a type of railway that takes passengers up and down steep inclines using carriages pulled by cables. Last week, the steepest in the world, the Stoos Bahn tram, opened in Switzerland.

The $52.6 million project, which took 14 years of planning and construction, runs from Schwyz to the mountain village of Stoos. As it travels up the mile-long track, the circular cars rotate to keep the 34 passengers standing upright. At its steepest point, the funicular climbs at a gradient of 110 percent.

The track starts at 1840 feet above sea level and finishes at an altitude of 4284 feet. Moving at 22 miles per hour, it takes four minutes to complete a journey that consists of three tunnels and a bridge measuring 1640 feet in length.

Most of the track is designed for just one train, but it splits into two at a certain point to allow the carriages to pass each other.

Financial and engineering problems meant the funicular opened two years later than planned. It breaks the previous steepness record held by the Gelmerbahn at Bern, Switzerland, which has a gradient of 106 percent.

The railway replaces a previous funicular that has been in operation since 1933. "After 14 years of planning and building, everyone is very proud of this train," said Ivan Steiner, a spokesman for the railway.

Children regularly use cable cars to get to school in Switzerland. In addition to attracting tourists, the new funicular will be used to connect communities, reports the BBC’s Imogen Foulkes.

Permalink to story.

 
Cool project in concept, an exercise in pointless waste in reality.
52.6 Million, 106 residents.
Just to improve upon a replace an existing tram. Government efficiency.
 
Cool project in concept, an exercise in pointless waste in reality.
52.6 Million, 106 residents.
Just to improve upon a replace an existing tram. Government efficiency.
They're a wealthy nation so they have that option. Are their children struggling in poverty and not getting good education?
 
Nope
The steepest one is on the inclined tower of Montreal's Olympic stadium. It gradually gets steeper as it climbs, to reach almost 80 degrees at the top.
 
I'll bet, at least as anticipated, it was thrifty. Building a road, including three tunnels and the long, long bridge, would have been much more expensive. This saves the schoolchildren from having to travel whatever long roundabout route it would otherwise be. I don't think the Swiss would have done it if it weren't practical.
 
I'm confused. Wouldn't 90° be straight vertical? Does this mean this railroad starts to invert?!

110% grade.. not 110 degree grade. Percent grade is the rise divided by the run and expressed as percent. (So a 45 degree slope I.e. 1 unit run over 1 unit rise, is 100% grade) ... this train angles to about 49 degrees. Edit... this may not seem like much, but most very steep roads rarely exceed 15-18% grade. If you drove down a 20% grade road (11 degrees) it would feel much like a 45 degree (or 100% grade slope). These passengers probably feel as though they are going nearly straight up, even though they are just barely past 45 degrees. We get this phenomenon due to our relationship with gravity... right up to 45 degrees it still feels like gravity is mostly helping us stay on the surface, for each degree past 45 it feels as if gravity is now actually helping us fall away from or leave the ground (and it is) right up until we completely lose contact at 90 degrees.
 
Last edited:
Nope
The steepest one is on the inclined tower of Montreal's Olympic stadium. It gradually gets steeper as it climbs, to reach almost 80 degrees at the top.

I'm sure they are referring to trains that traverse natural landscapes, not buildings. You could call an elevator a funicular and say that it goes straight up with an infinite grade.
 
110% grade.. not 110 degree grade. Percent grade is the rise divided by the run and expressed as percent. (So a 45 degree slope I.e. 1 unit run over 1 unit rise, is 100% grade) ... this train angles to about 49 degrees.
Thank you for the explanation. I was fixing to have to look this up.
 
110% grade.. not 110 degree grade. Percent grade is the rise divided by the run and expressed as percent. (So a 45 degree slope I.e. 1 unit run over 1 unit rise, is 100% grade) ... this train angles to about 49 degrees.
I have never heard of % being used before. I was just about to ask about that lol
 
Cool project in concept, an exercise in pointless waste in reality.
52.6 Million, 106 residents.
Just to improve upon a replace an existing tram. Government efficiency.
50,000 per resident nearly. The residents tax portion would be close to 3500 a year for 14 years to fund this project properly. This is why tax dollars and budgets are important. What got sacrificed for this project.
 
I'm confused. Wouldn't 90° be straight vertical? Does this mean this railroad starts to invert?!
No,. They have the incline percentage tricked up against distance climbed versus distance traveled.

In other words, if you travel 100' horizontally, and go 110' vertically over that same distance, then the grade is "110%".

If we want to talk about "angle of climb", you really can't go past 90 degrees, or dead vertical. After that, it becomes a question of whether you choose to measure the obtuse angle, which in this case would be "99 degrees", or the acute angle, which would be "81 degrees".

OK, how about if we try it in 3 dimensions, as with aircraft? There, the speed of the aircraft, (time), affects the other variables, angle of climb, and distance traveled The only fixed value would be the rate of climb, generally expressed in "feet per minute", (FPM).
 
Last edited:
110% grade.. not 110 degree grade. Percent grade is the rise divided by the run and expressed as percent. (So a 45 degree slope I.e. 1 unit run over 1 unit rise, is 100% grade) ... this train angles to about 49 degrees.
Close, but not quite

Degrees are measured from the center of a circle. If you extend a perpendicular line from 100' from the center of the circle toward vertical, you'd find that the 45 degree line falls short of the right angle coming from the radius line.

Basically, when calculating degrees you're projecting a 200' circle (degrees) into a 200' foot square box when attempting to measure the linear, "percentage rise".

This should help you visualize what I'm saying:

image-20160309-13717-1c27vzw.png


Draw a line from the lower left corner of the square to the upper right. That gives you 45 degree angle, relative to the base of the square. .

In any case let's make the square 1 foot. Pythagoras says: "the sum of the squares of the sides of a right triangle, is equal to the square of the hypotenuse".

OK, so 1 squared is 1, and 1 & 1 is 2 < (2 is the square of the hypotenuse). The square root of 2 is 1.41 (and some change).

Now multiply all my answers by 100, and you'll see that @ a 45% grade, the train would travel 141 feet, against 100 feet of forward travel. So the answer is absolutely not, a 49 degree grade..
 
Last edited:
Close, but not quite

Degrees are measured from the center of a circle. If you extend a perpendicular line from 100' from the center of the circle toward vertical, you'd find that the 45 degree line falls short of the right angle coming from the radius line.

Basically, when calculating degrees you're projecting a 200' circle (degrees) into a 200' foot square box when attempting to measure the linear, "percentage rise".

[snip image]

Draw a line from the lower left corner of the square to the upper right. That gives you 45 degree angle, relative to the base of the square. .

In any case let's make the square 1 foot. Pythagoras says: "the sum of the squares of the sides of a right triangle, is equal to the square of the hypotenuse".

OK, so 1 squared is 1, and 1 & 1 is 2 < (2 is the square of the hypotenuse). The square root of 2 is 1.41 (and some change).

Now multiply all my answers by 100, and you'll see that @ a 45% grade, the train would travel 141 feet, against 100 feet of forward travel. So the answer is absolutely not, a 49 degree grade..

You are clearly interested in appearing very smart, but not doing great at reading comprehension. I never said a 45% grade was a 49 degree grade. I said a 110% grade was roughly a 49 degree grade. which it is... (sorry 111% grade is 48 degrees, my bad at my horrible mental maths being off by one degree.) Also, my example was using rise over run to calculate percent, so you don't need to introduce tangent (which is what you are using in your example, without naming it) when going directly from rise over run. You only use tangent when trying to go straight from degrees to percent without the rise and run, since tangent gives you the rise/run ratio.
 
Back