Gelmerbahn funicular in Switzerland has the steepest grade in the world; 106%!!!
Here it is:
Earl
|
Replies sorted oldest to newest
Ummm... wouldn't a 106% grade mean the car and track is actually inverted/ starting to go upside down??
I think he is referring to 106% which is a little steeper than 45° which would be a 100% grade (1 foot elevation in 1 foot of travel). Grade callouts can be easily confused.
Neil
Grade is rise over run. So a 106% grade means that if the train moved 100 ft horizontally, the train would rise 106 feet.
Earl
Wikipedia strikes again!!!
I think they initiated the reference to a 106% grade (which is impossible as Train Nut pointed out, but which is repeated in various places on the web).
45 degrees is just over 70 percent grade. The run is measured along the hypotenuse of the triangle, not horizontally. For grades in the 1 - 5 percent range it doesn't matter much how the run is measured, but for steep grades it must be the actual distance traveled (the hypotenuse), not the horizontal distance.
A 100% grade is 90 degrees - straight up!
According to the same source (!) the rise is 1470 feet over a run of 3373 feet which is about 44 percent. That is the average over the whole route so parts are probably steeper. Another (more reliable?) source lists the max angle as being in the 50 degree range.
Still impressive!
Jim
Jim,
Here grade is rise/run and run is the horizontal distance. Here is an illustration from Wikipedia:
A 45 deg angle is a 100% grade because rise=run. The posts above that said the train would be upside down are confusing grade with angle.
The 106% grade calculates to a 46.7 deg angle. Arctan (1.06)=46.668... degrees.
Earl
Earl,
You are correct in that the angle of inclination is the arctan of delta h over d. But "d" is not the distance used in a % grade calculation.
To figure the % of the grade when talking about the slope of roadways, railroads (including models), hiking trails, etc., you use the "l" in your diagram not the "d".
The "d" is used in certain specialized construction terms, such as in a "1 in 4 pitch roof", for example,....but not when the grade is a route being traveled and is to be expressed as a percent.
In basic math, the calculation of the term "slope" also is delta h over d, but again that is different from the meaning of percent grade as used in roadway engineering.
As I said, usually roadways and railroads have gradual grades where there is not much difference between "l" and "d". But, in the case of very steep grades it becomes critical to use the actual distance traveled (l) in a percent grade calculation.
When describing hiking trails or highways, you often hear them described as rising 1500 ft in 5 miles or 2000 ft in 5 miles where the 5 miles is obviously the actual distance traveled (l). This is the number used in a percent grade calculation.
One reason this convention in the measuring of "run" was chosen was from a simple engineering viewpoint. Grades aren't always straight. It is much easier to calculate a "run" by traveling along a route with various curves, etc. and clock the mileage than it would be to survey the horizontal distances.
Another way to look at it is to do some calculations for % grade at angles over 80 degrees or so. Using your way, the percent would rise astronomically approaching 90 degrees - not a very useful calculation and making comparison of steep grades by % very difficult. The purpose of using the percent grade in the first place is to be able to compare steepness easily.
Wikipedia is sometimes misleading in its explanations (to put it kindly ).
Jim
Another explanation is that they are using a 180 degrees as if up to 90 it goes straight up and then after ninety it begins its descent so that 106 degrees would be 6 less than 90 or 84 degrees if it was facing the other way. It doesn't use as much math and matches the pictured angle.
However, I went to public school, so please bear with me.
EML
Access to this requires an OGR Forum Supporting Membership