It has been years since I had a job in transportation and I goofed in another post. If you could show me where I am going wrong it would be appreciated.
Lets say I would like to raise track from level to 5 inches high over a 48 inch run.
The grade for a 1 inch rise over 100 inches is 1%. 5 inches over 100 is 5%. To figure that into a 48 inch run I first divide 100 by 48 and come up with 2.083333.
Then I multiply my 5% by 2.083333 and get a 10.41665% grade? (sounds reasonable.)
To turn that 10.41665 into something I can use a miter saw to cut I press the Tangent button on my calculator and get 0.1838.
So according to this if I cut a series of slightly less than 2% angled supports I will be in business?
If I plug a 48 inch base and a 5 inch height into my visualgeometrycalculator app on the cell it tells me the angle of the low corner is 5.946 degrees.
Well your 10 plus per cent grade is right but you won’t be.pulling many cars up it. As far as the angle for the top of the piers you have a right triangle of 5" high by 48" long so the angle where the slope begins can be found from the tangent of the angle which is 5/48. You can them look up the angle in a chart that coverts tangents to angles. Should be one on the net somewhere.
when you start with the angle in degrees and you want to calculate the tangent or grade (not as percentage) you have to use the Tangent button. Only after multiplying with 100 you get the grade as a percentage.
However when you start with the tangent and want to calculate the angle, you’ll need the reverse procedure!
You will probably need to push two buttons at the very same time: SHIFT + TANGENT (or INV+TANGENT or 2nd + TANGENT). On some machines it is: TAN-1.
With the tangent=0,1 (grade of 10%) the angle is almost 6 degrees, completely in line with your findings.
BTW grades above 4 percent are called an attraction in itself (John Armstrong). Not calculating extra drag for curves, not taking sufficient easements into consideration.
Slope is rise over run. In your case it is 5/48 = 0.104 move the decimal point 2 places to the right for percent, in this case 10.4%. A couple of notes, the units need to be the same. i.e 5 inches of rise over 48 inches of run. Everything after the third decimal place is more precision then we need.
Degrees is the angle measurement. The best way to get this from the slope is to use a free online calculator such as here In this case 10.4% converts to 5.94 degrees.
And these calculations do not account for any transition at the ends to ease into and out of the grade; just a sudden change which no locomotive or rolling stock could tolerate.
Cacole said it, but if I could be permitted to guild the lily a bit, what it means is that while you are fashioning the necessary vertical curves that get your locomotive and its couplers into and out of the grade proper with the trailing car(s) still attached, you only have what is left between those transitions on either end of the grade to get your height. This necessarily means you’ll end up with a grade between the transitions closer to 11.5%, and good luck with that one.
Well, the final number is correct (although you can calculate it much more simply, the % grade is just the slope = rise / run, multiplied by 100 to make it a percentage) but I don’t know about “reasonable”. Not much is going to go up a 10.4% grade.
