being new to the lingo…is a two percent grade the same as two degrees?
Hi, and welcome. The grade is expressed as a percentage of units of linear movement in elevation divided by the linear movement along a horizonal axis. For example, if you move 8 feet toward a direction, but also rise 3 inches, you get 8’ X12" (to get common units between the rise in elevation and the “run” horizontally) equaling 96" and divide your rise of 3" by the figure of 96. That comes to 0.03125, or moving the decimal two places to get hundredths, 3.1%. For the distance travelled horizontally, in other words, the elevation change is positive 3.1%. That’s a substantial grade, even for model railroads.
Degrees are an expression of angle. We use them sometimes when we talk about a curve, but not in the modeling world. The big railroads use degrees when plotting out curves as they survey.
thank you for the info. the formula helps explan the difference and now i can calculate my own thanks
dug
Nope. A 2% grade has an enclosed angle of 1.15º.
I can explain how to calculate that figure if you have a calculator with the following: square, square root, reciprical (1/X), and tangant/cotangent functions.
If you want a real handy tool, go to Sears and buy their digital level/elevation/pitch tool, I think it sells for around $30, it is one handy MRR tool for laying elevated track. I love it, got it for Christmas last year from honorable son #3…(hint, hint).
Save the $30, and build your own tool from scrap parts. If you know what your grade is going to be, or what you want it to be, make one on a level place like the floor. Then make a wedge using a short piece of wood and a 3" nail. Just place the nail across the wood and slide it along until it creates a mini version of the grade from nail to the edge of that wood. Tape the nail in place, and place a short 6" level on it from the nail to the same edge of the wood.
Now when you place the wood block on a slope, if the level’s bubble is centred, you have the same grade slope as originally made. Most model RR grades are less than 3%; aim for about 2-1/4% maximum, less is better. Steeper grades will really limit what a loco can pull up it. If there is a curve on the grade, the extra drag will reduce the loco’s pulling capacity even more.
Have fun, George.
Anyone used the 5-12 dollar grade gauge tool that Micro Mark sells? It seemed like a good tool at a really good price. I’ve thought about getting one. I don’t have a catalog handy for a part number, sorry.
Isn’t the simple/ basic explanation for a grade: how many inches of climb over 100 inches of travel?
a 2" climb in 100 inches equals a 2% grade?
Thanks.
Angle Gauge pn #82280 sale price $5.50 and the Laser Level #83041 for $9.20
Just not sure how acurate the angle gauge would be… but then $6.00 don’t even get you a happy meal…
Hi, and welcome aboard!
As previously mentioned, percentage and degrees for inclines can both be used. I do find that in the planning process, using “percentages” works much easier for me. In example, my layout uses 2 percent grades to go from the staging tracks to the main level, and on up to the outer level that surrounds the layout. In building this, it broke down to a 2 inch raise in 100 inches of linear track, or about 1/2 inch raise in 25 inches of linear track (I rounded to two feet). With a good level, this was pretty easy to build the grades, even with a good portion being curved trackage.
Hey, ENJOY !!!
Mobilman44
Capt. Grimek, yes,…if your grade’s length just happens, by happy coincidence, to be 100" long. Mine are much longer, so that a 2" rise over my longer distance would turn out to be much less than 2%.
Since the rise you desire and the run you desire are likely to be dependant on available space (and let’s not forget the transition into and out of the grade, which reduces the steady grade run), the formula that works generically anywhere, and for any scale, is rise divided by run using the same units for the numerator and the denominator. For example, my rise is 8" (yeowww!!!), but I allowed 264" to achieve that altitude on my layout. Dividing the 8" by the 264" yields a figure of 0.031, or 3.1%
To answer your question without going into detail as others have done, yes a two precent grade is the same as a two degree grade.
Bill
Awk! No, it most definitely is not! You can only have up to 90 degrees from horizontal to absolute vertical, whereas a percentage, by definition, is out of a “per centum”, Latin for “by 100”. Ninety and one-hundred are not the same thing, so degrees and grade percentages do not equate. That is what trignonometry tells us. See R.T. Poteet’s response above…he alludes to the fact.
-Crandell
In most cases grade is expressed as a % and implies vertical change, in most cases “degrees” are used to express curvature or horizontal change.
A quick and dirty grade “calculator” is to laminate 1/8" material in s a stair step fasion with each step about 1/2" deep. If you use a 2 ft level, each step represents 1/2 % increments of grade.
To determine an existing grade, put the level on the grade and put the step under the low end, moving the level up one step at a time until its level, then see when step you are ont to measure the grade.
To set a grade, put the step of the grade you want under the low end of the level and then adjust the roadbed up or down until the level is level.
You can make one for about 25 cents and 30 minutes of time out of 1/8 masonite, balsa or other stock.
Dave H.
Hi!
I have that Micromart “grade gauge” and - IMHO - it is not practical to use for grades of 2 or 3 percent, as the gauge markings are way too small to accurately indicate the lower numbers. If you are doing something over 5 %, I would say it would be a help. Mine will go up for auction come wintertime.
Please note: Micromart has some fabulous tools and supplies and I’ve bought a lot of neat stuff there over the years. The gauge in question is a “high quality item”, but not applicable - IMHO - for the smaller grades of a typical model railroad.
Mobilman44
Welcome Dug [#welcome], and as you have already noticed I am sure, there is plenty of help here for any question/s you may have.
I will not reiterate what others have said so well in most cases, so I will just say this: Take your start point on your layout where the rise will begin and mark the spot. Then mark the end point where you want the track to end it’s rise. Measure the VERTICLE rise at this end point. Now measure the HORIZONTAL distance from the Start mark you made, to the End mark you made. Divide the Verticle number by the Horizontal number, this will be your grade elevation. As an example, lets say that your Vertical measurement is 2" and your horizontal measurement is 48". 2"/96" = 0.021. Simply multiply this number by 100 to get % grade elevation, in this case 2.1% . This would be an acceptable grade and would take an entire 8 foot bench run to achieve.
Hope this helps you,
In other words, a 45 degree grade would equal a 100% grade, right?
Exactly. And I embarrassed myself by stating, some months back, that a vertical grade, perpendicular to a horizontal surface, was 100%, and I was corrected in short order. We all have brains poots now and then. [(-D]
-Crandell
Your 48" horizontal measurement should be 96".
I don’t know who corrected you because you were absolutely correct. Grade is always measured along the hypotenuse of a right triangle. If the opposite–the side reflecting how much the right triangle is going to diverge off of the adjacent–is a factor of one and the hypotenuse is a factor of one then the adjacent has to be a factor of zero–1 squared minus 1 squared equals zero-- and your opposite and your hypotenuse have to be in coincidence; if you want a 100% grade you have to go straight up in the air to get it. That, Mr Crandell, ain’t 45º–its 90º.
A 45º angle will give the following:
1 squared plus 1 squared equals two. The square root of two is 1.414. Divide that into one and you come up with .7071 and when you multiply that by 100 you get 70.71. A 45º angle renders a gradient of 70.71%.
For those of you who may have had electronics training the figures 1.414 and .707 should be familiar.
R.T. Poteet,
It was me who corrected Crandell.
As you know, a 45 degree angle means that both the horizontal (run) and the vertical (rise) sides of the triangle are of egual lengths. Let’s just say that both are 100" long.
So, with the formula being : (rise) / (run) x 100 = % grade… 100" / 100" x 100 = 100 %
Now, let’s say the angle is 90 degrees. That makes the vertical (rise) 100" and the horizontal (run) 0" (as there is no horizontal movement).
Again, with the formula being : (rise) / (run) x 100 = % grade…