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
dug wrote:being new to the lingo.....is a two percent grade the same as two degrees?
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.
From the far, far reaches of the wild, wild west I am: rtpoteet
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Now when you place the wood block on a slope, if the level's bubble is centred, you have the same grade slope as originally made.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.
Raised on the Erie Lackawanna Mainline- Supt. of the Black River Transfer & Terminal R.R.
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
ENJOY !
Living in southeast Texas, formerly modeling the "postwar" Santa Fe and Illinois Central
Capt. Grimek wrote: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.
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%
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.
Dave H. Painted side goes up. My website : wnbranch.com
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.
Welcome Dug , 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,
- Harry
HarryHotspur wrote: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.
grayfox1119 wrote: Welcome Dug , 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,
Your 48" horizontal measurement should be 96".
Jay
C-415 Build: https://imageshack.com/a/tShC/1
Other builds: https://imageshack.com/my/albums
selector wrote: HarryHotspur wrote: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. -Crandell
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 wrote: selector wrote: HarryHotspur wrote: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. 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.
selector wrote: HarryHotspur wrote: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.
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... 100" / 0" x 100 = ....
WHOA !!! Unless the rules of arithmatic have changed in the 20+ years since I took a math class in high school, it is impossible to divide by 0.
So, as you can clearly see, a 45 degree angle is a 100% grade.
Grade is always measured along the hypotenuse of a right triangle.
Actually, the grade is the measurement of the adjacent angle (between the hypotenuse and the horizontal (run) side of the right triangle) expressed as a percentage (rather than degrees). The length of the hypotenuse has no bearing on the grade being calculated. What does have a bearing on the grade are the lengths of the adjacent side (horizontal run) and the opposite side (vertical rise).
R.T. Poteet, I'm not sure where you got your formula. Are you using the theory that the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. If so, why do you divide 1.414 into 1 (or 1 / 1.414) ? In any case, you are not using the correct formula, so you won't get a correct answer.
I'm sorry, but I don't think any of you guys are on the right track trying to relate percent grade to an angle. Are you really trying to convince me that a 45 degree angle is a 100 percent grade because the rise (100) divided by the run (100) multiplied by 100 percent equals a 100 percent grade? I think not.
What if the rise is 200 and the run is 100? Are you going to tell me that the grade is now 200 percent? Come on now!
{edit: okay, I fess up. I did a little more looking and I guess that it is possible to have a grade more that 100 percent. For purposes of figuring things on a model railroad, I still say it is unnecessary confusion!!!!}
The angle you guys are defining by dividing the rise (side opposite the angle) by the run (side adjacent to the angle) is the tangent of the angle. Once you do this math, you can go to a table where you can determine the angle. It so happens that when the rise is equal to the run, the value of the angle tangent is 1. A table will tell you that the angle with a tangent of 1 happens to be 45 degrees. In case anyone is interested, the tangent of 90 degrees doesn't have a value because it is infinitely large; however, the tangent of 89.99 degrees is about 5,729. I'm not making this up. See http://hometown.aol.com/mathobservations/tan.html if you're interested.
Now, you can certainly determine the angle you have if you divide the rise by the run, but the only correct way to easily determine the percent grade is the original formula given in the posts above. Any attempt to throw angles into any of this is throwing effulent at a piece of rotating equipment.
Regards!!!!!
This may help using a visual aid.
When the rise equals the run, you will always have a 45 degree angle, and this equates to a 100% grade because you are rising in elevation the same distance as the horizontal run. You can understand this better by considering a perfect square. If you draw a diagonal, the angle will be 45 degrees.
chateauricher, let us repair back to school!
There are three sides to a right triangle: OPPOSITE--except in some special occasions this is usually designated as the shortest side diverging from the apex of the right angle and usually points upward; ADJACENT--the second--usually longest--side diverging from the apex of the right angle; HYPOTENUSE--the side directly opposite the apex of the right angle. The right angle in a right-angled triangle is, of course, 90º with the other two angles split equally into angles totalling 90º. To achieve 45º angles the OPPOSITE and the ADJACENT must be equal in length. In any right-angled triangle the HYPOTENUSE must always be greater than either the OPPOSITE or the ADJACENT.
Grade is always measured along the HYPOTENUSE of a right-angled triangle; rise will be reflected by the distance of the OPPOSITE. Grade is computed as the rise divided by the distance traveled along that HYPOTENUSE expressed as a percentage.
To find the distance traveled along the HYPOTENUSE we must resort to the PYTHAGOREAN THEOREM which states:
the square of the length of the HYPOTENUSE of a right triangle equals the sum of the squares of the lengths of the other two sides
i.e. our OPPOSITE and ADJACENT. Remember, to achieve a 45º angle the OPPOSITE and ADJACENT must be equal in length. Assuming a 100' length of our OPPOSITE--our rise-- then the length of our ADJACENT must also be 100'. Fixing these numbers into the PYTHAGOREAN THEOREM we add the square of 100' to the square of the other 100' and we come up with 20,000 square feet the square root of which is 141.42 feet. The length of the HYPOTENUSE when the OPPOSITE of a 45º angle is 100' is 141.42 feet. Now to the nitty-gritty; divide 100' by 141.42 feet and multiply by 100 to create a percentage and you come up with a 70.71% grade!!!!
This should be enough said but I sincerely doubt if it is. What ADJACENT-HYPOTENUSE angle must we have in order to create a 100% grade? The square of the HYPOTENUSE minus the square of the OPPOSITE equals the square of the ADJACENT: 100' square minus 100' square equals zero feet square; the square root of zero is zero. The only way possible to have an ADJACENT equal to zero is to have coincidence between our OPPOSITE and our HYPOTENUSE. If these lengths are coincident then our OPPOSITE-HYPOTENUSE angle is going to be 0º and our ADJACENT-HYPOTENUSE angle is going to be 90º and if we have a 90º angle then our HYPOTENUSE has to be pointed straight up in the bloomin' air coincident with the OPPOSITE. That, friend , is a 100% grade which is what I told crandell!
Yeah, all these numbers flying around have made me dizzy too, and I know all this stuff well. Grade is based on 100. A 2% grade is 2"rise in 100", 2' in 100', 2miles in 100 miles, you get the picture. To be able to get accurate numbers for any distance, you will either have to get friendly with a calculator capable of trig, or get a grade tool from micromark. If you have a regular old calculator, you can use these numbers;
1%grade---- .01 X distance = rise or .12" rise per foot (almost 1/8")
2%grade---- .02 X distance = rise or .24" rise per foot (almost 1/4")
2.5%grade-- .025 X distance= rise or .3" rise per foot ( almost 5/16")
easy huh. Just remember to do the distance in inches. The trig stuff comes in handy for other things that can be used for trackwork but are not really necessary.
If you know how high (rise) you need to end up, you would need to know one other variable (grade OR run) You can just substitute your numbers into the formula to get the answer
If you know what grade you want and the height you need, then the run is equal to rise/grade*
(*you must divide grade by 100 first. For 2% grade, use .02, 3%=.03 etc. )
If you know the run and height, then the grade is equal to rise / run X 100.
These simple formulas will figure out what you need for any height and any distance with any grade you would use or even dream of.
Hope this didn't muddy your mind any more
FWIW, I bought one of those grade meter things (probably from Micromark) and it's pretty handy, even though I have no trouble calculating the grade using rise over run. I like the gradometer (or whatever it's called) because you can measure the grade on any section of track that 5" long or so. Without the meter, there are many places on my layout where it's virtually impossible to measure the elevation of the track.
Also, you can build one for almost nothing. It works on gravity, so you only need a weight with a pointy end pointing down, and the other end held up by a thread just in front of a backboard. On the backboard, mark the degrees (-5 to +5 or whatever you need) using a grade school protractor. That's it.
Can we agree that the adjacent side of the right triangle is the horizontal side, or the run ?
Can we also agree that the opposite side of the right triangle is the verticle side, or the rise ?
That makes the hypotenuse (the longest side) the hill we are trying to climb.
What we want to know is : what is the angle between the horizontal run (adjacent side) and the hypotenuse (hill), expressed as a percentage ?
The accepted formula for calculating this is : (rise) / (run) * 100 = % grade
Let's say the grade is 100 %, and the rise (verticle change in elevation) is 100", what is the run (horizontal distance travelled) ? x = run ; y = rise; z = % grade
This creates a right triangle with two sides 100" long. You will agree that such a right triangle (with both non-hypotenuse sides being the same length) has a 90 degree angle, and two 45 degree angles.
Therefore, a 45 degree angle is a 100% grade.
It appears that you are confusing the "run" with the hypotenuse. The rise is always verticle. The run is always horizontal. The hypotenuse is neither horizontal nor verticle.
One must remember a couple of things when thinking or working with all of this.
1. There is NOT a "linear" relationship between "Degrees" and "Percent of Grade". The Percent of Grade for 45-degrees is 100% --- but that does not mean that the Percent of Grade for one half of 45-degrees is 50% --- it is, instead 41.42%. You cannot say that for each Degree change that there will be a linear change in the Percent of Grade. (It should also be noted that "Percent of Grade" is not the same as "Grads" --- There are 100 Grads in a quarter circle and there is a linear relationship between Degrees and Grads.)
2. When standing on the surface of the earth and measuring a distance, most often we lay the measuring device on the surface we are measuring and that distance would be the hypotenuse of a right triangle when measuring for Percent of Grade. That is the wrong place to be measuring the "RUN". We always draw the diagram with the slope on the top of the triangle representing "Grade" and that is upside-down. If you want to measure the grade of your RR track in your garden, you should place a level on the track and raise the LOW end up until the bubble shows level and then you measure how high you lifted that end... the length of the level is now the "RUN" and the distance the low end was lifted is the "RISE", and the track is the "Hypotenuse" of the right triangle and will be longer than the length of the level used for the base of the triangle... and note the "RUN" is on top of the triangle.
Semper Vaporo
Pkgs.