jpwhawaii wrote:Hello All, Great discusssion here, even the mistakes help to make this clearer when explained. Something no one has mentioned yet, is there a curve in your grade? Your level tool will not work. It will show a higher grade than you actually have. To calculate the length of your curve you need to use 2 X pie (3.14) X Radius.A 30" curve has a circumference of 188.4" (distance around the complete circle)2 X 3.14 x 30 = 188.4"How much of that curve are you laying? Half circle is 50% X 188.4" = 94.2" of distanceQuarter circle is 25% X 188.4" = 47.1" of distanceEighth of a circle is 12.5% X 188.4" = 27.55" Just my two cents.
Hello All,
Great discusssion here, even the mistakes help to make this clearer when explained. Something no one has mentioned yet, is there a curve in your grade? Your level tool will not work. It will show a higher grade than you actually have. To calculate the length of your curve you need to use 2 X pie (3.14) X Radius.
A 30" curve has a circumference of 188.4" (distance around the complete circle)
2 X 3.14 x 30 = 188.4"
How much of that curve are you laying?
Half circle is 50% X 188.4" = 94.2" of distance
Quarter circle is 25% X 188.4" = 47.1" of distance
Eighth of a circle is 12.5% X 188.4" = 27.55"
Just my two cents.
I never knew you could multiply by a pastry. I did know that you could multiply by pi though.
Sorry, I couldn't help myself. =D
--Jake
You can find the answer on the Wikipedia site listing under "grade" and looking under the heading for grade as in slope. This section describes the grade of a slope as being rise in units over the horizontal run in units. Thus, a two percent grade means the track climbs two units in one hundred horizontal units. A grade of 100% would result in a rise of 100 units in 100 horizontal units. Grades steeper than 100% are possible, but of only academic interest unless you are planning the side slope of a solid rock cut.
Incidentally, if the factor of adhesion of a locomotive is 25% and all the wheels are driven and all are clean with perfect motion, the steepest grade the locomotive can climb, pulling no cars and no tender, would be 25%, very steep indeed.
Five percent was a lot of trouble on the real railroads, but grades in this range are more manageable on our models simply because we are used to running short trains with relatively large lashups of motive power. If you want to model a helper district, a grade in this range would be about right. Otherwise there isn't enough contrast between the train length in the helper district and the more level part of the railroad. For more normal operation, grades of less than 2 percent would be preferred if you don't want train length to suffer too much.
I have a little brass Mogul that will pull twenty five cars on level track, but less than half that number on a one percent grade. On a four percent grade the locomotive is down to three or four cars.
JulesB wrote: When laying sanitary pipe lines we used to use batter boards and a transit. All pipe slopes were in percentages. Used the transit for line and drove a nail into the batter board and hung a plumb bob for line. The batter boards were place at what ever legnth the pipe sections were, usually 8'. If the slope was 3.1% and devided by 100 = .031 x 8 = .25 = the rise per pipe section, close enough for a run which usually was 200'-250' at most. Used the rod for grade. Laying line at 1/2% grade were a problem before lasers came out. I had one of, if not the first. It used a micrometer to adjust the grade. Later units were tube and self leveling lasers.Production went up about 90% Jules
When laying sanitary pipe lines we used to use batter boards and a transit. All pipe slopes were in percentages. Used the transit for line and drove a nail into the batter board and hung a plumb bob for line. The batter boards were place at what ever legnth the pipe sections were, usually 8'. If the slope was 3.1% and devided by 100 = .031 x 8 = .25 = the rise per pipe section, close enough for a run which usually was 200'-250' at most. Used the rod for grade. Laying line at 1/2% grade were a problem before lasers came out. I had one of, if not the first. It used a micrometer to adjust the grade. Later units were tube and self leveling lasers.
Production went up about 90%
Jules
That's sounds good to me. Only thing I can say is "poo rolls downhill", trains must go up.
REGARDLESS, if your locomotives slip their drivers, remove cars from, or add locomotives to, the train!
Mark
PS -- I think I need a long ocean cruise to Mexico. Doesn't the Norwegian Sun leave for Mexico next Monday from San Francisco?
R. T. POTEET wrote:Charts, maps, diagrams, and plans may well be annotated in percentage but the surveyor, to achieve that percentage, must set his transit or theodolite with the proper angle to achieve that grade (slope) in a specified chain length.
Not the way we were taught. We never measured vertical distances as an angle. They were always done by measuring the vertical change from level using a level and a rod. All of the railroad engineering activites I have been around have used a level and a rod to check grade.
Regardless, it is still way quicker to run a model railroad grade using a level and some sort of vertical measure. For example putting a 1/4' drill bit under a 2 ft level gives you essentially a 1% grade.
Dave H.
Dave H. Painted side goes up. My website : wnbranch.com
markpierce wrote: Semper Vaporo wrote: jpwhawaii wrote: fwright wrote: jpwhawaii wrote:I think your logic is backwards on the last post. Resistance through a curve is increased dramatically on a real railroad but very little on a model railroad. So, the distance over the curve will help to lower your grade considerably.Au contraire, my friend. The tapered wheel treads on the prototype keep the friction on a curve very low until the wheel has to start sliding to make up the increased distance on the outer rail and/or the flange starts rubbing on the rail. Then friction shoots up dramatically, and the prototype employs devices such as flange oilers to keep friction in check.NMRA RP25 does not require tapered treads, although most manufacturers do taper their wheel treads for ease of manufacture. Depending on how much of the tread width and flange root radius is allowed in the calculation, the radius at which a typical HO 33" wheel set starts to slip instead of roll is in the 20s inches (assumes an NMRA maximum tapered tread). By theory, sharper than that radius will see a significant increase in friction.John Allen proposed a formula (pre-RP25 with no taper and knife edge flanges) that added the result of 31/radius to the grade for the effective grade. An example would be a 2% actual grade on a 20" radius curve. The formula says to add 1.5% for an effective grade of 3.5%. I never found a basis for Allen's formula, but it's certainly in the ballpark empirically. I believe the RP25 tapered wheelsets cause less added friction than the formula advertises for larger radii. Once the wheel starts having to slide, there would be little difference in friction between modern and older design wheels.my thoughts, your choicesFred W Thusly, following this concept, there is no way for the average modeller to acurately calculate a grade if there is any curve, and there usually is.So the best, most accurate, "estimation" would be to use a level measurement on tangent and perhaps the curvature calculation throughout the length of the entire grade. Top and bottom transitions would increase the overall grade, unless..., you begin your calculation past and before those points.Considering that a model locomotive will pull considerably less standard cars up a 4% grade than a 2% grade I do believe that these calculations are very important in the planning stages as opposed to finding that out after the wood is cut and laid to the best guess.One more very important point, and here I think we'll all agree, swap out all of your wheelsets to metal wheels with needlepoint axles.Well, you will have to define "average modeler"... but to measure the average grade through a set of curves you measure the amount of "Rise" (the total height of the hill) and use the Pythagorean Theorem to solve for the "Run", i.e.: Calculate the "run" by using the rail length as the hypotenuse of the right triangle formed by the height, run and hypotenuse.Calculated Run = Sqrt((Length of rail * Length of rail) - (Rise * Rise))% Grade = (Rise / Calculated Run) * 100I figure the average modeler can do that with a calculator in hand.Actually, considering the fact that for really low % of grade values, the length of the rail will be really close to the "run", so rather than do all the key pokin' on the calculator, just use the length of the rail. Fer instance: Suppose the length of the rail is 36-inches and the rise is 2-inches. The real RUN = Sqrt((36 * 36) - (2 * 2)) = 35.9444-incheswhich is just 0.055598 inches shorter than the length of the rail.The % grade using 35.9444-inches would be 2 / 35.9444 = 5.564%The % grade using 36 inches would be 2 / 36 = 5.5555%A difference of 0.009 between using the "accurate" (calculated) run and the "estimated" (but really easy to measure value) run.Just use the length of the rail... the grade will probably have some undulations in it as well as the "lead-in" and "lead-out" so it is just an average value anyway. And just how accurately can most folk measure the Rise or Run or the length of the rail?Then, just how much do you add to the effective grade to account for the curve resistance?Semper is "torturing" me.Mark
Semper Vaporo wrote: jpwhawaii wrote: fwright wrote: jpwhawaii wrote:I think your logic is backwards on the last post. Resistance through a curve is increased dramatically on a real railroad but very little on a model railroad. So, the distance over the curve will help to lower your grade considerably.Au contraire, my friend. The tapered wheel treads on the prototype keep the friction on a curve very low until the wheel has to start sliding to make up the increased distance on the outer rail and/or the flange starts rubbing on the rail. Then friction shoots up dramatically, and the prototype employs devices such as flange oilers to keep friction in check.NMRA RP25 does not require tapered treads, although most manufacturers do taper their wheel treads for ease of manufacture. Depending on how much of the tread width and flange root radius is allowed in the calculation, the radius at which a typical HO 33" wheel set starts to slip instead of roll is in the 20s inches (assumes an NMRA maximum tapered tread). By theory, sharper than that radius will see a significant increase in friction.John Allen proposed a formula (pre-RP25 with no taper and knife edge flanges) that added the result of 31/radius to the grade for the effective grade. An example would be a 2% actual grade on a 20" radius curve. The formula says to add 1.5% for an effective grade of 3.5%. I never found a basis for Allen's formula, but it's certainly in the ballpark empirically. I believe the RP25 tapered wheelsets cause less added friction than the formula advertises for larger radii. Once the wheel starts having to slide, there would be little difference in friction between modern and older design wheels.my thoughts, your choicesFred W Thusly, following this concept, there is no way for the average modeller to acurately calculate a grade if there is any curve, and there usually is.So the best, most accurate, "estimation" would be to use a level measurement on tangent and perhaps the curvature calculation throughout the length of the entire grade. Top and bottom transitions would increase the overall grade, unless..., you begin your calculation past and before those points.Considering that a model locomotive will pull considerably less standard cars up a 4% grade than a 2% grade I do believe that these calculations are very important in the planning stages as opposed to finding that out after the wood is cut and laid to the best guess.One more very important point, and here I think we'll all agree, swap out all of your wheelsets to metal wheels with needlepoint axles.Well, you will have to define "average modeler"... but to measure the average grade through a set of curves you measure the amount of "Rise" (the total height of the hill) and use the Pythagorean Theorem to solve for the "Run", i.e.: Calculate the "run" by using the rail length as the hypotenuse of the right triangle formed by the height, run and hypotenuse.Calculated Run = Sqrt((Length of rail * Length of rail) - (Rise * Rise))% Grade = (Rise / Calculated Run) * 100I figure the average modeler can do that with a calculator in hand.Actually, considering the fact that for really low % of grade values, the length of the rail will be really close to the "run", so rather than do all the key pokin' on the calculator, just use the length of the rail. Fer instance: Suppose the length of the rail is 36-inches and the rise is 2-inches. The real RUN = Sqrt((36 * 36) - (2 * 2)) = 35.9444-incheswhich is just 0.055598 inches shorter than the length of the rail.The % grade using 35.9444-inches would be 2 / 35.9444 = 5.564%The % grade using 36 inches would be 2 / 36 = 5.5555%A difference of 0.009 between using the "accurate" (calculated) run and the "estimated" (but really easy to measure value) run.Just use the length of the rail... the grade will probably have some undulations in it as well as the "lead-in" and "lead-out" so it is just an average value anyway. And just how accurately can most folk measure the Rise or Run or the length of the rail?Then, just how much do you add to the effective grade to account for the curve resistance?
jpwhawaii wrote: fwright wrote: jpwhawaii wrote:I think your logic is backwards on the last post. Resistance through a curve is increased dramatically on a real railroad but very little on a model railroad. So, the distance over the curve will help to lower your grade considerably.Au contraire, my friend. The tapered wheel treads on the prototype keep the friction on a curve very low until the wheel has to start sliding to make up the increased distance on the outer rail and/or the flange starts rubbing on the rail. Then friction shoots up dramatically, and the prototype employs devices such as flange oilers to keep friction in check.NMRA RP25 does not require tapered treads, although most manufacturers do taper their wheel treads for ease of manufacture. Depending on how much of the tread width and flange root radius is allowed in the calculation, the radius at which a typical HO 33" wheel set starts to slip instead of roll is in the 20s inches (assumes an NMRA maximum tapered tread). By theory, sharper than that radius will see a significant increase in friction.John Allen proposed a formula (pre-RP25 with no taper and knife edge flanges) that added the result of 31/radius to the grade for the effective grade. An example would be a 2% actual grade on a 20" radius curve. The formula says to add 1.5% for an effective grade of 3.5%. I never found a basis for Allen's formula, but it's certainly in the ballpark empirically. I believe the RP25 tapered wheelsets cause less added friction than the formula advertises for larger radii. Once the wheel starts having to slide, there would be little difference in friction between modern and older design wheels.my thoughts, your choicesFred W Thusly, following this concept, there is no way for the average modeller to acurately calculate a grade if there is any curve, and there usually is.So the best, most accurate, "estimation" would be to use a level measurement on tangent and perhaps the curvature calculation throughout the length of the entire grade. Top and bottom transitions would increase the overall grade, unless..., you begin your calculation past and before those points.Considering that a model locomotive will pull considerably less standard cars up a 4% grade than a 2% grade I do believe that these calculations are very important in the planning stages as opposed to finding that out after the wood is cut and laid to the best guess.One more very important point, and here I think we'll all agree, swap out all of your wheelsets to metal wheels with needlepoint axles.
fwright wrote: jpwhawaii wrote:I think your logic is backwards on the last post. Resistance through a curve is increased dramatically on a real railroad but very little on a model railroad. So, the distance over the curve will help to lower your grade considerably.Au contraire, my friend. The tapered wheel treads on the prototype keep the friction on a curve very low until the wheel has to start sliding to make up the increased distance on the outer rail and/or the flange starts rubbing on the rail. Then friction shoots up dramatically, and the prototype employs devices such as flange oilers to keep friction in check.NMRA RP25 does not require tapered treads, although most manufacturers do taper their wheel treads for ease of manufacture. Depending on how much of the tread width and flange root radius is allowed in the calculation, the radius at which a typical HO 33" wheel set starts to slip instead of roll is in the 20s inches (assumes an NMRA maximum tapered tread). By theory, sharper than that radius will see a significant increase in friction.John Allen proposed a formula (pre-RP25 with no taper and knife edge flanges) that added the result of 31/radius to the grade for the effective grade. An example would be a 2% actual grade on a 20" radius curve. The formula says to add 1.5% for an effective grade of 3.5%. I never found a basis for Allen's formula, but it's certainly in the ballpark empirically. I believe the RP25 tapered wheelsets cause less added friction than the formula advertises for larger radii. Once the wheel starts having to slide, there would be little difference in friction between modern and older design wheels.my thoughts, your choicesFred W
jpwhawaii wrote:I think your logic is backwards on the last post. Resistance through a curve is increased dramatically on a real railroad but very little on a model railroad. So, the distance over the curve will help to lower your grade considerably.
Au contraire, my friend. The tapered wheel treads on the prototype keep the friction on a curve very low until the wheel has to start sliding to make up the increased distance on the outer rail and/or the flange starts rubbing on the rail. Then friction shoots up dramatically, and the prototype employs devices such as flange oilers to keep friction in check.
NMRA RP25 does not require tapered treads, although most manufacturers do taper their wheel treads for ease of manufacture. Depending on how much of the tread width and flange root radius is allowed in the calculation, the radius at which a typical HO 33" wheel set starts to slip instead of roll is in the 20s inches (assumes an NMRA maximum tapered tread). By theory, sharper than that radius will see a significant increase in friction.
John Allen proposed a formula (pre-RP25 with no taper and knife edge flanges) that added the result of 31/radius to the grade for the effective grade. An example would be a 2% actual grade on a 20" radius curve. The formula says to add 1.5% for an effective grade of 3.5%. I never found a basis for Allen's formula, but it's certainly in the ballpark empirically. I believe the RP25 tapered wheelsets cause less added friction than the formula advertises for larger radii. Once the wheel starts having to slide, there would be little difference in friction between modern and older design wheels.
my thoughts, your choices
Fred W
Thusly, following this concept, there is no way for the average modeller to acurately calculate a grade if there is any curve, and there usually is.
So the best, most accurate, "estimation" would be to use a level measurement on tangent and perhaps the curvature calculation throughout the length of the entire grade. Top and bottom transitions would increase the overall grade, unless..., you begin your calculation past and before those points.
Considering that a model locomotive will pull considerably less standard cars up a 4% grade than a 2% grade I do believe that these calculations are very important in the planning stages as opposed to finding that out after the wood is cut and laid to the best guess.
One more very important point, and here I think we'll all agree, swap out all of your wheelsets to metal wheels with needlepoint axles.
Well, you will have to define "average modeler"... but to measure the average grade through a set of curves you measure the amount of "Rise" (the total height of the hill) and use the Pythagorean Theorem to solve for the "Run", i.e.: Calculate the "run" by using the rail length as the hypotenuse of the right triangle formed by the height, run and hypotenuse.
Calculated Run = Sqrt((Length of rail * Length of rail) - (Rise * Rise))
% Grade = (Rise / Calculated Run) * 100
I figure the average modeler can do that with a calculator in hand.
Actually, considering the fact that for really low % of grade values, the length of the rail will be really close to the "run", so rather than do all the key pokin' on the calculator, just use the length of the rail. Fer instance: Suppose the length of the rail is 36-inches and the rise is 2-inches.
The real RUN = Sqrt((36 * 36) - (2 * 2)) = 35.9444-inches
which is just 0.055598 inches shorter than the length of the rail.
The % grade using 35.9444-inches would be 2 / 35.9444 = 5.564%
The % grade using 36 inches would be 2 / 36 = 5.5555%
A difference of 0.009 between using the "accurate" (calculated) run and the "estimated" (but really easy to measure value) run.
Just use the length of the rail... the grade will probably have some undulations in it as well as the "lead-in" and "lead-out" so it is just an average value anyway. And just how accurately can most folk measure the Rise or Run or the length of the rail?
Then, just how much do you add to the effective grade to account for the curve resistance?
Semper is "torturing" me.
Humblest of apologies... I actually was only trying to show my "proof" for deciding that all you need is the length of rail and the rise to calculate the percent of grade, which I didn't really figure out for sure until I was most of the way through the calculations. I has suspected it to be true, but not until I had actually typed the answers did I convince myself of the fact.
As for "torturing" you... well, I am ashamed to admit that when read your comment I giggled. So apparently it does entertain us (well, "me" anyway) to confuse people!
"Sorry, sorry, sorry!"
Read that the way Stephanie Vanderkellen (Julia Duffy), the ditsy "rich" maid, spoke it in the Bob Newhart comedy program in which he ran the Stratford Inn in Vermont. If you don't know that show you will have no idea the "sincerity" with which it is offered, and you will have missed the absolute funniest comedy gag EVER perpetrated... it took two completely different TV "Series" (over many years, 6 seasons for the 1st series and 8 for the 2nd) to reach the punch line/scene in the very last show. See: http://en.wikipedia.org/wiki/Newhart#.22The_Last_Newhart.22
Semper Vaporo
Pkgs.
dehusman wrote: R. T. POTEET wrote:Oh yes they do! Have you ever heard of a transit or a theodolite? Have you ever heard of a surveyor? Have you ever heard of a survey stake? Have you ever heard of a surveying chain?--although I do think these things have a more specific name I can't think of it at the moment.Yes I have. I have a degree in Civil Engineering. And every prototype railroad chart, map, diagram and plan I have ever seen in the last 30 years has grade marked in %.You run a grade with a level and rod. Done it. (modern surveyors use very sophisticated GPS equipment instead)Horizontal curves are measured in degrees and degree of curvature.Dave H.
R. T. POTEET wrote:Oh yes they do! Have you ever heard of a transit or a theodolite? Have you ever heard of a surveyor? Have you ever heard of a survey stake? Have you ever heard of a surveying chain?--although I do think these things have a more specific name I can't think of it at the moment.
Yes I have. I have a degree in Civil Engineering. And every prototype railroad chart, map, diagram and plan I have ever seen in the last 30 years has grade marked in %.
You run a grade with a level and rod. Done it. (modern surveyors use very sophisticated GPS equipment instead)
Horizontal curves are measured in degrees and degree of curvature.
Charts, maps, diagrams, and plans may well be annotated in percentage but the surveyor, to achieve that percentage, must set his transit or theodolite with the proper angle to achieve that grade (slope) in a specified chain length.
From the far, far reaches of the wild, wild west I am: rtpoteet
"It is actually not such a simple concept, now is it. " border="0" width="25" height="20" />"
It appears so for some.
"As for not trying to confuse you again.... what'd we do for fun then? " border="0" width="15" height="15" />"
Please continue in your efforts to keep me from being bored.
markpierce wrote: Man! How did such simple concepts seem to get so complicated? You guys have my head spinning. Regardless, 1 unit forward and one unit upward is a 1-to-1 ratio (or 100% grade, because 100% is equal to 1, and 1 divided by 1 equals 1), and that makes for a 45-degree angle. Please don't try to confuse me again. Mark
Man! How did such simple concepts seem to get so complicated? You guys have my head spinning. Regardless, 1 unit forward and one unit upward is a 1-to-1 ratio (or 100% grade, because 100% is equal to 1, and 1 divided by 1 equals 1), and that makes for a 45-degree angle. Please don't try to confuse me again.
It is actually not such a simple concept, now is it.
As for not trying to confuse you again.... what'd we do for fun then?
The outside wheel on the curve is moves toward the outside rail. The wheel diameter close to the flange has a greater diameter, so the riding surface of the outside wheel has a slightly larger diamer than the inside wheel. This helps to compensate for the fact that the outside wheel has to travel a greater distance than the inside wheel.
Surveyers deal with vertical and horizontal directions and distance. So in measuring the steepness of the grade, the surveyer will note the measured or observed vertical angle in degrees and the difference in height of the instrument and the height of the target being observed to calculate the true degree of angle.
Railroaders use the percent of grade (the ratio of the vertical change for a given horizontal distance) to measure the steepness of the grade. The data to calculate percent of grade (change in elevation versus distance) is obtained from the surveyers' sightings.
dehusman wrote:. . . . . . . . . . In N America, no railroad measures grades in angles for any operational reason . . . . . . . . . .
Oh yes they do! Have you ever heard of a transit or a theodolite? Have you ever heard of a surveyor? Have you ever heard of a survey stake? Have you ever heard of a surveying chain?--although I do think these things have a more specific name I can't think of it at the moment.
In 2005 I stopped up in the Texas Panhandle and watched BNSF on their double tracking; the railroad sure as heck had a surveyor working that day!
Semper Vaporo wrote: Oh wow, you got the Moor Powher blue Ray laser level? But that "Benford" tract-o-meter is the Elbownian knock-off of the "Binford" model... the spring will rust on ya real quick.
Oh wow, you got the Moor Powher blue Ray laser level? But that "Benford" tract-o-meter is the Elbownian knock-off of the "Binford" model... the spring will rust on ya real quick.
That's what happens when you have inferior tools. My spring is already rusted solid. And I have failed...I came in 2nd in the best toys race....
Yes, a curve increased the distance traveled which reduces the actual grade, but the fact that it is a curve increases the rolling resistance and that is usually counted as increasing the effective grade. The two values do not cancel each other out as the amount of increase in rolling resistance is much more than the reduction in actual grade.
EDIT: but then, thinking about that a bit more, that is why the RR have so many curves in places where it might look like they could just go straight. They are reducing the grade by snaking up the hill. Measuring the grade has to be done on small segments, not the beginning and end of the track (bottom of hill to top of hill) without taking into account the length of the track over that same "as the crow flies" distance.
fwright wrote: Semper Vaporo wrote: AHHHh! But BRAGGING RIGHTS are at stake here!"MY loco will pull 10 XYZ cars up a 4 PERCENT hill!""GOLLY, MY loco can't seem to pull 5 XYZ cars up my 3 DEGREE hill and I got ball bearing wheels! HOW COME?"That's because your 3 DEGREE hill is over 5 PERCENT grade! You are right. I forgot that the other justification for the grade calculations (besides being anal ) is bragging rights. You can't brag properly unless you have measured your grade to 3 decimal places using the latest and greatest Moor Powher blue ray laser level with super computer and 9 line 64K color LCD screen attached. Of course, you also need the Benford super tract-o-meter (aka spring scale) to accurately measure locomotive pull to the nearest .01 grams. Model railroading is fun.
Semper Vaporo wrote: AHHHh! But BRAGGING RIGHTS are at stake here!"MY loco will pull 10 XYZ cars up a 4 PERCENT hill!""GOLLY, MY loco can't seem to pull 5 XYZ cars up my 3 DEGREE hill and I got ball bearing wheels! HOW COME?"That's because your 3 DEGREE hill is over 5 PERCENT grade!
AHHHh! But BRAGGING RIGHTS are at stake here!
"MY loco will pull 10 XYZ cars up a 4 PERCENT hill!"
"GOLLY, MY loco can't seem to pull 5 XYZ cars up my 3 DEGREE hill and I got ball bearing wheels! HOW COME?"
That's because your 3 DEGREE hill is over 5 PERCENT grade!
You are right. I forgot that the other justification for the grade calculations (besides being anal ) is bragging rights. You can't brag properly unless you have measured your grade to 3 decimal places using the latest and greatest Moor Powher blue ray laser level with super computer and 9 line 64K color LCD screen attached. Of course, you also need the Benford super tract-o-meter (aka spring scale) to accurately measure locomotive pull to the nearest .01 grams. Model railroading is fun.
Even more to the point - why are you even trying to measure a grade? The prototype measures their grades because they have to use load charts and calculations to assign enough power to handle the desired train over the grades. Even then, they don't always get it right.
In the model world, our grades are usually constrained by the available space. The grade will be what it will be to achieve the desired track plan. And that grade will be acceptable for operations and appearance - or not. Where the steepness of the grade is truly optional, does it matter whether the grade is actually 2.85% or 1.63%? Again, all that really matters is acceptablity for appearance and operations, which are fairly subjective.
Despite my engineering degrees and well-trained desires to calculate everything to the nth degree, I have concluded that anything more than a +/- 1% estimation of a grade during the planning phase is a waste of time. If you use a construction method that requires bending material to form the grade (cookie cutter, spline), you will automatically have some vertical curve transition to the grade. This transition automatically means the real grade is steeper in places than calculated on the plan. But we all know decent vertical transitions are critical to smooth operations on grades.
Again, the vertical bending radius of the roadbed/subroadbed and space available are going to mandate a given grade profile - does it matter whether it's 2.8% or 2 degrees? All that really matters is your locomotive(s) pulls an adequate (how ever many cars that may be) train up the grade, and you are satisfied with the appearance. And again, predicting the grade +/- 1% is probably good enough to make the determination of adequacy for the desired operations in advance.
I have learned the same things about horizontal curves and/or software track plans, too. We tend to have an over-infatuation with calculation instead of performance and appearance.
selector wrote: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
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
Another thing to note, there is no defined grade that can be perfectly vertical. A vertical grade would technically be undefined. 90 degrees = vertical, with an undefined slope, a 100% grade would be a 45* angle, a 200% grade = 67.5* angle, and so on.
EDIT: My bad, simply scrolling down the first page would have shown me that this has already been better explained by other members. Either way, the point still stands.
Easier way. COMPLETELY forget all the stuff with angles. In N America, no railroad measures grades in angles for any operational reason.
Just use the % method. Its waaaaaaaaay easier to measure, its waaaaaaaaay easier to build. All you need is a level and a tape measure.
Vertical distance/horizontal distance x 100 = percent grade.