grading question

3"rise per 100" run equals 3%

A foot rise per hundred foot run is 1%. Have to add a bit for the approach and exit of the grade. Hope this helps

rise divided by run (distance) = grade expressed as a percentage

 

So as stated above; a 1" rise over a 100 " run would equal a 1% grade

 

If you know the height you need to rise and you know the distance you have available, divide one by the other and you will have the resulting grade needed.

 

As an alternative, if you know the rise and you know the maximum grade desired, you can solve for the distance needed to make the climb. So rise/grade = Run (1"/.01=100")

 

I also like to use a shortcut which is a 1" rise over 8 feet (only 96 inches) = 1% grade. It's not exactly 1% but close enough and makes the math a lot easier.

Great question.  I love this question.  When my kids and other young people tell me the math they get in school has no real application in life I show them how I use Algebra in my Real Estate Business, at my store, on my layout and at the baseball field.  I showed kids how to use the Pythagorean theorem to solve for the distance between first and third base or the catcher's throw from home to second to catch a runner stealing.

Unfortunately with the space and track I have I need to raise a section so I can run the rest of the track under it to close the loop.

If you are short on space, you can raise one track and lower the other to reduce the amount of linear distance needed to raise the track to the desired height.

 

You might be interested in Woodland Scenics' SubTerrain Lightweight Layout System, which features readymade 2%, 3% and 4% inclined foam pieces.

 

-John

To add an empirical thought to the calculation for the basic grade %...

 

As your grades get steeper (e.g., >2%), you'll need to add extra distance to the 'run' to allow for gradual vertical transitions at the bottom and top of the grades.  That is to say, the transitions from level-to-grade, grade-to-level must allow for the equipment you're operating.  For instance, steam engines may encounter shorting problems as the engine enters an abrupt transition at the bottom of a grade.  In a like manner, the same engine might lose traction on the grade as it reaches a sharp transition at the top, since the leading drivers will lose contact with the rails.

 

Diesels can have similar problems, but having separate trucks can minimize issues steamers encounter.......as long as the trucks have some amount of vertical swivel.

 

No rules of thumb come to mind on this transition 'thing', but I've had to make accommodation on at least a couple occasions for this phenom.  It adds to the total length of a grade when planning a layout.

 

BTW, another grade issue is re curves.  Besides reducing traction if they're part of the grade, entering/departing a curve at the top or bottom of a grade can also be problematic....another reason for ensuring smooth transitions.

 

Happy New Year!

 

KD

Here's another easy way to measure grades.

 

Get yourself a 2-foot long bubble level. Place it on the track on the grade. Raise the LOWER END until the bubble shows level. Every 1/4" you have to raise the lower end to make it level equals 1%. If the low end is 3/4" off the track to make it level, that's a 3% grade.

Is the 3% grade criteria for "visible grades" for aesthetics, or is there a performance issue that can damage locos?

Reason I ask is that I have 2 hidden grades, that I "eye balled" to make.

The locos seem fine with it, i.e. with a DCS Handheld set to 15-20 the locos seem fine going up them.

Anything over 2% doesn't look so good.  Steeper grades and how the engine handles them depends on other things like length of train, number of motors, traction tires, cleanliness of track.

.....

Dennis

% Ascending Grade = (Rise/ Run) x 100

 

% Descending Grade = (Drop / Run) x 100

 

In both cases:

• Keep Rise, Run & Drop in the same units (inches or feet, or meters). Do not mix the units;
• The parenthesis surrounding the Rise / Run and the Drop / Run are not mathematically necessary, but are shown to give a better insight into the mechanics of the calculation;
• Rise is the increase in height measured 90 degrees from the Starting Surface;
• Run is the horizontal distance you have to accomplish the Rise or Drop;
• Drop is the decrease in height measured 90 degrees from the Starting Surface;
• Multiplying by 100 converts the result to a percentage.

Example:

The train needs to climb 6 inches up from the bench top or floor in a distance of 12 feet.

Solution:

• The rise is 6 inches;
• The run is 12 feet, but needs to be inches, so multiply 12 feet x 12 to get 144 inches;
• % Grade = (6 / 144) x 100 = 600/144 = 4.17 % Grade.\

 

That's ironic to me.  I'm the President of our local little league and can't tell you how many times I've showed the kids (in the dirt) how to use the Pythagorean theorem to determine the distance from home to 2B or how to explain that when you take a lead from 1B you do so in a straight line and not from the back off the bag (shorter distance to 2B).   I just wish math teachers would use more real life applications when teaching - kids might be a little more attentive...lol

 

--Greg

Anything over 2% doesn't look so good.

As a reference, Lionel and MTH's trestle systems will give you about a 5% grade, so a 2% grade would be more than half as steep over the same run.

 

-John

To answer this question I have always used to go up 4" you need 8 running feet for a 4% grade.  2% 4" is 16 feet.  If you remember these numbers you can caculate percent grade versus how many feet you need to get up and down.  Always remember it takes the same amount of footage to get down as itm takes to get up.

If you have a run that you will only be going one direction on (rare but it does happen on some layouts), the descent can be steeper (within reason) without issues.

Thank you guys this has been very helpful. Hopefully I can do this right the first time and avoid making a mess of it.