N scale grade ?

[Anonymous]

im wondering what the grade formula is like: 4x100=?
i cant remember the formula.'
im tryin to go up 5" in 5ft

[lonewoof]

Formyula is : grade % = feet rise / 100 feet run. So multiply your 5ft run by 20 (100ft) and your 5" rise by 20 (100") and you get 8.333'/100' = 8.3%. Pretty steep grade...

[cheese3]

the formula is x amount of rise over 100 and that will give you the % needed. is this what you mean?

[cheese3]

sorry i did not see lonewolf's reply

[jkeaton]

Basically, grades are the same in all scales - and lonewolf has given you the formula. 5" rise in 5 feet run is basically not doable - if your locomotives get up it, they will certainly be running very hard and hot - and N scale locomotives seem to overheat more readily than bigger scales. In the real world, few railways had anything like an 8% grade - even Interstate highways rarely have grades that steep. Railways that needed grades like that usually adopted a rack system, where a gear on the locomotive engaged a rack of gear teeth between the rails. It's less a question of getting up the grade (if you can't get up then you'll spin your wheels), then of being able to safely get down (sliding wheels don't stop you very well).

Tell us a bit more about the layout and the grade and some of the bright minds around here will have suggestions, I'm sure.

Jim

[Anonymous]

For most any model RR 4% is the absolute maximum grade possible, unless you're willing to do a logging railroad with Shays. Most people prefer keeping down to 2% or 3%, as that allows most locomotives to pull decents trains up and over the hills. This limitation is pretty much scale independant.

To go up 5" at a 3% grade will allow: 100/3*5 = 167 inches = 13.9 feet

Mark in Utah

[7h9h_]

i like N scale

[chateauricher]

QUOTE: Originally posted by lonewoof
Formyula is : grade % = feet rise / 100 feet run. So multiply your 5ft run by 20 (100ft) and your 5" rise by 20 (100") and you get 8.333'/100' = 8.3%. Pretty steep grade...

I don't know why you're multiplying both the rise (5in) and the run (5ft) by 20. I'm not sure how you did it; but you did get the correct answer of 8.3%.

The formula is ...
(vertical) rise ÷ (horizontal) run x 100 = % grade
(make sure both rise and run are measured in the same
units : feet, inches, cm, mm, etc.)

The grade equivalent of a 5in rise in a 5ft run is calculated this way ...

1. Convert both the rise and run into the same measuring units (eg: inches, in this case) :

rise = 5in
run = 60in (5ft x 12)

2. Divide the rise by the run :

rise ÷ run = 5 ÷ 60 = 0.8333

3. Multiply the result by 100 in order to get a percentage instead of a decimal :

0.8333 x 100 = 8.333%
A grade of 8.333% is extremely excessive. IF your locomotive can climb it, it probably wouldn't be able to pull much of a load. And descending that grade would be dangerous.