Just another question on how to figure grade on a new layoyt. I know that rise/run is good for straight track, but do I use the same formula for inclines on curves? Also can someone tell me is it better to use glue to hold track down or shoud I use track nails?
Thanks
Welcome to the forums.
Rise run is the same on straight or curves, it is how much you climb in a given distance.
As for attaching track, you will get a number of answers. I use cheap latex caulk and have had success, as have others. Some use Liquid Nails for Projects if you are using extruded foam (pink/blue board). Be sure to “for Projects” as some of the Liquid Nails products will eat foam. If yu are working on curves you might want to use some T pins to hold it in place while the caulk sets up and weight on the track holds it down on the adhesive until it sets. Overnight was all mine needed.
Good luck,
There are several ways. There is a mathematical way using the radius and the length of a straight line from one point of the curve to another (but I don’t have the formula). One can lay out a chord of small rope along the track, marking the chord where you want to measure the distance, and then straighten out the chord to measure. One can add short, straight measurements which will slightly understate the length. Once the length is determined, one can calculate the amount of rise is needed for a specific grade percentage. Or buy a device like a Sears digital torpedo level (the 8" length is handy) that will read out the grade digitally.
Curved track adds resistance to a train, so for a given grade a locomotive can’t pull as much on a curve up a grade compared to straight track. I don’t know if anyone has scientifically determined how much of a grade reduction on a curve is needed to compensate for this. There are variables, such as the scale of the models, the radius in question, etc. If guessing, for HO I’d say 15%, so a 2% grade should be reduced to 1.70% on a moderate curve. The sharper the curve, the greater the compensation needed. It will be coincidental if the compensation is perfect, but a slight reduction of grade on the curve should reduce the differing performance.
Mark
While the basic formula is the same, both I and the prototype compensate for curvature by easing the grade. My technique is to use a 24 inch long carpenter’s level raised 1/4" per percent of grade at the uphill end. I level along 24 inch chords to establish riser height. How much this reduces the grade is largely determined by the sharpness of the curve - quite a bit for 16" radius on my sharply-twisted mountain-climbing branch, considerably less on my 30" radius mainline grade.
I have had excellent results using latex caulk to anchor flex track in an un-climate-controlled garage in the Dessicated Desert - a trick I learned by reading it on these forums.
I won’t tell you my grades, to protect the guilty. Suffice it to say that it takes two 0-6-0Ts to get five four wheel cars up the mountain!
Chuck (Modeling Central Japan in September, 1964)
if you know the radius of the curve, multiply x 2, then multiply that figure x 3.14. that will give you the circumference of the circle. divide that figure by 360 and multiply the result by the degrees of curvature.
example find the run of a 90 degree curve of 36 inch radius track. 36x2=72. 72x3.14=225.08 or 225" is close enough. divide 225 by 360 = .625" multiply .625x90=56.25". got it? the formula is twice the radius x pi (3.14) = circumference of the circle.
actually, twice the radius times 3 is close enough. then figure a half circle or quarter circle etc.
basic algebra but i’m sure a real teacher could explain it in clearer terms.
grizlump
Due to the difference between arc definition curves and chord definition curves, the answer to your curve question can become quite complicated. But as an over simplification the arc length must be determined.
If any two elements of a curve are known, all other elements can be calculated. As is the case with most model curves, the radius will be one known element. The Chord length (straight line distance from the beginning to the end of the curve) can be physically measured giving a second element. Then the arc length can be calculated from these figures. The central angle of the curve ( I ) is expressed in degrees must be known, and is difficult but not impossible to measure on a model curve. This is what grizlump9 is mistakenly refering to as the degree of curve. The chord distance is equal to the sin of one half of the interior angle multiplied by twice the radius: C = (Rx2) x sin 1/2 I = chord, thus if the radius and chord distance are known the following formula can be used to determine the central angle: I = (arcsin C / 2R) x 2. Now that you have the interior angle the following formula can be used to determine arc length: arc = (I / 360) x 2piR. Without a scientific calculator this can become quite a chore.
Then your grade can be calculated using the usual formula along the centerline of the curve.
Good luck…Jim J
Try this on for size:
(1/(2RX3.1416/100))X360
The result of this computation will return the number of degrees it will take to achieve a run of 100" along the arc of this circle.
I did all the math when planning, but when it came to the laying of the spline I used a tailors cloth measuring tape to decide what height my risers should exactly be. The tape would follow the curved spline and give me a good precise measurement. I layed one spline along the risers and after I double checked the length of spline to the next riser I did a final height adjustment. Then I added more spline. If you can see it in the photo I kept the grade quite low on the curve in the front and increased it to just under 2 percent the rest of the way. Also I purposely change the grade throughout the climb as I thought it would be a little more realistic looking rather than having one set angle. It is suppose to be a long Journey through the Rockies after all. I never went over 2 percent though. The curve in the foreground is a 31" radius and the 6" climb is about 40’ long. Good luck.
Brent
I came across these handy little charts a few years ago.
As to securing track, latex caulk, the cheap stuff, is the way to go!
If making a helix you need to hold the rise steady and chage the radius to make an acceptable grade. Why? Because you need 3-4" clearance from the top of the rails to the bottom of the substrate so the train has room to run. So if you use 4" separation to be safe and 100" of track you have a 4% grade but what radius yields 100" of track. the formula for a circle is pi x d or 3.1416 x the diameter. So dividing 100 by 3.1416 we get 31.86" diameter or 15+ inchs radius (1/2D) for a four percent grade which is pretty steep. So what radius is needed for a 2% grade. That would be four inch rise in 200". 200"/3.1416= 31.8" radius. That’s why helixs are only for people who have the room to spare in my opinion.
I’m not sure how this chart is interpreted. Does it mean that a 30" radius curve creates the drag equal to a 1.1% grade? So, is a helix with a 2% grade and a 30" radius equal to a grade of 3.1% on the straight? If the train is twice as long as the curve, is then a 30" radius curve equal to a 0.55% grade? The magnitude of curved track on pulling power is larger than I have sensed, but then, I haven’t performed scientific experiments. I’d like to hear your opinions, experiences, or authoritative conclusions, please.
Mark
I use track nails. Predrill the holes so it’s a press fit. That way you can push them in almost snug without kinking the tie. It also makes it easier to salvage the rail for changes or the next layout.
Enjoy
Paul
Mark
The equivalent grade for curved track came from John Allen. I have never found any evidence beyond it being Allen’s hypothesis for a formula for the extra drag created by curved track. It would have some basis on Allen’s experience in HO scale using trucks before the days of RP25 wheels and free-rolling wheel sets. A practical experiment (pure speculation on my part as to Allen’s conduct of any supporting experiments) would be to check how many cars a given locomotive can pull up a known straight grade as can compared to how many cars can be pulled around a known radius curve.
As a photographer, John Allen would likely have been familiar with 1/<
Yeah! What he said!
I actually don’t know how true it is in HO since I model N. I just thought it was an interesting chart done by an authoritative person. I know there have been many advances since Mr. Allen’s time, so anything better then the chart you could look at as icing on the cake!
Is that like 36000/(2R*3.1416) ?