Although I could probably find an internet tutorial and figure this out, I thought that it would be more fun to pose the question here.
I’m redesigning the interchange/yard area of may layout and breaking it in half to have sort of an east facing yard and a west facing yard. In the middle will be the main runaround (they are also the ladders). The runaround will be in the shape of a rhombus, if you remember geometry class.
I’m building it with 4 LH PECO #8 code 83 North American turnouts. All 4 running East/West, two facing North, and two facing South. While the shape will be symetrical, it looks like the shallow frog angle will have to give the rhombus sort of an “offset” shape. PECO says the frog angle on its #8 is 7.15 degrees
I want to keep the diverging tracks perfectly straight, used as ladders.
Edited for more clarity (I hope):
Height of the rhombus along the veritcle axis is 9 inches and the lenght along the horizontal axis is about 98 inches.
What is the distance between the SW and NE frogs…as the crow flies?
What is the total distance of the two legs that connect the SW and NE frogs? …that will be the linear length of the runaround track (s)
No cookies available for prizes, just my appreciation. Enjoy.
Why are you using #8 turnouts in a yard. You’d gain longer tracks if you used #6s. #8s are great for mainline crossovers but overkill for the slower speeds in a yard. Peco turnouts have sharper rail closure radius than other brands, but I’d still bet the #6 closure radius is greater than your minimum radius.
Even with my knowledge of geometry, a picture, or sketch would go a long way here.
The rhombus shape is a common approach to build yards with all tracks being nearly equal in length, but understanding how that intergrates with you other trackage is not well described.
Many times either drawing myself (pencil-paper) or using someones software… the trackplan never matches what the track actual does (can do). Meaning, for instance, Atlas’s Right Track Software you would think would match Atlas track angles- degrees- bends- etc. etc. It doesnt. And others are the same - for whatever reason.
When this happens I usually grab the pieces of track in question and lay them out just as if i were actually putting them down permanantly. Then I take measurements… realign this or that, even sometimes come up with a better idea. Or, and this has happened… DANG IT! It just wont fit!!! Arrrgh!
You may want to try this and forgo the head scratching - hair pulling (for those of us that still have it) - math. “Model Railroading is FUN!” Math is for school, and school was never fun! Not wise to mix the two.
As an added incentive, once the track is ‘pre-laid’, measured, and or re-arranged how you like it… its a simple matter of marking your centerline so when you do lay the track permanant like… you already have marks in place that are dead on the way you want them and where you want them.
This is what I’m building, but its a lot flatter. The SW and NE angles on this diagram are about 60 degrees, whereas the LH turnout frog would make them about 7.1 degrees The NW and SE angles shown here would be the inverse of 7.1 (so that all 4 angles added up to the required 360 degrees of course), with the “mainlines” (actually some are switching leads to industries, but that’s not relevant) extending the top and bottom tangent lines to the east and west. All four sides make up the runaround
The two verticle lines are the ladders, with the angles of the yard tracks coming off at 7.1 also…but that’s not relevant to my question.
And imagine squishing the rhombus flatter…the NE and SW corners would be pushed farther apart as this rhomus was flattened.
Its very flat, but all track is straight, so I’m thinking that basic geometric principals and formulas should be able to tell me how far apart the SW and NE frogs will be diagnally. The height of the rhombus is 9 inches and its lenght is about 98 inches.
This is sounding like a geometry test question, LOL.
I understand that there are easier ways to do this. I just thought that it would be fun to maybe think about how it was done in the old days.
I don’t like the way that 73 foot long centerbeam flats and wood chip hoppers look going through #6 frogs. All cars look better going through the shallower angles.
Everything I see on google satellite these days looks higher than #6 frogs, just about everywhere.
And I only need to hold about 20 cars on a 12 foot long by 1 foot deep yard. Really, each yard is going to be about 2 tracks. Both yards terminate into backdrops at either end of the space, giving the illusion of longer tracks. Less clutter. I have the space.
And this is the short coming of the PECO turnout, with its “compact” footprint.
With Atlas turnouts you just join the turnouts to each other on the straight route and you get diverging routes spaced at 2" track centers.
And I still don’t understand which direction the yard leads enter and leave? Not that it matters that much.
And I’m still not sure I understand the question? Or did I cover it in my second sentance - you need to know how big of a spacer to add between the turnouts?
With Atlas it would be zero for 2" track centers.
Just another reason I don’t like PECO turnouts, short spacers on yard leads.
As a professional draftsman, who has done my share of track planning, the easy answer to this question is to draw the parallel yard tracks and then cross them with a line at the frog angle.
Yes, the prototype uses turnouts larger than #6 in yards, more like #8 or #10 or even bigger.
Thanks y’all for indulging my question, its amittedly sort of academic. And my questions are poorly described. The questions are:
What is the distance between the SW and NE frogs…as the crow flies?
What is the total distance of the two legs that connect the SW and NE frogs …that will be the linear length of the runaround track (s)?
I neglected to give some dimensions (but that’s okay because nobody really asked about the math) . I edited my comments above to give a veritcal height of 9 inches and a horizontal length of (about) 98 inches from left to right (if it were a rectangle), and the angles are 7.15 degrees (and 172.85 degrees for the other two angles).
I laid this out on the benchwork and realized that fussing around trying to get things “square” would be a lot easier if I had some precise measurments to use. Since all tracks in this rhombus are suppossed to be perfectly straight (with no kinks, small curves, or bows; unlike a traditional runaround) and the angles are precise, geometry formulas and math should give me the precise answers to shoot for. It also tells me runaround length (not including the narrow portions that approach the frogs)
Its a way of knowing the answers without employing any type of track planning process. Its really a geometry question. It may not be answerable on this forum.
Yes, laying it out on grid paper is the way to do this, then measuring the different lengths according to scale, AFTER its laid out.
It dawned on me that since I will be laying out a perfectly symetrical geometric shape, we can know the exact dimensions and runaround length we will have BEFORE I start laying out lines on the benchwork or grid paper.
Its really no different than forming a siding with tracks 2 inches apart, but where using two LH turnouts to do that makes the mainline shift to flow through a diverging path of a turnout(most sidings are built with a LH and a RH turnout). I’m essentially extending the crossover length to make 9 inch center spacing between the “main” and the siding/runaround, which pushes total length of the siding to be longer (the distance between the two frogs). (I just happen to be using two more turnouts to return each track back to parallel). In this case, the crossover portions are becoming a functional part of the runaround since its becoming longer. (They will also serves as the ladders)
I think that when making the crossovers taller, I’m gaining some functional length to the runaround compared to how much the total horizontal length of the siding also has to grow. But I’m not sure about that.
Its just something that we could know before we start laying anything out on paper.
If anybody wants to try, I’ll skip the track planning part and focus strictly on the geometry and math. Here is the picture (albeit taller and not as long as actual…this is a “wracked square” and I’m describing a “wracked rectangle”)
You can draw this at home if you like: Assuming 2 sets of angles at 7.15 and 172.85, and the total height of the figure is 9 inches, and its length from SW point to NE point along a horizonatal axis is (about) 98 inches ( if we dropped a plumb off the NE tip to extend the bottom line to that point) …What is the length of each of the four sides? (of course, 2 sets of 2 equal length sides).
I think this is enough information for a geometry mathmetician to calculate.
Based on the angles you gave (7.15 degrees and 172.85 degrees) and the length projected onto the x-axis (98"), there would be four sides, each 49.19" long. The 9" height does not fit. (actually 6.12")
Holding the 98" projected length and the 9" offset, then the angles become 10.49 degrees and 169.51 degrees and the four sides become 49.41" long.
Robert, I know you did what it sounded like he was explaining, but that is not what he is looking for.
He does not mean the x-axis thru the rhombus.
He means draw a rhombus using the angles he indicated, a long skinny rhombus where the two horizontal sides are 9" apart.
The left and right sides of the rhombus will become the “C” side of a right triangle scribed inside or outside the tip of the rhombus.
That triangle “A” side, plus length of the horizontal side of the rhombus = 98"
How long is horizontal side of the rhombus? How long is the left/right side of the rhombus?
What I want to know is why 9"?
If I was laying out track I would be using center lines on 2" centers so the width would be an even number. OR, if I wanted to map the space that would be used it would sill be an even number, 2" “spsce” for each track.
But to answer the question, the horizontal sides would be 26.2542" long, the angled sides sloping up from horizontal at 7.15 degrees would be 72.3080" long.
I would draw a picture, but I don’t really know how to do that quickly on here.
I agree. This is why I try to keep my nose out of these sort of threads. But I can’t seem to help myself . . . . kinda like watching a car wreck . . .
My question is why a rhombus?
Yes, exactly. I calculated the same. You suggested this simple solution a few posts earlier where this could be solved to a pretty close tolerance by carefully laying out two parallel lines 9" apart and slicing another line across at a 7.15 degree angle and then measuring the results. No need to fidget with rhombuses (rhombii ??).
I know how to draw one quickly, but it is a pain in the neck to get the image up on this forum.
