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.
- Douglas
If only there were existing software that would allow you to draw this and see exactly your dimensions/options are. Oh wait, there is.
I use 3rdPlanit (trackplanning.com) to do my tinkering in the virtual space.
Onewolf42 If only there were existing software that would allow you to draw this and see exactly your dimensions/options are. Oh wait, there is. I use 3rdPlanit (trackplanning.com) to do my tinkering in the virtual space.
I may try that, but that would require the internet research I was avoiding for the fun of it.
No, 3rd planit comes with templates for different commercial turnouts built-in. You just select what you're using and plop it down.
Hmmm. The thread isn't as much fun as I thought it would be.
Doughless Hmmm. The thread isn't as much fun as I thought it would be.
Rich
Alton Junction
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.
Ray
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.
I'm with Ray, why #8's in a yard?
And I don't have any PECO turnouts as a template....
Sheldon
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.
Easy peasy, no math thats sleazy!
Clear Ahead!
Douglas #2
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.
Colorado Ray 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. Ray
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 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.
I always build yards, or passing sidings with perfectly parallel tracks, never willy nilly.
Again, on paper or on the bench work, if you lay out the parallel tracks, the leads will cross them at the frog angle
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.
I would still like a sketch of the whole area.
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.
LINK to SNSR Blog
ROBERT PETRICK 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.
ATLANTIC CENTRAL Robert, I know you did what it sounded like he was explaining, but that is not what he is looking for.
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 . . .
ATLANTIC CENTRAL What I want to know is why 9"?
My question is why a rhombus?
ATLANTIC CENTRAL 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.
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 ??).
ATLANTIC CENTRAL I would draw a picture, but I don't really know how to do that quickly on here.
I know how to draw one quickly, but it is a pain in the neck to get the image up on this forum.
Robert
Thank you Robert. But the length is the variable in my case and not the height.
At a 9 inch height, what length would fit? And then what would each leg length be.
NVM, see below.
Thanks for your help.
ATLANTIC CENTRAL ROBERT PETRICK 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. Sheldon
Sheldon's right. Here's a scaled sketch based on what was told.
Each track would hold about two 73' cars.
Colorado Ray ATLANTIC CENTRAL ROBERT PETRICK 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. Sheldon Sheldon's right. Here's a scaled sketch based on what was told. Each track would hold about two 73' cars. Ray
Thanks for the sketch. Actully the long sides and short sides are flipped, making the runaround having a less diagnal look to it, but the 4 lengths, but the 4 sides comprise the runaround tracks not the storage tracks. The sorage tracks will come off the two long sides, which will serve as ladders.
Including what Sheldon calculated, 72.30 inches for the long sides, the runaround length adds up to just over 98 inches...then deducting clearance needed for the loco to runaround the cars, which is considerable given the 7.15 angle of the #8s, give me a functional runaround/train length of probably around 76 inches? (we...er...you..could use math to tell me how much length I lose to provide clearance for a standard width HO car on a track angled at 7.15 degrees...at both ends of course).
The yard tracks would come off of the longer sides of the runaraound at each direction. Coming off the bottom of the long side on the right would be longer than 26 inches, especially given that I have more space available to the right beyond the diagram.
However, if we increase the height (or depth as laying on the bench) to 9.5 or 10 inches, that extends the length of the long sides and moves the NE corner to the right (the SW corner is a fixed point), and then I may run out of room for a proper tail track/ lead track at the NE.
Why am I doing this? As you can see, the diagonal shape puts the runaround more towards the back drop as the train moves from left to right.
With #8s, its tough to have a conventional runaround that runs parallel to the bench work edge, and still have enough room to swing any track to be close to the backdrop for any amount of distance. Now I've got 4 tracks coming off the corners of the runaround to maximize the amount of straight trackage in the space, which is important to me for modern open space railroading with long centerbeam flats.
And it would still be so much simpler to see the proposed track diagram.
I have used track arrangements similar to this which take advantage of the specific angle of a turnout for decades now.
And the prototype does the same thing all the time.
Still not sure what the point of all this?
ATLANTIC CENTRAL And it would still be so much simpler to see the proposed track diagram. I have used track arrangements similar to this which take advantage of the specific angle of a turnout for decades now. And the prototype does the same thing all the time. Still not sure what the point of all this? Sheldon
You have the diagram in Ray's post, Edit: just swap out the short sides and the long sides to make a really short ladders and room for only two yard tracks off each ladder. If I could have drawn it and posted it, it would have saved a lot of confusion. I admit that.
I could explain further as to how much total space I have for everything, but this runaround section is the critical place.
Now we can adjust the height a bit to see what it does to the length. If anybody wanted to play along. I understand if people have better things to do.
The point is that I usually plan by laying the track by eyeball with some straight pencil lines writtne on the benchwork. But this seciton I want perfect from all corners, so knowing that the length of a short side is 26.25 inches and not 26.5 inches, for example, matters to keep things "square" so to speak.
I don't trust myself to lay out the lines at 7.1 degree angles, not having a chalk line to snap, and keeping everything perfect. Knowing the lengths of the 4 sides beforehand helps to know that the turnouts are being placed perfectly.
I knew you'd come through though, thanks.
So I am assuming the storage tracks run the long dimension? That may seem obvious to you, but not out here with us.
Once you get the first one right, the rest are parellel lines.
A chalk line is not eaxpensive, can I send you one?
Doughless Now we can adjust the height a bit to see what it does to the length. If anybody wanted to play along. I understand if people have better things to do.
For every inch you increase the "height", you increase the length 8 inches. That's kinda where the No 8 turnout designation comes from. Exactly fits the 7.15 degree angle offset.
I'm not sure how this is at all useful because he's asking about FROG to FROG and any measurements for the track are taken from the centerline of the track and point of intersection of the centerlines of the track.
So even if you calculate the distance from the PI on the SW switch to the PI on the NE switch, its not going to give you the distance between the frogs and it's not going to give you the length of the run around and it's not going to give you the capacity or length of a yard track, it just gives you a dimension that is not really useful for any design purpose.
Dave H. Painted side goes up. My website : wnbranch.com
You do realize that PECO offeres free downloadable switch templates. You can print them off, put two pair of switches together, and then position them exactly as you want them and measure the clearance with actual cars.
dehusman I'm not sure how this is at all useful because he's asking about FROG to FROG and any measurements for the track are taken from the centerline of the track and point of intersection of the centerlines of the track. So even if you calculate the distance from the PI on the SW switch to the PI on the NE switch, its not going to give you the distance between the frogs and it's not going to give you the length of the run around and it's not going to give you the capacity or length of a yard track, it just gives you a dimension that is not really useful for any design purpose.
Yes Dave, I originally asked frog to frog thinking it would help know the exact distance but it isn't needed.
Better than having the PECO templates, I already have PECO #8s to play with.
The exercise was to understand what the lengths of the pair of sides would be before I started laying out lines or track.