Figure 8 using E-Z track

[train man]

Keep playing with AnyRail.  You'll learn a lot, and it's a lot of fun.    

OK Joe. I set up the figure 8 in anyrail using the 11.25 inch curves and 30 deg crossing. Then I worked out the math to see what I would need in the way of straight sections. It seems things are even worse for this case than for the 90 deg crossing case.

I find that the distance from the center of the crossing to the end of each curved section is about 3.0144 inches. This value is the tangent of 15 degrees times the radius. Subtracting half of the 2 19/32 inch crossing length from this gives the needed straight section length as about 1.71753 inches. Unfortunately, none of the E-Z track straight sections shown in anyrail is very close to this. The closest straight lengths I see are 1 29/32 inches (part 44841-1) and 1 1/8 inches (part 44829-3). Considering both sides of the cross, putting in two of the first parts comes out too big by 0.38 inches, and putting in two of the second parts comes out too small by 1.185 inches. If there is another E-Z track straight section that I missed that would work better please let me know.

This is what motivated my original question. Even though it is possible to work out the math of what's needed, that doesn't mean the sections exist that will get it to fit properly.

Of course I may have not worked this out correctly. Please let me know if you come up with something different.

[train man]

I think I may have been able to answer my own question again Joe. I had not noticed that way down on the bottom of the track sections shown in anyrail (down below where the crossings and switches are shown) there are more straight sections. Part number 44899-2 is 7/8 inches in length. Two of these give a length of 1.75 inches, which is very close to the 1.7153 inch length that is needed, assuming my math is right (and I hope you will comment on that).

I'll bet even the 90 deg crossing case will work out using those special short lengths shown at the bottom of the E-Z track library!

[train man]

OK. I took another look at the 90 deg crossing case. For that case I came up with a straight length needed between the end of the crossing and the start of each curve as being 10.6875 inches. And combining straights of lengths 4 15/16 + 4 1/2 + 1 1/4, I get exactly the needed length! The part numbers for those sections are 44811, 44829-1, and 44899-3, respectively.

So, assuming you agree with the results Joe, it indeed looks like both the 30-deg and 90-deg figure 8's can be made just fine using E-Z track.

[James in FL]

I'm not liking AnyRails numbers if these are the dimensions you are listing.
When I measure 7 sections of the 5 in straight my average is 5.038.
The average of the 1 1/8 is 1.131.
30 degree crossing (1 sample measured 7 times) averages 2.582.
The 90 = 1.128.
The 1 29/32 section (only sold with the 90 degree crossing) averages 1.9095.
The 1 ½ = 1.507

The seven sample average is what we were taught way back when we learned "no two identical parts are exactly alike".
There is no 4 15/16 straight that I am aware of, number listed refers to 5 in straight section.
We could cross over into standard deviation here, but let's not.

Let us not forget that we are playing with toy trains here.
In this industry, the most we can expect is tolerances in the +/- .010 range maybe a bit more.
I no longer have access to a precision optical measuring device but I would fully expect the stated degree and radius of the curve track has at least as much deviation.
So, if you want to use Bachmann EZ track, (or for that matter, that from any other manufacturer) know before hand, your trackwork may not be precise and may have some slight kinks and waves to it. It's good enough and will work.
If you don't want to settle for "good enough" and "will work", and desire a more precise fit, you may want to consider a combination of flex and hand laying, or hand laying exclusively.
Even then, know we are not perfect, neither is the 1:1 real thing.
What works out perfectly on paper does not translate exactly into real world circumstance.
Build it with what you have, and enjoy it.
Good luck

@ Joe

Thank you for your time to look in here on us.
Getting back to the question of the overall length of the rectangle...

What is the proper formula to figure this out?
Is it in fact, Pythagorean Theorem or am I way off track?
Always willing to be educated.
School me please.
In reply to your question (discounting AnyRail numbers) my measurements = straight leg equals 1.7233.
What does Anyrail say?

[train man]

I'm not liking AnyRails numbers if these are the dimensions you are listing.

You are right James. The values I listed are all from anyrail. And I agree, I am not looking for exact precision. Just want to get values that work.

I realize you asked Joe about the length calculation. But the trig is sitting in my notes, so let me tell you how I am calculating it. The distance from the center of the cross to the top of the loop is the sum of 3 terms:

1)The radius of the loop
PLUS
2) The radius of the loop times the sine of half of the crossing angle times the tangent of half the crossing angle
PLUS
3) The radius of the loop times the cosine of half of the crossing angle.

The full length is double the result of that sum.

The above calculation gives the length from track center to track center. So the width of the roadbed must also be added to the final result.

[James in FL]

1)The radius of the loop
PLUS
2) The radius of the loop times the sine of half of the crossing angle times the tangent of half the crossing angle
PLUS
3) The radius of the loop times the cosine of half of the crossing angle.

I'll buy into that, again your numbers look good.

Thanks TM

P.S. I think we have lost many lurking, I hear something frying besides me.

[train man]

P.S. I think we have lost many lurking, I hear something frying.

I think you're probably right. But please, let me confess. In my other life, in addition to doing trains, I am also a physicist. I use trig all the time. And I would like to thank both you and Joe for the insight into how the Figure 8 works. I don't know if I would have ever realized that the sum of the angle of the loop plus the crossing angle is 360 degrees. That is the key result that makes everything come out right. With that clue, working out the details of the Figure 8 was not hard. Without that clue, I did not know where to start.

Thanks again!

[train man]

Hopefully this is not beating a dead horse, but it is possible to simplify the length calculation by applying some trig theorems to the 3-terms sum I gave previously. An exactly equivalent expression (that is, one that gives exactly the same result as the longer expression) involves the sum of just two terms:

1) The radius of the loop
PLUS
2) The radius of the loop times the secant of half the crossing angle.

Again the result of this calculation needs to be doubled, then the width of the roadbed added to it, to get the overall length.

Sorry if this is a bit too much, but I enjoy doing this kind of math.

[Joe Satnik]

Isn't trig fun?

What I came up with:

Given

crossing degrees "CRDG",

crossing length "CRL"

curve radius "R",

and track bed width "TBW",

a mathematically perfect Figure-8 has the following formulas:

Overall Width = 2R + TBW

Straight length between opposing curves = 2R x Tangent(CRDG/2)  (Good for an up-and-over Figure-8.)

Straight length between crossing and curve = R x Tangent(CRDG/2) - CRL/2     (There are 4 of these straights per Figure-8.)

Overall Length = 2R x [1+Secant(CRDG/2)] + TBW, where Secant is the same as 1/Cosine

If you are running fence-to-fence on your table top, add another TBW to both Overall Length and Overall Width for loco and rolling stock overhang.

It is possible to create an HO 18"R and 22"R "Double-Cross-Demolition-Derby" Figure-8 using  four 90 degree crossings (and their included 2" pieces).

Try it out on Anyrail.

22"R and 26"R (actually, any 4" difference in radius) would work in HO too, but would be incredibly long and wide.  

If anyone claims I did, I will categorically deny that I have ever walked around wearing a hat in the shape of a Figure-8.

Hope this helps.  

Sincerely,

Joe Satnik

Edit: Added HO references.

[train man]

Isn't trig fun?

Overall Length = 2R x [1+Secant(CRDG/2)] + TBW, where Secant is the same as 1/Cosine

Yes Joe, trig is fun! Very glad to see you got the same result for the overall length as I did. It's good that we are on the same page with the math.