Figure 8 using E-Z track

[train man]

Is it possible to make a figure 8 using E-Z track? If so, what track sections do I need to get?

[James in FL]

Hi train man,

Absolutely it's possible.

Tell us what the radius of curves you plan to use is, and the degree of crossing, we can then give you the formula to figure it out.

[train man]

Thanks for the reply James. Just to be clear, I want a figure 8 that uses a crossing, not an up and over. As for the radius, I do not have a specific value in mind. I just want the setup to be as compact as possible, and to use only standard sections, not flex track. It is n scale.

[James in FL]

Yes, as compact as possible using N scale EZ track.
Not over and under, but rather level crossing.
Got it.

I was going to take the easy way out and refer you to another previous post I was involved in.
Those posts seem to be missing thanks to the recent site upgrade.

Give me a bit of time as I have to cut the lawn before dark.

Wifey says so.

BRB

[train man]

Thanks for your help.

[James in FL]

Ok, to build the most compact Figure 8 from EZ track you will need the smallest radius curves and the largest degree crossing Bachmann makes.
Smallest radius curves presently available are 11.25r (30 degrees) Item # 44801).
Largest crossing 90 degrees (Item # 44841).
For straight sections you will need; (for each leg of crossing)
(1)   5in. straight (Item 44811) 1 ea. per leg total 4
(2)   4.5 straight (Item 44829) 1ea. per leg total 4 (you will need 2 packs of these)
(3)   1.125 straight (in same pack as above) 1ea. per leg total 4
The straights are the radius minus half the crossing length.
Crossing is 1.125
The calculation comes out to 10.6875 per leg.
This will be .125 short but the EZ track is forgiving and it will work out fine.


Curves
360 - 90 = 270
270/30 = 9 per side.


Welcome back to trigonometry 101.

Good luck

[train man]

Thanks James. It is interesting that there is not an exact solution. As I feared, this indicates that Bachmann did not design the sections so a figure 8 comes out just right. But I understand what you say about the track being forgiving.

Do you know what the maximum width and length (to the outer edges of the roadbed) that this design will produce for the figure 8?

[train man]

Welcome back to trigonometry 101.

OK James. After working through the trig, I may have been able to answer my own question. If the figure 8 is enclosed in a rectangle that runs between the limiting track centers, the width of that rectangle is just twice the track radius, or 22.5 inches. The length of the rectangle is two plus twice the square root of two times the radius, which gives about 53.3198 inches. Do you agree?

Of course, to get the length and width to the edge of the roadbed, I would have to add the total width of the roadbed to these two results. I think that is an extra 1.5 inches that must be added to the length and the width.

Is that about right?

[James in FL]

If you're not dead on with your calculation, you're close enough.


I'm not totally confident in my formula.
Twice the square root of twice the radius, I get 9.486.
Then I add twice the diameter (45) and I get 54.48.
Then I subtracted half the crossing and I get 53.917.
I'm leaning this is correct but again not totally sure if my formula is.

If I subtract the entire crossing I get 53.355.

Perhaps Mr. Satnik will weigh in here, and correct me.

I measure the width of the roadbed to be 1.0625

55.125 x 23.5

[train man]

I'm not totally confident in my formula.
Twice the square root of twice the radius, I get 9.486.

Hmm. I think we are using different formulas. From what you say it sounds like you are using a square root of the radius. The formula I worked out for the length of the rectangle, from track center to track center, looks like:

(2 +2 sqrt(2)) X radius

Or approximately 4.8284 X radius = 4.8284 X 11.25 = 54.3198 inches

And oops, I see I accidentally typed 53.3198 yesterday.

As for the roadbed, I don't have a piece of track with me. I am basing the 1.5 inch width on the fact that the description of the Thunder Valley train set says that the supplied track produces a circle of 2 foot diameter. Assuming a track radius of 11.25 inches, the diameter center-to-center would be 22.5 inches, so it appears the roadbed adds 1.5 more inches.

[James in FL]

Yes, formula I was attempting to use was that of a right angle extending 10.68 on each leg. Then a 30 degree arc, 11.25r  (tangent) connecting the two and calculating to the center of the arc, and multiplying by 2
I must admit I did not take the missing .125 in to consideration, yet I subtracted ½ the crossing, I think therein lies my error.
Last time I sat in a trig class was 40+ years ago, so that's where my doubt in my formula comes to play.
I'll PM Joe and ask him to look in here.
Maybe he can get me on track.

Your calculation looks good to me @ 54.3.

I think it's a sin / cos thing and at this point my brain begins to boil

[Joe Satnik]

Greetings all.

Busy-ness got me away from the board for a few days.

You can play with and figure out Figure-8's on AnyRail.com (free download).

Smallest crossing degrees gives the smallest overall length. 

In this case, it is 30 degrees.

AnyRail shows the 30 Degree Crossing Item #44840 has a length of 2.598".

Half that crossing length is 1.299" (for use in calculating the straight leg length).

The math is shown half-way down the first page of this thread:

http://www.bachmanntrains.com/home-usa/board/index.php/topic,11761.0.html

Helpful Hint: Tangent of 15 degrees = 0.2679 (for use in calculating the straight leg length).

See how close your straight leg length calculations are to what I found by playing around on Anyrail, which is between 1.5" and 2". 

Hope this helps.

Sincerely,

Joe Satnik

[train man]

Thanks for the heads up on AnyRail, Joe. The 30 deg crossing may minimize the length, but it causes the width to blossom out. No problem though, since the software will let me try out different arrangements.

[Joe Satnik]

TM,

Not sure what you mean by "blossom out", but an 11.25" radius circle is still the same width no matter which degree crossing is attached to it,

as long as the crossing length is short enough.  (Case in point, the HO thirty degree crossing is actually too long for 18"R circles.)

The number of curve sections needed varies with the degrees of the crossing, as in this formula:

(360 - crossing degrees) / degrees per curve section = number of curve sections needed per side. 

In this '30 degree crossing' and '30 degree per curve section' case, (360-30)/30 = 11 sections per side, times 2 = 22 curve sections total. 

In the '90 degree crossing' case, (360 - 90) / 30 = 9 sections per side, times 2 = 18 curve sections total.

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

Hope this helps. 

Sincerely,

Joe Satnik

[train man]

TM,

Not sure what you mean by "blossom out", but an 11.25" radius circle is still the same width no matter which degree crossing is attached to it,

What  I had meant Joe was that I had thought it would be necessary to go to a larger radius curve to use the 30 deg crossing. But from what you say, I see I was wrong. So I'll give the 30 deg crossing and 11.25 radius a try in the software. I just hope things fit better than they did with the 90 deg crossing. In that case, it seems you wind up with quarter inch gaps on the straight sections.