Is it possible to make a figure 8 using E-Z track? If so, what track sections do I need to get?
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.
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.
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
Thanks for your help.
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
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?
Quote from: James in FL on August 26, 2014, 11:35:02 PM
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?
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
Quote from: James in FL on August 27, 2014, 10:39:22 PM
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.
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
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
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.
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
Quote from: Joe Satnik on August 29, 2014, 10:04:35 PM
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.
Quote from: Joe Satnik on August 29, 2014, 10:04:35 PM
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.
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!
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.
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?
Quote from: James in FL on August 30, 2014, 06:22:20 PM
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.
Quote1)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.
Quote from: James in FL on August 30, 2014, 10:28:58 PM
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!
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.
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.
Quote from: Joe Satnik on August 31, 2014, 06:10:21 PM
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.