ONLINE
STORE
"ASK THE BACH MAN"
FORUM
PARTS, SERVICE,
& INFORMATION
CATALOGS AND
BROCHURES

Welcome, Guest. Please login or register.
Did you miss your activation email?
September 19, 2018, 07:01:07 PM
Home Help Search Login Register
News: Check out the photo gallery link above or >click here< to see photos of recently announced products!
+  Bachmann Message Board
|-+  Discussion Boards
| |-+  N
| | |-+  Bachmann N 30, 45, 60 and 90 degree crossing lengths.
« previous next »
Pages: [1] 2 Print
Author Topic: Bachmann N 30, 45, 60 and 90 degree crossing lengths.  (Read 7242 times)
Joe Satnik


View Profile WWW
« on: January 05, 2010, 04:45:56 PM »

Dear All,

Do any of you own, or have access to, any of the four crossings listed?

(Bachmann N 30, 45, 60 or 90 degree crossings.)

Please measure their length, vertical face of roabed to opposite face, to the nearest 1/16" or mm.

Measure the length of any included fitter straights, also. 

Let us know how packaged, e.g. display box, blister card, etc.

Thanks. 

Sincerely,

Joe Satnik   
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #1 on: January 06, 2010, 05:44:26 PM »

Post edit to omit misinformation.

90° crossing

Measuring with calipers in inches;

The crossing measures 1.125 (+.005, -.000) across all four surfaces, for all intents and purposes it is 1⅛” and square.
In millimeters it measures 28.575 (+.127, -.000).
There is no embossed part number on the crossing.

The four filler pieces that are included with the 90° crossing are embossed as 2 29/32” ST.
They are not embossed with a part number.

Here’s the weird part, they actually measure 1 29/32”.

With calipers they measure from a low of 1.906 to a high of 1.913.
For planning 1 29/32’ will do.



Logged
James in FL

View Profile
« Reply #2 on: January 06, 2010, 06:16:16 PM »

Joe

Sometime ago, I purchased the Bachmann assorted straight section track pack.
Included in the pack are two 4½” straight pieces (N4829A-1), two 2¼” pieces (N4829A-2) and two 1⅛” pieces (N4829A-3).
With calipers they check out as advertised.

I am just wondering out loud, why the “oddball” 1 29/32” pieces are included with the 90° crossing?

It will be interesting to see what the filler pieces included with the 3 other crossings measure out to.

« Last Edit: January 06, 2010, 06:22:40 PM by James in FL » Logged
Joe Satnik


View Profile WWW
« Reply #3 on: January 06, 2010, 07:17:54 PM »

James,

Thanks for all your hard work.   1-1/8" crossing, plus four 1-29/32" fitter straights. 

Here's the one picture that I trust:   

http://www.normstrainworld.com/images/BACH44841.jpg

All the rest are scans of a Bachmann catalog picture (of an HO crossing) re-labeled as the 44841 N crossing.

To build a 90 degree Figure 8, the four straight legs are each "radius minus half the crossing length".

So, 11-1/4"R - 9/16" = 10-11/16".

Combine available straights and fitters to get as close to 10-11/16" as possible. 

For a 90 Degree x 19" Radius Figure 8,

19"R - 9/16" = 18-7/16"

Combine available straights and fitters to get as close to 18-7/16" as possible.

Hope this helps.   

Sincerely,

Joe Satnik
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #4 on: January 07, 2010, 05:30:13 PM »

60° crossing

With calipers the crossing measures 1.552” or 39.43mm across opposite vertical faces.
In layman’s terms that’s .010” short of 1 9/16”.

For planning let’s call it 1 9/16 (1.5625)”.

The embossed part number on the bottom of the crossing is N4842-B.

The four filler pieces included with the crossing are embossed N 1⅛ ST.
The P/N is N4829A-3.
They measure out as advertised 1⅛” (1.125 + .004 - .000).

Bachmann has re-released this crossing as having longer legs whereas the filler pieces roadbed will not need to be trimmed by the modeler.
The new ones do not come with the filler pieces.

I do not have the newer version.





« Last Edit: January 07, 2010, 09:20:54 PM by James in FL » Logged
Joe Satnik


View Profile WWW
« Reply #5 on: January 07, 2010, 08:51:12 PM »

Dear James,

Again, thanks for your work and dedication. 

Calculating the straight lengths for the 60 degree Figure 8s requires a different (more complicated) formula.   

I gotta run, will get back to you with it later.

Joe   
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
Joe Satnik


View Profile WWW
« Reply #6 on: January 07, 2010, 09:47:52 PM »

Dear James.

Math:

xdeg = crossing angle degrees

xlength = crossing length

R = curve radius

strleg = straight leg (1 of 4) length

strleg = R X [tangent (xdeg/2)] - xlength/2

In the 60 degree crossing case, 60/2 = 30,  and tan(30) = 0.57735

Degrees of curves needed for each side = 360 - xdeg = 360 - 60 = 300 degrees per side

300 degrees/30 degrees per section = 10 sections per side, or 20 total 11.25" radius curves. 

300 degrees/15 degrees per section = 20 sections per side or 40 total 19" radius curves.

Hope this helps.

Sincerely,

Joe Satnik

Edit:  added brackets [ ] to equation, corrected tan (30) value to 0.57735





« Last Edit: January 09, 2010, 02:21:50 AM by Joe Satnik » Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #7 on: January 08, 2010, 07:36:44 PM »

Joe,

Please double check my math.
11.25r x .57735 = 6.49518 - .78125 = 5.71393
And
19r x .57735 = 10.96965 - .78125 = 10.1884.

I maybe wrong yet again.

Thanks for all your input to this forum, the next time a figure 8 question comes up, maybe we can direct them here, to this thread.

I think I've got it but please correct me if I don't

Sincerely,

James in fl

« Last Edit: January 09, 2010, 09:11:03 PM by James in FL » Logged
Joe Satnik


View Profile WWW
« Reply #8 on: January 09, 2010, 02:18:53 AM »

Dear James,

My humble apologies.  You are correct, typo error on my part.   I'm glad you caught it.  I will go back and correct it. 

Sincerely,

Joe Satnik
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #9 on: March 04, 2010, 01:04:54 AM »

The 30 degree crossing measures 2.582 across the opposite faces.
No extra track filler pieces are included with the crossing.

In the 30 degree crossing case, 30/2 = 15, and tan(15) = 0.26794

Degrees of curves needed for each side = 360 - xdeg = 360 - 30 = 330 degrees per side

330 degrees/30 degrees per section = 11 sections per side, or 22 total 11.25" radius curves. 

330 degrees/15 degrees per section = 22 sections per side or 44 total 19" radius curves.

Strleg 11.25r X .26794 = 3.01432 - 1.291 = 1.72332

Strleg 19r X .26794 = 5.09086 - 1.291 = 3.79986



The 45 degree crossing measures 1.744, across the opposite faces. let’s call it 1.750.
No extra track filler pieces are included with the crossing.

In the 45 degree crossing case, 45/2 = 22.5, and tan(22.5) = 0.41421

Degrees of curves needed for each side = 360 - xdeg = 360 - 45 = 315 degrees per side

315 degrees/30 degrees per section = 10.5 sections per side, or 21 total 11.25" radius curves. 

315 degrees/15 degrees per section = 21 sections per side or 42 total 19" radius curves.

Strleg 11.25r X .41421 = 4.65986 - xlength/2 (.8750) = 3.78486

Strleg 19r X .41421 = 7.86999 - xlength/2 = (.8750) = 6.99499


Someone, (Joe?) please double check the math.
« Last Edit: March 04, 2010, 01:14:27 AM by James in FL » Logged
Joe Satnik


View Profile WWW
« Reply #10 on: March 04, 2010, 01:38:24 PM »

Dear James,

Thanks for the crossing measurements.
 
Your calculations look good.

A couple of notes or exceptions: 

1a.) Unless you want to cut an 11.25"R - 30 degree curve in half, the 45 degree crossing would only be good for the 19"R curves.

1b.) You could add a single 19"R curve to the middle of each 11.25"R loop, but that would throw off the straight leg calculations a bit.  (The next trigonometry challenge?...)
 
2.)  Now comes the hard part:  Finding fitter straights that add up to leg lengths you calculated.....Fortunately loop ends tolerate straight-leg slop pretty well.

Hope this helps.

Sincerely,

Joe Satnik
 
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #11 on: March 04, 2010, 05:20:30 PM »

Some thoughts…

Several posts back I questioned the “oddball” sections included with the 90° crossing. Now I see where they can come in handy, for instance with the 45° crossing using 19r curves.
A 5in straight plus the “oddball” 1.90in. straight gets it pretty close.
Using eight of them with the 30° and 19r also works out pretty close, closer than any other combination.
Problem there is that they are only available with the 90° crossing and you’d have to buy 2 crossings to get the eight pieces.
Also you will need at least two (four would be better) assorted track packs as they include 2ea.of the smaller sizes.
Going that route, there is still going to be some slop on most all possible combinations of filler pieces with all 4 crossings.

So let’s say $50 for four assorted track packs plus another $30 for two 90° crossings… that puts the cost at a good quality dremel tool with ~ 50 accessories.
Or you could spend 3 bucks for a competitors track saw and spend the rest on fodder track pieces which could be cut much closer than using manufactured short lengths.

Just thinking out loud…

P.S. Thanks again Joe for all your input into the forums and the trig lessons.

Good Luck

James in fl
Logged
Joe Satnik


View Profile WWW
« Reply #12 on: March 04, 2010, 07:01:06 PM »

Dear James,

The mixed radii recipe mentioned in 1b.) above (mostly 11.25"R x 45 degree Figure 8 ) calculates to 4-7/8" straight leg length. 

1 ea 45 degree crossing,
4-7/8" straight or fitter straights
5 ea. 11.25" x 30 degree curves
1 ea. 19" x 15 degree curve
5 ea. 11.25" x 30 degree curves
4-7/8" long straight or fitter straights
Attach end to crossing.
repeat for other side.

Recipe may work with 5" straights.   

Hope this helps.

Sincerely,

Joe Satnik 
Logged

If your loco is too heavy to lift, you'd better be able to ride in, on or behind it.
James in FL

View Profile
« Reply #13 on: March 04, 2010, 09:44:49 PM »

Joe,

Yep, that’s (1b) yet another case where having 8 pieces of the 1.906 section would get you closer. (x2 + 1⅛ = 4.937) leaving you only .062 over, rather than .125 over using the 5 in. straight.

I knew I needed to go back to the LHS for something.  Smiley

Eventually I’d like to build and run on each scenario we have discussed above.


« Last Edit: March 04, 2010, 11:01:47 PM by James in FL » Logged
James in FL

View Profile
« Reply #14 on: September 21, 2010, 05:12:06 PM »

Figure 8 configurations using the new 17.5r 15° curves.

90° crossing;
360° - 90° = 270° per side.
270 / 15 = 18pcs. per side.
Strleg  = 17.5 - .5625 = 16.9375. (4x).

60° crossing;
360° - 60° = 300° per side.
300 / 15 = 20pcs, per side.
Strleg =  17.5 x .57735 = 10.103625 - .78125 =  9.322375. (4x).

45° crossing;
360° - 45° = 315° per side.
315 / 15 =  21 pcs. per side.
Strleg = 17.5 x .41421 =  7.248675 - .8750 = 6.373675. (4x).

30° crossing;
360° - 30° = 330° per side.
330 /15 =  22 pcs. per side.
Strleg = 17.5 x .26794 = 4.68895 – 1.291 =3.39795. (4x).

Someone please confirm/refute my trigonometry calculations.
 Joe ?

« Last Edit: September 21, 2010, 05:17:44 PM by James in FL » Logged
Pages: [1] 2 Print 
« previous next »
Jump to:  
Powered by SMF | SMF © 2015, Simple Machines Valid XHTML 1.0! Valid CSS!