Trains.com

Lionel 42" circle is really ___ wide???

3854 views
13 replies
1 rating 2 rating 3 rating 4 rating 5 rating
  • Member since
    August 2004
  • From: St. Paul, Minnesota
  • 2,116 posts
Lionel 42" circle is really ___ wide???
Posted by Boyd on Thursday, April 12, 2007 1:52 AM
Don't laugh at me,,, I don't have any Lionel 42" circle sitting around. The upper turnaround loop on my bedroom layout I'm building looks to be too small to run 48" Fastrack on. Its 48" wide from the wall to the edge of the plywood. Fastrack seems to be 48" center rail to center rail. Of course I will need clearance room away from the wall. 36" Fastrack is 39" edge of roadbed  to edge of roadbed. So I might end up going with 42" tubular Lionel track. Whatever curve track I use on this loop I want to be able to add switches in the circle down the line when I have more $$$. A majority of the layout will have Fastrack 48, 60  and 72 curve for the main line, but there is some K-line 72 curve track in the 027 profile. Money is tight so I have to work with what I have. The different ways of describing a  train tracks curve width is a bit confusing.

Modeling the "Fargo Area Rapid Transit" in O scale 3 rail.

  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Thursday, April 12, 2007 8:52 AM

Tubular track is usually described by the (approximate) diameter of a circle of track, to the outside ends of the ties.  You could put together a circle and measure the diameter, but to get any kind of accuracy, you would have to make two measurements of diameters at right angles to each other, without disturbing the track between measurements, and average them.

There is a simpler way, using a single curved piece.  Just measure the chord length between the ends of the center rail.  Then divide half of this by the sine of half the angle that that piece would occupy in a circle, to get the radius to the center rail.  For example, with 8 pieces to a circle, divide half the chord by the sine of half of 45 degrees, which is 22.5 degrees, or .382683.  With 12 pieces to a circle, divide by the sine of half of 30 degrees, which is 15 degrees, or .258819.  With 16 pieces to a circle, divide by the sine of half of 22.5 degrees, which is 11.25 degrees, or .19509.

Be sure to measure the chord exactly between the centers of the ends of the railheads (excluding any track pins, of course).  This method is not very sensitive to whether the track has been bent a little to another curvature, especially for the gentler curves--it will tell you the original or intended radius.  But, if you have any worry about that, pretend that you have divided the piece into two equal parts and measure each part separately, using the sign of half the halved angle, then average those two results.

Here's an example:  Suppose we have a piece of ordinary O27 curve.  We measure the chord and find that it is 9 9/16 inches, or 9.57625.  We divide that by 2 to get 4.78125, then by .382683 to get 12.494, which we round to 12 1/2 inches, the actual radius of the track, to the center rail.  If we want the nominal diameter, we double that radius to 25 inches and add the length of a tie, 2 inches, to get 27 inches, q.e.d..

Bob Nelson

  • Member since
    January 2005
  • 1,991 posts
Posted by Frank53 on Thursday, April 12, 2007 9:11 AM
 lionelsoni wrote:

There is a simpler way, using a single curved piece.

Just measure the chord length between the ends of the center rail.  Then divide half of this by the sine of half the angle that that piece would occupy in a circle, to get the radius to the center rail.  For example, with 8 pieces to a circle, divide half the chord by the sine of half of 45 degrees, which is 22.5 degrees, or .382683.  With 12 pieces to a circle, divide by the sine of half of 30 degrees, which is 15 degrees, or .258819.  With 16 pieces to a circle, divide by the sine of half of 22.5 degrees, which is 11.25 degrees, or .19509.

Be sure to measure the chord exactly between the centers of the ends of the railheads (excluding any track pins, of course).  This method is not very sensitive to whether the track has been bent a little to another curvature, especially for the gentler curves--it will tell you the original or intended radius.  But, if you have any worry about that, pretend that you have divided the piece into two equal parts and measure each part separately, using the sign of half the halved angle, then average those two results.

Here's an example:  Suppose we have a piece of ordinary O27 curve.  We measure the chord and find that it is 9 9/16 inches, or 9.57625.  We divide that by 2 to get 4.78125, then by .382683 to get 12.494, which we round to 12 1/2 inches, the actual radius of the track, to the center rail.  If we want the nominal diameter, we double that radius to 25 inches and add the length of a tie, 2 inches, to get 27 inches, q.e.d..

Bob:

Could you describe the simplier way now? Whistling [:-^]

  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Thursday, April 12, 2007 9:19 AM

Okay:  Measure the chord and multiply it by the magic number.  The magic number is

1.306563 for 8 pieces to a circle,

1.931852 for 12 pieces to a circle, and

2.562915 for 16 pieces to a circle.

Simplier, no?

Bob Nelson

  • Member since
    October 2004
  • From: S.E. Ohio
  • 5,434 posts
Posted by Blueberryhill RR on Thursday, April 12, 2007 9:22 AM
 Frank53 wrote:
 lionelsoni wrote:

There is a simpler way, using a single curved piece.

Just measure the chord length between the ends of the center rail.  Then divide half of this by the sine of half the angle that that piece would occupy in a circle, to get the radius to the center rail.  For example, with 8 pieces to a circle, divide half the chord by the sine of half of 45 degrees, which is 22.5 degrees, or .382683.  With 12 pieces to a circle, divide by the sine of half of 30 degrees, which is 15 degrees, or .258819.  With 16 pieces to a circle, divide by the sine of half of 22.5 degrees, which is 11.25 degrees, or .19509.

Be sure to measure the chord exactly between the centers of the ends of the railheads (excluding any track pins, of course).  This method is not very sensitive to whether the track has been bent a little to another curvature, especially for the gentler curves--it will tell you the original or intended radius.  But, if you have any worry about that, pretend that you have divided the piece into two equal parts and measure each part separately, using the sign of half the halved angle, then average those two results.

Here's an example:  Suppose we have a piece of ordinary O27 curve.  We measure the chord and find that it is 9 9/16 inches, or 9.57625.  We divide that by 2 to get 4.78125, then by .382683 to get 12.494, which we round to 12 1/2 inches, the actual radius of the track, to the center rail.  If we want the nominal diameter, we double that radius to 25 inches and add the length of a tie, 2 inches, to get 27 inches, q.e.d..

Bob:

Could you describe the simplier way now? Whistling [:-^]

Let's see here.......I'm a 10 year old kid trying to figure out if my curved track are going to fit on the 4 x 8 sheet of plywood, that my Dad bought me. I take 4 sections of curved 0 guage track and put them together. They fit. And I have about 8 1/2 inches on either side for my farm and station. If I do this to both ends, I can put two long straight tracks together and have an oval. Wow. I did it.

Chuck # 3 I found my thrill on Blueberryhill !!
  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Thursday, April 12, 2007 9:33 AM
That's it, Boyd.  Just buy a circle of every kind of track that you think might work for you and try it out...;-)

Bob Nelson

  • Member since
    May 2005
  • 382 posts
Posted by trigtrax on Thursday, April 12, 2007 10:26 AM

Tubular O-42,54,72 is measured at the center rail. (Note tubular O-27 and O gauge does not follow this rule)

Using O-42 the diameter of the center rail is 42 inches. Ties are 2 1/4" so the outside diameter requires 44 1/4

Likewise O-54 requires 56 1/4, O-72 requires 74 1/4.

These are minimums to fit track.. More clearance is required for swing.

  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Thursday, April 12, 2007 11:33 AM

I measure the radius to the center rail and diameter to the ends of the ties of my O27-profile track as:

Marx and Lionel O27, 12 1/2 and 27

Marx O34, 15 3/4 and 33 1/2

K-Line O42, 20 1/4 and 42 1/2

K-Line O54, 26 3/8 and 54 3/4

K-Line O72, 35 1/4 and 72 1/2

Bob Nelson

  • Member since
    February 2002
  • From: Mpls/St.Paul
  • 13,892 posts
Posted by wjstix on Thursday, April 12, 2007 12:39 PM
IIRC K-Line was the first to produce O-42 curves about 25 years ago, they chose that as being the largest possible curve you could completely fit onto a 4'x8' sheet of plywood. I assume Lionel O-42 is the same(??)
Stix
  • Member since
    August 2004
  • From: St. Paul, Minnesota
  • 2,116 posts
Posted by Boyd on Friday, April 13, 2007 12:08 AM

Lionelsoni,,, did you help write federal tax code?

I would go out and buy circles of all the various brands of track on the market but I deliver pizza for a living and thus I don't make a lot of money. Plus the tips go down quite a bit in the spring.

I'm still a bit confused. Lionel 027 profile old style tubular track in 42" circle IS _____ inches wide from the outside edge of the metal tie to the outside edge of the metal tie, not center rail to center rail. 

Modeling the "Fargo Area Rapid Transit" in O scale 3 rail.

  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Friday, April 13, 2007 8:52 AM
I don't have a single piece of Lionel O42 in either profile.  But the diameter of K-Line O27-profile to the ends of the ties is 42 1/2 inches, as I posted.  If, as William suggests, Lionel copied K-Line, then Lionel's might be the same.  If so, the chord length should be 10 1/2 inches.  Perhaps someone who has some would measure the chord for us.

Bob Nelson

  • Member since
    August 2004
  • From: St. Paul, Minnesota
  • 2,116 posts
Posted by Boyd on Saturday, April 14, 2007 12:28 AM
I stopped at the hobby shop and got some Lionel 42" circle track and its the biggest circle that will work in that space. I got 1/4 circle of 54" too.  Sorry I didn't see your post that 42" track is 44-1/4" wide.

Modeling the "Fargo Area Rapid Transit" in O scale 3 rail.

  • Member since
    March 2005
  • From: Southwest Georgia
  • 5,028 posts
Posted by dwiemer on Saturday, April 14, 2007 4:47 AM

Another option.  You can add different size curve pieces to make the bend.....If you have Fastrack, you can use a mix of O36 with O60, etc.  You can check with your hobby shop and see if you can exchange some pieces.  You can do this also with the other style track as well.  I suggest getting a couple of the different pieces and trying it.  I know guys like Bob can figure this out in their heads, but for me, I would need to see it.  Just a thought, it is good to think outside the box now and then.

Dennis

TCA#09-63805

 

Charter BTTs.jpg

  • Member since
    March 2004
  • From: Jelloway Creek, OH - Elv. 1100
  • 7,578 posts
Posted by Buckeye Riveter on Saturday, April 14, 2007 6:16 AM

If I took this problem to my LHS, more than likely they would pull out several boxes of different sized curved track and there on the top of the counter, we would put together something that would work.

Boyd, I didn't know that pizza orders slowed down in the Spring.  Makes perfect sense.  Football is not on TV.

Celebrating 18 years on the CTT Forum. Smile, Wink & Grin

Buckeye Riveter......... OTTS Charter Member, a Roseyville Raider and a member of the CTT Forum since 2004..

Jelloway Creek, OH - ELV 1,100 - Home of the Baltimore, Ohio & Wabash RR

TCA 09-64284

Join our Community!

Our community is FREE to join. To participate you must either login or register for an account.

Search the Community

FREE EMAIL NEWSLETTER

Get the Classic Toy Trains newsletter delivered to your inbox twice a month