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1954-90cross

 

 

This is a screen-capture from a 1954 Lionel pamphlet.  Sadly, I noted nothing more than the year when I saved it.

 

Note the top right suggestion. The puzzle to to figure out what they were thinking.

 

On the face they suggest using 3xO72 curves to accomplish 90deg of turn, but we know that O72 requires 4x22.5deg curves to do that.

 

But O42 curves do turn 30deg so they would work for the 90deg requirement.

 

In either case, the two 10" straights are not correct for connection to the curves.

 

I've played with this in RRTrack and worked out solutions (using O31 and O27), so I'm only wondering about how this suggestion got published.

 

Imagine the guy who bought track using this schematic and his surprise when he got down on the carpet.

 

  --Joe

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  • 1954-90cross
Last edited by Rail Reading
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I believe there is only one 072 curve involved, and it is the one that the arrow points to. The pieces of track immediately above and below the 072 curve are 45 degree arc pieces that are each cut down to create arc pieces of 33.75 degrees. The two pieces of 33.75 degrees and the one at 22.5 degrees all add up to 90 degrees and thus complete the loop. Rich

You guys are too used to software, and Fastrack "perfection.

Bet its a bit of a force, & you might need to file/trim the center & outer rails a little, but after playing on software, I see it as "close enough" for tubular. 

 

Build all but the 0-72. Now align the 0-72 radial center line to horizontal & move it close, but don't "lock it" to either end just look at it.

 If you lock it, that gap you see now, will really be split among a few joints (like 2 or more pairs).

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