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Incidentally this whole bent stick thing is a modelers adaptation of reality. 

 

To arrive at an easement length, consider taking one half of any curve's radius and center that over the X spot.

 

Half of a 74" radius is 37". Which equals L.  Half of L is 18.5".

 

Stop the fixed curve 18.5 inches before X and run the easement 18.5" into the tangent after X and bend the stick.

 

Now here is where many mess up.  As I mentioned in an earlier post above, I found it is very important to make sure the marking side of the bent stick is at the mid point of the X dimension.  In your case, using my wet thumb, 3/8".

 

I have had to correct many curves where bent sticks were used.  Bent sticks frequently yield exaggerated  center lines near their ends.

 

Great expressions of easements are far more attractive than any tangent lost imo.

 

My own RR has less than 5% tangent track.

Last edited by Tom Tee

Gentlemen

 

All those equations you are citing are simply trying to describe what Mother Nature gives you with a bent stick.    The key is to make sure your stick can naturally get to an equilibrium state.   I found I needed to constrain the stick at both ends by at least a foot.  I also found the offset did not matter that much from the point of view of either appearance or operation.  I used about 3/4" for all my curves,  if I recall.  But if you need something concrete I would defer to the master and follow John Armstrong's advice.

 

 

 

 

Last edited by John Sethian
Originally Posted by bob2:

Should I expect noticeable improvement in appearance as the train eases in gradually?  Or am I going to lose 40" of tangent track without much advantage?

I have no idea if you would notice it but I certainly would. But, it's not just about appearance, it also has to do with smooth, reliable operation. Think about how you drive your car. Not having easements is like jerking the wheel over instead of gradually rotating it into a curve. 

 

Ed Kelly wrote:

The article in OST can be found at
http://www.oscalemag.com/docs/ost_14.pdf
Scroll down to page 48.

Ed, thanks for posting the link.  I had not realized that this was on line, and I had only a few issues.  I did read it, although I did not verify the various tables as yet.  Problem was,  I was caught by his first sentence, that his procedure reflected railroad practice.  That is just not so (I'll be specific later).  In any case, the railroads turned 10-chord spirals using a transit at the TS; in some cases a table of corrections had to be applied to these angles.  This is not a procedure that can be applied on the layout table, obviously.  The most striking thing about the author's recommendations is the great length of the spirals (not that recommending spirals on yard tracks isn't a close runner-up).

It is like he feels this feature must not only be 1:48 of the real one, but even more on flatter curves.  That makes no sense, particularly the even more.  I can assure you the real railroads do not use a fraction of the radius.  The fact is, a 60"R curve is a 24-degree curve (very closely so), which is twice as sharp as the sharpest of real mainline curves (about a 12-degree curve).  The model is compressed as to radius.  I have to agree with Tom Tee that compression is needed in spiral lengths as well.

The author states that a curve which decreases uniformly in radius is a railroad spiral.  Not so.  A railroad spiral increases uniformly in curvature.  The two are not the same.  They are not even particularly close.  Just remember that the radius of a spiral at the TS is infinite.  The trackworker takes his 62-foot stringline, sets it on the gage line with his gaging tools, and checks the rail at the center (31-feet) of that line, with his folding carpenters rule.  Offset to rail of 4-1/4 inches is 4 degrees and 15 minutes of curvature.  There are reasons the degree is used in place of radius.  But is it not useful on a layout table, so if spirals are used, one must become used to converting back and forth between curvature and radius, is all.
Just divide scale feet of radius into 5730 feet to get degree of curvature.

Railroads use mostly 200-foot spirals with 20-chords; you may see 25- and 30-chords.  But I would compress this in the model by half to 100-foot spirals.
I do realize there can be a problem with steam with sprung axles but no working equalization, but halving superelevation and/or running the super changes beyond each end of the spiral is a way to deal with this.

I found my book of tables, all 209 pages on spirals, including the RR 10-chord.  It is an easy way to get exactly what you want (no failure of the stick to curve enough at the sharply curved end...).  Ten pages of it gives enough data to tackle the model; no copyright problems-- it's a 1940 government publication, accurate to 1/8" in 1:1.  I'll make up an example or so of its use,  certainly the 60"R and 48"R curves, together with the lines of table data used.  I'm thinking x & y's off the tangent from TS to SC.  Certainly quarter points can be read right off the tables.  --Frank

Next time, I might use a stick, anchored as John says.  

 

I currently have superelevation of 1/8" measured at the outside of the ties, with roughly one foot of lead-in on the 70 and 74" loops, and have no problems with rigid wheelbase steamers of four or fewer driving axles.  I have only one fully equalized Northern, but never noticed any real difference in performance.

 

All my steam models with five or more driving axles on a single frame are sprung.  I have no idea whether I could get away with such longer mechanisms unsprung.  It is my longer passenger cars that have problems - trucks need to be laterally free, and spacing between cars must be slightly increased, even on the 74" radius.  Spiral easements would have no effect on these sorts of problems - but if all it takes is two inches, I am in.  I just never bothered before.

Bob2,

 

When laying out curves it can be very easy to mark the right of way.  Simply make a template for the basic radii you will be using, place the template and draw your line.  Each curve can be laid out in under 5 minutes.

 

When working a tight area it can be helpful to have multiple templates of the same radius to slide them trombone style to work around obstructions until  C/Ls at each end are met.

 

I use a variety of templates of many different radii and select  track spacing.  The nice thing about templates is that they can also be used with a pattern router for sub road bed cutting.  Run the bearing directly on the template for a single track or guide the router flange along the edge for a double track sub road bed.

 

For making a template I use a new length of 1 1/2" X 1/4" lath clear.  Then I fasten a machined piece of wood on the tangent and a series of finish nails 1/4" apart 16" along the fixed radius.

 

For real tight radii I use 1 1/2" strips of tempered Masonite instead of clear lath.

 

Four of these templates are made with 6 mm underlayment & lath skirting.

 

The brown ones are 1/2" multi ply sign board.

 

There is 18" of tangent on each template.

 

The two short templates are 5' fixed radius.

IMG_6919

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Last edited by Tom Tee
         
         
         
         
         
         
         
         
         
         
         
         
         

At the top of the sample table line entries below are three lines of control and reference.  For all that,

my first attempt at a table has been a bit ragged.  The tabs in my Notepad draft seem to conflict with

the block boundaries in the table window here; then WYSIWYG wrapped everything.  So it wasn't

WYSIWYG for tables.  A direct loading without preview avoided line wrap.  I leave it for now.


Anyway, I wanted to include the full width and info of this Table II as it exists.  Note that directly it
is a table for a length of spiral one foot long.  I will explain how the total spiral deflection angle theta
is gotten.  For a 60" R curve, which is closely a 24-degree curve modelled, and a 100-foot spiral, the
total spiral deflection angle delta is 12-degrees for a 24-degree curve [ie, (theta-sub-s of 12-deg) =
(L-sub-s of 100 feet /200) x (D-sub-c of 24-degrees)].  The example given is for simplicity at this
time a 24-degree curve.  A 60"R is actually a 23-7/8 -degree curve.  My recommendation (after
study) for simplicity, toaccount for this, is to subtract 1/16th-inch (or 1 part in 400) from the distance
from TS to POT opposite original PC (that point determined by "k", at which offset "p" is measured);
while making the SC also a PCC where the curvature decreases by 1/8-degree (or 1 part in 480).  
The offset "p" itself would have to decrease by 1/4th of a 16th (cosine 12-degrees or ~.96 x 7/16th =
less by ~1/4 of a 16th).  I'll cover this procedure in later post, although it just amounts to sliding a
template of a 100-foot spiral made for a 24-deg curve into place, mentally.


Below are six lines taken from my tables:  T. II - Functions of Transition for L-sub-s = 1 [foot].
Abbreviations:  theta-,x, & y- all sub-s:  values are at SC; L.T. & S.T.  long & short tangents of the
spiral, L.C. long chord of the spiral; **degenerative case of no curvature, only for which case the
general statements of PC at half length of spiral-- and S.T. = 1/3 L of spiral, LT = 2/3 L-- are true.
For dimensions of L spiral = 100, move decimal point two digits to the right.

These tables were prepared under direction of Joseph Barnett of the Public Roads Administration,
and published in 1940 by the Government Printing Office, and are thus in the public domain.

------------------------------------------------------------------------------------------------------------
Alignment cols & tabs from Notepad...          HEADERS & RANGE OF VALUES SEEN IN TABLE II
0             1              2               3              4              5              6               7               8
123456789012345678901234567890123456789012345678901234567890123456789012345678901
T0          T1          T2         T3          T4          T5          T6          T7          T8---Tabs
theta-sub-s              k=opp PC                                                                      theta-sub-s
      [Ls]     p=offset               x-sub-s     y-sub-s     L.T.        S.T.        L.C.       
-------------------------------------------------------------------------------------------------------------
00.0 L000   .00000    .50000    1.00000    0.00000    .66667    .33333    1.00000    00.0**
10.0          .01453    .49949    0.99696    0.05805    .66773    .33430    0.99865    10.0
20.0          .02896    .49798    0.98788    0.11535    .67097    .33725    0.99157    20.0
30.0          .04321    .49546    0.97293    0.17114    .67650    .34229    0.98787    30.0
40.0          .05718    ,49199    0,95235    0.22473    .68452    .34962    0.97851    40.0
50.0          .07078    .48757    0.92649    0.27544    .69536    .35957    0.96656    50.0
---------------------------------------------------------------------------------------------------------------------------------

            
WORKED EXAMPLE:  100' SPIRAL TO 24-degree CURVE; Total SPIRAL ANGLE=12 degrees.
---------------------------------------------------------------------------------------------------------------------------------
An example (described in the introduction) follows:  D-sub-c = 24  L-sub-s = 100 Theta-sub-s = 12
12.0 L100   .01743    .49927    0.99562    0.06959    .66821    .33473    0.99805    12.0~60"R
06.8                                    0.99859    0.03952
06.75  75      Interpolation...   >.99861    >.03923    ...when    req'd.                    3/4 point
06.7                                    0.99863    0.03894   
03.0    50                             0.99973    0.01745                                             1/2 point
00.8                                    0.99998    0.00465
00.75  25      Interpolation...   >.99998    >.00436    Read up                              1/4 point
00.7                                    0.99998    0.00407    Looking    from TS    to SC...
00.0    00     .00000    .50000  1.00000    0.00000    .66667    .33333    1.00000    00.0** TS
            
---------------------------------------------------------------------------------------------------------------------------------
Multiplying the 5 sets of table II values by 100, 75, 50, 25, & 0 gives x,y at quarter points of spiral:
12.0 L100     01.743    49.927  99.562      6.959       66.821    33.473    99.805      12.0~60"R
06.75  75                             74.896     2.942                                                  3/4 point
03.0    50                             49.986     0.872                                                  1/2 point
00.75  25                             25.000     0.109                                                  1/4 point
00.0    00                             00.000     0.000                                                  00.0 TS
Layout:  Use the 40 scale on an engineer's triangular scale as 4 scale, with smallest division = 0.1 ft.
Or, use the 1/4" scale on an architect's triangular scale. with smallest division = 1 inch, by changing
the decimal part above into inches (i.e. 0.083 = 1 inch, etc.-- process not shown here).  Continuing:
---------------------------------------------------------------------------------------------------------------------------------
Now convert real feet above to inches on the 1:48 scale layout:  times 12, divide by 48= divide by 4.
12.0 L100  00.436     12.482    24.890     1.740        16.705    08.368     24.951      12.0~60"R
06.75  75                             18.724     0.736                                                   3/4 point
03.0    50                             12.496     0.218                                                  1/2 point
00.75  25                             06.250     0.027                                                  1/4 point
00.0    00                             00.000     0.000                                                  00.0 TS
---------------------------------------------------------------------------------------------------------------------------------
Now convert the decimal inches into inches & 32nds:  fractional part times 32 = full size on layout:
12.0 L100  0-7/16     12-15/3   24-7/8     1-3/4        16-23/32 08-3/8    24-15/16     12.0~60"R
06.75  75                             18-23/32  0-3/4                                                    3/4 point
03.0    50                             12-1/2     0-7/32                                                  1/2 point
00.75  25                             06-1/4     0 -1/32                                                  1/4 point
00.0    00                             00-nil      0-nil                                                     00.0 TS
---------------------------------------------------------------------------------------------------------------------------------
NOTE:
Explanation of the selection of theta value at quarter points will have to be in a later post.  It is only
that the values are related as the square of their distance along the spiral.  Thus from TS thru quarter
points 1/4, 1/2, 3/4 and SC the values are 0, 1/16, 4/16, 9/16, and 16/16 times 12-degrees,
---------------------------------------------------------------------------------------------------------------------------------
     
Last edited by F Maguire

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