[TheForge] Helical Railing Equations

[Bruce Freeman]

Okay!  Here come the promised equations.  Remember, these are untested.

I.  First, assume we have an enormous shop and an overhead crane and
scaffolding, and we choose to make a perfect helical railing.

Say you want to make a helix of diameter D that advances along its axis
by height H per revolution.  The expression for the length of the
railing per revolution is

L^2 =  (pi * d)^2 + H^2     or L = sqrt[ (pi * d)^2 + H^2 ]

Where:
" ^2" means "squared"
"sqrt[...]" means "square root of the amount in square brackets"
pi = 3.14159... , and 
"*" means "times"

Now, if we are to roll this on the flat then "stretch" it vertically
into a helix, the diameter to which we have to bend it would be D (which
will be greater than d):

D = L / pi = sqrt[ d^2 + (H / pi)^2 ]

Where:
"/" means "divided by"

The stretching will twist the railing.  It would be difficult to convey
in words the DIRECTION of twisting, so try this:  Take a short piece of
rubber hose, and make a circle of it on the flat.  Mark the top edge. 
Now stretch it vertically to turn it into a helix OF THE PROPER
DIRECTION OF ROTATION.  (You do NOT want to make a railing for the
mirror image of your stairway!!!)  Observe which direction the hose
twists.  You will either want to pre-twist the railing in the OPPOSITE
direction of this (which I'm not sure will work correctly), OR you will
simply tweak the railing back to horizontal (in which case you probably
don't need the following equation, but may be interested in it anyway):

I *believe* the twist will be 
T = arctan[ H/ (pi * d) ].

For those who are not familiar with trigonometric functions, calculate
the number inside the brackets on your pocket calculator, then look up
the "arc tangent" or "arctan" in a trig table, OR just hit the "[shift]
[tan]" keys on the (scientific) calculator.  In the latter case, be sure
that you've selected angles to be represented as degrees (typically
indicated by "deg" or "D" in the display, sometimes by the superscript
"o"), or you'll be reading radians or grads, which most of us don't
relate to very well.

EXAMPLE:

Suppose your helix will be 100" in diameter and 100" in height per
revolution.  d=100, h=100.  Then you'll need railing of length L per
revolution:
L=sqrt[(100*pi)^2+100^2] = 329.7" = 27.5' per revolution.  Already you
can see that this coudl be a bit rough to handle!

You'll need to bend this railing to diameter D:
D = L / pi = 104.9"  (a little more than the final diameter, d)

If you then stretch this such that the "coils" are separated by 100",
you'll have a perfect helix, except that the railing will be twisted. 
You'll need to untwist it by:

T = arctan[ 100/ (pi * 100) ] = arctan(1/pi) = 17.66 degrees PER
revolution of the helix.

MAYBE you could countertwist the railing either before or after bending
it to diameter D, but I'm not ready to swear to that yet.

II.  The Approximation.

My (untested) assertion is that you could make up four (my guess) FLAT
arcs of the railing material and assemble them into a complete helix. 
The shape would not be a perfect helix, but I believe it may be
indistinguishable to the eye.  As I've said before, the question is not
WHETHER this could be done, but HOW MANY fractions the rotation would
have to be cut into before the reassembled approximate helix would in
fact be indistinguishable from a perfect helix.  I believe four would
suffice.   (I'd rather it were three.) But this part is untested...

Now, the equations that apply are identical to the above, but we must
divide them by n=4, the number of arcs per rotation, for these purposes.
 I'll use "subscript" n or 4:

Length of railing to make the fractional or quarter arc:

Ln = 1/n * sqrt[ (pi * d)^2 + H^2 ]                              L4 =
1/4 * sqrt[ (pi * d)^2 + H^2 ]         

Dn = L / (n* pi) = 1/n * sqrt[ d^2 + (H / pi)^2 ]             D4 = L /
(4* pi) = 1/4 * sqrt[ d^2 + (H / pi)^2 ] 

Tn = 1/n * arctan[ H/ (pi * d) ]                                     T4
= 1/4 * arctan[ H/ (pi * d) ]

EXAMPLE:

Suppose your helix will be 100" in diameter and 100" in height per
revolution.  d=100, h=100.  Suppose you choose to make quarter-arc
segments as I suggest:

Then you'll need railing of length L per segment:
L4 = 1/4 * sqrt[(100*pi)^2+100^2] = 82.4" = 6.86' per revolution.  This
will be easy to handle!

You'll need to bend this railing to same diameter D calculated above:
D = L / pi = 104.9"  (a little more than the final diameter, d)
However, this will be more useful if expressed as the radius
R = D/2 = 52.47"

MAYBE you could countertwist the railing either before  bending it to
diameter D, but I'm not ready to swear to that yet.
However, you certainly can "countertwist" it before assembly:

T4 = 1/4 * arctan[ 100/ (pi * 100) ] = 1/4 * arctan(1/pi) = 4.42
degrees from one end of the segment to the other.  More specifically,
the ENDS of the segment must be twisted with respect to the middle to
give this overall twist.  Hence, one end must be twisted -2.21 degrees
relative to the middle, and the other +2.21 degrees.  This is NOT much
twist, so tweaking the railing with wrenches may indeed be the most
practical way of achieving it.

Okay.  There you have it.

Bruce
NJ