I am looking for a formula that I can use in Excel to determine the
correct radius of a handrail for a curved staircase.

Example; Staircase has a run of 12.00" and a rise of 7.375"
         Staircase is against a wall that is curved with a consistent radius 
           of 174' 11.75"

         Problem; What is the correct radius to bend the handrail to?
         If the handrail stayed level with the ground I would need to bend it 
         to the 174' 11.75" radius.  But because there is an incline the radius 
         must change to compensate for the pitch of the stairs.


I want to use Excel once I have the formula to figure this out.
I want to be able to input the "Run"  the  "Rise"  the horizontal "Radius"
  then have the "inclined radius" calculated automatically.

I want the formual to be able to work with differnt staircases.
When I adjust any of the inputs the incline Radius should change automatically.


Generally the only information that I am provided with to do the job is the:
  Run, Rise, horizontal radius.

Thanks for the help.
Fabman101

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Request for Question Clarification by redhoss-ga on 20 Nov 2003 05:25 PST

Does the staircase have both an inside and outside wall (fully
enclosed) or is it open to the inside (is the handrail really at the
outside wall). What type of material are you forming the handrail from
and what forming process do you plan to use. Your problem is similar
to a product I have made before.

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Clarification of Question by fabman101-ga on 20 Nov 2003 18:01 PST

Clarification of question:
Thanks for the input so far.
Perhaps this will help clarify;
This is not a spiral staircase,  this handrail is only 20 feet in overall length.
The staircase does not make much of a turn, it is similar to what you
might see outside a public library entrance.

The material to be used is 1.50?  OD  steel round tubing with a heavy wall.


Lets look at the spring for example.
If I am looking at the spring from the top view I will not see it as a
spring or a spiral but as a true circle.
When that spring is manufactured the machine must roll it to a certain
radius and also the pitch that the spiral of the spring needs when
finished.
The machine cannot roll it to the same radius you would see when
looking at it from the top or it would be to tight when it spiraled.
So what is that radius the machine rolls the spring to?

Most of the handrails I am concerned with do not make a complete
spiral or circle, most of them have a very large radius with a mild
curve.

I do want to try to figure out the spiral formula as well but I
thought I would start with what seemed to be the simpler problem
first. Maybe they are the same problem?

Hope this helps.
Fabman 101

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Request for Question Clarification by mathtalk-ga on 22 Nov 2003 19:10 PST

Hi, fabman101-ga:

I've verified the formula proposed by racecar-ga and checked it
against a couple of simple alternatives.  For the data given in your
problem, racecar-ga's formula give a radius that is independent of the
actual length of the handrail.  This could be computed from the total
rise and run, but evidently the figures you've given (7.375 and 12
inches, resp.) are only meant as "characteristic" rise and run values.

Assuming 20 feet as the (straight line) distance between ends of the
handrail, here are the respective values for radius computed according
to racecar's approach and two more:

racecar's way:  241.0710155 feet

3pt. circle:    241.031743  feet

2tang. circle:  240.9926596 feet

All this depends on a couple of assumptions, so before posting an
Answer, let's check these by clarifying a couple of things:

1.  Are the stairs themselves running parallel to each other?  (Then
the stairs will lie in an inclined plane, and the elliptical curve
suggested by racecar is the appropriate one.)

2.  Is the handrail symmetric with respect to the curve of the side
wall?  (It would be possible for the handrail to positioned so that it
extends further up or further down from the widest point of the wall's
bulge.)

regards, mathtalk-ga

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Clarification of Question by fabman101-ga on 24 Nov 2003 08:12 PST

I am not 100% on my math terminoligy but I will give it a try;


I am not sure about your question if the stairs are running parrell to
each other.   This may help,  visualize one of those large round oil
storage tanks and the stairway that goes up the side of it. The inside
handrail rests right up againts the tank wall.



The second part of your question;  I would say that yes the handrail
is symetrical to the curved wall.  And the wall is a syemetric curved
wall.
The radius of the wall stays the same all the way when looked at in a
flat plane it is not an elipse.

Let me know if you need more information.

Thanks
Fabman

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Request for Question Clarification by mathtalk-ga on 24 Nov 2003 08:37 PST

Thanks for the clarification, fabman101-ga.  In the case of the stairs
running around the outside of a tank, the steps are not in parallel,
ie. like they would be in front of most public libraries, but rather
the directions of the steps turn as they goes around the wall.

In theory the helical spiral is the right model for this, where an
elliptical arc would be more appropriate in the "public library"
scenario.  However the curvature given in your example is truly
slight.  Did you realize that a straight handrail (given the 20 feet
length assumption from a previous clarification, please note my
question about computing length from rise & run measurements) would
only be about 2 inches "out" at the midpoint?

If all the curves you are dealing with are so slight (see racecar's
calculation, a radius of turning of slightly over 80 yards for a
section of only 20 feet in length), then the difference between
helical and elliptical will hardly matter.  An installer would easily
"bend" the ends to fit without especially noticing that this was being
done.

I guess it doesn't make a big difference, but your new picture of a
handrail running on the outer curvature of a tank (rather than the
inner curvature of a wall) affects how a "conservative" design would
be chosen.  We need to avoid having the handrail go "through" the
wall, but depending on which way the wall is curving (in or out) we'd
be careful to have either too much or too little curvature in the
design.

So, please clarify where the length of the handrail will be given,
either as a separate measurement, derived from rise & run, or not
known at all; and specify whether an inner curved surface or outer
curved surface (e.g. tank) is the object.

regards, mathtalk-ga

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Clarification of Question by fabman101-ga on 25 Nov 2003 14:08 PST

The information that I am normally given is;

The radius of the wall,  as in a flat plan

The Rise and Run,  or the degree of the rise from horizantal.
   ( Same thing to me I can convert this either way)

And the desired legnth of the handrail.
  ( Not sure that this matters,  my material is only 20 foot max so if 
    handrail is longer they get it in several peices)

Also,  the radius ranges from very large to maybe down to 48.00"
Most of the time they are more in the large range like 240" and up.

Let me know if you need more info thanks for all the brainstorming so far.

PS: If you think I need to go with more than one formula let me know
and we can work from there.

Thanks
Fabman

Hi, fabman101-ga:

The sections of the handrail for a spiral staircase are not circular
arcs.  If they were, they would lie flat (in a plane), and a moments
reflection should convince you this is not the case.  So describing
the handrail geometry is not a matter of altering the radius of
circular pieces in some way that accounts for the rise, e.g. by
increasing the radius.

It is still correct to speak of the radius of curvature of the
handrail in the same fashion as the radius of the cylinder which
"contains" the handrail.  What remains to be specified to a
fabrication shop is the rise in height per degrees of turning.  For
example, if the handrail is to rise through 10 feet and in that height
make one complete revolution (360 degrees around the spiral
staircase), then there's a rise in height of one foot per 36 degrees
of turning.

To fabricate such a piece, you might start with a flat circular
pieces, say of 180 degrees or semi-circular shap.  The two ends are
then displaced perpendicularly to the plane of the original circle. 
Assuming the length of the handrail were to remain constant, this also
has the effect of reducing the radius of turning.  [Consider for
example the limiting case in which the two ends are fully extended,
and the semi-circle becomes a straight segment.]  In this sense there
is a calculation to be done, to find the proper radius of a circular
piece to start with, so that it winds up with the proper radius after
stretching.

regards, mathtalk-ga

Mathtalk is right: the handrail will not lie in a plane--it has the
shape of a helix, like part of a spring.  Anyway there is a GA theorem
that says that mathtalk is always right.

However, if the staircase makes much less than a full revolution (say
20 degrees or less) the railing can be approximated by a circular arc.
 It sounds likely that with a 174 foot radius, you aren't going to be
turning very many degrees.  If this is the case, then the railing will
have approximately the shape of the 'flat' part of the ellipse which
is formed when you cut a cylinder of radius R (the radius of the wall)
at an angle A ( arctan(rise/run) ).  The radius of curvature of the
ellipse at the 'flattest' point is R * sec^2 (A).  sec^2 (A) = (
rise/run )^2 + 1, so...

For a staircase that only turns through a small number of degrees, the
radius of curvature of the railing is:

R * ( M^2 + 1 )

where R is the radius of curvature of the wall, and M = rise/run.

For the numbers in your example this gives a radius of 241' 0.85".

Here is something that might be of interest.
http://www.dennisallenassociates.com/awards_article_7.htm

Mathtalk-ga,   are you still working on this problem or do you have
any suggestions?

Thanks
Fabman

Let a be the radius of the staircase, b= the run, c= the rise. This is
a helix with parametric equation r=(a*cos t, a*sin t, a*c*t/b) because
when the run is b which is a*t (so t=b/a) then the rise is a*c/b * b/a
which is c. Its derivative is r'= (-a*sin t, a*cos t, c/b). For
s=arclength ,ds/dt= length of dr/dt =a*sqr(1+(c/b)^2). The unit
tangent vector  T to the curve is dr/dt / ds/dt which is (-sin t, cos
t, c/b )/sqr(1 + (c/b)^2). Its derivative is (- cos t, - sin t,
0)/sqr(1 + (c/b)^2). The curvature kappa = length of dT/ds / ds/dt
which = 1/( a (1 + (c/b)^2 )). The "correct radius" is the radius of
the osculating circle, which is 1/ kappa. So "correct radius" = a (1 +
(c/b)^2). This is a very simple formula for Excel. When a=174' 11.75",
b=12", c=7.375" then "correct radius" = 2892.85" = 241' .85".

themathstatstutor

Hi, fabman101-ga:

I've been busy with non-GA things over the Thanksgiving holiday, sorry
about the delay.

I've been polishing my presentation: playing with a spreadsheet
implementation of the three formulas (for approximating the ellipse by
a circle), while trying to work in elegantly the comparison between
helix and ellipse.

For your original example, each "end" of the handrail on an ellipse
would be only 1/40 of an inch "out" from the true helix.  Tighten the
radius of the staircase down to 48 feet, and the effect is only
slightly worse, 1/3 of an inch "sprung" at either end.

So for the range of curvatures that you are dealing with, the
"racecar" formula is very good.  What I wanted to do with the
spreadsheet is to find the "worst case" for the range of slopes and
radii you'll be working with.  The error is clearly maximized with the
smallest radius, so that part is easy.  But the slope is "interesting"
(to a mathematician!).  When the slope is zero, then the handrail is
truly circular and there's no error.  If the slope were "straight up"
then the handrail would become a straight line, and again no error. 
So the maximum error will occur at some intermediate slope.

Of course from a practical point of view the slope will not vary too
far from the roughly 60% grade that you used in your example.

I'll post an answer this evening, and again, sorry for the delay.

regards, mathtalk-ga

No problem, Those holidays are always busy.

Take your time,  I was just checking in.

Thanks
Fabman