A Helical Stairway Project

Tom Farmer

This is a real-world problem. It’s what I understand from a friendly

exchange of several e-mails during a recent semester in which I was teaching

a multivariable calculus class. The first message sent to me began with, “I

am a carpenter with a problem,” and it turned out that the problem involved

parameterizing lines and curves in space and finding intersections between

lines and planes – exactly some of the topics in multivariable calculus. My

students got a kick out of the developing story.

By way of several messages back and forth, I learned about the plans

for a 10-story hospital (now completed) that has an unconventional stairway

system including, between each pair of floors, a free-standing, curved flight

of stairs that follows an arc of a helix. The stairway system is an important

visual element in this building. From the bottom floor, a visitor can look up

and see the entire system including the 9 helical flights. Thus, these curved

sections must look elegant as well as function properly.

The entire system is enclosed within a quarter cylinder of radius 21 feet,

and the stairway from one floor to the next has the following parts (see

Figures 1 and 2):

- a main landing adjoining the axis of the cylinder; then (rising clockwise

around Figure 1),

- a conventional flight of steps out from the axis to a landing bounded

by the wall of the cylinder; then

- the curved flight of steps that rise along a helical arc to another landing

along the cylindrical wall; and, finally,

- another straight flight of steps back toward the axis and rising to the

main landing on the next floor.

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Figure 1: Stairway system.

2

Figure 2: One flight showing stringer and concrete beam.

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Questions from the carpenter

In order to pour the concrete for the helical staircase, the carpenter envi-

sioned a temporary framework consisting of two main beams, AB and CD

in Figure 1, surmounted by a number of horizontal, radial beams that meet

the cylindrical surface along a curve. Upon this framework would be con-

structed the ramps of plywood and the forms (the temporary walls) that

would contain the concrete. A sideview of the finished product (Figure 2)

shows the zigzag outline of the steps on top of an 8-inch thick band that

forms the stringer supporting the steps. Below the stringer is a thicker con-

crete beam. Although the steps, the stringer, and the beam are all poured at

once, it is convenient to speak of them as separate objects. The ideal tempo-

rary framework is one that could be dismantled and reused, and this would

help to guarantee consistent results from one floor to the next. In fact, more

important than its structural role, the beam AB would serve as a template

in preparing to pour the staircase on each floor. The points A and B are

essentially the bottom and top points of the helical curve that the bottom

of the concrete beam is to follow. The points C and D are constructed from

A and B by drawing horizontal lines toward the axis of the cylinder and

measuring a distance equal to the long dimension of each step.

If AB is a straight beam (rather than having a curved profile) then it turns

out that the curve we get on the cylindrical wall is only an approximation of

a helix. This may not be obvious, but we work out the formulas below. So

here is the initial question raised in planning for this temporary framework:

What curved profile for the beam AB will cause the curve traced out on

the cylinder to be a true helix? A related question is: If a straight beam is

used instead of a beam with the proper curved profile, will the constructed

curve be a sufficiently good approximation of a helix? In other words, what

problems will stem from using this non-helical curve?

A primary consideration in building a stairway is that the rise (riser

height) and run (tread width) of each step must remain constant – if the

rise or run varies from one step to the next, then climbers might stumble.

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A defining characteristic of a true helix is that it has constant slope along a

(right circular) cylinder, and this allows constant rise and run. On the other

hand, if the stairway were to follow a curve that is not quite a helix, then

the slope along the outer cylindrical surface would not be constant. Thus, if

the risers were held at constant height, then the treads would have to vary in

width. Similarly, if the treads were of constant width, then the risers would

have to vary in height. Either situation would mean trouble for the climbers.

Of course, even with a true helix, the tread width varies along the length

of the tread. Each tread in a helical section is bounded between the outer

cylinder of radius 21 feet and an inner cylinder of radius 16.5 feet (in this

application the tread length is 4.5 feet). The other boundaries of a tread are

along radial lines. Thus, two climbers walking side by side will face different

tread widths. But, as long as a person maintains a constant distance from the

axis of the cylinders, the tread width remains constant and climbing should

be no problem. By the way, as can be seen in Figure 2, the cylindrical walls

in this construction are imaginary and the stairway system is actually in an

open space alongside a cylindrical wall of glass.

Imagine that in preparing for a helical flight, we have drawn on the outer

cylindrical wall (supposing it existed as a solid wall) the outline of the steps

and the helical arc to be followed by the top surface of the stringer. The

outline is determined by the specifications of the rise and run of each step,

the desired number of steps in the flight, and the location of the lower landing.

But now let’s talk about the bottom surface of the concrete beam that is

to support the flight. Ideally, the bottom surface would follow a helix parallel

to the one formed by the steps. That’s why we would like for the beam AB to

have a profile that matches the helix. If, instead, AB were a straight beam,

then the bottom surface of the concrete beam would have the wrong shape.

Of course, even so, we could make the top surface of the stringer have the

correct shape simply by allowing the concrete beam or the stringer to have

varying thickness. Would that work?

Well, because the sides of all the stringers and the concrete beams are

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visible from the stairwell, allowing them to have varying thickness turns

out to be unacceptable. This is so even though the necessary variation in

thickness is quite small (as we shall see).

A mathematical reply

The main goal is to discover the correct profile for a beam AB so that hori-

zontal rays from the axis of the cylinder and passing through points on the

top surface of the beam meet the cylindrical wall in a helix. We choose co-

ordinates so that the (vertical) z-axis is the axis of the cylinder with radius

R = 252 inches (21 feet) and let z = 0 represent ground level. Although the

actual stair system rises in the clockwise direction, we follow mathematical

convention in considering a helix that rises counter-clockwise. Also, let the x-

axis and y-axis be such that the first octant of space contains the entire stair-

way system as suggested by Figure 1. Then the coordinates of the points on

the desired helix for the first helical flight of stairs are x = R cos t, y = R sin t,

and z = ct, where t ∈ [α, β]. Here c is a constant determined by the slope

of the helix and [α, β] is the interval of polar rays that contains the helical

flight. The values of c, α, and β are derived from the building plans.

Any point P = (R cos t, R sin t, ct) of the helix uniquely determines a hor-

izontal line segment QP , where Q = (0, 0, ct) is on the axis of the cylinder.

Where does this line meet the vertical plane containing A and B? We can

take the direction vector of the line QP to be 〈cos t, sin t, 0〉, so using s as

the parameter, we find the line is given byx = s cos t

y = s sin t

z = ct.

(1)

Since A = (R cosα,R sinα, cα) and B = (R cos β,R sin β, cβ), the verti-

cal plane ΠAB containing these points has equation

(sinα− sin β)(x−R cosα) + (cos β − cosα)(y −R sinα) = 0. (2)

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To see this, we just verify that A and B satisfy the equation and note that

the equation does not depend on the z-coordinate of a point.

Substituting (??) into (??), we find that the desired point of intersection

U is x =

[ R sin(β − α)

sin(β − t) + sin(t− α)

]cos t

y =[ R sin(β − α)

sin(β − t) + sin(t− α)

]sin t

z = ct.

Because the line QP radiates from the axis of the cylinder, we call U the

radial projection of the point P of the helix onto the plane ΠAB.

Next we need to compare U with the point V where the line AB hits the

plane ΠP containing P and the z-axis. The equation of this plane is simply

(− sin t)x+ (cos t)y = 0.

And, in parametric form the line AB isx = R cosα + v(R cos β −R cosα)

y = R sinα + v(R sin β −R sinα)

z = cα + v(cβ − cα),

using−→AB as the direction vector. It follows that V is the point given by

x = R[cosα +

[ sin(t− α)

sin(β − t) + sin(t− α)

](cos β − cosα)

]y = R

[sinα +

[ sin(t− α)

sin(β − t) + sin(t− α)

](sin β − sinα)

]z = c

[α +

[ sin(t− α)

sin(β − t) + sin(t− α)

](β − α)

].

While it may not seem likely from the formulas, the x- and y-coordinates

of U must agree with those of V . After all, both points are on the line of

intersection of the vertical planes ΠAB and ΠP . To verify that they actually

do agree just requires the use of the sine difference formula or a computer

algebra system.

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Hence, we focus on the difference in the z-coordinates of V and U . This

is exactly the vertical separation between a point on a straight beam AB

and the corresponding point on a beam with a profile that matches the true

helix. This difference leads to the error in the thickness of the stringer if a

straight beam were used instead of a beam with the proper curved profile.

Thus, in the end, we need to look at the function determined by the difference

between the z-coordinates of V (t) and U(t) as the point P (t) varies on the

helix. This function is given by

f(t) = c[α +

[ sin(t− α)

sin(β − t) + sin(t− α)

](β − α)− t

]= c

[(α− t) sin(β − t) + (β − t) sin(t− α)

sin(β − t) + sin(t− α)

].

Now we put in some data

In order to get a clearer sense of what the function f says about the beam AB,

we put in some data. The carpenter provided information from the building

plans for the lower floors of the building. While the helical staircases for the

upper six floors use 16 risers as shown in Figure 6, the lower floors use 11.

So for the lower floors, 11 risers times 6 916

inches per riser yields an overall

vertical rise of 72.1875 inches. Also, α = 0.3560 and β = 0.9857, so c =

72.1875/(β − α) ≈ 114.64 inches. Using this data, we numerically evaluate

f . In fact, we could derive expressions for the maximum and minimum values

of f in terms of the parameters c, α, and β, but they are quite complicated

and do not seem to shed light on the design of the beam AB. So instead,

using MATLAB, we estimate the maximum and minimum values of f to be

about ±0.4684 inches.

Note that the f values are vertical differences rather than measurements

perpendicular to the line AB. Since AB has an angle of inclination of

θ ≈ 0.4332 radians (computed from the coordinates of A and B), a vertical

difference f(t) corresponds to a value f(t) cos θ perpendicular to the line AB.

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So the maximum and minimum values of f , namely ±0.4684 inches, corre-

spond to values of f cos θ ≈ ±0.425 inches. This means that using a straight

beam AB would cause the concrete beam to vary in thickness by about 0.85

inches in order for the top of the stringer, and therefore the steps them-

selves, to follow the path of the true helix. This variation, being noticable,

was deemed unacceptable.

Figure 3: Notch the main beam to position the radial beams.

On the other hand, such small adjustments in the profile of the beam –

plus or minus a half inch along a beam that is about 172 inches long – could

be accomplished by cutting notches of varying depth in a straight beam in

order to position the radial beams at the proper heights. That is, we wouldn’t

need to carve the entire beam; just let the notch depths match the desired

curved profile. Figure 2 illustrates the idea. The top of the main beam is

located a little below the line AB so that the radial beams can all fit into

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a notch of some positive depth. Each radial beam is topped by a triangular

wedge to provide a surface for attaching the plywood ramp. Of course, many

such issues belong in the domain of the carpenter, not the mathematician.

In order to make a picture of the curved profile of the beam, we want

to view f as a function of linear distance along the line segment AB rather

than a function of the polar angle t. The distance from A along AB is given

by

172v(t) =172 sin(t− α)

sin(β − t) + sin(t− α),

where 172 inches is the distance from A to B. With this we can use MAT-

Figure 4: Vertical displacement along AB.

LAB to plot f(t) versus 172v(t) as in Figure 4. Obviously, the vertical and

horizontal scales here differ from each other in order to accentuate the fact

that the beam is not straight.

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Back to reality

Back in the real world, the carpenter actually avoided the problem that he

posed – he used a different approach entirely. In the photograph in Figure

5, we see how the radial beams are supported independently by vertical

columns. What they did was to build a temporary deck at the level of each

main landing. Then, vertical columns standing on this deck were used to

support the framework and the forms into which the concrete was poured.

They did not use a main beam AB. Nonetheless, I give the carpenter credit

for seeing and caring about the underlying mathematical question that he

raised.

Acknowledgment. I wish to thank the carpenter, Bob Nelson, for bringing

this problem to my attention. The helical stairway now stands in American

Family Children’s Hospital, University of Wisconsin (Madison), Builder: J

H Findorff; Carpenters: Bob Nelson, Jamie Smith and Brian Heble. I also

thank the referee for suggestions that improved the clarity of the paper.

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Figure 5: Actual support structure. (Photo by Bob Nelson.)

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Figure 6: Nearly finished stair system. (Photo by Bob Nelson.)

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