Roller Door problem
By Murray Bourne, 28 Nov 2012
In an earlier article, I discussed the Length of an Archimedean Spiral. We saw we can find the length of such a spiral using integration, once we know the equation of the spiral and the beginning and end points. The equation for an Archimedean Spiral is in polar form, and is determined by an angle θ, the amount of turn of the spiral.
[Image source]
Brent, an engineer with a roller door company, asked an interesting question in the comments for that article. He posed a "reallife" issue he faced.
His problem is the reverse. He knows length of the spiral already (it's the height of the roller door, plus the bit that's wrapped around the drum wheel), but he needs to know the number of turns of the drum for the given height of the door and radius of the drum.
This is not such a straightforward problem.
I suggested he could produce a table for already known door heights and number of turns. and then use interpolation to find the required number of turns (at least approximately).
Brent also turned to PhysicsForums to ask for help. He wrote:
Hi,
I'm an engineer designing a spring system for a garage roller door. I need to know the number of turns of the door for all the size combinations.
I've found this page which gives a good equation for finding the length if you know the number of turns, starting radius and gap between spirals:
The equation of the spiral is r = x + yθ, with x = starting radius, y = gap/(2π), and to find L we're taking the integral from a = 0 to b = 2πn (where n = turns).
When you know n, this is straightfoward, and even I could work that out. But it's been a decade since I've done anything like this, so I was wondering if anyone could help me solve for n in this integral:
My way of solving this is to find L for n=1,2,3,4,5 etc, graph it in Excel and use "find trendline" to get an equation. Any help appreciated, thanks.
The response there was not very helpful.
Brent provided his Excel spreadsheet so I could see what he was doing.
Specific Example
Brent's roller door has a barrel radius of 127.5 cm and the thickness of the door slats (called the "curtain") is 4.0 cm. This means the radius will increase by 4.0 cm for each turn of the barrel.
So x = 127.5 and y = 4/(2π)=0.6366 in this case.
The following daigram (not to scale) indicates our situation:
Applying the integration formula above, we have the length of the curtain wrapped around the barrel for 1 turn:
Similarly, for 2 turns (that is, 0 to 4π), we have:
Continuing on for 3 up to 6 turns, we obtain the following values:
Turns  Length 

0

0

1

814

2

1654

3

2516

4

3406

5

4320

6

5259

Next, we use Excel to plot these values and to draw a trendline through them. We choose "polynomial of degree 2" because a consideration of the integral and an inspection of the Excel graph (when extrapolating beyond 6 turns) shows it is parabolic.
Excel can tell us the equation for the graph produced, and it is included on the chart.
We need to translate the "y" and "x" variables on Excel's chart for our own situation. The vertical scale is the length of the curtain, and the horizontal scale is the number of turns. Also, the "0.1667" value is a rounding error, which we can discard.
So in our case, we would write the Length of the curtain (L) for the number of turns, n as:
Solving for n (using the quadratic formula) gives us 2 solutions. The first one is:
The second solution gives a negative result for n, so we discard it.
We now check our work (by comparing a known value).
If the length is 3406 cm, then the number of turns given by our new formula is:
This is certainly close to the 4 turns we expected.
So Brent could now use this formula to find the number of turns for any roller door's curtain length given a barrel radius of 127.5 cm and a curtain thickness of 4 cm.
Linear Solution
The Excel graph for the range of turns 0 to 6 is very nearly linear. Using the "Linear" trendline option in Excel gives the model for the Length given the number of turns as:
L = 876.5n  62.4
Solving for n gives:
Applying this simpler formula to some of our known values gives:
For length L = 814, number of turns, n = 0.99989. (expected 1)
For length L = 1654, number of turns, n = 1.9582. (expected 2)
For length L = 5529, number of turns, n = 6.0712. (expected 6)
So for lengths of the roller door curtain less than about 10 m, this simpler linear model gives acceptable results.
Other solutions
Neil provided another solution in the comments.
He proposed:
where
n = number of turns
OL = overall length
ID = inner diameter of the spiral
OD = outer diameter of the spiral
However, Brent pointed out this can't be used because he doesn't know the outer diameter until he knows the number of turns.
Jacob then weighed in with another proposal, which involved taking an average of the inner and outer radius and then applying the circumference of a circle to that average.
Jacob suggests if we don't know the number of turns, we can use this formula to find it:
Applying that formula to our example above, with y = 4 and ID = 127.5×2 = 255 and our known case where L = 3406 gives:
This is very close to the 4 turns we expected.
Applying the formula for another known case, L = 5259, we obtain 6.08 turns, nicely close to the expected value of 6.
Conclusion
Brent has a choice of formulas to use, all of which give acceptable results. The linear model is the easiest.
Parabolic model:
Linear model:
The model suggested by Neil and refined by Jacob:
Here is a table that shows the values from the given formulas.
Length (via calculus) 
Turns (original) 
Turns (parabolic model) 
Turns (linear model) 
Turns (Neil/Jacob model) 

0  0  0  0  0 
814  1  1.0001  0.9999  1.0158 
1654  2  2.0014  1.9582  2.0318 
2516  3  2.9992  2.9417  3.0431 
3406  4  4.0004  3.9571  4.0571 
4320  5  5.0005  4.9999  5.069 
5259  6  6.0006  6.0712  6.0802 
See the 9 Comments below.