# Length of an Archimedean Spiral

By Murray Bourne, 21 Sep 2011

Reader Anantha recently wrote asking the following interesting question:

To find the total length of a flat spiral having outer end radius = 15.5 units, inner radius = 5 units & the increase in radius per turn = 0.81 unit, the total No. of turns in the spiral is 7.5.

This is an example of an **Archimedean Spiral**, otherwise known as an **arithmetic spiral**, where the arms get bigger by a constant amount for each turn.

We can see Archimedean Spirals in the spring mechanism of clocks and in vinyl records (used by the recording industry before CDs and MP3s); and in tightly coiled rope.

Watch mechanism [Image source: Hamilton Clock]

Coiled rope [Image source]

## However, her question can’t be solved

This was another example of an impossible-to-solve reader question. (Some of them are not possible to solve either because there is not enough information given, the algebra is unreadable, or some key vocabulary is not used correctly.)

In this case, it’s not possible because it has a fundamental flaw.

If the inner radius is 5 units and the increase in radius per turn is 0.81 units, then 7.5 turns will give us an outer radius of:

5 + 0.81 × 7.5 = 11.075

So the outer radius cannot be 15.5 units if the increase per turn is 0.81.

Here’s what the spiral with 0.81 radius increase per turn looks like:

Archimedean spiral, 0.81 units between each arm;

outer radius is 11.075

On the other hand, if is it important to have the outer radius being 15.5 units, then the increase per turn would need to be:

(15.5 − 5)/7.5 = 10.5/7.5 = 1.4

Here’s the spiral with inner radius 5, outer radius 15.5 after 7.5 turns:

Archimedean spiral, inner radius 5, outer radius 15.5;

distance between each arm is 1.4 units

The increase per turn is 1.4 units.

## Finding the Length of the Spiral

Before we can find the length of the spiral, we need to know its equation.

An Archimedean Spiral has general equation in **polar coordinates**:

*r = a + bθ,*

where *r* is the distance from the origin, *a* is the start point of the spiral and *b* affects the distance between each arm. (The total angle turned by the spiral in radians is given by 2π*b*, so the distance between each arm is 2π*b* divided by the number of turns.)

For both spirals given above, *a*= 5, since the curve starts at 5.

For the first spiral,

2π*b = *0.81, giving us *b* = 0.12892

So the formula for the first spiral is

*r = *5* + *0.12892*θ*

Using the same process, the formula for the second spiral is:

*r = *5* + *0.22282*θ*

## Length of the first spiral

We’ll use the formula for the Arc Length of a Curve in Polar Coordinates to find the length.

The starting value for θ is *a = *0 and after 7.5 turns, the end point is *b * = 7.5 × 2π = 15π = 47.12389.

The derivative under the square root is straightforward:

Substituting everything into our formula gives us:

You can see where that final answer comes from in Wolfram|Alpha. It may be possible to find the actual integral on paper (it involves hyperbolic functions), but why waste our lives doing so? We are interested in the length, not pages of algebra!

## Length of the second spiral

I’ve solved the second one as an example in the section Arc Length of a Curve in Polar Coordinates.

So Anantha, I hope that helps to answer your question.

See the 23 Comments below.

21 Sep 2011 at 5:23 pm [Comment permalink]

Excellent .

These plane geometry equations are simple and beautiful .

22 Sep 2011 at 1:10 pm [Comment permalink]

The explanation is very clear.

30 Apr 2012 at 8:54 am [Comment permalink]

Hi, thanks for this explanation.

I’m trying to work out the number of turns for a given length of spiral, (I’m an engineer for a garage door company, need to know turns to specify springs), I don’t suppose you could give me some pointers getting N when I know L? I could use Solver in excel but I need to do it hundreds of times.

Brent.

1 May 2012 at 8:59 pm [Comment permalink]

@Brent: You can’t solve it by any straightforward algebraic or calculus process – not that I can see, anyway.

Is your distance between each spiral always constant?

If so, I’m wondering if you can do it by getting a good approximation, then do a few actual calculations to zero in on the number of turns required.

For example, in the example in my post, the “average” radius of all the spirals is around 10 (1/2 way between 5 and 15). We know the length is 379, so the number of turns would be about 379 / (2 * pi * 10) = 6. This misses by quite a bit.

But a better “average” radius is 8 (obtained by working backwards from the known information), and this gives us the number of turns 379 / (2 * pi * 8 ) = 7.5.

Let’s try 2 other lengths and see how we go, and I’m assuming our distance between spirals is constant at 1.4.

(1) Length 250 units. The average radius will be less – let’s guess (for now) 7.5.

The number of turns will be 250 / (2 * pi * 7.5) = 5.3

Using this and the integration formula, gives 238, which is not bad for a reasonable guess. A small tweak or 2 would get you there.

(2) Going the other way, let’s see what happens with length 500.

The average radius will be more, say 8.5.

Number of turns: 500 / (2 * pi * 8.5) = 9.4 turns

Applying the length formula, gives 520, which is once again pretty close. A few tweaks and we’ll be very close.

Now, I don’t know whether that would save you any work compared to what you are doing now. But maybe if you created a table with a bunch of “good” guesses and the resulting number of turns, and then used some kind of linear (or probably non-linear) interpolation, you would get close each time for a given length.

Does that help or am I describing what you’re already doing?

2 May 2012 at 8:26 am [Comment permalink]

Thanks for your reply Murray,

I’ll describe what I ended up doing. I looked at the integral on the wolfram link in your blog post, saw it was a pretty regular shape – looks like a rectangle of width 2.pi.n and height average of top and bottom points. I put all this into an Excel table and used a data table to calculate length for n=1,2..10.

Then I graphed n-L and found that there’s a quadratic trendline with and r^2=1 fit. So I used the linest function in excel to estimate the constants in the quadratic equations based on my table of n and L values, and used that to drive the rest of the excel table which calculates the required springs. (I used the quadratic equation formulae to find the roots of the quadratic. 5 years at uni learning maths, used it once now in 9 years of engineering…) I needed to have it in spreadsheet because I have to do this for every height and width combination of the door.

I couldn’t have done any of that without your blog post. Thanks again. Brent.

2 May 2012 at 12:01 pm [Comment permalink]

Glad you got it to work, Brent.

And you’re not the first engineer who has pointed out the lack of utility for much of the math learning at university. Of course, some engineers have got to know some of it, but maybe we need to re-think what is contained in the compulsory math components of engineering degrees.

2 May 2012 at 2:16 pm [Comment permalink]

The thing about learning all the maths that you do in an engineering degree is not that you’ll necessarily do all the things in your career, but that some of the subjects you’ll do at uni require it. If you don’t do the maths that lead up to it, then you’re never going to be able to cope with thermodynamics or fluid mechanics. And you do need to be able to go through some of the maths in those subjects to truly say you understand the basic concepts.

The trouble is that there are simply so many fields that an engineer will go into, and you need to give an understanding of the fundamentals of just about everything: thermodynamics, fluid mechanics, structures, kinematics, control, electrical, materials etc etc etc. I might not have used the higher level maths much since leaving uni, but I’ve definitely used my understanding of all those subjects on a regular basis, and I gained my understanding by going through the maths.

What I would say to my 21 yr old self if I could: keep better track of your notes and text books, because it might be a decade before you need that information again but you will kick yourself if it’s lost forever.

4 May 2012 at 7:50 am [Comment permalink]

Brent,

I hope this helps. It is a couple days late. If you know your overall length (OL), ID of the spiral, and OD of the spiral then the equation for the number of turns (N) is:

N = OL/(PI*(ID+OD)).

Also, on the original posting with the spacing of .81 units, the OD of the spiral should have been 11.075 units not 10.67 units. I think you solved for 7 turns, not 7.5 turns.

5 May 2012 at 12:54 pm [Comment permalink]

Thanks for your input, Neil.

Hmmm – seems a lot simpler that what Brent and I were talking about!

Thanks for spotting the error – I have amended the post.

7 May 2012 at 7:19 am [Comment permalink]

Neil,

Thanks a lot for the information.

As it turns out for a garage door I don’t know the OD – it’s going to change depending on N.

8 Nov 2012 at 6:08 pm [Comment permalink]

This is a bit late, but new readers may still find it usefull.

Another way to make a quick approximation is to take the average For the first spiral in the article, inner radius is 5, and outer radius is ~ 11. The average radius is then (5+11)/2 = 8. The circumference of a circle with radius 8 is 2*pi*8 = 50.27. 7.5 turns of that length is 8*50.27 = 377.25. This, IMHO at least, is reasonably close to the ‘exact’ 378.8.

No matter how you look at it, you need three of ID, OD, N and L to compute the fourth – you cannot ‘just’ deduce N from L if you dont know ID and OD. But knowing ID, OD and L, you can quickly make a good estimate based on circles with diameter (ID+OD)/2.

It is a little different if you have the increase per turn. Say you have a roll of tape and know ID, tape thickness t and tape length L. You can then use the following formula to compute the number of turns (or layers, if you will):

N = (t – ID + sqrt( (ID-t)^2+(4*t*L/pi) )/(2*t)

and from this you can also compute the outer diameter OD

OD = 2*N*t+ID

Note, that both of the above formulae are approximations based on concentric circles, rather than a true spiral. However, the end result is reasonably close. Also note that it is diameters, not radii, you need to plug in!

9 Nov 2012 at 11:36 am [Comment permalink]

@Jacob – Thanks! We get closer and closer to a good approximation with each new mail.

25 Jan 2013 at 9:40 pm [Comment permalink]

Hi thanks a lot for your highly explanatory blog-entry! I still tried to solve the integral and after petty substitution I ended up with an integral of the form sqrt(x^2+1) … which can be solved relatively easily once you know the trick (see http://www.matheboard.de/archive/30730/thread.html for example) … however, in the blogpost from “wolfram” the integral is given by a much more complex formula (http://mathworld.wolfram.com/ArchimedesSpiral.html ) … now I was wondering if you have any idea where my error lies with the relatively simple method? best regards and please keep up the good and informative posts

15 Apr 2013 at 1:58 am [Comment permalink]

Ok I’m trying to solve something for a fiction story and it’s been years since I’ve done calculus. So maybe someone can solve this for me. If you wanted to travel to the galactic core from Earth you would want to travel down the spiral arm and not through the less dense regions if you needed to resupply. So Earth is about 7900Kpc from the galactic core. The start of the spiral arm at the end of the galactic bar is about 3000Kpc from the core. I do not know the distance between the arms. What I’m trying to figure out is how far you would travel down the spiral arm to get to the galactic bar. Any help would be greatly appreciated. Thanks.

16 Apr 2013 at 12:10 pm [Comment permalink]

Hi Ray

I think the dotted magenta distance is what you mean from your description. Before we go any further, is it correct?

[Image source: UniverseToday]

16 Apr 2013 at 12:13 pm [Comment permalink]

I think for something with the precision like this you could use the pythagorean approximation just fine.

27 Apr 2013 at 11:13 pm [Comment permalink]

Hello,

I’m not a mathemetician, but I have an archimedian spiral question too.

I would like to create a plastic spiral for developing long lengths of cine film. If the film length is 15 metres (15000mm), and the inner radius of my spiral should be 20mm, and the outer radius maybe 150mm. Ideally I would like about 2mm spacing between the winds. How many turns would there be?

The spiral will always need to be 15metres long, but I might like to experiment with different radii, spacing and number of turns etc. is there a formula that helps me do the maths for this please?

With the answer, I should be able to draw the correct spiral in Adobe Illustrator, and from there, create the plans needed to have the spiral made out of plastic…

Many thanks if anyone can help.

4 Oct 2013 at 11:09 pm [Comment permalink]

If spiral equation is r = a + b*Theta, where a is start radius

Spiral length is L = (1/b)Integral(sqr(r^2 + b^2))dr

Solution:

L = .5*[r*sqr(r^2 + b^2) + b^2 * log(r + sqr(r^2 + b^2))]/b

In Excel, in the VB code pages add a module and enter following code:

Option Explicit

Const cPi As Double = 3.14159265358979

Public Function LSpiral(ByVal Theta As Double, Pitch As Double, Start As Double)

Dim b As Double

Dim r As Double

b = 0.5 * Pitch / cPi

r = Start + b * Theta

LSpiral = Integral(r, b) – Integral(Start, b)

End Function

Private Function Integral(x As Double, b As Double)

Dim Tmp As Double

Tmp = Sqr(x ^ 2 + b ^ 2)

Integral = 0.5 * (x * Tmp + b ^ 2 * Log(x + Tmp)) / b

End Function

Then in the spreadsheet have 3 cells for Theta, Pitch and Start – you could have a cell for Number of turns and get theta from Turns * 2Pi

You can play with values to get the length you want

6 Oct 2013 at 11:11 am [Comment permalink]

@Robert – thanks for the Excel-based approach.

10 Jan 2014 at 11:41 pm [Comment permalink]

hi

I am a co op student and i am trying to make a roll of scotch tape in NX 7.5. I am attempting to use Law Curve to create a spiral line for an extrude to follow. how does the math (r = a + b?) equation or formula flow to create this? the roll needs to be roughly 15 mm thick and the total radius is 32.2 mm.

13 Jan 2014 at 5:16 pm [Comment permalink]

?Hello R.A.

This sounds like an interesting problem. Just to clarify a few things:

(1) What does “NX 7.5″ mean?

(2) Does that mean the inner radius is 17.2 mm (since the roll itself is 15 mm thick and the total radius is 32.2?)

(3) How thick is the scotch tape? (I’ll assume 0.07 mm, which I found here: http://www.tedpella.com/tape_html/tape.htm)

With these assumptions, we have

a = 17.2 (the inner radius)

b = (0.07 * number of turns)/ (2 pi)

So all together, the formula for your spiral for an angle theta would be:

r = 17.2 + ( (0.07 * number of turns)/ (2 pi)) theta

Does that help to get you started?

23 Apr 2014 at 7:45 pm [Comment permalink]

Hello Mr. Murray,

R.A means nx7.5 software. My query is similar. I want to draw a arithmatic spiral with the help of law curve in nx7.5. Pl. help. Thanks.

24 Apr 2014 at 10:56 am [Comment permalink]

@Singh: Thanks for the heads up about NX7.5. It looks like great software.

You just need to use the same formula, r = a + b θ, as explained in the article. You can use t instead of θ.