Skip to main content
Search IntMath

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.

Archimedean Spiral
Watch mechanism [Image source: Hamilton Clock]

Archimedean Spiral
Coiled rope [Image source: Delderfield]

24x7 Tutor Chat

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
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
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,


r is the distance from the origin,
a is the start point of the spiral and
b affects the distance between each arm.
(b is the distance between each arm.)

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

For the first spiral,

b = 0.81, giving us b = 0.12892

So the formula for the first spiral is

r = 5 + 0.12892θ

Using the same process,

b = 1.4, giving us b = 1.4/(2π) = 0.22282

So the formula for the second spiral is:

r = 5 + 0.22282θ

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.

24x7 Tutor Chat

See the 60 Comments below.

Leave a comment

Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

  1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
    `a^2 = sqrt(b^2 + c^2)`
    (See more on ASCIIMath syntax); or
  2. Use simple LaTeX in the following format. Surround your math with \( and \).
    \( \int g dx = \sqrt{\frac{a}{b}} \)
    (This is standard simple LaTeX.)

NOTE: You can mix both types of math entry in your comment.


Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.