# How to calculate curve length of a helix with linearly variable pitch?

I have designed a 3D helix which has a variable pitch $$P(z)$$ which is defined over the axial axis $$z$$ and has the form $$P(z)=a*z+b$$, $$a$$ and $$b$$ are constants. The larger the pitch, the sparser the helix should be at that part. It's like a wave which has variable wavelength. The radius of the helix is $$r$$. The total helix height (in $$z$$ direction) is $$L$$.

Since the helix has a linearly varying pitch, this spring-like helix will become more and more "dense" when $$z$$ increases if pitch decreases with $$z$$.

The question is how to express the helix length using all the aforementioned parameters?

I have a preliminary form, but this does not meet the result that I measured in the 3D CAD software.

• In your picture $x$ is the direction of the amplitude, but only the frequency increases (or the wavelength decreases reciprocally), as $z$ increases. Can you please clarify what you mean. For example, draw a longer segment. Commented Dec 12, 2018 at 15:05
• @JyrkiLahtonen Sorry for the confusion, it's indeed not clear. If using the coordinate system in the figure, the pitch function will be $P(z)=az+b$. And this shape is a 3D helix (like a spring) which becomes more and more "dense" when $z$ increases. This "density" is controlled by pitch $P(z)$. I will edit the question to clarify. Commented Dec 12, 2018 at 15:12
• Is your pitch the distance between turns (the usual definition) or the inverse, the number of turns per unit distance?. You talk about the spring getting denser as $z$ increases, while in the standard definition it should get less dense. Commented Dec 12, 2018 at 15:37
• @RossMillikan The pitch has the former definition. That is it's similar to a wavelength. And I think whether it's becoming denser or sparser depends on the sign of $a$ in the definition of $P(z)$. Commented Dec 12, 2018 at 16:06

Parametrize the helix as

$$\begin{eqnarray} x(t) &=& r \cos (2\pi t) \\ y(t) &=& r \sin (2\pi t) \\ z(t) &=& at + b \end{eqnarray}$$

The length of the helix is

$$L = \int_{t_1}^{t_2}\left[ \left(\frac{{\rm d}x}{{\rm d}t} \right)^2 + \left(\frac{{\rm d}y}{{\rm d}t} \right)^2 + \left(\frac{{\rm d}z}{{\rm d}t} \right)^2\right]^{1/2}{\rm d}t = \int_{t_1}^{t_2}[b^2 + 4\pi^2 r^2]^{1/2}{\rm d}t = \sqrt{b^2 + 4\pi^2 r^2}(t_2 - t_1)$$

• The way I read the question gave me the impression that the "pitch", i.e. the frequency, varies linearly. So your "constant pitch" $=2\pi$ should actually increase linearly as a function of $t$. Admittedly the OP is unclear, because the say that the pitch should vary as a function of $x$, but in their image it varies as a function of $z$. Commented Dec 12, 2018 at 15:02
• @JyrkiLahtonen You're right, it is a bit confusing. I will delete the answer if this is not what the OP had in mind Commented Dec 12, 2018 at 15:03
• Hi, thanks for the reply. I have difficulty in understanding $z(t)=at+b$. It seems to me that $z$ is related to pitch $P(x)$ through integration: $\Delta t*P(z)=\Delta z$. Would you elaborate on it a little bit? Commented Dec 12, 2018 at 15:06
• Your solution is for a constant pitch.
– user65203
Commented Dec 12, 2018 at 16:36

Just served as some inspirations for others, I posted my derivation process. Maybe someone can help to review the process? I hope this is counted as en eligible answer.

My derivation process is like the following:

$$\frac{dz}{d\theta}=\frac{P(z)}{2\pi}$$ so that: $$\theta=2\pi\int_0^z\frac{1}{P(z)}dz+const$$

Consider $$\theta_{z=0}=0$$:

$$\theta=2\pi\int_0^z\frac{1}{P(z)}dz=\frac{2\pi}{a}\log(az+b)$$

Thus $$z$$ can be expressed by $$\theta$$:

$$z=\frac{e^{\frac{a\theta}{2\pi}}-b}{a}$$

For $$x$$ and $$y$$:

$$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$

And the helix length is:

$$\begin{equation*} L_{helix}=\int_{\theta_1}^{\theta_2}\left[\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2+\left(\frac{dz}{d\theta}\right)^2\right]^\frac{1}{2}d\theta \end{equation*}$$

The upper and lower boundary of $$\theta$$ can be decided by plugging boundary conditions for $$z=0$$ and $$z=L$$ into $$\theta=\frac{2\pi}{a}\log(az+b)$$

So the final form before solving the integral is:

$$\begin{equation*} L_{helix}=\int_{\frac{2\pi}{a}\log{b}}^{\frac{2\pi}{a}\log{(aL+b)}}\left[r^2+\left(\frac{1}{2\pi}e^{\frac{a\theta}{2\pi}}\right)^2\right]^\frac{1}{2}d\theta \end{equation*}$$

This can be solved in some software. To me, I think this first equation $$\frac{dz}{d\theta}=\frac{P(z)}{2\pi}$$ is the key one. I'm not sure its correctness. I get it from intuition.

The pitch is the derivative of the $$z$$ position on the angle parameter, hence the element of arc is

$$\sqrt{(at+b)^2+r^2}\,dt,$$

which integrates as follows:

https://www.wolframalpha.com/input/?i=integrate+sqrt((az%2Bb)%5E2%2Br%5E2)+dz

• Hi, I don't quite get the pitch is the derivative of the z position part. You mean it's the derivative of $z$ in terms of which variable? Would you elaborate on that? Commented Dec 12, 2018 at 16:09
• @ZhangZe: I was wrong, it's in terms of $t$ where $t$ represents the rotation angle. The $z$ value is a quadratic function of $t$.
– user65203
Commented Dec 12, 2018 at 16:32