Jtdylktuy

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.

Parametrize the helix as

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

The length of the helix is

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

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:

$$frac{dz}{dtheta}=frac{P(z)}{2pi}$$
so that:
$$theta=2piint_0^zfrac{1}{P(z)}dz+const$$

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

$$theta=2piint_0^zfrac{1}{P(z)}dz=frac{2pi}{a}log(az+b)$$

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

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

For $x$ and $y$:

$$x=rcos{theta}$$
$$y=rsin{theta}$$

And the helix length is:

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

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

So the final form before solving the integral is:

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

This can be solved in some software. To me, I think this first equation $frac{dz}{dtheta}=frac{P(z)}{2pi}$ 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