how to define a circular helix in terms of the Frenet Frame Suppose $\alpha: I \rightarrow \mathbb{R}^3$ is a regular curve with unit speed which is also a circular helix. How can we define this carefully in terms of the standard $T,N,B$ frame along the curve ?
I have a few ideas, but I want a second or third opinion. This is to help with a particular homework problem I'm working through in Barrett O'Neill's Elementary Differential Geometry text. A different definition in terms of curvature and torsion is given. I'm not interested in that definition as the whole point of the problem is to show the constant curvature and torsion imply circularity of the given helix. 
Thanks in advance for your help, I hope check this in a couple hours.
 A: Following the carifying comments above, let's go with the second definition. The idea there is that the curve's a circular helix if there's an axis ($P + td$) such that the projection of the curve onto the plane perpendicular to the axis traverses a circle of radius $A$ with speed $AB$ in that plane. That seems to capture circular helix-ness. 
Now suppose you have a curve with constant curvature and torsion. To prove it matches the definition, you need to come up with the axis and the constants $A$ and $B$. Well, if you think of a model circular helix around the $z$-axis, say, the normal vector always lies in the $xy$-plane. So you could "discover" the axis by taking any two nearby normals and computing their cross-product. Of course, the more nearby they are, the shorter the cross-product vector will be, so to get a nice unit vector for defining the axis, I'm implicitly suggesting that you say this:
Let $$\vec{d} = \lim_{h \to 0} \frac{1}{h} N(h) \times N(0)$$.
You can rewrite that as 
\begin{align}
\vec{d} &= \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \times N(0) \\
 &= 
\left( \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \right)\times N(0) \\
&= N'(0) \times N(0)
\end{align}
Now you can use the Frenet-Serret formulas to work out an explicit value for the vector $\vec{d}$. (Note: The result won't generally be a unit vector, but it'll have nonzero length, and then you can make it a unit vector; without the division by $h$, you'd end up with a zero vector.)
Going back to the model of a circular helix around the $z$-axis, you should be able to infer the radius of the circle from the constant curvature and torsion. (Should be 1/curvature, I believe, but I'd have to check whether torsion somehow sneaks in there). That lets you pick a point $P$ at which to start the axis: 
$$
P = \gamma(0) + r N(0)
$$
where $\gamma$ is your curve, and $r$ is the radius you computed. For vectors $\vec{e}$ and $\vec{f}$, I'd recommend $N(0)$ and $\vec{d} \times N(0)$. Now all you have to do is show that everything matches up as needed. 
Since you didn't ask for a complete proof, but instead for a definition of circular helix that would allow you to prove that constant-curvature-and-torsion curves were indeed circular helices, I think I've given you what you asked for. I can fill in details if necessary, but that might take away the fun from you. 
Following comments and discussion
Here's a more complete answer that depends only on the curve $\alpha$, and its Frenet frame. Since comments reveal that you're teaching (or taught) a DG course, I'll let you do the computations here, and just post the main results. 


*

*The helical curve 
$$
H(s) = (r \cos us, r \sin us, v)
$$
has tangent vector 
$$
T(s) = (-ru \sin us, ru \cos us, v)
$$
which is a unit vector only if 
$$
r^2 u^2 + v^2 = 1.
$$ 
So I'll consider only triples $(r, u, v)$ that satisfy that condition. I'll also restrict to $r > 0$. 


The derivative of $T$ is 
$$
T'(s) = (-ru^2 \cos us, -ru^2 \sin us, 0)
$$
whose length is $ru^2$; by the Frenet formulas, this is the curvature, $\kappa$, of the helix, and the normal vector to the helix is 
$$
N(s) = (-\cos us , \sin us, 0).
$$
The derivative of the normal is 
$$
N'(s) = u(\sin us, -\cos us, 0). 
$$
The binormal is
$$
B(s) = T \times B = (v \sin ut, -v \cos ut, ru).
$$
The torsion, by Frenet, is $\tau = N' \cdot B = uv$.
So for such helical curves, we have
\begin{align}
ru^2 &= \kappa \\
r^2u^2 + v^2 &= 1 \\
uv &= \tau.
\end{align}
We can solve these for $r, u, v$ to get
\begin{align}
u &= \sqrt{\kappa^2 + \tau^2}\\
v &= \frac{\tau}{\sqrt{\kappa^2 + \tau^2}} \\
r &= \frac{\kappa}{\kappa^2 + \tau^2}
\end{align}
Thus for any positive $\kappa$ and any $\tau$, we can find a unit-speed helical curve $s \mapsto H_{\kappa, \tau}(s)$ by using the values of $u, v, r$ above in $H$. 
Now let's look at the constant-speed, constant torsion curve $\alpha$. Let $Q = \alpha(0)$, and let $u, v, r$ be derived from the known curvature and torsion as above.
$$
\newcommand{bT}{\mathbf T}
\newcommand{bN}{\mathbf N}
\newcommand{bB}{\mathbf B}
\newcommand{Tb}{ T_\beta}
\newcommand{Nb}{ N_\beta}
\newcommand{Bb}{ B_\beta}
\newcommand{bD}{\mathbf D}
\newcommand{bE}{\mathbf E}
$$ 
Let $\bT, \bN, \bB$ denote the unit tangent, normal, and binormal of $\alpha$ at $s = 0$. 
Define
\begin{align}
\beta(s) &= (Q + r\bN) - r\cos(us)\bN + r\sin(us) \bE + vs \bD, \text{ where}\\
\bD &= \frac{1}{\|\bN \times N'(0)\|} \bN \times N'(0)\\
&= \frac{1}{\|\bN \times (-\kappa\bT + \tau \bB)\|} \bN \times  (-\kappa\bT + \tau \bB)\\
&= \frac{1}{\|\kappa\bB + \tau \bT)\|}  (\kappa\bB + \tau \bT)\\
&= \frac{1}{\sqrt{\kappa^2+ \tau^2}}   (\kappa\bB + \tau \bT), \text{ and}\\
\bE = \bN \times \bD,
\end{align}
which you can work out in terms of $\bT, \bN,$ and $\bB$ similarly. 
Then direct computation shows that 


*

*$\beta$ and $\alpha$ start at the same point. 

*$\beta$ has constant curvature $\kappa$.
3, $\beta$ has constant torsion $\tau$. 
By the fundamental theorem of curves, $\beta$ must equal $\alpha$. Bur $\beta$ is evidently a helix with centerline $\bD$. (Indeed, it's the the result of rotating the "standard helix" $H(\kappa, \tau)$ by the matrix whose columns are $\bT, \bN, \bB$, and then translating by  $(Q + r\bN)$.)
A: I know that the question has already been answered, but I found a definition in John Oprea's book which might be useful. You have your curve $\alpha: I \to \Bbb R^3$, parametrized by arc-length, say. If $\alpha$ is a cylindrical helix, exists a constant unit vector $\bf v$ such that $\langle {\bf T}_\alpha(s), {\bf v} \rangle = \cos \theta $. Define $\gamma: I \to \Bbb R^3$ by: $$\gamma(s) = \alpha(s) - \langle \alpha(s) - \alpha(s_0), {\bf v}\rangle {\bf v}$$
for some $s_0 \in I$. Notice that $\gamma$ need not be parametrized by arc-length, despite its parameter being $s$. The curve $\gamma$ is the projection of $\alpha$ into the plane orthogonal to $\bf v$ passing through $\alpha(s_0)$. You can prove, for example, that $\kappa_\gamma = \kappa_\alpha/\sin^2\theta$. We say that $\alpha$ is a circular helix if $\gamma$ is a circle. 
A: I want to go out on a limb here, and make some observations for which I don't have proofs, but which seem likely to be true, and seem more like what James was looking for. 
Go back to the "standard helix" I defined in the first answer (which I'll now call $\alpha$). Suppose at each "time" $s$, you look at the frame $T(s), N(s), B(s)$ at the origin. What are the coordinates of $\alpha(s)$ in this (ever changing) affine coordinate frame? [Credit to Tom Banchoff for teaching me to ask this question!]
First of all, the $N$ coordinate is constant! (it's exactly $-r$). 
Second, the $B$ and $T$ coordinates are linear functions of $s$, with the ratio of slopes being $\tau:\kappa$, where $\tau$ and $\kappa$ are the torsion and curvature of the curve $\alpha$, of course. 
I suspect (and will perhaps try to prove) that if for an arclength-paramaterized curve $\alpha$, there is a point $P$ amd constants $c_1, c_2, c_3$ such that 
$$
\alpha(s) - P = c_1 s T(s) + c_2 s B(s) + c_3 N(s)
$$
then $\alpha$ is a helix in the sense I described earlier, with $P$ being the point of the axis closest to $\alpha(0)$ and with direction vector $N'(0) \times N(0)$, and with constant curvature and torsion in the same proportion as $c_2 : c_1$, and with radius $-c_3$. 
$$ 
\newcommand{nothing}[1]{}
\nothing{abc}
$$
$$\nothing{To do so, let me follow an idea Tom Banchoff taught me: that expressing a curve in the ever-changing Frenet coordinate system actually might be useful, and that once you've done this, you should differentiate your heart out. I'm sorry it took so long for me to realize it might have some use in this problem. 
Well,
\begin{align}
\alpha(s) - P &= c_1 s T(s) + c_2 s B(s) + c_3 N(s) \text{, so} \\
T(s) = \alpha'(s) 
&= c_1 T(s) + c_2 B(s) + s(c_1 T'(s) + c_2 B'(s)) + c_3 N'(s) \\
&= c_1 T(s) + c_2 B(s) + s(c_1 \kappa N(s) - c_2 \tau T(s)) + c_3 (-\kappa T(s) + \tau B(s)) \\
&= (c_1 -sc_2 \tau -\kappa c_3) T(s) + sc_1 \kappa N(s) + (c_2 + c_3 \tau) B(s) 
\end{align}
}
$$
The proof is actually surprisingly simple: differentiate the formula for $\alpha(s)$:
\begin{align}
T(s) &=\alpha'(s) \\
&= c_1 T(s) + c_2 B(s) + s(c_1 T'(s) + c_2 B'(s)) + c_3 N'(s) \\
&= c_1 T(s) + c_2 B(s) + s(c_1 \kappa N(s) - c_2 \tau N(s)) + c_3 (-\kappa T(s) + \tau B(s)) \\
&= (c_1  -\kappa c_3) T(s) + s(c_1\kappa-c_2 \tau) \kappa N(s) + (c_2 + c_3 \tau) B(s) 
\end{align}
Both the left and right hand sides are vectors written in the $TNB$ basis, so we can equate coefficients, to get
\begin{align}
c_1 - \kappa c_3 &= 1 \\
s(c_1\kappa - c_2\tau) &= 0 \text{ for all $s$}\\
c_2 + c_3 \tau &= 0.
\end{align}
The first equation means that $\kappa$ is constant; from the second, we can then conclude that $\tau$ is constant, and that $c_1:c_2 = \tau:\kappa$. 
From the proof given above, in my earlier answer, we can therefore conclude that the curve is a circular helix. 
In short: 
The conjecture above characterizing helical curves is correct, so a curve $\alpha$ is a circular helix if and only iff there's a point $P$ ad constants $c_1, c_2, c_3$ such that 
$$
\alpha(s) - P = c_1 s T(s) + c_2 s B(s) + c_3 N(s).
$$
The remaining details about $P$'s relation to $\alpha(0)$, and the relationship of the axis to $N'(t) \times N(t)$, I leave to you. 
Thanks for pressing for a purely-Frenet-like characterization of helices -- this was a lot of fun. 
A: For any smooth space curve $\gamma$ with nonvanishing curvature and parametrized by arc length $s$ one has Frenet's formulas
$$\left[\matrix{\dot{\bf t}\cr\dot{\bf n}\cr\dot{\bf b}\cr}\right]=
\left[\matrix{0&\kappa(s)&0\cr-\kappa(s)&0&\tau(s)\cr0&-\tau(s)&0\cr}\right]\ 
\left[\matrix{{\bf t}\cr{\bf n}\cr{\bf b}\cr}\right]\ ,\tag{1}$$
where $t\mapsto \kappa(t)>0$ and $t\mapsto \tau(t)$ are the  curvature and the torsion along $\gamma$. 
If $\gamma$ is a circular helix then for fixed $h\in{\mathbb R}$ one has $$\gamma(s+h)=T_h\gamma(s)\qquad(-\infty<s<\infty)\ ,$$
where $T_h:\>{\mathbb R}^3\to{\mathbb R}^3$ is a euclidean motion. $T_h$ is called a "Schraubung" in German, because you are rigidly screwing the helix along itself. In technical terms, $T_h=R_a\circ S_a$, where $S_a$ is a shift along a certain axis $a$, and $R_a$ is a rotation around the same axis. This implies that $\kappa$ and $\tau$ are constant for such a helix, so that $(1)$ is in fact a constant coefficient system of ODEs:
$$\left[\matrix{\dot{\bf t}\cr\dot{\bf n}\cr\dot{\bf b}\cr}\right]=
\left[\matrix{0&\kappa&0\cr-\kappa&0&\tau\cr0&-\tau&0\cr}\right]\ 
\left[\matrix{{\bf t}\cr{\bf n}\cr{\bf b}\cr}\right]\ .\tag{2}$$
Conversely: Any such system $(2)$ with given $\kappa>0$ and $\tau\ne0$ defines a circular helix up to a rigid motion. This can be seen as follows: Assume that a $\kappa>0$ and a $\tau\ne0$ are given. Then set up a helix in the form
$$\gamma:\quad t\mapsto \bigl(r\cos t,r\sin t, c\>t)\tag{3}$$
with parameters $r>0$ and $c\ne0$ to be chosen appropriately. Then compute $\kappa$ and $\tau$ as functions of $r$ and $c$ according to the well known formulae, and finally adjust $r$ and $c$ such that the given $\kappa$ and $\tau$ result. The helix $\gamma$ then satisfies $(2)$, and any other curve satisfying $(2)$ for the given $\kappa$ and $\tau$ will differ from $\gamma$ by a rigid motion.
To sum it all up: A curve $s\mapsto \gamma(s)$ in ${\mathbb R}^3$ is a helix iff the curvature $\kappa>0$ and the torsion $\tau\ne0$ are constant along $\gamma$.
A: Show the constant curvature and torsion imply circularity of the given helix solution".
It is not (generally) so!!
Even among geodesics on surfaces of revolution there are many choices, circular helices are not the only solution. They are given as examples because they are the easiest to model as one principal curvature is zero and also perhaps as a well known example in everyday life.
A great circle on a sphere has constant (zero) (geodesic) torsion and constant (normal) curvature.
EDIT1: Asymptotic lines on constant (zero) (normal) curvature and constant (geodesic) torsion.
EDIT2: I have placed the words (geodesic) and ( normal) in parentheses because we are talking about two things at the same time.
The Frenet-Serret frame you refer to is meant for stand alone curves in space without regard to the surface on which the curve sits. The same space curve can be made to sit on many surfaces by adjusting/choosing any normal curvature and any geodesic torsion arising from Mohr's circle of curvature. The circular helix you gave as an example is one choice among them where normal curvature and torsion are both constant. I hope I could make myself clear. 
" 
EDIT3: Toroidal Loxodromes with constant Kappa and Torsion:
