7
$\begingroup$

This is a question about understanding the concept of curvature. Firstly, what exactly is curvature of a curve (not the formula, what does it actually mean conceptually)? Secondly, I am confused about how one can figure out when a curve would have the maximum curvature. I was told it is when the derivative of the curvature function (k(x)) = 0. But, I don't understand the concept/reason behind it.

In the book I am using there are more than one definitions for curvature. Some of them are:

Number One:

$$k\left(x\right)=\frac{\left|f''\left(x\right)\right|}{\left[1+\left(f'\left(x\right)\right)^2\right]^{\frac{3}{2}}}$$

Number 2:

$$k\left(t\right)=\frac{\left|\vec{r}'\left(t\right)\:\times \:\:\vec{r}''\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|^3}$$

Number 3:

$$k\left(t\right)=\frac{\left|\vec{T}'\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|}$$

$\endgroup$
  • 1
    $\begingroup$ Well regarding the last part usually things have zero derivative when they're at a maximum... are you unfamiliar with calculus or did it just not connect right away? $\endgroup$ – spaceisdarkgreen Oct 2 '17 at 19:45
  • $\begingroup$ I am familiar with calculus, it is hard for me to connect all of the concepts conceptually. $\endgroup$ – user485656 Oct 2 '17 at 19:58
  • $\begingroup$ All the equations are essentially the same in that they all measure the same quantity. For these it boils down to application. $\endgroup$ – Triatticus Oct 2 '17 at 20:03
  • $\begingroup$ For definition $3$, what is $T$? Also, definition $1$ is a special case of $2$, if you think of number $1$ as $2$ applied to a parametrisation of the type $r(t)=(t,f(t))$. $\endgroup$ – Arthur Oct 2 '17 at 20:07
  • $\begingroup$ From the title, I was at first thinking the "why" was asking "why do we care about the maximum curvature"? To which a possible answer would have been: if you bend a pipe cleaner to that curve then the point of maximum curvature is typically the point where the stress is greatest and it's most likely to break. $\endgroup$ – Daniel Schepler Oct 2 '17 at 22:20
3
$\begingroup$

Curvature at a point is what it sounds like - a measure of how "curvy" a curve is. How sharply it's bending at that point, if you will. There are many (and well varied) notions of curvature, but by the sounds of your question, this is what you are talking about.

In order to understand your second question, you need to be a bit more precise in telling us what definition of curvature you're using, I think.

EDIT:

From various comments and your edit to your post, my only remark about why the curvature is maximum at the point where the derivative = 0 is that if you imagine the curvature being plotted on a graph (a plot of curvature vs time $t$), there probably be a place where the curvature reaches some local maximum. This will be like a peak - it'll look like the top of a hill. For this reason the derivative with respect to time there will be zero. It depends on the nature of your curve as to whether this is a local maximum or a global maximum.

$\endgroup$
  • $\begingroup$ In the book I am using, there are many different formulas for the curvature. 1. |f ''(x)| / (1 + f '(x)^2)^(3/2) 2. k(t) = |T'(t)| / |r'(t)|. Those are exampled of two of them $\endgroup$ – user485656 Oct 2 '17 at 19:46
  • $\begingroup$ I've edited the post to answer the question that I think you're answering. $\endgroup$ – Matt Oct 2 '17 at 20:19
11
$\begingroup$

Imagine you have a circle that tangent the inside of the curve. If the radius of the circle is small, then it fits, but if the radius gets to be too big, it doesn't fit into a tight bend.

The largest circle the fits the curve at any point is called the osculating or "kissing" circle.

The radius of curvature is the radius of the osculating circle.

Curvature is the reciprocal of the radius of curvature.

Once you have a formula that describes curvature, you find the maximum curvature (or minimum radius) the same way you find the extrema of any smooth function.

$\endgroup$
  • $\begingroup$ I am trying to do this numerically. Would that be a separate question ? I have a python script that creates a contour out of so many points available. Now you are saying that I need to fit a circle to that contour and then then the inverse of the radius of curvature is the curvature of that point. So essentially I am going to have create a circle at each point of the contour am I right ? $\endgroup$ – gansub Oct 4 '17 at 2:34
  • $\begingroup$ if you have a smooth curve, you can use the formula in the original post. If you calculate vectors normal to your curve. The point where nearby vectors intersect, will be at the center of said circle, and then the radius and curvature will neatly fall into place. $\endgroup$ – Doug M Oct 4 '17 at 16:08
2
$\begingroup$

As you walk along a curve you turn. At any given point, the rate at which you turn (compared to how fast you walk) is the same as if you were going along the circumference of a circle of some radius. The reciprocal of that radius is the curvature. So when walking through a point in the curve where the curvature is $1$, it will feel like a circle of radius $1$, while curvature of $2$ corresponds to a circle with radius $0.5$, and so on. (At least, that is one definition of curvature.)

As for when to find the maximum, differentiating and finding series works the same way it does in any other application: at the point where the curvature is as large as it's going to be, it goes from increasing to decreasing, which means that the derivative goes from positive to negative. It must be $0$ in the middle.

$\endgroup$
2
$\begingroup$

The slope of a curve $\gamma$ at a point $p \in \gamma$ equals the slope of the line $L$ through $p$ that is a "best fit" for the curve to first order, which means: $\gamma$ and $L$ both pass through $p$ and they have the same first derivative at $p$. That line $L$ is called the tangent to $\gamma$ at $p$.

The curvature of $\gamma$ at a point $p$ equals $1/r$ where $r$ is the radius of the circle $C$ through $p$ that is a "best fit" for the curve to second order, which means: $\gamma$ and $C$ both pass through $p$, they have the same first derivative at $p$, and they have the same second derivative at $p$. That circle $C$ is called the osculating circle to $\gamma$ at $p$.

Using this definition of curvature, one can actually derive any of those formulas by using calculus and analytic geometry.

$\endgroup$
0
$\begingroup$

The curvature is what makes the difference between a straight line and a curve, i.e. a measure of "non-straightness". And it is intuitive that a curve of constant curvature is a circle.

For curves that are not circles, the curvature must be defined locally, i.e. it varies from place to place. To get an estimate, you take three points in a small neighborhood and construct the circle through them.

This can be handled by differential calculus and one can show that when you shrink the neighborhood, the circle tends to a limit, and the inverse of its radius is given by the formulas that you cited.


Now like any non-constant function, the curvature can have minima and maxima. Beware that the sign of the curvature distinguishes between left turns and right turns, so that the extrema correspond to places where the radius is the smallest, irrespective of the direction. The curvature can even become infinite, forming a so-called angular point (the local circle degenerates to a single point).

It is also interesting to consider the zeroes of the curvature, where the curve becomes locally straight. These are called inflection points.

$\endgroup$
0
$\begingroup$

The most important definition and the one that gives the best explanation for the curvature is actually missing from your list. If a curve $\alpha(s)$ is parametrized by arclength (meaning that $|\alpha'(s)|=1$ for all $s$) then the curvature is simply the norm of the second derivative: $$k(s)=|\alpha''(s)|.$$ Thus curvature measures how "fast" the (unit) velocity vector is turning. All the other formulas are consequences of this one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.