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Wiki says that the curvature of a circle is 1/r, but if we consider the radius in meters or centimeters or other we get different results.

How do we know the right value?

user157860
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  • $1$ centimeter = $0.01$ meter. Are you confused by the conversion from centimeters to meters or something else? – Axion004 Oct 21 '19 at 12:57
  • @Axion004, of course not.if we consider the meter the curvature is 0.01, if we consider the centimeter the curvature is one, if the millimeter than it is ten. Shouldn't we have an absolute value, like we do with Kelvins? – user157860 Oct 21 '19 at 13:04
  • Temperature is just absolute in the sense that its zero is well-defined; you can still have different systems with their zero at absolute zero like Kelvin and Rankine. Anyway, the usual definition of curvature is dimensional. If it were not, one could not distinguish the relative curvature of different circles. – Ian Oct 21 '19 at 13:13
  • By the way you should note that in principle slope is also dimensional unless you require both axes to have the same units. – Ian Oct 21 '19 at 13:14
  • @Ian, the slope 15% gives me a real clue of the hardness I will face. All circles have same curvature, then you are just telling me what the radius of the curve is. – user157860 Oct 21 '19 at 14:17
  • It's telling you the radius of the osculating circle, which isn't the same at every point along the curve. But if you want something more specific, then you have to compare the curvature to another length scale to nondimensionalize it. For example you might compare it to the distance that you will go in a given unit of time, in which case the curvature is telling you something about how fast you need to turn per unit time (in radians per time, so length has been scaled out now). – Ian Oct 21 '19 at 14:21
  • Another possibility is that if you have a polar curve then you can compare the curvature to the arclength per angle, say $s$; then $\kappa s = 1 + \frac{r'^2 - r r''}{r^2+r'^2}$, where ' denotes differentiation with respect to the angle. This sort of measures the local deviation between the curve and a circle. – Ian Oct 21 '19 at 14:26
  • @Ian, that's interesting, can you expand it in a proper answer with example?I'd love to accept your answer – user157860 Oct 22 '19 at 07:34

2 Answers2

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The curvature $\kappa$ is "tangent direction change per length increment", and has dimension ${1\over{\rm length}}$. It follows that for a physical curve in our ambient space the numerical value of $\kappa$ depends on the chosen length unit.

When you have a circle drawn on a paper with length unit marked in the figure (say, on the coordinate axes) then it is agreed that the unit circle has $\kappa=1$ for all students in the class, independently of the sizes of their figures.

Note that you have the analogous problem already with lengths. It has nothing to do with the sophisticated notion of curvature.

Christian Blatter
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  • for lengths we know exactly what is a meter, now and before. If I have a length of 10 meters ten I know it is 1/3.10^7 C. If I am taking a bend on the highway, a sign tells me the slope, and also I need to know the curvature of the road, if you give me such curvature, you are simply indirectly telling me the radius of the curve – user157860 Oct 21 '19 at 13:10
  • @user157860 You are being told the radius of the so-called osculating circle, which changes at each point of the path. – Ian Oct 21 '19 at 13:18
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You'll get the right answer no matter which units you use.

Suppose you have a circle whose radius is 100 inches, which is the same as 254 centimeters or 2.54 meters. Then the curvature of the circle is

$$\frac{1}{100\ \mathrm{inches}} = \frac{1}{100}\ \mathrm{inches}^{-1},$$

but the curvature of the circle is also

$$\frac{1}{254\ \mathrm{cm}} = \frac{1}{254}\ \mathrm{cm}^{-1},$$

and the curvature of the circle is also

$$\frac{1}{2.54\ \mathrm{m}} = \frac{50}{127}\ \mathrm{m}^{-1}.$$

All of these answers are correct, because $\frac{1}{100}\ \mathrm{inches}^{-1}$, $\frac{1}{254}\ \mathrm{cm}^{-1}$, and $\frac{50}{127}\ \mathrm{m}^{-1}$ are all the same quantity.

Tanner Swett
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  • in this way you are just re-telling me, in an indirect way what is the radius of the circle What is the point, the difference?what info does that add? difference between a – user157860 Oct 21 '19 at 14:12
  • @user157860 That's right, the curvature does just say, in an indirect way, what the radius of the circle is. It doesn't add any information. But curvature shows up in a couple of formulas. For example, you can find the turn rate of a vehicle by multiplying its speed by the curvature of its path. Another useful fact is that a straight line doesn't have a radius, but it does have a curvature: the curvature of a line is $0$. – Tanner Swett Oct 21 '19 at 17:56