Radius of Curvature Derivation Help

In summary, the conversation is about understanding the derivation of the radius of curvature from a book. The equations for tangent and normal unit vectors are given, and the parametric equation of a circle is mentioned. The conversation also touches on a different way of expressing the concept of curvature.
  • #1
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Homework Statement


This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.


Homework Equations


Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:

[tex]\textbf{u}=\frac{d\textbf{r}}{ds}[/tex]

and

[tex]\frac{d\textbf{u}}{ds}=k\textbf{n}[/tex]

where k is called the curvature.

We can expand r(s) around a certain point:

[tex]\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...[/tex]

using the equations for u and n:

[tex]\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...[/tex]

which is the same as:

[tex]\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...[/tex]

And from here they conclude that this is the equation of a circle with radius [tex]\frac{1}{k}[/tex], which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.

Thanks to advance to whoever is able to help.
 
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  • #2
The parametric equation of a circle is ( a+ Rcosu, b+Rsinu).
 
  • #3
Grand said:

Homework Statement


This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.


Homework Equations


Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:

[tex]\textbf{u}=\frac{d\textbf{r}}{ds}[/tex]

and

[tex]\frac{d\textbf{u}}{ds}=k\textbf{n}[/tex]

where k is called the curvature.

We can expand r(s) around a certain point:

[tex]\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...[/tex]

using the equations for u and n:

[tex]\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...[/tex]

which is the same as:

[tex]\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...[/tex]

And from here they conclude that this is the equation of a circle with radius [tex]\frac{1}{k}[/tex], which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.

Thanks to advance to whoever is able to help.

I'm wondering if your latter equation is well derived. Did you take it from your book?
 
  • #4
Yes, it is from the book and I was able to derive it by myself.

@Eynstone, that accounts only for the sin term, not for the -cos term
 
  • #6
Oh wait! it's not expressed differently, it's just rearranged so that you can manage to get the radius of curvature, check that first website I handled, and enjoy up to point (31). ;)
 

What is the radius of curvature and why is it important in science?

The radius of curvature is a measure of how curved a curve or surface is at a given point. It is important in science because it helps us understand and analyze the shapes and movements of objects in the physical world.

How is the radius of curvature derived?

The radius of curvature is derived using the formula R = (1 + (dy/dx)^2)^3/2 / (d^2y/dx^2), where dy/dx represents the slope of the curve at a given point and d^2y/dx^2 represents the rate of change of slope.

What are the applications of the radius of curvature in different fields of science?

The radius of curvature has various applications in physics, engineering, and astronomy. It is used to analyze the shape of lenses and mirrors in optics, calculate the turning radius of objects in motion, and determine the curvature of space-time in general relativity.

Can the radius of curvature be negative?

Yes, the radius of curvature can be negative. A negative radius of curvature indicates that the curve or surface is concave, meaning it curves inward. A positive radius of curvature indicates a convex curve or surface, which curves outward.

Are there any limitations to using the radius of curvature as a measure of curvature?

While the radius of curvature is a useful measure of curvature, it is limited to measuring only the curvature at a single point on a curve or surface. It does not provide information about changes in curvature along the length of the curve. Additionally, it may not accurately represent the overall shape of complex curves or surfaces.

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