# What is the radius of curvature formula for an ellipse at slope = 1?

1. Aug 24, 2013

### jjredfish

What is the radius of curvature formula for an ellipse at slope = 1?

I have found b^2/a, and a^2/b for the major and minor axis, but nothing for slope = 1.

Thanks.

2. Aug 24, 2013

### MrAnchovy

MathWorld has a comprehensive entry on ellipses. Note that the measure of curvature given in that article, as is usual, is the reciprocal of the radius of curvature.

3. Aug 24, 2013

### jjredfish

I read that page before I came here, but it didn't help with this specific problem. I don't need the radius from the center, or the focus. I need the radius of the line that is normal/perpendicular to slope =1. So, basically, a 45 degree angle from the edge of the ellipse to its center of curvature. I need the formula that describes this for any ellipse. I have come up with (b sqrt2)/((a^2/(b^2))+1) but it isn't accurate.

4. Aug 24, 2013

### MrAnchovy

The equation you need is referenced as (59) on that page.

That doesn't make any sense.

And nor does that, but I think I know what you mean. You want the radius of curvature (given by the reciprocal of (59)) where dy/dx = ±1. Use the parametric equations for y and x in terms of t to find the right value of t and substitute it into (59).

Edit: or do you want the radius of curvature where y=x? again you can find the appropriate value of t and use in (59).

Last edited: Aug 24, 2013
5. Aug 24, 2013

### jjredfish

Sorry, I'm not a mathematician (obviously) so I may be explaining it wrong. Let me try it with the help of a diagram.

I would like to know the formula that would give the length of the radius of curvature, for any ellipse, where the slope is at a 45 degree angle. In this diagram, it would be the formula for the length of the red line (if that line actually was the length of the radius). Please note that the radius may not (and most likely does not) end at the axis.

I have been out of school for many decades, so this is not homework. And I am not asking for a specific answer. I am asking for the formula, in standard ellipse terms of a, b, c, e, etc. Thanks.

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6. Aug 24, 2013

### jjredfish

R=b^2/a, and R=a^2/b describe the equivalent radius where the major and minor axis intersect with the ellipse. I would like to know the same sort of formula, but where the slope of the ellipse is at 45 degrees (instead of zero degrees, and 90 degrees).

(59) may contain the formula I need, somehow, but I don't have the math knowledge to distill it down to what I need from that. I also don't know what k or t are, and I don't know how something can use an exponent that is a fraction. I am probably working on a junior high math level at this point, which is why I came here for help. I understand R=b^2/a, and R=a^2/b, and want to know what their 45 degree equivalent would be, in the same simple nomenclature, if possible.

Last edited: Aug 24, 2013
7. Aug 24, 2013

### MrAnchovy

8. Aug 24, 2013

### janhaa

I don't know if this helps
the curvature is
$$\kappa=\frac{|x' y^" - y' x^" |}{(x'^2+y'^2)^{3/2}}$$

for x = x(t) and y = y(t)

$$R = {1\over \kappa}$$

9. Aug 24, 2013

### jjredfish

To janhaa: Sorry, that is completely over my head, and not in the form that I need it in (for my non-math brain).

To anyone who may be able to help: The best way I know to describe what I am seeking is from my description above: R=b^2/a, and R=a^2/b are well-know, common, and standard formulas for the radii where the major axis, and minor axis, intersect with the ellipse. I am seeking the equivalent formula, but where the slope of the ellipse is at 45 degrees, instead of zero degrees, or 90 degrees. It does seem like it should be possible, don't you think?

Thanks to anyone and everyone for any help.

Last edited: Aug 24, 2013
10. Aug 24, 2013

### janhaa

If the blue point has coordinate: (x, 0)
The start of the red line has coordinate: (c, 0)
theta, $$\theta$$is the obtuse angle
r is the radius (the red line)
then
$$\cos(\theta)=\frac{x-c}{r}$$

11. Aug 24, 2013

### jjredfish

Hi janhaa, thank you for your interest. Sorry, that was the best diagram I could find, but the blue point should be ignored. It is not part of the issue.

The issue is: if the red line (not drawn to scale) is the length of a radius, with its end point being on the ellipse (at the point where the ellipse has a slope of 45 degrees), and that same end is scribing the ellipse's radius of curvature (at slope = 45 degrees), how long is the red line? In other words, how long is the "red line" radius? The red line almost certainly will not end on the axis as shown in the diagram, that is just a coincidence.

Thanks

Last edited: Aug 24, 2013
12. Aug 24, 2013

### jjredfish

Hi MrAnchovy,

It is for part of a software program I am working on. I am 48 years old, so it is not for homework, if that is your concern.

Have I described the problem well enough, in a way that is understandable/makes mathematical sense?

Thanks

Last edited: Aug 24, 2013
13. Aug 24, 2013

### MrAnchovy

The answer is going to lie somewhere between $\frac{a^2}{b}$ and $\frac{b^2}{a}$, it would take about a dozen lines of maths to get it, and is almost certainly going to include terms such as $(a^2+b^2)^{3/2}$ which you don't understand, so it would be a waste of time.

On top of that, I can't believe that in any program involving ellipses this is the only thing you are going to need. I suggest you find someone to work with on an ongoing basis - the underlying maths is not particularly hard but you won't achieve anything by writing code to solve equations without understanding it.

14. Aug 24, 2013

### MrAnchovy

And yes this is an adequate description of the problem:

although to be clear you should probably ask for the solution in terms of the semi-axes of the ellipse a and b.

15. Aug 25, 2013

### jjredfish

Hi MrAnchovy, thanks for your reply. Actually, the formula for the ellipse is the only complicated math in the entire program (the program itself is not about math), and it is the only remaining component keeping me from finishing so I would really appreciate if you would at least give it a try. I am not a genius, or a mathematician, but I am not as stupid as I must seem to those who are, either. And though I may not be able to work through the dozen lines of math it takes to get there on my own, I would be able to understand the final formula enough to incorporate it into the program, especially if it is in terms of a and b, as you suggest. Would you please at least give it a try? Thank you again.

Last edited: Aug 25, 2013
16. Aug 25, 2013

### rcgldr

Instead of using the parametric form the radius of curvature versus x =

r(x) = | ( 1+(y')^(2) ) ^(3/2) / (y'') |

starting with equation for ellipse, and hoping I do the math right:

(x^2/a^2) + (y^2/b^2) = 1

y = (b/a) ± sqrt*(a^2 - x^2)

y' = ± (b/a) x / sqrt(a^2 - x^2)

y'' = ± a b / (a^2 - x^2)^(3/2)

r(x) = | (1 + (b/a)^2 x^2 / (a^2 - x^2) )^(3/2) / (a b / (a^2 - x^2)^(3/2)) |

y' = 1 when x = ± a^2 / (sqrt(a^2 + b^2)

plug this into r(x), which is a mess, but you'll have a formula.

17. Aug 25, 2013

### chingel

As was said, $R=|\frac{(1+(y')^2)^{3/2}}{y''}|$
You want $(y')^2=1$
As was in the previous post, $y''=\frac{ab}{(a^2-x^2)^{3/2}}$ and $x^2=\frac{a^4}{a^2+b^2}$
Then $y''=\frac{(a^2+b^2)^{3/2}}{a^2 b^2}$
So the result is $R= a^2 b^2 (\frac{2}{a^2+b^2})^{3/2}= a^2 b^2 \frac{2}{a^2+b^2} \sqrt{\frac{2}{a^2+b^2}}$
I wrote the square root out separately if you don't like fractional exponents.

18. Aug 25, 2013

### jjredfish

To: rcgldr and chingel, thank you so much!! You guys rock! Your posts are very helpful. It really helps that you simplified it for me and put it in terms I can understand and relate to the things I know.

So, for the fractional exponent "3/2", the denominator is actually the root... so "2" is the square root in this case. And the numerator "3" is just the normal exponent. Is that correct?

19. Aug 25, 2013

### chingel

Yes you can think of the exponent 3/2 as the square root raised to the third power.

20. Aug 25, 2013

### jjredfish

Thank you, chingel, I finally know how to use fractional exponents now!

And thanks to you I know the formula for the length of the radius at slope=1.

One last question, if you don't mind... is there any way to know where either end of that radius is located in relation to the center of the ellipse (a=0,b=0)?

Thanks again so much for your help.

Last edited: Aug 25, 2013
21. Aug 26, 2013

### rcgldr

If I did the math correctly, r(x) can be further simplfied:
$$r(x) = \big | \frac{(a^4-a^2 x^2 + b^2 x^2)^{3/2}}{a^4 b} \big |$$
$$r(x) = \big | \frac{(a^4 + x^2 (b^2 -a^2))^{3/2}}{a^4 b} \big |$$$$r(0) = \frac{a^2}{b}$$
$$r(a) = \frac{b^2}{a}$$

Choose a point on the ellipse and imagine a circle with radius = curvature of radius tangent to that point. One end of the radius is on that point on the ellipse, the other end is at the center of the circle. If you plot the path of the center of these circles, you get the evolute. The evolute of an ellipse is a stretched astroid. Links to articles:

Evolute - Wikipedia, the free encyclopedia

Evolute -- from Wolfram MathWorld

Ellipse Evolute -- from Wolfram MathWorld

Osculating circle - Wikipedia, the free encyclopedia

Last edited: Aug 26, 2013
22. Aug 26, 2013

### MrAnchovy

Well no actually, to raise something to the power 1.5 you do exactly that, you don't raise it to the power 3 and take the square root.

I knew there would be more questions! For a similar amount of effort it takes to derive equations for the single case you have asked for, any undergraduate mathematician could write a class (or extend the class you are using) in whatever high-level language you are using to do all the calculations you want, including those you don't know you want yet. And because it would use standard formulae it would be much more likely to be error-free.

And to answer your question, yes the centre of curvature is found using the expressions on this page. You will see they use the same t parameter as the expression I gave for the radius of curvature, so can easily be added to a generalised solution. However deriving an expression for the single point where $\frac{dy}{dx} = -1$ is harder and more error prone.

You are like a starving man standing by the ocean begging for fish: what you really need is a net.

Last edited: Aug 26, 2013
23. Aug 26, 2013

### jjredfish

To MrAnchovy: I really don't want to argue. I appreciate what you are saying, but I don't have the time, ability, or money to buy/use a commercial fishing net. I need one or two fish, and may never go fishing again. Thanks.

Last edited: Aug 26, 2013
24. Aug 26, 2013

### jjredfish

So I replace the x with the slope, which = 1, in my case?

25. Aug 26, 2013

### jjredfish

I tried using "x=a cos t", and "y=b sin t" to get the coordinates for the center of the radius (where slope=1), but I am getting nonsensical answers, so I am obviously doing it wrong.