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

[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.

 

[pbuk]

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.

 

[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.

 

[pbuk]

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.

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

I need the radius of the line that is normal/perpendicular to slope =1.

That doesn't make any sense.

So, basically, a 45 degree angle from the edge of the ellipse to its center of curvature.

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).

 

[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.

 

[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.

Thanks for any help you can offer.

 

[pbuk]

Can I ask why you are interested in this answer?

 

[janhaa]

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.

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)

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

R : the radius

 

[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.

 

[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
$$\cos(\theta)=\frac{x-c}{r}$$

 

[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

 

[jjredfish]

Can I ask why you are interested in this answer?

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

 

[pbuk]

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.

 

[pbuk]

And yes this is an adequate description of the problem:

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

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

 

[jjredfish]

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.

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.

 

[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.

 

[chingel]

To help you out a little more:

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.

 

[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?

 

[chingel]

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

 

[jjredfish]

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

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.

 

[rcgldr]

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

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)?

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

 

[pbuk]

I finally know how to use fractional exponents now!

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.

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)?

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.

 

[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.

 

[jjredfish]

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

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

 

[jjredfish]

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

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.

 

[rcgldr]

If I did the math correctly, r(x) can be further simplfied: ... equations ...

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

No, you replace x with the x coordinate value that corresponds to a slope of 1. This gives you the effective length of the radius. You also need an equation for the normal (a line perpendicular to the tangent), to determine the direction of the radius, or use the evolute formula.

Using parametric equation format, for a given point on the ellipse, (x = a cos t, y = b sin t), the outer end of the radius is at (x,y), and the inner end is at the point ((a^2-b^2) cos^3(t) / a, (b^2-a^2) sin^3(t) / b). The distance between those points will be the radius of curvature, r(x), which was already shown above. Given a point on the ellipse (x,y), you'll have to solve for t, t = arccos(x/a) or t = arcsin(y/b), and you'll have to compensate for which quadrant the point is in.

 

[chingel]

Using the previously found relation for when the slope is +-1: $x^2=a^2 cos^2 (t)=\frac{a^4}{a^2+b^2}$
Simply substituting the cosine into the formula on the Wolfram page on the ellipse evolute, the result is:
$$x_e=\frac{a^2-b^2}{a}(\frac{a^2}{a^2+b^2})^{3/2}$$
$$y_e=\frac{b^2-a^2}{b}(\frac{b^2}{a^2+b^2})^{3/2}$$
The second endpoint of the radius is of course on the ellipse where
$$x=\frac{a^2}{\sqrt{a^2+b^2}}$$
$$y=\frac{b^2}{\sqrt{a^2+b^2}}$$
This is for the endpoints of the radius when the slope is -1 in the first quadrant.

 

[pbuk]

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. Rcgldr and chingel are helping me out with my cane pole and worm. That said, if you want to take the time to explain the whole net to me, slowly, in plain english, I will take the time to try to understand it.

Also, it seems to me, when one knows the length of a radius, it is not far-fetched to ask about its location, as both seem necessary to define a radius. That is just asking about both ends of the same fish.

OK, it is actually much easier to do that than to provide specific formulas giving the answers to your questions, which I suppose is what I could have said in the first place.

The points on an ellipse centred at $(x, y) = (0, 0)$ satisfy the equations $x = a \cos t, y = b \sin t$ where t is a parameter varying from 0 to $2\pi$ radians (or 0° to 360°) and a and b are the semimajor and semiminor axes. We can therefore define any point on a given ellipse simply by determining a value for $t$.

Now the radius of curvature at any point on an ellipse so defined is given by (I'll work in pseudocode)

 ((b * cos(t))^2 + (a * sin(t))^2)^1.5 / (a * b)

and the centre of curvature (called the evolute) by

x_evolute = (a*a - b*b) / a * (cos(t))^3
y_evolute = (b*b - a*a) / b * (sin(t))^3

- these equations are taken straight from the Mathworld pages already linked in this thread.

So you can find the x and y coordinates and the radius and centre of curvature for any point on an ellipse (which are the things you have asked for), as long as you have $t$ - the only thing missing is that you need to know how to find $t$. Now we need a bit of maths: the gradient (slope) of the ellipse is given by $\frac{dy}{dx} = \frac{y'}{x'} = \frac{b \cos t}{- a \sin t} = \frac{-b}{a \tan t}$, so if we know the gradient $g$ we can use this code to find t:

 t = arctan(-b / (a * g))

.

Putting it all together:

// define the ellipse
a = semimajor_axis
b = semiminor_axis

// set the gradient at the point we are interested in
g = -1

// work out the 't' parameter for the given gradient
t = arctan(-b / (a * g))

// find the position on the ellipse
x = a * cos(t)
y = b * sin (t)

// find the evolute (centre of curvature)
x_evolute = (a*a - b*b) / a * (cos(t))^3
y_evolute = (b*b - a*a) / b * (sin(t))^3

// find the radius of curvature
r = ((b * cos(t))^2 + (a * sin(t))^2)^1.5 / (a * b)

// note that abs(r - sqrt((x - x_evolute)^2 + (y - y_evolute)^2)) should be small - use this in unit testing

 

[jjredfish]

Thanks for taking the time to explain all this MrAnchovy.

I will try to understand it, and let you know if I have any questions.

 

[vendetta2115]

The formula for an ellipse is x^2/a^2+y^2/b^2=1. The formula for the slope of an ellipse is y'(x)= - (x/a^2)/(y/b^2). Setting y'(x)=1, we have (x/a^2)/(y/b^2)=-1. Solving these two equations yields the solutions:

x = - a^2/sqrt(a^2+b^2), y= +b^2/sqrt(a^2+b^2),
and
x = +a^2/sqrt(a^2+b^2), y= - b^2/sqrt(a^2+b^2)

The radius of an ellipse at any point (x,y) is given by the formula R=sqrt(x^2+y^2). Substituting in the expressions for x and y and simplifying, we find that the radius from the center of the ellipse to the point where the slope of the tangent line is 1 is conveniently simple:

R=sqrt[(a^4+b^4)/(a^2+b^2)]

The radius of curvature of an ellipse is given by the equation R_c = [[a^2sin^2(t)+b^2cos^2(t)]^(3/2)]/(ab), where t is the angle from the x-axis to the point (x,y). Since tan(t)=y/x, then t=arctan(-b^2/a^2). So finally we have

R_c=[[a^2sin^2(arctan(-b^2/a^2))+b^2cos^2(arctan(-b^2/a^2))]^(3/2)]/(ab).

Since a>0 and b>0, this simplifies to:R_c=a^2 b^2 ((a^2+b^2)/(a^4+b^4))^(3/2)This equation gives you the local radius of curvature at the point (x,y) where the slope y'(x)=(+/-)1 for any ellipse of the form x^2/a^2+y^2/b^2=1.

For example, an ellipse where a=2.5, b=5 has a tangent line equal to 1 at x=1.11803, y= - 4.47214, which occurs at an angle t=75.96 degrees from the x-axis. At that point, the ellipse has a radius R=4.60977, and a radius of curvature R_c=1.59508. Note that this ellipse will also have tangent lines equal to (+/-)1 at x=(+/-)1.11803, y=(+/-)4.47214.