Equation to graph a 180 degree curve comprised of a radius and an ellipse


by JimEd
Tags: curvature, ellipse, equation, graph, radius
JimEd
JimEd is offline
#1
Nov30-12, 02:30 PM
P: 46
Hi All, I'm not a math guy so I am coming to you for help.

I am trying to come up with an equation to graph any 180 degree curve that is comprised of: a 135 degree radius, and a 45 degree ellipse (135 + 45 = 180). The two curves being the same curvature (slope?) where they meet.

The portion of the ellipse I want to use would be the part between line B (the vertical one) and line A (the red one), on the drawing below. The radius portion would start at A and go to the right.

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Given: the length of the semi-major axis of the ellipse, the length of the semi-minor axis of the ellipse, and the radius. Use any example numbers you like (but remember the curvature (slope?) of the radius and the curvature (slope?) of the ellipse have to be the same where they meet, so I believe that means the value of the radius is dependent on the values of the ellipse).

Please don't take this wrong, but I'm an older guy, and this isn't homework, so ideally I would like someone to solve it (if it is possible), and then briefly explain it to me. As opposed to someone leading me around to try to solve it for myself - which would be extremely painful for both of us!

Thanks very much.
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mfb
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#2
Nov30-12, 03:46 PM
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(but remember the curvature (slope?) of the radius and the curvature (slope?) of the ellipse have to be the same where they meet, so I believe that means the value of the radius is dependent on the values of the ellipse)
Only if you add more requirements. Does the end of the ellipse-part (and end of the 180-curve) have to be some special point of the ellipse?

Some helpful equations:
An ellipse aligned with the coordinate axes can be expressed as ##\frac{x^2}{a^2}+\frac{y^2}{b^2}=1## or ##y(x)=\pm b \sqrt{1-\frac{x^2}{b^2}}##

Curve radius is given by $$R = \frac{(1+y'^2)^{\frac{3}{2}}}{y''}$$ where ' is the derivative with respect to x. The curve radius of a circle is the regular radius of the circle, of course.
45 away from the intersections of the major axes and the ellipse are points with y'=1. This simplifies curvature to ##R=\frac{\sqrt{8}}{y''}##, but it applies to a special case of your construction only.
JimEd
JimEd is offline
#3
Nov30-12, 04:09 PM
P: 46
Quote Quote by mfb View Post
Does the end of the ellipse-part (and end of the 180-curve) have to be some special point of the ellipse?
Hi mfb,

Thanks, you are right, I should have included that. I will add it, and drawing to my original question/post.

The end of the ellipse portion (at the end of the 180 degree curve) would correspond to the middle of the major axis (where the ellipse intersects line B on the drawing, above).

Thanks.

JimEd
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#4
Dec1-12, 09:28 AM
P: 46

Equation to graph a 180 degree curve comprised of a radius and an ellipse


(below)
mfb
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#5
Dec1-12, 12:57 PM
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P: 10,809
Quote Quote by mfb View Post
where ' is the derivative with respect to x.
'' is the second derivative.

If you know a and b and the 45-part of the ellipse is like in the sketch, with the origin of the coordinate system in its center:
Derive y and solve for y'=-1 (it has a nice general solution) to find the corresponding x-value of point P where ellipse and circle meet. Use this x-value in y(x) to get the corresponding y-value for P. Use some tool of your choice to draw y(x) between the calculated x-value and x=a.

Derive y' to get y'', insert your calculated x-value and plug that in the formula for the curve radius to get the curve radius R.
The center of your circle is now ##\frac{R}{\sqrt{2}}## to the left and below P. Its center is M(c,d) and the circle equation is ##(x-c)^2 + (y-d)^2 = R^2## or ##y(x)=d+\sqrt{R^2-(x-c)^2}##. Draw this for x=c-R to point P.

##y(x)=\pm b \sqrt{1-\frac{x^2}{b^2}}##
That has an error, it should be ##y(x)=\pm b \sqrt{1-\frac{x^2}{a^2}}##
JimEd
JimEd is offline
#6
Dec1-12, 02:14 PM
P: 46
Quote Quote by mfb View Post
'' is the second derivative.

If you know a and b and the 45-part of the ellipse is like in the sketch, with the origin of the coordinate system in its center:
Derive y and solve for y'=-1 (it has a nice general solution) to find the corresponding x-value of point P where ellipse and circle meet. Use this x-value in y(x) to get the corresponding y-value for P. Use some tool of your choice to draw y(x) between the calculated x-value and x=a.

Derive y' to get y'', insert your calculated x-value and plug that in the formula for the curve radius to get the curve radius R.
The center of your circle is now ##\frac{R}{\sqrt{2}}## to the left and below P. Its center is M(c,d) and the circle equation is ##(x-c)^2 + (y-d)^2 = R^2## or ##y(x)=d+\sqrt{R^2-(x-c)^2}##. Draw this for x=c-R to point P.
Thank you so much for this, it looks brilliant, but it is way over my head. I am using the online graphing calculator at https://www.desmos.com/calculator and I can't get any of these equations to show up as... anything. Is there any chance someone could make a graph of it? If so, thank you, I know it is asking a lot. I'm a little better with visual learning.

I'm probably misunderstanding you, but, from P to the center of the circular portion (as opposed to the ellipse portion) should just be R (the radius), not R / sqr 2. Correct?
mfb
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#7
Dec2-12, 07:48 AM
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P: 10,809
The total distance is R, but it is composed of the distance in x- and y-direction, which have the same magnitude.
JimEd
JimEd is offline
#8
Dec2-12, 08:59 AM
P: 46
Can anyone make a drawing or a graph of this, please?

I know this is a math forum, and I do appreciate math, but I am not very good at it. That's why I am here. I don't know how to "derive" things, so I need a more basic explanation. Pretend I'm an average high school freshman and you'd probably be pretty close to my level of understanding.

I get that R / sqr 2 = the length of the x and y paths from P to M, now.

Thanks
JimEd
JimEd is offline
#9
Dec3-12, 12:06 PM
P: 46
Anyone?


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