Tool Path Equation 2

[sreid]

This is really a math question but I don't know where else to ask it.

I need the equation (if there is one)of the curve for the path of the center of a circle which is rolling around an ellipse. Any suggestions for math forums would also be appreciated.

 

[MintJulep]

More than you could ever want to know about an elipse at:

http://mathworld.wolfram.com/Ellipse.html

Using equation 10 from that site, simply add the r of your circle to both a and b. (I think)

 

[sreid]

I do like math world. But the problem appears simple but may not be. One can easily find a point on an ellipse. And it's easy to find the tangent and normal at the point. And then separate the the X and Y components from the circle radius and add them to the point on the ellipse X and Y. What is not clear is how to find the inverse.

 

[Lcubed]

sreid,
Isn't it the equation for an ellipse, larger-or smaller-by the radius of the circle? Am I missing something?
Regards,
Lcubed

 

[sreid]

A visualization. Make the ellipse thin and roll the circle around the ellipse. The trace of the circle center is sort of like two straight line with circular arcs at the ends. It's some kind of oval but clearly not an ellipse.

 

[IRstuff]

Not sure if you really need an inverse function, per se.

If you know where the major and minor widths are, subtracting the diameter of the circle leaves you with the two axes of the ellipse itself.

TTFN

 

[zekeman]

You start out with the fact that you are mapping the new coordinates from the x-y coordinates of the ellipse. The slope at the new points are dy'/dx' which are equal to dy/dx at the corresponding points of the ellipse. If the eq of the ellipse is:
x^2/a^2+y^2/b^2=1 then
dy/dx=-x/y*b^2/a^2 and the normal to the ellipse which passes thru the centre of the circle is -dx/dy=y/x*a^2/b^2.
Therefore you can write the slope eq of the ellipse in terms of the new coordinates as
dy'/dx'=(-x'-rho*cos(thet))/(y'-rho*sin(thet))*b^2/a^2
where thet is the arctan of dx'/dy' and rho the radius of circle and therefore
sin(thet=dx'/dy'/sqrt(1+dx'/dy')^2)
cos(thet)=1/sqrt(1+dx'/dy')^2)
Now you have a difficult diff eq that probably doesn't have a closed solution, but may be more useful than mapping a solution.

 

[gearguru]

I did not check zekeman's calculation, but what you are looking for is the mathematical problem, covered in the "differential geometry" : You are looking for the ENVELOPE created by the circle with the diameter = radius of the tool, when it rolls around the ellipse.
If zekeman's method does not work, search on the Web for the mathematical solution.

 

[sreid]

Thanks to all. The magic term is "Parallel Curve."

 

[gearguru]

We used the term "equidistant";
"parallel" was used for straight lines only.
But it was on another continent, long time ago and not in English language. Mathematically it is solved as envelope, in CAD we call it "offset" - all of it the same.

 

[MintJulep]

If you roll a circle around a circle, you get a circle.

A circle is just a special ellipse where both axis are equal.

Why wouldn't this also work for the general case?

sreid, why wouldn't it be an ellipse?

Going back to mathworld, you are just changing the constant a to a different constant, a+r. Seems like it should get you a larger ellipse.

 

[amorrison]

Recommend you check that circle diameter is less than the "with" of the ellipse at the narrow endpoints.

Mathematically stated ?? The ellipse rate of curvature must be greater than the circles rate of curvature at all points.

 

[zekeman]

Good point. Stated another way, if you are thinking of cutting an ellipse with a milling cutter, you must have the radius of the cutter less than the smallest radius of curvature of the ellipse due to undercutting. Therefore the tool path you get by rolling does not translate into the reverse at points where the curvature criteria is not met.Any book on cams or gears for has an explanation of undercutting.

 

[EnglishMuffin]

MintJulep: You are incorrect. If you have AutoCAD, you can prove it to yourself very easily. Draw an ellipse, and then create a second curve by offsetting the ellipse by some arbitrary amount. Then overlay a second (unique) ellipse with the same major and minor axes as the offset curve. You will find there is a difference between the ellipse and the offset curve - although the offset curve (which is actually a piecewise approximate spline) will look just like an ellipse to the casual observer. I believe there is a closed form equation for this curve, albeit a very complicated one, although why anyone would need one in this day and age is beyond me. A numerical solution to any desired degree of accuracy (using the simultaneous Newton Raphson method for example) should be fairly straightforward, and of course there is always CAD and the tool path offset capability built into all modern NC machining software.
Although it sometimes works and can provide insight, I am afraid that if you make a habit of generalizing from the special to the general case you will make a lot of mistakes!

 

[Kapitan]

Finding the centre of a circle rolling around the outside of an ellipse is a simple task when given the points (Xe,Ye), on an ellipse (Ae and Be) (the semi major and minor axes of the ellipse) and the radius of the rolling circle Rc.

First thing is to find the angle of the normal to the ellipse at a given point (Xe,Ye). This is done first on a circle where Ac = Bc = Be,
with points (Xc,Yc) and the tangent of the normal is (Yc/Xc).

By considering the ellipse as a X scaled circle the tangent scales as well, so that at a point (Xe,Ye) we now have the normal angle ;

Theta = ArcTan ( Yc * Ae / Xc * Be )

Then positioning the roller centre (Xr,Yr) is adding the X and Y components of a triangle whose hypotenuse is the ellipse normal of length Rc.

So the coordinates of the rolling circle on the ellipse are :

Xr = Xe + ( Rc * Cos ( ArcTan ( Yc * Ae / Xc * Be ) ) )

Yr = Ye + ( Rc * Sin ( ArcTan ( Yc * Ae / Xc * Be ) ) )

If the circle were to roll inside the ellipse then the components would be subtracted from the ellipse points.

Plug some numbers into this if it seems too good to be true.
No need to plunge into differential geometry or the Newton-Raphson method, more an understanding of conics and coordinate geometry.

A constant offset from any ellipse does not give another ellipse, the new curve is an oval.

 

[EnglishMuffin]

Kapitan: As I understand it, sreid did not ask how to calculate the tool path parametrically, starting out with the coordinates of a known ellipse, which as you say is a relatively simple matter - he specifically asked for an equation for the tool path (ie a direct relationship between the x and y values of the curve). You inevitably end up with two interrelated equations, just like those you have written, which either have a very complicated closed form solution (which I believe does exist) or require a numerical solution , which (I repeat) can be achieved with a Newton Raphson iterative routine (such as that employed by TK solver for example). I again repeat - why sreid requires such a specific equation if he needs it only for machining purposes is beyond me, since one can compute the path in terms of the coordinates of the original ellipse by a number of means, including yours.

 

[zekeman]

Kapitan,
A small point. Your equations(in parametric form) are almost ok except that your terms in Ae/Be should be (Ae/Be)^2.

If you have a graphing calculator, the curve can be shown nicely, using the parametric feature.

 

[zekeman]

English muffin,Kapitan

There is nothing inherently wrong with the idea of presenting a solution in parametric form, and nothing in the question precludes such a solution.
However as a practical matter, if sreid intends to use the solution for machining an elliptical part with a milling cutter then he would be well served to use the parametric form which is slightly incorrect as presented. They should read:
Xr=x + Rc*cos(arctan(y/x*(a/b)^2)
Yr=y + Rc*sin(arctan(y/x*(a/b)^2

The ellipse being:

x^2/a^2+y^2/b^2=1
Again for machining, the solution is not valid for the undercutting condition,
Rho<Rc
where Rho= radius of curvature at x,y coordinate of ellipse.

 

[jbel]

Is this a toolpath for a machining center????

If so, why not just use a G41 or G42 command and the original path that you want the cutter to follow. If the control uses FANUC programming language all of this is a moot point.

 

[zekeman]

Correction:

My comment about undercutting applies only to cutting an internal ellipse where it is obvious you can't, if the radius of curvature at any point of the ellipse is less than the cutter radius.

No such limitation exists for the convex ellipse.