Ellipses, tilted circles, circle approximation

[THX-1138 13]

I have a machining problem that I am trying to solve, and it appears that math may be of assistance.

 

Basically, I have a 12mm cylinder and a piece of 10mm brass bar stock, and I want to cut a groove in the bar that approximates the curve of the cylinder fairly closely. Unfortunately, I don't have any tools that can work with a diameter as small as 12mm, so I thought that perhaps using a larger diameter circle, 'tilted' at an angle so it intersects the bar as an ellipse, might get me close enough.

 

(Of course, I never thought before about any formulae relating a 'tilted' circle to the defining particulars of its cross-sectional view based on the angle. Does viewing a circle at an angle result in a proper ellipse?)

 

That's probably incredibly unclear, so let me see if I can depict it. The attached images are, in order: the cylinder and the bar (with the desired groove marked in blue); the cutting tool (variable diameter, but not small enough); a top view of the tool and the bar; and an end-on view of the tool and the bar.

 

Obviously this method will only yield an approximation of the groove I need. The question is, however, what diameter and angle to use for the tool to get something 'good enough.' An ellipse with one focus 6mm from the end would probably do for starters..

 

Plus I learn a bit more about geometry..

[thedarkshade 76]

I wish I could help you but I'm feeling very well today:doh:. Anyways, a google search came up with this!

[THX-1138 13]

Yes, I studied the Wikipedia article, but either it didn't answer my questions or else I'm too thick to understand how it did answer them. (Actually, the section on Semi-latus rectum and polar coordinates makes it clear that a 'tilted circle' is, in fact, a proper ellipse. One down..)

 

Thinking about this some more.. Given a tool diameter d, the foci would be d - 12mm apart, in order to give a 6mm distance from the nearest focus to an endpoint of the major axis. How to turn that into an angle measurement is still something I've not figured out, although the same section gives a clew. Figure out the eccentricity and the angle is easily calculated.

 

(I'm probably not awake enough to figure out what I'm doing wrong, but I'm also deducing that the constant sum of the distances to the foci is always equal to the length of the major axis.)

 

I went back to the Wikipedia article after a massive caffeine infusion, and I think I got a bit closer.

 

If the cutting tool is 50mm in diameter, and I want the least distance from a focus to an endpoint of the major axis to be 6mm, then using the formulae in the article we get:

 

1. [math]\varepsilon = \frac{c}{a} = \frac{(\frac{50}{2} - 6)}{25} = \frac{19}{25} = 0.76[/math]
    
   
2. [math]\varepsilon = \sin{\psi} = 0.76[/math]
    
   
3. [math]\psi = \arcsin{\varepsilon} = \arcsin{0.76} \approx 49.46\,^{\circ}[/math]
    
   
4. [math]b = \sqrt{a^{2} \cdot (1 - \varepsilon^{2})} = \sqrt{625 \cdot (1 - 0.5776)} = \sqrt{264} \approx 16.25mm[/math]
   

 

Which is a nice step along the way. However, the above will result in the semi-latus rectum (who says mathematicians don't have senses of humour?) being

 

[math]l = a \cdot (1 - \varepsilon^{2}) = 10.56mm[/math]

 

so the width of the groove if it is 6mm deep will be 21.12mm. That's much too wide; the cylinder will roll around in it like cavemen carrying an egg (see the film Caveman ). Obviously I need to futz around with the diameter of the tool, the depth of the groove, and the eccentricity of the ellipse. (Since the cylinder is going to be soldered into the groove, it needn't mate perfectly -- just 'good enough.')

 

 

Apologies for 'thinking out loud' here, but perhaps this will be useful to someone else in the future.

[jgalaz 10]

This thread was useful for me since I was trying to figure the same question on.

More specifically, I wanted to know how the minor axis of the projected ellipse varies with tilting angle Phi.

I'll take a look at the Wikipedia website here referenced to see whether the equations there are of any use (If not, I'll have to figure out the relationship myself).

I'm intuitively guessing that if (x^2)/a^2 + (y^2)/b^2=1, all you have to do is divide by cos(phi) which ever axis is reduced upon tilting, if tilting parallel to either of the ellipse axes that is.

For example, if the ellipse is ORIGINALLY A CIRCLE on a plane P, it's shape is:

(x^2)/R^2 + (y^2)/R^2=1

If tilting parallel to the y axis, then the circle will start becoming an ellipse by reducing its radius along the x direction:

(x^2)/(R*cos[phi])^2 + (y^2)/R^2=1

 

Which is the same as:

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

After the following variable substitution:

a=R*cos[phi]

b=R