Finding the focal point of a parabola inside of a sketch...

[texaspete]

Hi all, I am working on WF2.0. I have a sketch it contains only a parabola. It is dimension in a way that does not identify the location of the focal point. The design is a reflector for a tail lamp. I need to find the focal point of the parabola. the RHo value is .50. I believe that is a parabola. Does anyone have any suggestions; I'm validating the reflector / parabola to a focal point location. Thanks in advance - texaspete
I believe the formula is y=(1/4)*(Focal distance)*(X^2).

 

[Heckler]

How was the sketch created? using an equation defined in a graph function?

Here is some information on parabolas

http://en.wikipedia.org/wiki/Parabola

http://en.wikipedia.org/wiki/Parabola#Derivation_of_the_focus

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2005 SP 5.0 & Pro/E 2001
Dell Precision 370
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XP Pro SP2.0
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_`\(,_
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Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

 

[texaspete]

Heckler, The sketch (I wish I could show you) was developed by a one of my employees. He did not used the techique that PTC Knowledge base suggest. He created the curve (in the sketch) using relationship and equations.
ex. sd2=sd1*2 sd0=90(degrees) sd3=100 (It appears that this is x value) and some other sd#=SD#; Y value must be calculated.

Still open for other technique to check curve and focal point.

I guess I will have to make a parabolic curve the way I know how and compare the geometry.
I thought I could pick a point on the curve, put an AXIS or a line tangent to the curve at that point. Then make a line normal (90degrees) to the line mention last. Then mirror a line that the light ray would be parallel to and go throught the point mentioned earlier. Do this 2 times and the lines that are mirrored (the lines that represent the light rays) should intersect at the focal point.

Any comments - texaspete

 

[Heckler]

Are you trying to input this data in a ray tracing program? We often do this for some of our optics stuff but the program is a simple 2D ray trace program.

I would have used the graph function. Without seeing all the data it's hard for me to determine....other than crunching the numbers using the standard equations for a parabola.

let us know what your solution ends up being.

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2006 SP 5.0 & Pro/E 2001
Dell Precision 370
P4 3.6 GHz, 1GB RAM
XP Pro SP2.0
NVIDIA Quadro FX 1400
o
_`\(,_
(_)/ (_)

Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

 

[jeff4136]

Glad you asked. I've wanted to examine and understand the curves for some time. Now, after some sperimentin, I know I don't. `;^)

For a full (sym tan angles relative chord, rho 0.5) parabolic curve; simply reversing the equal length focus to directrix and directrix -> curve -> axis lines layout procedure is easy to do. Two equal length pairs nails it.

For less than a full curve; I don't know. I'll post an example file (WF2) on mcadcentral under the subject line "Focal Point?" shortly if you want to take a gander.

 

[jeff4136]

> ... post an example file (WF2) ...

I got in a last minute dabble before posting the file and it appears that for the partial curve (as might be used in a revolved feature section) IF the edge tangent direction is properly defined the standard layout will indicate the focus.

Interested to hear how it might compare with your findings (I ~think~ my assumptions are valid, not sure).

 

[jeff4136]

Pete, not sure from your response on mcadcentral if you
wanted to pursue this further but thought I'd tie up a
couple of loose ends.

Looking at your test.bmp; the curve should be good to go if
sd46 = atan( sd17 / (sd21 * 2)) -and- rho = 0.5.
That tan angle definition is what I didn't understand (as
seen in my example Sketch_1). If it's not correct the conic
axis is rotated and is no longer coincident with the conic
arc end point. (I'd seen the effect looking at curvature
graphs but had never put the pieces together so I could get
my noodle around what was actually happening.)

Once the curve is determined to be 'good', you can find the
focus by either the reverse layout or reflected rays. To
reflect the rays (I'm not sure if test.bmp is supposed to
be an example?) I would ...
_ Put a couple of points on the conic reference projection
and lock them down with dimensions.
_ Create centerlines coincident with them, perpendicular to
the reference projection.
_ Create rays parallel to axis, coincident to points.
_ Mirror the rays about the centerlines.