checking the focal point on a parabola

[mmccurdy8P45Z]

I have created a solar dish using the conical command and revolve, based on the required parameters.  I need to double check the focal point. Is there a way to do this in Fusion 360? 

Diameter = 170" (this size is required for other machining later in the process)

Distance from the apex to the rim = 22.5

Focus = 81" (This is the critical number to maintain)

When using the conic command, I clicked on the two outside points representing the diameter, then a point 22.5" down, representing the apex. When I clicked on that point, the bottom of the conical curve (parabola), was up a bit from the 22.5. A number appears (.75). I'm not sure what that number does. I guess I have more than one question.

Mark McCurdy  

[SaeedHamza]

Could you add some pictures to make things clearer?

Regards

[etfrench · Accepted solution]

The mysterious number is the Rho value

A little Fusion 360 documentation.

And a list of ray tracing software as I don't think Fusion 360 can do this with conic curves.

Here's one way to draw an accurate parabola:

[TheCADWhisperer]

A parabola is a Conic Section.

A Conic section can be created in any 3D CAD app.

Slice the appropriate cone with appropriate workplane.

Project the cut edge.

Revolve.

[mmccurdy8P45Z]

Thank you for the help. I will redraw using your method. The rhino value is also very helpful and necessary to know. 

My only question still remains, can the final model be tested in Fusion, maybe rendering or animation, to check the focal point?

I appreciate all the help and information.

Mark McCurdy 

[etfrench]

I don't think Fusion 360 can test the focal point.  I added the link to ray tracing software for this purpose.  The Rho value is absolutely necessary to set the focal point.  The link @TheCADWhisperer supplied has the formulas necessary for calculating this.

[TheCADWhisperer]

It can be solved with geometry.

[untrustedone]

As I wanted to find the focus of a parabola, and i had just used some random numbers, here is my method for determining it. It is precise but not accurate, as you have to eyeball a tangency to the parabola. 

A parabola is that line where the distance from a directrix line and the distance from the focus are equal. Funnily enough, a circle has the same radius all the way around. So we need to find a circle which is tangent to the directrix and the symmetry line and whose center is on the parabola.

Create a sketch using the same plane that you split the cone. Project the edges of the parabola.

Using construction lines, (don't really need the profiles), construct a line that the parabola is symmetrical.

Create a perpendicular line to the symmetry line, on the outside of the parabola. This will be your directrix line.

create another line parallel to this line, and manually make it tangent to the vertex of the parabola.(this is the part that loses you the precision[feature enhancement:: tangency to conic lines] Once that is where you want it, fix it in place.

Create a circle somewhere that is not otherwise constrained. Create a line parallel to the directrix whose endpoints are on that circle and make the line coincident to the circle center. 

Make the circle tangent to the directrix and the symmetry line. Make the circle center coincident to the parabola.

(almost there)

Make a line from the circle diameter line to the fixed parabola tangent line. Make a line from the directrix to the parabola tangent line. add perpendicular constraints to those lines and the tangency line. 

Now make those last two lines equal. The circle's diameter line and the symmetry line intersection is your focus.

You can verify this by drawing any circle tangent to the directrix and whose center is on the parabola. It's diameter will pass through the focus.

[MichaelT_123]

Hi Fellows,

This is the problem middle ages astronomers with a help of goose feathers had grappled with

... successfully.

Consider finding three points on the curve and solve underlying set of equations.

More points and a curve fitting algorithm might result in the better approximation.

No goose chasing is necessary... Python scripting is at your fingers tips!

Regards

MichaelT

[untrustedone]

Hi,

Yeah, Python and me don't get along. I suppose I could spend several weeks learning, but my past brushes with Python were dismal failures. It took me half an hour to come up with the method, and a lot of it was searching how to make a parabola in Fusion.  

Just wanted to contribute a method for those of us that are mere mortals. 

Just a minor rant: having to use a solid to get back to a sketch is counterintuitive. The math is there, just not in 2D sketch land

[MichaelT_123]

All good 🙂, Mr. Untrustedone.

It is a mater of the purpose, the solution's scope and the efforts required.

Regards

MichaelT

[untrustedone]

To make accurate, you could make a sketch on a place tangent to the cone, create a line and project that on to your focus calculating sketch. Now you can make coincident the tangent line to the point.

[r.moss]

Hi, try this!

It is only 2D - a 3D version would be more of a challenge.

All you need to do is create a normal to the parabola and mirror your ray in it.  Fusion used to let you create a tangent to a conic curve at any point but it seems to be broken at the moment (you get the tangent to one of the end points instead).  As a work-around you can generate your own tangent or normal from the gradient.

It's easiest to first create some User Parameters (Modify - Change Parameters menu).  Let's define a mirror with axis along the y-axis and passing through the origin.  Some people like to think of a parabola as y = k*x^2, others as y = x^2/(2*R) where R is the radius of curvature at the origin (follows from the general equation for radius of curvature, R = ((1 + (y')^2)^(3/2))/y'' which simplifies to R = 1/y'' if y' = 0.  Then differentiating y = x^2/(2R) twice gives y'' = 1 / (2/(2R)) = R, I'll believe that!).

All parabolas are the same shape, in the same way that all circles are the same shape - the difference in size is just down to the k or R value used.  So a y = (x)^2 parabola could be enlarged by a factor of 2R in both the x and y directions if written as y = (2R) * (x/(2R))^2 = x^2/(2R).

I'll use R rather than k because we know that the focal length should come out as 2R.

So let's create R, X, Y  e.g. R = 2 m, X = 0.5 m, Y = X^2/(2*R).

I'll also create x as the ordinate where my ray hits the curve.  (It would be nice to be able to drag it but I don't think you can retrieve the resulting position).

It's easiest to position the control points for a symmetrical parabola.  On a sketch, place three points and add dimensions or constraints so they are at (X,Y), (-X,Y) and (0, -Y).  The last is the control point - when using rho = 0.5 the curve will pull down half-way towards i.e. it hits (0,0) as intended.

Now draw a conic curve through these and set rho = 0.5.

Now let's draw the normal.   The gradient of the tangent at our ray position x is y' = 2x/(2R) = x/R (tangent of angle to x-axis).  The atan function in F360 returns angles from -90 to +90 but negative values cannot be used in dimensions, so let's add 90 degrees and define theta = atan(x/R) + 90

That's the first screencast:

.

Unfortunately the sketch does not like the coincident constraint for the point on the curve and won't automatically redraw if you change the "x" parameter.    I think one can only have one screencast - see my next post!

Roger

[r.moss]

In this screencast I remove the troublesome coincident relation and replace it with (x,y) for the point that places it on the parabola.  I have also re-dimensioned the x-ordinate from a datum off to the left so that it never has to be negative.  You can see the reflected ray always cuts the y-axis at the same point (indicated loosely by the line I drew as a marker there).

Enjoy!

 

[r.moss]

Whoops, did I say focal length = 2R?  I meant R/2, sorry...