Inventor – 3D Axis Calcs Part 2

When we left off the last part of this session , we had discussed the cone plane intersection, where the cone angle had to be calculated from a non radial intersection.  This wasn’t the end of the world, we had one angle to deal with, the pocket angle.  We even discussed how to do the job with less trig, and more Inventor Work Features.

This time we will cover throwing 2 axial rotations at the cone intersection:

The Insert Angle and the Axial Rake Angle

Last time we discussed having the reference point to axis relationship vary as it rotated on a plane.  What happens when we rotate the plane they are on?

We will cover the following topics:

• Another Angle
• More Trig
• The Results

Another Angle

Take a piece of paper, and stand it up in front of you, bottom edge flat on a table, arms out in front. Now rotate the paper to the left, keeping some edge or corner of the paper touching the table (that was our Insert Angle from last week).  The upper left corner’s height off the table has dropped.  Hold that position.

Last week that height was the key to solving our cone angle. 

Now rotate the paper away from you, while maintaining the side angle.

The height of the upper left corner off the table is continuing to drop.  Inevitably, when the angle away from you reaches 90° flat on the table, that corner height will be 0.  Keep in mind that the known Insert Angle from last week has not changed, yet H is now 0°.

 

Inventor – 3D Axis Calcs Part 1

I often get wrapped up with ‘having to know the solution’, and I’m a sucker for a design challenge. My wife just loves it. If I just learned some equations, I wouldn’t understand the process.  That’s only useful in 1 instance, and not adaptable.  I gotta know how and why. Like when I was figuring center of gravity while calculating the direct material volume replacement for differing densities to achieve a target combined mass. That was the pinewood derby, using Inventor. (I established the equation using linear vector intersection,  and yes it is linear)

That’s what brings us to our current discussion, “arriving at the resulting angle of the edge of a cone as it passes through a plane that is neither radially or axially orthogonal to the centerline axis of the cone”. The farther off plane you get, the narrower the angle becomes.

We will cover the following topics:

• The Angle
• The Facts
• Model References
• The Trig

The Angle

I need to cut a truncated cone out of the head to create a skirt.  it needs to lie exactly on the edge of the dashed construction geometry line at the base of the pocket. 

 

We know that the Insert is angled at 30°, so why not feed that to it. So we will.

Inventor – Projected Geometry and Non-Orthogonal Planes

Due to the fact that this is the first 2010 article, I wanted to do something useful to most everyone, especially newbies. There is not much new in the way of Projected Geometry and Cut Edges in 2010.  I did run across some new issues however.

I spend some of my time redesigning parts and especially tooling so that the design intent can be achieved, on more limited machinery than the original design permitted.  In this case I was working with an indexable endmill using square inserts, greater pitches, and limited by a 3-Axis mill.  The next few articles will be pulled from what I learned on this particular part.

What I was doing

During the redesign, I decided to approach it from an axis orthogonal view, using trig to offset sketch planes.  In this case the key factors are the Flute plane, and the Insert Pocket plane.  The angular relationship is variable to the Z-Axis, both Radially, and Axially.

The offset sketch planes would receive the projected geometry to calculate an edge, and then I would hinge the respective Flute and Pocket planes at those projected edges (the axial relationship).  Then more sketching and cut extrusions.

The Problem

The problem lies with Projecting Geometry onto a sketch that lies on plane that varies in it’s orthogonal relationship to the plane the geometry was projected from.  It’s fine until you change the angular relationship.  You won’t notice it until something is dependent on that projection.  THEN it becomes a problem.  If you don’t vary the angular relationship until after building off that projected edge, and it’s continued dependencies, it becomes a disaster.  The whole thing lights up like a Christmas Tree.

Oddly enough, when you return the plane to it’s original angle, everything goes back together.

Above: The flute plane is Orthogonal and aligned to the Z-Axis.  No problems in the browser.