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°. ### The Trig We actually have 2 triangles to fight,  The Insert angle, and the Axial angle.  No sense in this being easy.

First we need to the height.  Lets work this 1 angle at a time.

Insert Angle – Get the height of the insert corner as the insert turns on its plane.  Cosine of the Insert angle x Side Length of the insert.

Ht1=cos(InsertAng)*SideLen = 4.480

Axial Rake angle – The height of the corner begins to drop as the insert plane is laid backward.  Imagine if it were laid down on it’s back.  The corner wouldn’t be very high at all. Cosine of the Axial Rake angle x The Height the corner is currently at.

Ht2=cos(AxialRakeAng)*Ht1 = 4.473

Next we need the width of the triangle, represented as the difference between the reference dimensions Inventor provided us.

Ddif = D1-D2 = 2.147

Now, the angular result.

From 90° we will subtract the arc tangent of the Ddif divided by the Ht2 height

AngRes = 90-atan(Ddif-Ht2) = 25.636°

You can load these equations in 1 at a time in the user parameters, or throw the whole thing at the feature.  I highly suggest that you step through them carefully, and then when you get a stabilized result, prepare a composite equation on another user parameter, until it stabilizes.  Then apply it to the feature (I used Copy/Paste, and then tossed the test parameter).  Troubleshooting is much easier 1 piece at a time.

This is the composite:

AngRes = 90° – atan((D2-D1)/(cos(AxialRakeAng) * (cos(InsertAng) * SideLen)))

### Results I have no room for mismatch in my designs, so I go ape in order to figure out why something just doesn’t seem to line up.  We could have over compensated for the mismatch by recording key offsets as we changed both angles from min to max, and then applied an angle that did not interfere with the design. Why?

Next time this comes up, ask yourself…..Is there a result that will work?

If the answer is yes, the it can be calculated!! Both by Inventor work features/planning, and by equations.

The trick is just looking at the problem from what the FACTS of the requirement must do in order to be accomplished, in relation to what you have now.  Don’t concentrate on the feature.  The Feature will emerge from the result.