Theoretical sphere location

[[Da...]]

I am trying to find a true position of a theoretical sphere placed in a partial cone. How the inspection is described is-- that you place a .4375" calibrated sphere and then calculate position in y and Z axis on a 20 degree offset. See attached inspection print for more context
All of this to say is there a way to mimic this using calypso. I currently have a theoretical sphere at the diameter 0.4375 at a formula-calculated x and y (same as a cone from the model). How do I find where the sphere lies tangent to the cone?

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[[Ke...]]

Construction->Tangent. You will put your sphere diameter in there.

[[Ri...]]

When you say partial, are there no two surfaces that are opposing?

If there are at least two opposing surfaces, you could trace two linear opposing lines up the wall of the Cone, and then use the Tangent construction to get what you want.

If you don't have opposing surfaces, you will have to use formulas to trig out the cosine of placing the ball up against the surface. I don't remember exactly how to do it because I only did it once years ago, and I had to have someone from Zeiss help me.

I'm certain that there are much smarter people than me in here that can help with the math.

[[Ri...]]

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You have to have two features to create the tangent with.

[[Da...]]

Either the length OR the diameter has to be a basic number. The variation would be on the other dimension.

Need more print to understand fully.

[[Jo...]]

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Had a similar print with a measure over pin distance. If Datum C is the Basic, you could offset it, then offset the angled line (the ball radius) to intersect. That would provide the missing dimension.

[[Da...]]

first off is that 20 degrees per side (40 degrees included) ?
How close are you holding the 20 degrees ? ( do you want the formula to adjust based on variation from the 20 degrees?)
How much of the cone is available to check ?

Calypso used to have trig functions in the formula drop down. Now it is gone (2019). I don't remember the the syntax for all the formula functions. I can write the formula and maybe someone can help with the syntax, you should be able to do this in formula...
Need the questions answered any ways.

[[Cl...]]

This is just me thinking out load to get some brainstorming started. I might be way off base here. Let me know.

Scan a line on the inside edge (center) of the cone, recall that lines points (if you can) into a plane. Move the plane into the cone .21875 (sphere radius).Create another theoretical plane at the moved plane location, and rotate it opposite whatever angle the first plane is?. Intersect the two planes. Recall the intersection into a theoretical .4375 sphere. Create a tangent from the theoretical sphere and scanned line, intersect the tangent with the scanned line. You should get a point at 1.1754 from -C-

I have not tried any of this so, not sure it it is right or will even work. Like I said, just thinking out load.

[[Da...]]

So... 10° half-angle, 20° full-included angle. approx. 160° of rotated cone available.
So what I did was I scanned the cone; and since the angle of interaction to the sphere has to be 10°, I determined that the diameter of the sphere at interaction had to be (=2*sphere_radius*cos10°). Using the cone addition function, I could find the Z value of the cone at the calculated value. Then I created the theoretical sphere with a calculated Z value (from Cone addition + (sphere_radius*sin 10°).
This seems to get the desired result for the sphere based on 1.279 to CL of the sphere. I'll try a couple of the other ideas to see if I can think through that logic and maybe get more reliable results.
Thank you for the input. Always want to learn more

*Credit to David Wilkens at Inspection Engineering for his included input as well