DRAFT ANGLE

[[Fe...]]

Good morning, how can I calculate the 10 degrees as shown in detail Y? Thank you.

 

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[[Ma...]]

You can either scan it like a cone or like a line.

For cone you then report half of it's angle.
For line you then report angle between features ( old Calypso has that close to profile callouts )

[[Jo...]]

Looks like the angle is relativ to the the extremly short area (1,924). That is in general not a good idea because minor deviations (form) on that short area will result in big deviations in the angle.

Theoretically the angle should be the same as to the area "11,14". But you sould have of course detailled knowledge about the usage/installation to just replace the the one reference of the angle.

I would measure both and compare to get a feeling for the situation.

[[Fe...]]

So can I scan the 99.5 circle and any circle of the cone and insert it into the cone half-angle feature?

[[Ma...]]

For a cone you need at least two circular paths or helix to obtain it's angle.

That 10° is meant from axis - not that little area.

[[Jo...]]

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How do you tell by the information given?

The line is clearly on the the "little area".

Because it would make sense or any hard tell?

[[Ma...]]

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There is a runout callout to A, then i assume that 10° is half of that cone or angle of line to A axis. I can not tell more since i don't see if it's full circle/cone or just a half.

[[Jo...]]

I share your assumption.

The drawing is kind of bad in that regard. They could have drawn the angel to the axis.

[[Fe...]]

[[Fe...]]

So they can be more understanding

[[Ch...]]

Ciao Frederico,

It would be best if you measured a cone with several lanes, then you have the cone angle.

[[Fe...]]

I need to calculate the angle between the Ø99.5 circle and the cone

[[Ma...]]

That circle is too short to obtain plausible result.

Edited yesterday at 09:31 AM

[[Ri...]]

From what I see you have about 34.576 of length to scan; I don't see the angle as only in the 1.924 reference dim.

I cone with 3 or 4 circular scans should be sufficient.