[SH...] Posted July 4, 2020 Share Posted July 4, 2020 How can I check these two holes PCD in calypso without using formula??IMG_20200704_172636.jpg Link to comment Share on other sites More sharing options...
[Da...] Posted July 5, 2020 Share Posted July 5, 2020 OK, I first want to be sure - is P.C.D. pitch circle diameter? I can't imagine the term right, what pitch? Anyway, Shabu why don't you want to use formula? Link to comment Share on other sites More sharing options...
[SH...] Posted July 5, 2020 Author Share Posted July 5, 2020 Please sign in to view this quote. Yes, it is pitch circle diameter,l don't know, in the drawing it mentioned like that. Using formula l can only calculate the diameter of individual holes with respect to centre. I want to calculate diameter including two holes. Or I have to calculate two diameter and add it then divided by two?? Link to comment Share on other sites More sharing options...
[Ja...] Posted July 5, 2020 Share Posted July 5, 2020 We can do it by recall option. First measure three circle. Then create a new circle from feature menu. In the option select recall and select three circle. It will create a circle passing from three circle. Link to comment Share on other sites More sharing options...
[SH...] Posted July 6, 2020 Author Share Posted July 6, 2020 I have only two circles Link to comment Share on other sites More sharing options...
[An...] Posted July 6, 2020 Share Posted July 6, 2020 To calculate a circle, Calypso needs at least 3 measuring points or three circle elements from a recall. without a formula, you won't be able to calculate it. it also makes no sense to specify polar coordinates in xy-axis and a pitch circle diameter Link to comment Share on other sites More sharing options...
[Ro...] Posted July 6, 2020 Share Posted July 6, 2020 Thats a terrible callout....... the only thing i can think of would be to cheat and create some 'added' data points (read false data points) by creating a couple theoretical circles where the X&Y are "(getActual (circle1),x)*-1", so that circle will appear in the opposite quadrant than the actual circle. You'll end up with 4 points of data, but i don't know if it will be any good. Link to comment Share on other sites More sharing options...
[An...] Posted July 6, 2020 Share Posted July 6, 2020 If you convert the polar coordinates into a circle and use the tolerance of 0.2 mm, you get a pitch circle diameter of Ø 76.532 mm with full (+) tolerance and a Ø 75.438 mm with full (-) tolerance. The tolerance of + - 0.2 mm for the specified pitch circle diameter can only work if the polar coordinates are precisely maintained. If I calculate the polar coordinate with a (+) tolerance and the other with a (-) tolerance, the dimension is still within the required tolerance of the pitch circle diameter! The tolerance of one polar coordinate may only be + - 0.1 mm, the other may only be + - 0.05 mm, only then can the tolerance of 0.2 mm of the pitch circle diameter be maintained! It would be better to provide the two polar coordinates with a tolerance of +- 0.05mm, then the pitch circle diameter is always within the required tolerance. Link to comment Share on other sites More sharing options...
[An...] Posted July 7, 2020 Share Posted July 7, 2020 Hello Shabu, I did the math. I have some pairings here. This mating would fit, but it's dangerous because the worker one polar coordinate with (+) tolerance, the other must machine with (-) tolerance. Calculated with full tolerance utilization. 18,80mm / 33,10mm = Ø 76,132mm pitch circle diameter 19,20mm / 32,70mm = Ø 75,840mm pitch circle diameter With this pairing, the tolerance of +- 0.2mm of the polar coordinates cannot be fully exploited, if one would do so, the tolerance of the pitch circle diameter will be exceeded. 18.94mm / 32.84mm = Ø 75.820mm pitch circle diameter 19,08mm / 32,98mm = Ø 76,202mm pitch circle diameter If the polar tolerance were set to + - 0.05mm, the tolerance of the pitch circle diameter would always be conform if the tolerance was fully utilized. Tolerance greatest against greatest 19.05mm / 32.95mm = Ø 76.120mm Tolerance smallest against smallest 18,95mm / 32,85mm = Ø 75,846mm Tolerance largest against smallest 19,05mm / 32,85mm = Ø 75,948mm Tolerance smallest against largest 18,95mm / 32,95mm = Ø 76,020mm I think you should talk to the designer about which dimensions are more important to him. Link to comment Share on other sites More sharing options...
[SH...] Posted July 8, 2020 Author Share Posted July 8, 2020 Please sign in to view this quote. Sorry for the late reply, thanks for your effort in my post, I got it what your telling but I don't think they will change the drawing. I went with formula.. Link to comment Share on other sites More sharing options...
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