Custom projection view in Creo

[IVsergiu]

Hello everyone!
I want to share the information how to create custom projection views for diferent scenarios.
It took me some time to get this done.
If you want to create ISO, Dimetric or custom projection views you can do it by using these formulas:

a= atan(sqrt(tan(α)*tan(β))/(tan(β))) - angle of rotation on vertical axis

b= asin(sqrt(tan(α)*tan(β))) - angle of rotation on horizontal axis

α and β - are the angles you want for projection view (ex. Dimetric view α = 7,  β = 42)
a and b are the input angles for rotation of the part, as a result you get α and β for the projection view.

The best thing is to add a mapkey to do it for you:
I've added this to my config file:

!Dimetric view
mapkey ww1 @MAPKEY_NAMECreate a dimetric view;@MAPKEY_LABELDimetric view 1 (ww1);\
mapkey(continued) `igToolbar_AncestorIGT_IGT_GRP_inh396938342.proe_win:casc340798662`;\
mapkey(continued) ~ Command `ProCmdViewOrient` ;\
mapkey(continued) ~ Select `orient` `SetupOptions` 1 `dynorient`;\
mapkey(continued) ~ Update `orient` `spinPH.YSpinBox` -20.268067;\
mapkey(continued) ~ Activate `orient` `spinPH.YSpinBox` -20.268067;\
mapkey(continued) ~ Update `orient` `spinPH.XSpinBox` 19.420537;\
mapkey(continued) ~ Activate `orient` `spinPH.XSpinBox` 19.420537;\
mapkey(continued) ~ Activate `orient` `panzoomPH.ZoomFitPB`;\
mapkey(continued) ~ Input `orient` `NameVw_IP` `08_dimetric_1`;\
mapkey(continued) ~ Update `orient` `NameVw_IP` `08_dimetric_1`;\
mapkey(continued) ~ Activate `orient` `NameVw_PB_Save`;~ Activate `orient` `OkPB`;

ww1 - is the shortcut to trigger the mapkey.
I hope this helps.

[tbraxton]

I think you may be off (slightly) on the values you are using for dimetric projection. In practice this would not be spotted in most applications. If you use something like Mathematica or a scientific calculator, I think you will find that you get the values below for the Dimetric angle equations rounded to 3 decimal places.

In DIMETRICAL projection the three sides have a proportion of 1:1:0.5

DIMETRICAL 3D views based upon a vert. angle of -20.705 (= ArcSin( Tan (ArcSin(1/3) ) ) degrees and a horiz. angle of 19.471 (= ArcSin(1/3) )

[IVsergiu]

The formulas I've posted are correct.
Bellow I've attached a picture from a drawing, where the calculated angles are almost the standard ones (it depends on the number of decimals for the input angles when the view is created)

 

[tbraxton · Accepted solution]

I am posting this for reference only. You will see that the angle values I posted above more closely match the target ratio of the projected lengths. One is left to determine what is good enough for themselves. Neither of the angle values result in the exact ratios of the projected lengths.