lamensterms Posted November 19, 2022 Posted November 19, 2022 Hey, Wasn't sure which forum was best to post this question... I've got a 3D model and a 2D isometric view (FLATSHOT) of the model. I'd like to reproduce the viewing angle (camera) of the 2D iso, in the 3D model. So I can create another 2D viewing from the exact same angle as the original 2D. Anyone know how to do this? Quote
lrm Posted November 19, 2022 Posted November 19, 2022 By "2D isometric view" do you mean one of the 8 potential isometric views or do you seek to duplicate the specific random axonometric projection of the 2D flatshot geometry? Also, although you use the term "camera" should it be assumed that you are dealing with a parallel projection. Camera views may yield a perspective projection. If a true isometric, just choose one of the predefined isometrics in the view command or you can rotate the model in 3D space. From a principal view, rotate the 3D model by + or - 45° about the vertical axis then + or - 35.2644° about the horizontal axis to yield an isometric orientation. If you need to reproduce a random axonometric project it would help to know 3 points that lie on a plane that is parallel to the viewing plan or 2 points that lie on a line perpendicular to the viewing plane. If you don't have these points then you would need to know the x, y, z coordinates of points in the 3D model and their corresponding u, v coordinates in the 2D projection. 1 Quote
ReMark Posted November 19, 2022 Posted November 19, 2022 What was the exact angle of the 2D isometric view? Quote
lrm Posted November 19, 2022 Posted November 19, 2022 I forget all about the Excel VBA "Trimetric View Calculator" program I wrote earlier this year. It might be of help to you. If you know the angles of two of the principal axes (assuming the third is vertical) it will calculate the two angles required for the VPOINT command. For example, if you know that in your trimetric view one principal axis is at an angle of 40° going up towards the right and the other 20° going up to the left then you would fill in the Excel worksheet as follows. Hitting the Calculate button would yield the results you see. You can enter these value into the vpoint command as follows: AutoCAD would then reorient your view to match the input requirements. The VBA programs uses numerical methods to find a solution so the time to compute an answer and the precision of the answer is a function of the input data. Note, I need to do a little more testing. The left and right axis input labels may need to be reversed. Trimetric view calculator VBA.zip 1 Quote
BIGAL Posted November 19, 2022 Posted November 19, 2022 Lrm could you just pass to (command "-VPOINT" X,Y,Z) You could just write the correct string to a cell and copy and paste to command line, no need to fill in dcl. 1 Quote
lrm Posted November 19, 2022 Posted November 19, 2022 Great suggestion @BIGAL I added the following to cell F17. Its contents can be copy and pasted to the command line. Make the top view active before using it. =CONCATENATE("(command -vpoint ",CHAR(34),CHAR(34)," ",CHAR(34),A17,",",B17,",",C17,CHAR(34),")") Quote
BIGAL Posted November 20, 2022 Posted November 20, 2022 Glad to help. Check what you have CHAR 34 -vpoint Char 34 Quote
lamensterms Posted November 21, 2022 Author Posted November 21, 2022 Hi lrm and everyone. Thanks for the replies. I have had a very quick look at the spreadsheet any I think it will work. Just FYI here are the angles of the view I am trying to reproduce Quote
lrm Posted November 21, 2022 Posted November 21, 2022 I fixed the -vpoint commad cell so all you need to do is copy/paste the contents of cell F17 to AutoCAD. Here are the results I got for your angles. Is this close enough? I'd like to improve the program and make it faster and more precise. Trimetric view calculator VBA.02.zip 1 Quote
BIGAL Posted November 21, 2022 Posted November 21, 2022 Just had a look at this (getvar 'Viewdir) try with some different views. Quote
SEANT Posted November 22, 2022 Posted November 22, 2022 This topic had come up on another forum. https://www.cadforum.cz/forum_en/forum_posts.asp?TID=12817&PN=1&title=ellipses-with-degree Sadly, the spreadsheet I linked to that post no longer appears available. It used some interesting geometric/trigonometric manipulation to acquire a direct computational result. If there's interest, I could recreate the spreadsheet as well as geometric proofs (.dwg files). Enjoy the holidays. Quote
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