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isometric / dimetric extrude

Mar 31, 2013 4:19 PM

Tags: #3d #rotate #extrude #isometric

Hi,

 

I'm trying to figure out a way to extrude an object at 26.565° from the left and right side of the object. The default isometric settings are for 30° isometric angles, but for the purpose of my project, that is too steep and I would like to use a typical 1:2 "isometric" ratio.

 

JETaImage has already provided a formula for creating this from the TOP of an object here http://forums.adobe.com/message/4535692, which involves rotating the object 45 degrees, and using extrude at 60°, 0°, 0°.  However, I'd like to be able to accomplish the same angle from the left or right side of the object.

 

 

This image shows the default isometric extrusion on the left (notice how it does not match the 26.565° (1:2) grid.  The box on the right is extruding from the top and fits the grid.

Screen Shot 2013-03-31 at 6.08.52 PM.png

 

Does anyone know the angles I need to extrude a shape from left / right, rather than the top?

 

Thanks!

 
Replies
  • Currently Being Moderated
    Mar 31, 2013 4:48 PM   in reply to growing

    First, are you, "growing", the same person as "not an artist 5" in the referenced thread? (Why are people so afraid to use their real names in this forum?)

     

    JET

     
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  • Currently Being Moderated
    Mar 31, 2013 5:15 PM   in reply to growing

    The box on the right is extruding from the top and fits the grid.

    What grid? Please show the grid you are refering to. I want to see the grid from which you arrived at the vertical dimension of the right box.

     

    JET

     
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  • Currently Being Moderated
    Mar 31, 2013 6:24 PM   in reply to growing

    But your requirements are ambiguous (and actually contradictory). On the one hand, you say you want to use the 1:2 so-called "isometric" oblique cheat and you want to create objects with 3D Effect that will fit the 1:2 cheat grid, which—as thoroughly explained in the other thread—is not an isometric construction and is not even an actual orthographic projection of a cube, so is therefore also not a dimetric construction.

     

    On the other hand, you say your box on the right—drawn with 3D Effect according to the method I described in the earlier thread, and which I assume is supposed to be a cube—is what you want. But if that is supposed to be a cube conforming to the oblique cheat, then its height is wrong for that grid.

     

    Is the box on the right supposed to be a cube? If so, is it supposed to be a cube drawn according to the oblique cheat? If so, then how did you arrive at its extrusion height?

     

    The answers to those questions have a fundamental bearing on the answer to your desire to be able to do extrusions along the right and left axes rather than the vertical axis.

     

    JET

     
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  • Currently Being Moderated
    Mar 31, 2013 6:51 PM   in reply to growing

    Then is the height proportion of everything in your drawing "just random?"

     

    Suppose your drawing needs to have a 1x1x1h cube and a 1x1x2h box in the same drawing. Don't you intend for those to be in proper proportion to each other? And do you intend for the heights to be in true orthographic proportion to the other dimensions, or do you intend for the heights to be proportioned in accordance with the oblique cheat grid?

     

    Said another way: Suppose your drawing has two cubes (1x1x1) in it. One of them is extruded along the vertical axis. The other is extruded along the right axis. Don't you intend for those to look like same-shaped cubes?

     

    Illustrator's 3D Effect only extrudes along one axis: The axis that is perpendicular to the plane on which the original path is drawn. Your extrusion depth (height of the box) may be "random" when the original path resides on the top plane, but its height is NOT going to be "random" when the original path resides on the left or right plane; the height in those cases is going to be determined by the original path.

     

    JET

     
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  • Currently Being Moderated
    Apr 1, 2013 8:23 PM   in reply to growing

    Growing,

     

    In this PDF, I am assuming that you really are not talking about the fake "isometric" convention but merely want to use 3D Effect with the left and right axes at 26.57°-from horizontal.

     

    Here's the thing: 3D Effect's interface is quite cumbersome because of the behavior of its roll, pitch, and yaw value fields, and the omission of numerical input fields for the proxy axis rotations. You can enter decimal values in the three fields, but the next time you re-open the already-applied effect from the Appearance palette, the value you entered automatically round to nearest integers and screw up the carefully applied values (and thereby screw up your drawing). And manually dragging the proxy cube axes results in herky-jerky redraw (especially in CS 5.5 which is extraordinarily frustrating.

     

    The PDF demonstrates two methods to do what you want, but if it were me, I'd use the second method, not the first. (Actually, I'd use neither; I'd just set up for dimetric drawing and not mess with 3D Effect.)

     

    First method (Page 1 in the PDF):
    Not as accurate as the second method. So if you need to "stack" or "abut" multiple objects, some ugly misalignment may become apparent, especially on long extrusions.

     

    Second method (Page 2 in the PDF):
    You should have less misalignment trouble using this method, because you enter just one value in the pitch field and it is an integer.You do all extrusions along the Y axis, as described in the 2012 thread. But instead of rotating the initial artwork 45°, you rotate it plus or minus 116.565°, store the result as a Symbol, and then rotate the Symbol Instance on the page. (This is so that the 3D Effect will rotate with it.) Do edits by editing the Symbols.

     

    JET

     
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