Drawing Cube in Dimetric Projection

[basty]

I would like to draw 1×1×1 cube in dimetric projection.

This is the standard rule of angle and length of dimetric projection drawing (see attached image called "dimetric theory" please).

Can anyone tell me please the length of OZ in order to draw 1×1×1 cube in dimetric projection? (see attached image called "my dimetric cube" please)

Thank you so much for anyone help.

 

 

 

Edited September 19 by basty

[lrm]

The length of OZ should be 0.9515.

 

To create the dimetric view you can rotate  a 3D cube by 45° about a vertical axis then tilt it upward by an angle alpha.   Given the angle of 96° in your drawing the angle alpha is:

alpha = asin (tan(90 - 96/2)) = 64.2112°

 

The scaling factor OZ = SQRT( 1 - [cos(45)*cos(alpha)]^2)   = 0.9515

 

The illustration below shows how I derived the scaling function.  The red lines are the projections of the cube edges to the viewing plane.

[basty]

i still don't understand. Can you explain please using another word and step-by-step instruction?

[basty]

How do I rotate it and tilt it upward since the object is 2D, not 3D? And can you tell me please how do you get that trigonometry formula?

Edited September 22 by basty

[lrm]

To understand the formulas and dimetric projections it is best to first think about how a 3d cube needs to be rotated relative to a viewing plane to generate the desired view. 

The cube below has an A on its front side, a B on its right side and C on top.

Rotating the cube by -45° about the vertical (Z) axis we have the following (the red box stays stationary and serves as a reference).

 

The front view looks likes this.

If the cube is rotated about a horizontal axis (relative to the screen) the cube will appear as a dimetric view.

For example, rotating the cube by 10° yields:

 

If the cube were rotated by 35.2644° instead of 10° we would have the following.

An isometric view!  Note that the angle 35.2644 is the atan(1/sqrt(2)).  The height of a unit cube divided by the diagonal of its base.

 

You posed the question what would be the length of the line if the projected angle was 42° (90 - 96/2) instead of 30°.

In the image below the cube has been tilted up by an angle alpha.  Since it is a unit cube the distances AB and BD are cos(alpha).  

Inspecting the drawing we can conclude that:

tan(90 - 96/2) = DC/AB = DC / cos(45)

or

DC = cos(45) * tan(90 - 96/2)

and

sin(alpha) = [cos(45) * tan(90 - 96/2)]/cos(45)

or alpha = asin(cos(45) * tan(90 - 96/2))/cos(45)

 

The projected length L, of AD becomes:

L = sqrt(1^2 - (cos(45) *cos(alpha))^2) = 0.9515

 

To duplicate the orientation of your original drawing the cube should be rotated by 132° about an axis perpendicular to the viewing plane.

 

 

The dimensions I have noted here are for a dimetric projection. In you original drawng it looks like you want a dimetric drawing. To make XO and OY equal to 1 the dimensions above should be scaled by (1.0 / 0.9515) or 1.0501 making OZ = 0.4351 * 1.0510 = 0.4572.

 

Dimetric projection:

 

Dimetric drawing.

 

 

 

 

 

 

 

Edited September 23 by lrm

[basty]

How do I rotate the cube by -45° about the, for example, vertical (Z) axis?

[lrm]

Use the rotate3d command. I find it much more useful than the 3drotate command.

[basty]

You said I should use the rotate3d command. But you said much more useful than the 3drotate command. I am really confuse here. Explain please.

[lrm]

Review the documentation for the rotate3d command and see what options are available. Note all the choices that you have for choosing the axis of rotation.

[basty]

Which one of these 3D views you're using?

 

 

[lrm]

Top.

[basty]

Your cube on 1st pic, in 5th post on this page doesn't look like a Top view. It's an isometric view. Why do you say it's a Top view in your previous post?

[lrm]

@basty

 

Here's a cube in the top view.

 

Rotate it about y axis by -45°,

Now rotate it about x by the angle alpha = 64.2112 yields a dimetric with the 96° angle of your drawing.

 

Rotate by 132 to duplicate the rotation of your original post.

Since your original post is a dimetric drawing and not a projection) it uses full size dimensions for two principal dimetric axes.  The drawing should be scale by 1/0.955 = 1.0510 yielding:

 So the answer to your oringinal question is to use a factor 0.4572 for the foreshortening.

 

 

 

Edited September 29 by lrm

[basty]

How do I rotate it about the y-axis by -45°?

 

At what base point I should specify?

[lrm]

Quote

How do I rotate it about the y-axis by -45°?

 

@basty Did you do your homework assignment from my Sept 23 post "Review the documentation for the rotate3d command and see what options are available."?

 

Quote

At what base point I should specify?

 

It doesn't really matter where the base point is.  Best to give it by the cube.

 

[basty]

That's the problem, I don't have the help files. If you don't mind, can you direct me please to the online help files?

[lrm]

google autocad rotate3d

[eldon]

@basty

 

Whilst you are google-ing autocad rotate3d, why don't you google dimetric projection?

 

The information that I found says that the dimetric projection is a three dimensional pictorial representation of an object. I understand that to mean that as long as dimensions in two planes are accurate, the dimension in the third plane is almost arbitrary. Just has to look reasonable.

 

If you want to draw a 3d cube in autocad, and rotate it, you are no longer drawing a projected view.

[basty]

I have this message when trying to do the rotate3d command:

 

 

What does it mean and what should I do?

 

[lrm]

@basty

Quote

What does it mean and what should I do?

 

Specify a point and then you will be prompted for an angle.  Specify 45 and see what happens!

 

Now undo and try again this time with a different point but still use 45 for the angle and look a the result  What do you see that is different from the two actions? What conclusion do you make about the affect that the point makes?