How to make a circle on an isometric drawing that is not on one of the three iso planes?

[vanrhyn_m]

I am trying to create an isometric drawing of an object which has a circle on a slanted surface. I know how to use iso circle to create circles on a left/right/top plane, but this surface is not one of those planes. I am sure there's a way to use ellipse to draw a circle in this kind of perspective, but it is escaping me so far!

[Patchy]

The hole is not perpendicular to the surface.

That figure is incomplete, if the centerline of the hole doesn't show parallel to the edge of the block then it needs another view showing the angle, you can't draw something that doesn't have enough information.

Look at the 2 red lines.

[Washingtonn]

Draw out the orthographic views as shown below to get the extents of the circles from which you can use the AutoCAD ellipse command

[Valentin_CAD]

@vanrhyn_m ,

Is much easier to draw this in a 3D Solid and use VIEWBASE to obtain all your views.

Not recommended, but you could also Explode a ViewBase generated from a 3D model to obtain a 2D view.

• How to Explode a ViewBase generated from a 3D model?

        

[Patchy]

Whoever drew this nonsense, it said here the hole 1 unit dia. is drilled thru, does it look like thru at all?

[Washingtonn]

Exercises, such as this one, are not necessarily drawn correctly but typically have sufficient dimensional information which if used can be drawn correctly – possibly having to rely on an assumption* along the way. It is an exercise for students to gain experience reading and understanding sketches, using a CAD program (or even manual drafting techniques), and preventing users from just tracing over a picture.

As for it being easier to make a 3D Solid and then use VIEWBASE to obtain all of your views… Yes, but ask the question - “what is the learning goal of the exercise?”  A new user just learning about orthographic and isometric drawings?

*Assumptions:

The centerline of the circles on the slanted face are midway along the 2.875 width dimension.

The dimensioning takes precedence over graphics.

[leeminardi]

@vanrhyn_m Keep in mind that lines in an isometric drawing that are not parallel to one of the principal axes should be drawn by transferring their principal axis dimension to components measures along the isometric axes (vertical, 30° or 150°).  It can be helpful to create an orthographic sketch of the inclined surface to determine measurement to use in the isometric.

For example, we know that the surface of interest has a slope of 2.375 : 1.5. The location of the centerline is 1.75 up the slope and the hole has a  1.5 dia.  From this we can determine the dimension to use in the isometric drawing.

Given these values we can create the following to locate the centerlines of the ellipse.

Note that the 0.9345 line is at an angle of 150°.

The top of the ellipse is determined as follows:

With the parallelogram for the ellipse it is easy to pick of the point for it major and minor axes.

[Kent1Cooper]

Suggestions already made will get you a long way toward a solution.  The tricky part is the accurate sizing/rotation of the eventual Ellipse so that its tangency points with the enclosing projected square land at the midpoints of the square's edges as they should.  You can't just use those midpoints as axis endpoints of the Ellipse, or the result will bulge outside the square edges.  Here, I used the upper left and lower right midpoints of the projected square as axis endpoints for an Ellipse, and PERpendicular Osnap to the other two edges to define the other axis, so at least the Ellipse is tangent to the long sides, though not at their midpoints.  The yellow is where the Ellipse bulges outboard.

Of course, the problem is that the axes of an Ellipse must be perpendicular, but the projected axes of the projected square in this situation certainly are not.  The "right" Ellipse must have axes that are not parallel with the square's edges.  You may need to eyeball aspects of it all.

See >this< about just this question.  They didn't have a way of defining the Ellipse precisely.  What they quoted [in red] from me is actually incorrect -- there's some true Ellipse that will do it, but not based on what you expect, except for its center.  See also the Wikipedia link at Message 12 there, if that helps [I haven't studied it carefully].

[leeminardi]

@Kent1Cooper you are correct that determining the ellipse that fills the parallelogram and is tangent to the midpoints of its sides is tricky!  The ellipse I drew does not meet these requirements and does slightly bulge.  In the image below, the green ellipse is the correct ellipse. It is tangent to the parallelogram at the midpoints.  I created it from a 3D solid model.  The yellow ellipse incorrectly uses the parallelogram centerlines as the major and minor axes.

[dany_rochefort]

@vanrhyn_m  Anybody who has attended an old school drafting course will tell you that there is only one way to do this correctly without faking it.  Just google : Ellipses in isometric perspective

https://www.peachpit.com/articles/article.aspx?p=2873372&seqNum=19#:~:text=The%20centerlines%20in%20...

 

[dbroad]

The angled line running through the ellipse is the major axis. I can't think of a way other than creating the 3d solid to conveniently determine either the ellipse rotation or minor ellipse distance.  The major axis length is 1.5 for the outer circle.

[leeminardi]

@dany_rochefort 

I took drafting way before there were computers and the manner in which we learned to draw the intersection of a cylinder and a plane was as you outline.  As no freehand drawing was allowed we used French Curves to ink the curved lines to get a clean results.  For the youngsters out there who never had manual drafting, here are some French Curve templates.

The ellipse you show is straight on to the plane of the ellipse, its true shape.  Unfortunately, this is not what needs to be draw in the OP's isometric drawing.  What is needed is the projection of the ellipse as viewed in an isometric drawing to the viewing plane.  As a result, the major and minor axes of the correct ellipse are not coincident with the centerlines of the ellipse's bounding parallelogram. As noted in my post (#9), the angle of the major axis from the edge line of the plane is about 13.9332°, the semi-major axis = 0.9185, semi-minor axis = 0.7316.

Since the deviation between the correct ellipse and the approximate ellipse you offer is small, it's close enough for an isometric drawing and most users would be satisfied with it.

Creating a 3D solid CAD model is the best way to go if you want a more precise and accurate isometric drawing.

To create an isometric drawing from a 3D model use rotate3d to rotate the model by + or - 45°about the y axis and the + or - 35.2644° about the x axis. Scale the result by 1.2247449  (i.e., sqrt(3)/sqrt(2)) and then use flatten.   

[Kent1Cooper]

---

@Kent1Cooper wrote:

....

See >this< about just this question.  ....  See also the Wikipedia link at Message 12 there, if that helps [I haven't studied it carefully].

---

But it really works!  I'm considering how that might be automated for any selected parallelogram Polyline.  Later....

[leeminardi]

@Kent1Cooper WOW!  The Rytz Construction does really work!  I applied it to the posted problem and it yielded the same ellipse that was created from the solid model.

Here's my construction using the nomenclature from the wiki reference.

That's quite the process.  How about  making a LISP program that constructs the ellipse given two sides of the parallelogram? 😁

[leeminardi]

I created the following program that uses Rytz's construction to create an ellipse given its center and two tangent points.  The program works on any three points in 3D and will create the ellipse on the plane defined by the three points.  No error checking is done (e.g., 3 points that are collinear).  

(defun c:tangentEllipse ( / ptC ptP ptQ uN ptPP r ptA ptB a b pt1 pt2)
; creates an ellipse that is centered at the specified 3D point
; and is tangent to two other specified 3D points.
; The ellipse is place in the plane defined by the three 3D points.
; The program uses Rytz's construction described at:
;  https://en.wikipedia.org/wiki/Rytz%27s_construction
; The program ends with the UCS set to World.  
;  Lee Minarid  9/29/2023
;
(command "_ucs" "w")  
(setq ptC (getpoint "\nPick ellipse center point: ")
      ptP (getpoint "\nPick ellipse first tangent point: ")
      ptQ (getpoint "\nPick ellipse second tangent point: ")
)
(setq uN     (uvec (cross (mapcar '- ptQ ptC) (mapcar '- ptP ptC)))
      ptPP   (mapcar '+
		     ptC
		     (cross (mapcar '- ptP ptC) uN)
	     )

      ptD    (mapcar '/ (mapcar '+ ptQ ptPP) '(2. 2. 2.))
      r	     (distance ptC ptD)
      ptA    (mapcar '+
		     ptD
		     (mapcar '*
			     (uvec (mapcar '- ptPP ptD))
			     (list r r r)
		     )
	     )
      ptB    (mapcar '+
		     ptD
		     (mapcar '*
			     (uvec (mapcar '- ptQ ptD))
			     (list r r r)
		     )
	     )
      a	     (distance ptQ ptA)
      b	     (distance ptQ ptB)
      pt1 (mapcar '+
		     ptC
		     (mapcar '*
			     (uvec (mapcar '- ptA ptC))
			     (list b b b)
		     )
	     )
      pt2 (mapcar '+
		     ptC
		     (mapcar '*
			     (uvec (mapcar '- ptB ptC))
			     (list a a a)
		     )
	     )
)
(command "_ucs" "za" '(0 0 0) uN)
(setq ptCucs	(trans ptC 0 1)
      pt1ucs	(trans pt1 0 1)
      pt2ucs	(trans pt2 0 1)
)
(command "_ellipse" "c" "_non"	ptCucs "_non" pt1ucs "_non" pt2ucs)
(command "ucs" "p")  
(princ)  
)  

; calculate unit vector of v1
(defun uvec (v1 / s)
  (setq s (distance '(0 0 0) v1))
  (setq s (mapcar '/ v1 (list s s s)))
)

;;; Compute the cross product of 2 vectors a and b
(defun cross (a b / crs)
  (setq	crs (list
	      (- (* (nth 1 a) (nth 2 b))
		 (* (nth 1 b) (nth 2 a))
	      )
	      (- (* (nth 0 b) (nth 2 a))
		 (* (nth 0 a) (nth 2 b))
	      )
	      (- (* (nth 0 a) (nth 1 b))
		 (* (nth 0 b) (nth 1 a))
	      )
	    )				;end list
  )					;end setq c
)					;end cross

[dany_rochefort]

@leeminardi i'm pretty sure my method is correct. Had you checked the link i provided you would of saw the rest of the procedure.

[leeminardi]

@dany_rochefort   The process you outline for determining an ellipse is correct for finding the intersection of a cylinder and a flat plane but that is not the full task here.  You need to take it one step further and determine the projection of an ellipse that is not parallel to any of the principal planes and then project it to the viewing plane for an isometric drawing.  

An ellipse on a plane skewed in space will not have its axes oriented the same as the ellipse project to the viewing plane.  The red line show the axes of the ellipse on the incline plane while the white line are of the ellipse that is projected to the viewing plane. 

In the images below your ellipse is the red one while the correct ellipse is in white.

Note how the red ellipse is not tangent to the midpoint nor within  the parallelogram containing the ellipse while the white ellipse is.