Tech drawing - ellipses in perspective

[DaveC426913]

I'm doing some 2D pencil diagrams (CUD - computer-unaided design).

So I've got a square in perspective (2 point perspective but 1 point perspective is good enough). Now I want to draw an ellipse within it, so that the ellipse is properly tangential to all four midpoints of the square. I have an ellipse template to help draw the ellipses.

I know that you can't just align the long axis and short axis of the ellipse to the diagonals of the square (especially since it will be impossible - the square's diagonals are not perpendicular).

I know you have to offset the angle to the ellipse so that it looks right (lies in the same plane as the square).

Is there a formula for what angle one would offset an ellipse's axes to get it to sit right? (A long calculation would not be much good - I'm looking for a shortcut.)

Any takers?

 

[haruspex]

I'm doing some 2D pencil diagrams (CUD - computer-unaided design).

So I've got a square in perspective (2 point perspective but 1 point perspective is good enough). Now I want to draw an ellipse within it, so that the ellipse is properly tangential to all four midpoints of the square. I have an ellipse template to help draw the ellipses.

I know that you can't just align the long axis and short axis of the ellipse to the diagonals of the square (especially since it will be impossible - the square's diagonals are not perpendicular).

I know you have to offset the angle to the ellipse so that it looks right (lies in the same plane as the square).

Is there a formula for what angle one would offset an ellipse's axes to get it to sit right? (A long calculation would not be much good - I'm looking for a shortcut.)

If four equally spaced points in a straight line are seen in projection, their cross-ratio is conserved. See http://en.wikipedia.org/wiki/Cross-ratio#Definition
If we take two vertices of the square and their midpoint as z1, z3, z2 respectively, and put z4 at infinity it gives a x-ratio of 2. In the projected form, z4 is the vanishing point of the line. This should allow you to plot the midpoints of the sides. Is that enough to fit the ellipse (whih I assume was a circle before projection)?

 

[DaveC426913]

You miss the point.

If four equally spaced points in a straight line are seen in projection, their cross-ratio is conserved. See http://en.wikipedia.org/wiki/Cross-ratio#Definition
If we take two vertices of the square and their midpoint as z1, z3, z2 respectively, and put z4 at infinity it gives a x-ratio of 2. In the projected form, z4 is the vanishing point of the line. This should allow you to plot the midpoints of the sides. Is that enough to fit the ellipse (whih I assume was a circle before projection)?

I've got the midpoints of the square. How do I fit the ellipse? Eyeball it?

 

[haruspex]

Since it's a square before projection, the inscribed shape was a circle before projection, right? After projection, the major axis of the ellipse will be parallel to the horizon line, no?
That doesn't look right in your thumbnail, but then the square doesn't either.

 

[DaveC426913]

Since it's a square before projection, the inscribed shape was a circle before projection, right

Right.

After projection, the major axis of the ellipse will be parallel to the horizon line, no?

No. Why would it?

That doesn't look right in your thumbnail, but then the square doesn't either.

Attached is a better pic.

There's no way an ellipse will fit into a perspective square without being rotated.

 

[haruspex]

Ah yes, they do tilt. Quite subtle: in a line of circles straight across left to right, the one in the middle is level, and they must tend back to level at infinity, but they tilt in between.
I tried this, but the algebra gets heavy:
- start with ax²+by²=1
- apply a rotation through theta
- apply a shift of origin; we now have 5 unkowns
- use some combination of the 8 equations based on knowing the ellipse passes through the midpoints of the square's sides and is tangential to the square at those places.
- solve for theta (!)

 

[I like Serena]

Yeah, I solved it for theta some time ago (using the matrix representation of an ellipse).
It gets heavy.
I couldn't find a clear construction method to find the long and short axes, or the angle.

So I recommend eyeballing it, using the fact that the ellipse has to "touch" the square at its midpoints.

At least a circle in perspective is an ellipse.

 

[haruspex]

Progress, perhaps.
I considered a simple example. Circle radius c at (a,b) in the XY plane. Observer at (0,0,z). Viewing pane at (*,1,*), i.e. parallel to XZ plane.
I get tan of the axis angle satisfies
t² - kt = 1
where
k = (z² - a² - b² - c²)/(za)

Postscript:
There's a sign wrong somewhere. Put negative z to view from above.
I plotted this up on a spreadsheet for 5 adjacent circles, i.e. stepping a by 2c. it looks right.