# Vieth-Müller Circles (Visual Depth Perception 8)

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This Demonstration shows that binocular disparity is constant on the blue circles through one distractor and the eye nodes. The sliders allow you to set a distance for the fixate point F away from the base point, and a distance for the distractor D beyond it. The scale is very small to exaggerate the angles in order to see what they measure at normal viewing distances.

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Contributed by: Keith Stroyan (March 2011)

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

One of the main goals of this set of Demonstrations, outlined in the first Demonstration, "Motion/Pursuit Law in 1D (Visual Depth Perception 1)" (see Related Links), is to show which geometric inputs to vision determine the depth of objects and especially the depth of D relative to F.

The fact that binocular disparity does not vary around the circles shown in this Demonstration implies that it is not a good measure of the distance of D beyond F unless D and F are on the axis.

Even when D and F are on the axis, binocular disparity alone does not determine depth. The Demonstration "Binocular Disparity versus Depth (Visual Depth Perception 9)" (see Related Links) allows you to vary the depth along a curve of constant binocular disparity. The Demonstration "Disparity, Convergence, and Depth (Visual Depth Perception 10)" (see Related Links) gives the depth formula.

The remainder of this notebook proves the geometric lemma that binocular disparity is constant on circles through the eye nodes.

The Circle Lemma

Following is a proof that convergence angles are the same for every point on a circle through both eye nodes and one point (either the fixate or the distraction). This means any distraction point on the circle through the fixate has retinal disparity zero. This is the Vieth-Müller circle, but more: any point on the circle through one distraction and both nodes has the same retinal disparity as any other distraction on this circle because they have the same convergence angle.

We will prove the theorem for half the convergence angle, taking a point F on the circle and the rays to one point L that corresponds to the left node and another point D on a horizontal diameter to G.

Our goal is to show the angle at G equals the angle at F. is just half of the convergence angle to both nodes and the diameter helps with the proof.

We add radial segments from the center to the points as shown in green and magenta. The resulting isosceles triangles show that the angles marked are equal and that occurs on the left as shown in green near L. The angles marked are equal because they are on opposite sides of crossing lines.

Now use the isosceles triangle LCF to see that for the angle shown next.

The sum of the angles of a triangle is , so and , so .

Combining the last two results and subtracting,

so and .

The convergence angles are always the same on circles through the two nodes and the point. Convergence angles are the same at every point of the blue circle, so the difference between convergence angles for points on the blue circle and the convergence angle of the fixate is the same: .

This constant disparity circle is shown interactively in the Demonstration.

NOTE: moving the fixate on the Veith–Müller circle (through the fixate) will change that circle slightly because the eye nodes move to follow the fixate.

For more information on the Vieth–Müller circle (or horopter) see the Wikipedia page.

## Permanent Citation