# Vertex distance

The vertex distance is the distance between the surface of your eye and the center back of your lens. As the lens moves further from the eye, the perceived strength of your lenses is altered.

• When a plus lens is moved away from the eye, it is perceived as being stronger
• When a minus lens is moved away from the eye, it is perceived as being weaker

This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes: to achieve a given correction for short-sightedness, the lenses of glasses needs to be stronger than the lenses of contact lenses.

## Practical Guidelines

• I have High Myopia. How does Vertex Distance affect me?
If you wore contact lenses before EndMyopia (EM) and continue to use contact lenses throughout your EM journey - Then vertex distance doesn't affect you.
If you wore glasses before EM and continue to wear glasses during EM then it is recommended to invest in a lens kit and choose your reduced lenses based on testing your visual acuity - Then vertex distance doesn't affect you. If you do not have access to a lens kit and you want to reduce in pre-defined diopter steps - Then vertex distance can affect you: For example, reducing from -14.0 dpt to -13.75 dpt at a vertex distance of 15mm results in an effective perceived diopter drop of only 0.17 dpt, further reducing to -13.5 dpt would result in an effective diopter drop of 0.34 dpt.
If you switch from glasses to contact lenses at high myopia or vice-versa you definitely will need to account for vertex distance before buying new corrections.
If you calculate your own correction based on blur distance from your eye, but intend to buy glasses, you will need to adjust for the vertex distance.
No. Your differential correction will not undercorrect you by a diopter margin where vertex distance plays a role.
• How do I measure my vertex distance?
Ask a second person to measure the distance from your closed eye to your glasses while you wear them. Or, as a literal "rule of thumb", try placing different fingers between your closed eye and the back of your glasses, then measure the thickness of your finger - This is an estimate for your vertex distance.
• I use the endmyopia.org Diopter Calculator App and my results differ from my manual centimeter measurements, why?
The Diopter Calculator App for Android (as for the June 2020) measures the distance from the screen to the tip of your nose. Since the distance from the nose to your eyeball is not taken into account, your focal length seems smaller (and your diopters higher) than with a manual measurement. You can however correct these values by applying the same formula as for the vertex distance, inserting the distance from the tip of your nose to your eyeball for ${\displaystyle x}$.

## Calculation

The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:

${\displaystyle D_{C}={\frac {1}{{\frac {1}{D}}-x}}}$,

where ${\displaystyle D_{C}}$ is the perceived diopter number, ${\displaystyle D}$ is the diopter strength of your lenses and ${\displaystyle x}$ is the vertex distance in meters. (More generally, ${\displaystyle x}$ is the distance the lens is being moved from its original position.) It is important to note here that this equation is sensitive to minus signs of your diopter strength and the direction of movement.

Example for a vertex distance of 15mm (=0.015m):

${\displaystyle +4.0dpt:D_{C}={\frac {1}{{\frac {1}{+4.0}}-0.015}}=+4.255dpt}$

${\displaystyle -4.0dpt:D_{C}={\frac {1}{{\frac {1}{-4.0}}-0.015}}=-3.774dpt}$

In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses, for distant objects. It can be seen that vertex distance appears to increase the strength of plus Lenses and decrease the strength of minus lenses. The effect is noticeable above 4.0 dpt and is mostly negligible for low myopia.

The expression looks less intimidating when you remember that (by definition) the diopter is the reciprocal of the focal length. So it's really just:

${\displaystyle f_{C}=f-x}$ where ${\displaystyle f_{C}={\frac {1}{D_{C}}}}$ and ${\displaystyle f={\frac {1}{D}}}$

Conceptually, the location of the image is changed because the lens has moved relative to the eye. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.

• for a minus lens, moving the virtual image further from the eye is equivalent to using a weaker minus lens
• for a plus lens, moving the (real) image towards the eye (so that it is not as far beyond the retina) is equivalent to using a stronger plus lens.

Note that the equivalence relationship holds only for distant source objects: the behaviour of glasses and contact lenses will be different for all closer objects, and the near point in particular will be different.