Vertex distance - Revision history

Divenal: /* Calculation */ add note that the equivalence is only for distant objects.

2020-07-13T08:47:53Z

Calculation: add note that the equivalence is only for distant objects.

Divenal: /* Calculation */ still hadn't convinced myself with the wording

2020-07-03T08:29:14Z

Calculation: still hadn't convinced myself with the wording

← Older revision		Revision as of 08:29, 3 July 2020
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	where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>		where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>

	Conceptually, the location of the [[Optics_related_math#Virtual_image\|Virtual image]] is changed because the lens has moved relative to the eye and source object.		Conceptually, the location of the [[Optics_related_math#Virtual_image\|Virtual image]] is changed because the lens has moved relative to the eye and source object. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.
	* a minus lens ~~creates~~ the virtual image of ~~a distant object near its focal point. When~~ the lens ~~is moved away from the eye, so too is the virtual image. This~~ is equivalent to using a weaker minus lens.		* for a minus lens, moving the virtual image further from original location of the lens is equivalent to using a weaker minus lens
	* a plus lens ~~creates~~ the ~~virtual~~ image ~~of a near object on the other side of its focal point. When the lens is moved~~ towards the ~~object,~~ the ~~virtual image is pushed further away from~~ lens~~, and therefore also the eye. This~~ is equivalent to using a stronger plus lens.		* for a plus lens, moving the (real) image towards the original location to the lens is equivalent to using a stronger plus lens.

	==References==		==References==

Divenal: /* Calculation */ add some words to describe what's going on in terms of virtual images.

2020-07-02T19:22:43Z

Calculation: add some words to describe what's going on in terms of virtual images.

← Older revision		Revision as of 19:22, 2 July 2020
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	where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>		where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>

	Conceptually, the ~~focus length is reduced (power~~ is ~~increased)~~ because it has moved ~~closer~~ to the source of the image.		Conceptually, the location of the [[Optics_related_math#Virtual_image\|Virtual image]] is changed because the lens has moved relative to the eye and source object.
			* a minus lens creates the virtual image of a distant object near its focal point. When the lens is moved away from the eye, so too is the virtual image. This is equivalent to using a weaker minus lens.
			* a plus lens creates the virtual image of a near object on the other side of its focal point. When the lens is moved towards the object, the virtual image is pushed further away from lens, and therefore also the eye. This is equivalent to using a stronger plus lens.

	==References==		==References==

Divenal: "corrected" is not the same as "perceived". Also add a few more words.

2020-07-02T18:56:40Z

"corrected" is not the same as "perceived". Also add a few more words.

← Older revision		Revision as of 18:56, 2 July 2020
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	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. This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes.		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==		==Practical Guidelines==
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	: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 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.

	* <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>		* <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>
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	<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,		<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,

	where <math>D_C</math> is the ~~corrected (~~perceived) diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.		where <math>D_C</math> is the perceived diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.

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

Phil3741: Fixed indent

2020-06-26T07:03:14Z

Fixed indent

← Older revision		Revision as of 07:03, 26 June 2020
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	* <div> '''I have ''High Myopia''. How does Vertex Distance affect me?'''</div>		* <div> '''I have ''High Myopia''. How does Vertex Distance affect me?'''</div>
	::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 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 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 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.

	* <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>		* <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>
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	:The [https://play.google.com/store/apps/details?id=org.endmyopia.calc 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 <math>x</math>.		:The [https://play.google.com/store/apps/details?id=org.endmyopia.calc 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 <math>x</math>.


	==Calculation==		==Calculation==

Deadpan: /* Calculation */ Minor correction.

2020-06-14T03:53:03Z

Calculation: Minor correction.

← Older revision		Revision as of 03:53, 14 June 2020
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	==Calculation==		==Calculation==
	{{Notice\|Advanced, this information ~~maybe~~ not apply. See above.}}		{{Notice\|Advanced, this information may not apply. See above.}}
	The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:		The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:

Viceroy Sam: /* Calculation */ typo

2020-06-12T12:11:37Z

Calculation: typo

← Older revision		Revision as of 12:11, 12 June 2020
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	==Calculation==		==Calculation==
	{{Notice\|~~Advance~~, this information maybe not apply. See above.}}		{{Notice\|Advanced, this information maybe not apply. See above.}}
	The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:		The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:

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	Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.		Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.


	==References==		==References==

Viceroy Sam: re-order and add notice.

2020-06-12T12:11:25Z

re-order and add notice.

@@ Line 1: / Line 1: @@
 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.  This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes.
 ==Calculation==
 The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:
@@ Line 24: / Line 45: @@
 Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.
 ==References==

Deadpan: Formatted the fractions using LaTeX

2020-06-09T16:53:25Z

Formatted the fractions using LaTeX

← Older revision		Revision as of 16:53, 9 June 2020
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	The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:		The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by:

	<math>D_C=1/(1/D-x)</math>,		<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,

	where <math>D_C</math> is the corrected (perceived) diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.		where <math>D_C</math> is the corrected (perceived) diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.
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	Example for a vertex distance of 15mm (=0.015m):		Example for a vertex distance of 15mm (=0.015m):

	<math>+4.0 dpt: D_C=1/(1/(+4.0)-0.015)=+4.255 dpt</math>		<math>+4.0 dpt: D_C=\frac{1}{\frac{1}{+4.0}-0.015}=+4.255 dpt</math>

	<math>-4.0 dpt: D_C=1/(1/(-4.0)-0.015)=-3.774 dpt</math>		<math>-4.0 dpt: D_C=\frac{1}{\frac{1}{-4.0}-0.015}=-3.774 dpt</math>

	In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses. It can be seen that vertex distance '''increases''' the strength of [[plus Lenses]] and '''decreases''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].		In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses. It can be seen that vertex distance '''increases''' the strength of [[plus Lenses]] and '''decreases''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].
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	<math>f_C = f - x</math>		<math>f_C = f - x</math>
	where <math>f_C = 1/D_C</math> and <math>f = 1/D</math>		where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>

	Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.		Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.

Phil3741: /* Practical Guidelines */

2020-06-08T21:06:22Z

Practical Guidelines

← Older revision		Revision as of 21:06, 8 June 2020
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	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 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 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 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.

← Older revision		Revision as of 08:47, 13 July 2020
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	<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,		<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,

	where <math>D_C</math> is the perceived diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.		where <math>D_C</math> is the perceived diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. (More generally, <math>x</math> 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):		Example for a vertex distance of 15mm (=0.015m):
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	<math>-4.0 dpt: D_C=\frac{1}{\frac{1}{-4.0}-0.015}=-3.774 dpt</math>		<math>-4.0 dpt: D_C=\frac{1}{\frac{1}{-4.0}-0.015}=-3.774 dpt</math>

	In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses. It can be seen that vertex distance '''~~increases~~''' the strength of [[plus Lenses]] and '''~~decreases~~''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].		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]].


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	where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>		where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>

	Conceptually, the location of the [[Optics_related_math#~~Virtual_image~~\|~~Virtual~~ image]] is changed because the lens has moved relative to the eye ~~and source object~~. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.		Conceptually, the location of the [[Optics_related_math#The_thin_lens_equation\|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 ~~original location of~~ the ~~lens~~ is equivalent to using a weaker minus 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 ~~original location to~~ the ~~lens~~ is equivalent to using a stronger plus 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.

	==References==		==References==