P.D. for reading and computer screen glasses

[Kyle]

Hi,

Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

[NOT Otis_Brown]

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

there is no formula. it is an objective measurement. the patient
looks at a target that is a specified distance away and then a
measurement is made. In general, near PD is about 3mm less than the
far PD. Intermediate PD (computer screen) would be about 1/2 that
value. How important it is to get the exact value depends on how
strong the reading glasses are. High prescriptions require more
accurate measurements.

[Kyle]

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

there is no formula.  it is an objective measurement.  the patient
looks at a target that is a specified distance away and then a
measurement is made.  In general, near PD is about 3mm less than the
far PD.  Intermediate PD (computer screen) would be about 1/2 that
value.  How important it is to get the exact value depends on how
strong the reading glasses are.  High prescriptions require more
accurate measurements.

In the United States. Can labs make eyeglasses optical centers exactly
the amount needed?
For example. If the PD is 63 and say 15mm from the frame. Can the labs
make this exactly
right? Or what's the tolerable allowance? In Asia, the centers can be
off 3mm in EACH SIDE and they consider it ok. If so. What's the use of
using 0.75mm per side if they are off by even 3mm.
What's the highest mm per side off optical center is considered ok in
the United States?

[NOT Otis_Brown]

Post by Kyle

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

there is no formula.  it is an objective measurement.  the patient
looks at a target that is a specified distance away and then a
measurement is made.  In general, near PD is about 3mm less than the
far PD.  Intermediate PD (computer screen) would be about 1/2 that
value.  How important it is to get the exact value depends on how
strong the reading glasses are.  High prescriptions require more
accurate measurements.

In the United States. Can labs make eyeglasses optical centers exactly
the amount needed?
For example. If the PD is 63 and say 15mm from the frame. Can the labs
make this exactly
right? Or what's the tolerable allowance? In Asia, the centers can be
off 3mm in EACH SIDE and they consider it ok. If so. What's the use of
using 0.75mm per side if they are off by even 3mm.
What's the highest mm per side off optical center is considered ok in
the United States?

the tolerance depends upon the strength of the prescription but
optical centers commonly are within 1mm of actual. If not then the
lab is not working carefully enough while blocking the lenses.

[retinula]

Post by Kyle

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

there is no formula.  it is an objective measurement.  the patient
looks at a target that is a specified distance away and then a
measurement is made.  In general, near PD is about 3mm less than the
far PD.  Intermediate PD (computer screen) would be about 1/2 that
value.  How important it is to get the exact value depends on how
strong the reading glasses are.  High prescriptions require more
accurate measurements.

In the United States. Can labs make eyeglasses optical centers exactly
the amount needed?
For example. If the PD is 63 and say 15mm from the frame. Can the labs
make this exactly
right? Or what's the tolerable allowance? In Asia, the centers can be
off 3mm in EACH SIDE and they consider it ok. If so. What's the use of
using 0.75mm per side if they are off by even 3mm.
What's the highest mm per side off optical center is considered ok in
the United States?

OC should be <1mm tolerance.

[Kyle]

Post by Kyle

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

there is no formula.  it is an objective measurement.  the patient
looks at a target that is a specified distance away and then a
measurement is made.  In general, near PD is about 3mm less than the
far PD.  Intermediate PD (computer screen) would be about 1/2 that
value.  How important it is to get the exact value depends on how
strong the reading glasses are.  High prescriptions require more
accurate measurements.

In the United States. Can labs make eyeglasses optical centers exactly
the amount needed?
For example. If the PD is 63 and say 15mm from the frame. Can the labs
make this exactly
right? Or what's the tolerable allowance? In Asia, the centers can be
off 3mm in EACH SIDE and they consider it ok. If so. What's the use of
using 0.75mm per side if they are off by even 3mm.
What's the highest mm per side off optical center is considered ok in
the United States?

OC should be <1mm tolerance.- Hide quoted text -
- Show quoted text -

Can anyone point to an eyeglass frame with adjustable PD such that you
can move
the frame toward and away from each other? I think I need this to get
precise because
after having different labs create glasses.. they always end up at
least 2mm off and
they say their machines are not accurate.

[Science_Research]

Hi Kyle

F. Y. I.

Here is a video on how to measure your P. D., and other information you will need to order Internet glasses.

Hope this helps you in your search for accurate lenses.

[y***@gmail.com]

Kyle:
"Can anyone point to an eyeglass frame with adjustable PD such that you can move the frame toward and away from each other?"

What on earth are you talking about? "...such that you can move the frame toward and away from each other" is utter nonsense.

[The Real Bev]

Post by y***@gmail.com
"Can anyone point to an eyeglass frame with adjustable PD such that you can move the frame toward and away from each other?"
What on earth are you talking about? "...such that you can move the frame toward and away from each other" is utter nonsense.

Not at all. They work sort of like binoculars. Plastic, two sets of
lenses. My MIL had some. Ask a local optometrist where to get them.

OTOH, if you mean moving a single set of lenses nearer and further from
your eye, safety glasses will generally do that. You might be able to
find prescription safety glasses via a large optical firm, or perhaps
you can have prescription lenses put in a safety frame.

BUT if you want to move the lenses nearer and farther from each other, I
haven't a clue. With bifocals the PD may be slightly different in the
segment, but not by much.

How big a change do you need?

[Robert Martellaro]

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

NPD = DPD - DPD/1+ W(1/s - F/1000)

W is the work distance in mm. S is the stop distance (the distance from the
eye's center of rotation to the corneal plane, plus the vertex distance,
typically 14mm and 13mm, respectively), and F is the lens power.

Assuming a 40cm work distance and no distance power, the near PD when the
distance PD is 63mm is 59mm.

But that doesn't mean we'll always use the near pd for reading, especially for
myopes and/or aspheric/atoric lenses.

Robert Martellaro
~~~~~~~~~~~~~~~~~~
Roberts Optical Ltd.
Wauwatosa Wi.
www.roberts-optical.com
~~~~~~~~~~~~~~~~~~
"Science is a way of trying not to fool yourself."
- Richard Feynman

[f***@yahoo.com]

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

NPD = DPD -  DPD/1+ W(1/s - F/1000)
W is the work distance in mm. S is the stop distance (the distance from the
eye's center of rotation to the corneal plane, plus the vertex distance,
typically 14mm and 13mm, respectively), and F is the lens power.
Assuming a 40cm work distance and no distance power, the near PD when the
distance PD is 63mm is 59mm.

What power do you assume for F to come out with exactly 59?
I'm calculating it and assuming F is -9 and S is 14+13=27.
The calculations is thus:

NPD = DPD - DPD/1+ W(1/s - F/1000)

=63 - 63/[1 + 400(1/27 - (-9)/1000)]
= 63 - 63/[1+ 400 (0.037 + 0.009)] = 63 - 63/(1 + 18.4) =
63 - 63/19.4 = 63 - 3.2 = 59.8

I don't get exactly 59. Any tips where I did wrong in the
calculations?

But that doesn't mean we'll always use the near pd for reading, especially for
myopes and/or aspheric/atoric lenses.

Why do you say that? Why doesn't it mean that myopes always use the
near
pd for reading?

Robert Martellaro
~~~~~~~~~~~~~~~~~~
Roberts Optical Ltd.
Wauwatosa Wi.www.roberts-optical.com
~~~~~~~~~~~~~~~~~~
"Science is a way of trying not to fool yourself."
- Richard Feynman

[Robert Martellaro]

Post by f***@yahoo.com

Post by Kyle
Hi,
Supposed your far Pupillary Distance (P.D.) is 63mm. When making
reading
and computer screen eyeglasses, how much must the P.D. be reduced?
Also what is
the exact formula to calculate this? Thanks.

NPD = DPD -  DPD/1+ W(1/s - F/1000)
W is the work distance in mm. S is the stop distance (the distance from the
eye's center of rotation to the corneal plane, plus the vertex distance,
typically 14mm and 13mm, respectively), and F is the lens power.
Assuming a 40cm work distance and no distance power, the near PD when the
distance PD is 63mm is 59mm.

What power do you assume for F to come out with exactly 59?
I'm calculating it and assuming F is -9 and S is 14+13=27.
NPD = DPD - DPD/1+ W(1/s - F/1000)
=63 - 63/[1 + 400(1/27 - (-9)/1000)]
= 63 - 63/[1+ 400 (0.037 + 0.009)] = 63 - 63/(1 + 18.4) =
63 - 63/19.4 = 63 - 3.2 = 59.8
I don't get exactly 59. Any tips where I did wrong in the
calculations?

Note that I assumed "no distance power". Your calculations were fine.

Post by f***@yahoo.com

But that doesn't mean we'll always use the near pd for reading, especially for
myopes and/or aspheric/atoric lenses.

Why do you say that? Why doesn't it mean that myopes always use the
near
pd for reading?

Myopes are accustomed to the base in prism induced when the eyes converge behind
lenses that have minus power in the horizontal meridian. It might not be a good
idea to take that away in one fell swoop.

Atorics should have the design pole placed center pupil on the horizontal
meridian, although position of wear optimized lenses may have more latitude
compared to semi-finished aspherics.

Robert Martellaro
~~~~~~~~~~~~~~~~~~
Roberts Optical Ltd.
Wauwatosa Wi.
www.roberts-optical.com
~~~~~~~~~~~~~~~~~~
"Science is a way of trying not to fool yourself."
- Richard Feynman