- [John] There, that's better.
(dynamic rock melody)
Hello, I would like to welcome you
to the Laramy-K OpticianWorks Training Center
where today we're continuing our series,
on "Let's Start to Finish."
The response to the first two has been overwhelming.
Thank you, so much.
I think it just proves that these concepts
are just key to being a good optician.
Today we're gonna continue those little,
tiny, baby steps, build on prior knowledge,
start looking at the same basic spheres:
simple horizontal decentration, but using monocular PDs.
And in our final example, we're gonna do
out decentration and a zero.
Example number one.
For this example, we'll have a right of plus 3.50,
a left of plus 3.75, customer PD of 29/28,
and a frame of 52-18.
That would be a common PD in powers
for a single-vision reader.
We have super frame, and super frame says it has
a boxing system of 52-18.
Now, couple of things are going to be different
than the last two videos.
One, our power is starting to creep up.
So instead of just taking my numbers of 52-18
at face value, adding them together,
and getting a 70 millimeters,
because the power's creeping up,
I'm actually going to double-check that.
I would actually measure my A exactly where
I'm supposed to according to the boxing system,
double-check that I have 52.
And then I would check my frame DBL,
distance between my lenses, and make sure that I had 18.
That checks out, so we're good to go.
We can use our 70.
The other difference between this and the last two,
although we're still doing baby steps
and very small changes, is that our PD
is given to us monocularly.
It isn't a single number, it's two numbers.
It's 29 and 28.
Meaning that this person is not perfectly symmetrical.
If you're looking at them, their right eye
is a tiny bit further away from the central line
of their nose than their left eye is.
Not uncommon at all.
Because our power's starting to creep up,
we'd want to consider doing that monocular PD.
So if I have two numbers for my PD,
I need two numbers to work with with my frame.
So I take my 70 millimeters and I divide it in half
and I get 35.
That's where that number comes from.
Where does 35 come from?
Splitting my frame, A and DBL,
or my distance between centers, in half, 35, 35.
Now, if I have a human PD of 29 in the right,
and I take my 35 minus 29,
and why do I use a calculator?
Because my brain doesn't work that good
and because I don't want to make a mistake and ruin a lens.
That gives me a difference of six millimeters, right,
and we're gonna do that in,
and 35 minus 28 gives me a difference
of seven millimeters in.
PD is narrower than my frame PD,
that would be in decentration,
I would be taking my lens in right
and moving it this, and over six millimeters,
and in my left, I'd be moving it over seven millimeters.
One more to account for the difference.
Here is what it would look like in the blocker.
We have got six millimeters of movement in our right,
so we'll go ahead and load up a block.
There, notch.
Good.
We're gonna move that over six.
We're gonna take our right lens
and move it right for in decentration.
That looks wonderful.
Notch up, tongue right, oh.
I knew there was something I was forgetting.
I'll get to it in just a second.
Then we move seven for the left, that's five,
that's six, that's seven.
Seven, that would be zero.
I'm moving that over seven millimeters.
And that looks wonderful.
The thing I meant to mention and I forgot.
These are just inherent in the design
of this particular blocker.
They do get in the way.
I know that makes it a little bit difficult sometimes
to see exactly what's going on behind them.
But those are not part of the marking device
or the grid system.
They're somehow in how everything's held together.
So, sorry, but that's the way that is.
Example number two.
And don't go anywhere if you're getting the hang of this
and you say, my god, I wish he would just hurry up
and move on with things.
Stick around for example number three
because I'm gonna do a zero and an out with that one.
For example two, we will have a right of a minus 5.50,
a left of minus 4.75, a customer PD of 28/30,
and a frame with a boxing system of 51-14.
I have got super frame.
Super frame has a boxing system measurement of 51-14.
My powers are creeping up.
I need to be a lot more careful with things
than I would be if that was a point 50 and a point 75.
So I'm gonna double-check this measurement.
If you're not familiar with the boxing system,
watch the video on the boxing system.
I'm gonna make sure that my A is 51.
And it is.
And I'm gonna make sure my DBL is 14, which it is.
So we're good to go there.
I have been given a monocular PD of 28/30.
Monocular meaning that this person,
the right eye is not symmetrical.
It has a different distance from the center of their nose
to the center of their pupil than it does in their left eye.
Because I have two numbers, 28 and 30,
I need to split my frame.
My 65 gets divided by two,
which gives me 32 and a half and 32 and a half.
When I have my 32 and a half, and I take away my 28,
I end up with a decentration amount
that I need to move this OC.
Four and a half millimeters in the right in
and two and a half millimeters in in the left.
Where did this come from?
Splitting my frame.
Where did these come from?
My monocular PD.
Once I get over to the blocker,
I would take my lens and because 28 is narrower
than 30 is, I'm going to have to displace
or decenter this lens more.
How much more?
I need to move this one four and a half millimeters over.
For my left, I only need to move it
two and a half millimeters over.
Let's see what that would look like in the blocker.
Example number two.
In my right, I have got four and a half.
That's four and a half.
Let's go ahead, load up a block.
Take my right, and right moves right for in decentration.
That would be on center, that's over four and a half.
And for my left, I have a little tiny two and a half.
There's two and a half.
That would be on center, over two and a half.
That would be my left.
Example number three.
Which I promised was gonna be a little bit different.
And it's gonna be a little different, too,
because I'm actually gonna kind of draw this out
and work it out in front of the camera.
So we may have a little fast motion here.
I'm not sure.
It may just be watching me, watching my back.
So let's get going.
For example three, we'll have a right of minus six,
a left of minus 6.50, customer PD of 33/32,
and a frame with a boxing system of 47-17.
I have super frame.
Super frame has a boxing system measurement of 47-17.
47 and 17 gives me a total of 64.
My powers are getting up there.
Gonna be super careful about what I do.
I'm gonna make sure that 47 and that 17 are correct.
Take out my PD stick.
The only time we use a PD stick.
Gonna check my A, which is 47.
And I would check my DBL.
A nice 17, so we're good there.
Let's see what we can fill in.
We've got 64 and even I think I can divide 64 in half.
So that would be 32 and 32, all right.
We have a monocular PD of 33/32.
This is going to be interesting.
Frame 32, 32.
Human 33, 32.
32 minus 33 is minus one.
And 32 minus 32 is zero.
Okay, we've got red flags, ding, ding, ding, ding.
Bells, whistles, red flags going off.
We've never seen a minus one before, have we?
What that tells us is that this person,
his or her PD, monocular PD for the right eye,
their pupil is actually wider,
it's further from the center of their nose out to the pupil,
than it is from the center of their nose
to the center of the eye wire opening.
For this person, their pupil for the right eye
is gonna be over here.
Look, it's extremely rare, but it does happen,
so be aware of it.
What does zero tell us?
Zero tells us that this lens is not gonna have
to be decentered at all.
They're pupil and the center
of the eye wire opening are equal.
So it's gonna be an overlap.
And we're gonna have that.
When I always need, the rule is for in decentration,
right moves right, left moves left,
and we'll look at this in the blocker in just a second.
For my right eye, it's not going to do that this time.
It's gonna go from here this way instead one millimeter.
For my left, I'm not gonna decenter that lens at all.
That center OC dot off of my lens meter
is gonna go right there.
The two outer ones don't matter at this point,
because we're doing a sphere.
Let's see what that would look like in the blocker.
For example number three, we have something
a little bit different going on.
Normally, if I saw one in the right,
I would go this way for in decentration,
but because we have out, I'm gonna move
my right lens to my left.
I'm gonna load up a block.
Set my right.
Exaggerate that just a little bit so we can see it.
Stick that on there.
And let's see here.
That would be the center of my block.
That would be my OC displaced out
towards the temple in my right
as if I was wearing a pair of glasses.
That is correct.
Looks great.
Something a little different.
Our left is a zero.
I'm gonna just set that on zero.
Gonna take my left lens and I am gonna put it on zero.
No decentration at all.
Put my block on there.
And everything should be lined up.
My X in the center of my dot.
Perfect, no decentration.
Human PD is equal to my frame PD.
There we go.
Thank you so much for watching.
If you are ready to get started finishing,
please, give us a call at Laramy-K.
We will be glad to help you out.
(dramatic musical flourish)