the frame diffrence

Frame Difference

The difference between the horizontal and vertical measurements for the lens area of a frame is known as the frame difference. For example, for the box measurement of the lens area for Figure 1112-13, we use the following formula:

Frame Difference = A – B

Frame Difference = 50 mm – 26 mm

Frame Difference = 24 mm

The greater the frame difference, the more rectangular the surrounding box looks.

Figure 1112-13 Frame Difference

Temple Length

The overall temple length is the measurement through the middle of the temple from the barrel to the end of the temple. See Figure 1112-14 below. For frames with a cable temple, the measurement is done by stretching the flexible cable out and measuring with a ruler. The overall temple length is the most commonly used measurement today.

Figure 1112-14 Overall Temple Length

A couple of older temple length measurements are not used much today. They are the:

  • length to bend – from the center barrel to the middle of the bend

  • front to bend – for turn-back endpieces, the measurement is made from the back of the frame front to the middle of the bend

Calculating the Geometric Center Distance (GCD)

The GCD can be calculated by using the following formula:

GCD = eyesize + DBL

For example:  a frame has an eyesize of 42 mm and a DBL of 22 mm.  What is the GCD?

GCD = 42 mm + 22 mm.
GCD = 64 mm.

If the GCD and the eyesize or the DBL are known, the unknown variable can also be calculated.

For example: a frame has a GCD of 72 mm and an eyesize of 60 mm.  What is the DBL?

72 mm = 60 mm + DBL.
DBL = 12 mm.

Figure 1112-15 Calculating the Geometric Center Distance (GCD)

Decentration

Binocular Decentration

Starting with the patient's binocular pupillary distance (PD), we must be able to determine the difference between the DBC and the PD for the purpose of decentration of the optical center of the lens. This is called "decentering the lens" because we must move the optical center away from the geometric center of the lens. The formula for decentering the lens for a binocular PD is as follows:

Example: If we have a frame with an A = 52 mm and DBL = 18 mm, we will have a DBC of 70 mm. The patient's binocular PD = 62 mm.

Since the DBC is greater than the PD, the decentration will be nasally or in by 4 mm. If the DBC was smaller than the PD, the optical center would move out or temporally.

Monocular Decentration

When using monocular PDs, we must also decenter each lens separately. To decenter each lens, we must divide the DBC by 2 and subtract each of the monocular PDs.

Example:

If we have a frame with an A = 50 mm and DBL = 16 mm, we would have a DBC of 66 mm; divided in half, the monocular DBC would be 33 mm. The patient's OD PD = 31 mm and OS = 33 mm.

OD Decentration = DBC/2 – PDOD

OD Decentration = 33 mm – 31 mm = 2 mm in

The DBC is greater than the PD, so the optical center is decentered in.

OS Decentration = 33 mm – 33 mm = 0

The OC will be at the frame GC.

Frame Difference

The difference between the horizontal and vertical measurements for the lens area of a frame is known as the frame difference. For example, for the box measurement of the lens area for Figure 1112-13, we use the following formula:
Frame Difference=A−B\text{Frame Difference} = A - B
Here, $A$ represents the horizontal measurement (also known as the eyesize), and $B$ represents the vertical measurement, or the deepest vertical dimension of the lens.

For example:
Frame Difference=50 mm−26 mm\text{Frame Difference} = 50 \text{ mm} - 26 \text{ mm}
Frame Difference=24 mm\text{Frame Difference} = 24 \text{ mm}
The greater the frame difference, the more rectangular the surrounding box looks.

Figure 1112-13 Frame Difference

Temple Length

The overall temple length is the measurement through the middle of the temple from the barrel to the end of the temple. See Figure 1112-14 below. For frames with a cable temple, the measurement is done by stretching the flexible cable out and measuring with a ruler. The overall temple length is the most commonly used measurement today.

Figure 1112-14 Overall Temple Length

A couple of older temple length measurements are not used much today. They are the:

  • length to bend

    • from the center barrel to the middle of the bend

  • front to bend

    • for turn-back endpieces, the measurement is made from the back of the frame front to the middle of the bend
      These older methods have largely been replaced by the overall temple length measurement due to advancements in frame design and manufacturing, as well as the need for a more standardized and direct measurement for fitting and ordering.

Calculating the Geometric Center Distance (GCD)

The GCD can be calculated by using the following formula:
GCD=eyesize+DBL\text{GCD} = \text{eyesize} + \text{DBL}
For example: a frame has an eyesize of 42 mm and a DBL of 22 mm. What is the GCD?
GCD=42 mm+22 mm\text{GCD} = 42 \text{ mm} + 22 \text{ mm}
GCD=64 mm\text{GCD} = 64 \text{ mm}
If the GCD and the eyesize or the DBL are known, the unknown variable can also be calculated.

For example: a frame has a GCD of 72 mm and an eyesize of 60 mm. What is the DBL?
72 mm=60 mm+DBL72 \text{ mm} = 60 \text{ mm} + \text{DBL}
DBL=12 mm\text{DBL} = 12 \text{ mm}

Figure 1112-15 Calculating the Geometric Center Distance (GCD)

Decentration
Binocular Decentration

Starting with the patient's binocular pupillary distance (PD), we must be able to determine the difference between the DBC and the PD for the purpose of decentration of the optical center of the lens. This is called "decentering the lens" because we must move the optical center away from the geometric center of the lens. The formula for decentering the lens for a binocular PD is as follows:
Binocular Decentration=DBC−Binocular PD\text{Binocular Decentration} = \text{DBC} - \text{Binocular PD}

Example: If we have a frame with an A = 52 mm and DBL = 18 mm, we will have a DBC of 70 mm. The patient's binocular PD = 62 mm.
Binocular Decentration=70 mm (DBC)−62 mm (PD)=8 mm\text{Binocular Decentration} = 70 \text{ mm (DBC)} - 62 \text{ mm (PD)} = 8 \text{ mm}
Since the DBC is greater than the PD, the decentration will be nasally or in by 4 mm per eye ($8 \text{ mm} / 2 = 4 \text{ mm}$). If the DBC was smaller than the PD, the optical center would move out or temporally.

Monocular Decentration

When using monocular PDs, we must also decenter each lens separately. To decenter each lens, we must divide the DBC by 2 and subtract each of the monocular PDs.

Example:
If we have a frame with an A = 50 mm and DBL = 16 mm, we would have a DBC of 66 mm; divided in half, the monocular DBC would be 33 mm. The patient's OD PD = 31 mm and OS = 33 mm.

OD Decentration = DBC/2 – PDOD
OD Decentration=33 mm−31 mm=2 mm in\text{OD Decentration} = 33 \text{ mm} - 31 \text{ mm} = 2 \text{ mm in}
The DBC is greater than the PD, so the optical center is decentered in.

OS Decentration = 33 mm – 33 mm = 0
OS Decentration=33 mm−33 mm=0 mm\text{OS Decentration} = 33 \text{ mm} - 33 \text{ mm} = 0 \text{ mm}
The OC will be at the frame GC.

Accurate decentration is crucial to ensure optimal optical performance and patient comfort, preventing prism-induced effects and visual fatigue.