Shanghai da Wei Optical Co., Ltd.

First, the definition:

Surface defects Standard: According to the U.S. military standard MIL-PRF-13830B with two sets of figures indicate the size of surface defects. For example, 40/20 (or 40-20) The former limit scratches size, which limits the size of pitting.

Michiko, bright roads are collectively referred to as scratches.

Spots, pits, ideas are called pitting.

Required length and width ratio greater than 4:1 for scratch; length and width ratio of less than 4:1 for pitting.

When the element in different regions is not the same surface finish, the equivalent diameter is calculated on areas: high surface quality requirements of the equivalent diameter of the area in which prevail within the region (such as the effective aperture area) of the outer lower surface quality area calculating the equivalent diameter of the whole device.

As the left component surface quality requirements are different, then the area A is determined to meet the requirements, they should take the calculated diameter of the inner circle. And B, then take the whole element diameter.

U.S. military standard requirements for non-circular element with a diameter equal to the area of ​​a circle diameter of the take.

For rectangular elements (when the length and width ratio ＜5:1) can use a simple formula: Equivalent Diameter = (length + width) / 2 calculation. As the left area equivalent diameter of components A (1 +3) / 2 = 2mm B area equivalent diameter of (2 +4) / 2 = 3mm

The length and width ratio ≥ 5:1 product is not a simple formula to calculate the equivalent diameter

Scratch:

In the U.S. military standard "MIL-O-13830" scratches the surface quality of a model as the number of levels compared to the standard scratches. (Note: U.S. standard unit of measure scratch unspecified determined that is the width and depth of the scratch, only in the observed pattern as the standard.)

Here is the usual scratches scratches progression numbers, standard templates are 10 #, 20 #, 40 #, 60 #, 80 # 5 level. SC-QA027 total of seven surface quality standard criteria scratches, the former 5 is the subject of U.S. military regulations, the latter two is the company's internal control standards. One by one to explain the following:

1, when an element exceeds scratches progression progression scratches surface quality, the component failed.

For example: surface quality requirements elements 60-40, represents the element of scratches must ≤ 60 #, if a component of> 60 # of scratches, the component failed.

2, when an element does not exceed the surface scratches progression progression quality requirements, but elements present maximum scratches, scratches all the maximum length of the element should not exceed the diameter of 1/4.

For example: a length of 10mm 30mm wide of components, the surface quality of 60-40, there are 2 60 # length of 3 mm scratches.

It is the equivalent diameter of 20mm

1/4D is 1/4 × 20 = 5mm

The maximum length and scratches:

3mm+3mm=6mm

6mm >5mmComponent exceeds the maximum element length and scratches diameter of 1/4. So element failed.

We look at the numbers on the left are two samples (not the actual picture is enlarged size)

Figure 20 # scratches, and the length> 1/4D. 20-10 For surface quality requirements of the product failed. However, this sample is not raised surface quality requirements, but to everyone their own decision to finish grade. Therefore, the product can be sentenced to 40 # scratches.

3, when the elements present maximum scratch, and the maximum length of the scratches not exceed 1/4D, requires that all stages of the scratch scratch length multiplied by the ratio of the diameter obtained with the component of the product and may not exceed the maximum scratch half progression.

Example: 30mm width 10mm with a long element, the element of the surface quality of 60-40, there are two elements 60 # scratches of 2mm, 4mm, 40 # 3 scratches.

It is the equivalent diameter of20mm

1/4D is 1/4×20=5mm

The maximum length and scratches:

2mm+2mm=4mm

4mm<5mm. Meet two.

However, all stages of the scratch scratch length multiplied by the ratio of the diameter and components of the product obtained as follows:

1,60 # scratch length of (2 +2); 40 # scratch length of (4 +4 +4 +4)

2,60 # scratch scratch length is multiplied by the element diameter ratio of 60 × (2 +2) / 20;

40 # scratch scratch length multiplied by the ratio of the diameter of the element 40 × (4 +4 +4 +4) / 20

3, all multiplied by the scratch scratch series length and diameter ratio of components and the product of the proceeds as follows:

[60×（2+2）/20]+[40×（4+4+4+4）/20=36

Component maximum scratch series of 60; 60 and half of 60/2 = 30) 36> 30; Thus, component failure.

As shown on the left of the product

Then 20 # scratches length 0.7mm. 2mm length 10 # scratch the surface quality of the product max 20-10 scratch length 1mm ＜1/5.

All series of scratches scratches length multiplied by the ratio of the diameter obtained with the component and is the product of:

20 ×0.7/5+10

×2/5=2.8+2=4.8 ＜ 10，The device can be sentenced to 20 # scratches

4, when an element does not exceed the surface scratches quality series of stages, and the element is no maximum scratches, scratches progression by multiplying all scratches diameter ratio of the length of the element and the sum of the products obtained shall not exceed maximum scratches progression.

Example ①: element is  10, 60-40 surface quality indicators, there are two 50 # scratches 2mm, 1 strip 40 # scratches 3mm, the other two 40 # scratches 2mm, 2 strip 20 # scratches 2mm, 10 # scratches Total length 10 mm.

All series of scratches scratches length multiplied by the ratio of the diameter obtained with the component and is the product of:

[50×（2+2）/10]+[40×(3+2+2)/10]+[20×(2+2)/10]+[10×10/10]=66

Scratch is the largest component60# 66>60；Therefore, the components failed.

On the left is that we look at the number three samples. (Picture is not the actual size of the amplified)

Figure 20 # scratch length 2.2mm. 10 # scratches the surface quality requirements for length of 7mm 20-10 This product failure. For surface quality requirements 40-20 series is multiplied by the product of all elements scratches length to diameter ratio and the sum of the products obtained as follows: 20×2.2/5+10×7/5=8.8+17=25.8

25.8 ＜ 40。Components meet40-20

5, when the indicator member scratches grade quality or better than this level is 20, the component surface is allowed to have dense scratches in any one of the element surface areas  6.35mm, allowed to have 4 or more 4 large equals 10 # scratches.

Example: Surface quality indicators20-10，20mm，

In Figure (1) shows the area or length of 1mm 2 of 0.5mm and a length of 2 10 # scratches. It is consistent with 5.1.1; 5.1.4 (without regard to 5.1.2; 5.1.3) but it does not comply with 5.1.5. Component failed. On the left is that we look at the number three samples. (Picture is enlarged not actual size)

Chart has five scratches, the element diameter is 6mm; 20-10 For surface quality requirements of the product failure. For surface quality requirements 40-20 series is multiplied by the product of all elements scratches length to diameter ratio and the sum of the products obtained as follows:

20×2.2/5+10×7/5=8.8+17=25.8

25.8 ＜ 40。Components meet40-20

6, the circular element not more than 20 # progression scratches with the same diameter. For square components are not allowed more than 20 # progression through element scratches.

7, when the interval between two or more less than 0.1mm scratch, scratch into a calculation, the combined length of the scratches scratches from the beginning to the end of scratches. Width at the outer edges of scratches.

Please judge finish?

Pitting:

According to the U.S. military standard MIL-PRF-13830B series pitting defects allowed to take the actual diameter, provides for 1/100mm as units of measurement. If pitting irregular shape. Maximum length and shall be taken as the average diameter of the maximum width. (Note: The U.S. military target different pitting and scratches, pitting is measurable ie pitting size is determined, ie, pitting 50 # diameter D = 0.5mm pitting)

"SC-QA027 surface quality standard" pitting criteria have six. The other five are U.S. standard requirements. A company internal control standards.

1, when an element exists over the surface quality requirements pitting progression, the component failed. Example: element is  20mm, quality indicators required for the 60-40, the element has a diameter of 0.5mm pitting.

50>40Component failed.

2, each 20mm diameter allows only a maximum pitting. Example: element is  20mm, quality indicators required for the 60-40, there are two components pitting diameter 0.4mm substandard components

3, each 20mm in diameter on the sum of all the pitting diameter must not exceed 2 times the maximum pitting example: 40mm × 40mm components, quality indicators required for the 60-40, one of the area  20mm 40 # pips a, 20 # point a, point 10 # a, while in another area  20mm, 20 # 2 bps, 10 bps 4 #, 5 # point 1.

A zone pitting diameter and:

=40+20+10=70 < 40×2=80

B, pitting diameter and:

=20×2+10×4+5×1=85> 80，

Area B failed. Therefore element failed.

Left component surface quality 60-40：

A district pitting diameter and is10+10+20=40＜80.

B, pitting diameter and is10 × 3+20 × 3=90＞80.

Area C pitting diameter and is20 ×2+10=50＜80.

Area B, the component failed failed.

Left is the second sample No. 11. (Picture is not the actual size of the amplified)

There is a 5 # idea # 1 3 pips. (But then these ideas might wipe will not see here only as an example to illustrate). This product if you do not consider the distance between two points can be determined pips series 5. Why? Because the product has only one maximum point 5 #, and all of the ideas was 0.05 + 0.03 = 0.08 ＜0.1 (5 # pips twice), all point can be determined for the series 5 #.

4, when the quality requirements of pitting is 10 or better level, the distance between any two points must be greater than hemp 1mm.

Example: element element 40mm × 40mm,

Quality indicators required for the 20-10, there are two 5 # of ideas,

The distance between the point0.8mm. 0.8mm＜1mm

Component failure

Left is the second sample No. 11. (Picture is not the actual size of the amplified)

There is a 5 # idea # 1 3 pips. Is the above example, it is determined just 5 #, 5 # series but for two spacing must be greater than 1mm, so components do not meet the requirements, all the components of ideas can only be sentenced to 20 #.

5, less than 2.5um pitting negligible.

6, as shown when there is intensive pitting pitting gathered when diameter of the outer circle pitting size.

Written two finish at the standard data format must be written, such as product ideas judged as 0 #, but scratches sentenced to 40 #, then the product finish for the 40-20 instead of 40-0. If the product idea and sentenced to 20 # 0 # scratches ruled, the product finish for the 40-20 instead of 0-20.

Bubbles and envelope: envelope as bubbles into account. According to the U.S. military standard MIL-PRF-13830B bubble series and pitting, as units are the same. Irregular bubble is taken straight average of the maximum width and length. The allowed values ​​and bubble pitting the same. But bubbles and pitting envelope must distinguish, separate evaluation. There are gray spots element is unqualified components.