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by：ChangHui
2020-05-11

Surface defect standard: according to the American military standard MIL -
O-
13830 b with two sets of figures show that the surface defect size.
For example, 40/20 (
Or 40 -
20)
The former size limit the scratch and the latter limit hemp dot size.
Line, bright road, injury, scratch called scratch.
Spots, pit points, ideas are called pitting.
Rules of length and width ratio is greater than 4:1 to scratch;
Long and wide as pitting than less than 4:1.
When components are different areas of the surface finish requirement is different, the calculation of equivalent diameter to the area of: high surface quality requirement within the area within the equivalent diameter of area shall prevail,
Such as effective aperture area)
Outside, the surface quality requirement of low area calculation are integral components of equivalent diameter.
Below components surface quality requirement is different, in the judgement area A is in accordance with the requirements, should take the circle diameter to calculate.
While B area take the entire element diameter Nick: in the United States military standard MIL -.
O-
13830 'the surface quality of scratches as few scratches at all levels than standard template.
(
Note: the standard is not sure of unspecified scratches scratches, width and depth of the template for the standard only in actual observation.
)
Here is often scratch scratch series number, standard template has 10 #, 20 #, 40 #, 60 #, 80 # 5 grade.
1.
When the components of the surface of the scratch series more than the quality requirements of scratch series, unqualified components.
For example: the surface of the element quality requirements for 60 -
40, the representative element scratches must be 60 #, or less if components have> 60 # scratches, the element is unqualified.
2.
When Nick series of components does not exceed surface quality requirements of the series, but when there is a biggest scratch element, the sum of all the length of the largest scratch should be no more than a quarter of the diameter of the element.
For example, has a 30 mm long 10 mm wide element, element of the surface quality requirement for 60 -
40, 60 # 2 to 3 mm long scratches.
The equivalent diameter of 20 mm 1/4 for 1/4 x 20 d = 5 mm scratch and the length of the biggest is: 3 mm + 3 mm = 6 mm 6 mm> 5 mm maximum row scar length and more than a quarter of the diameter of the element.
So the component is unqualified.
Below is a sample see (
Images are magnified not actual size)
The scratches in the graph of 20 #, and 1/4 length> d.
20 - to surface quality requirement
10 the product is unqualified.
But the sample did not surface quality request, but you yourself decided to finish level.
So the product scratches can be jailed for 40 #.
3.
When components exist largest scratch and scratch is not more than the sum of the length of the largest 1/4 d, require all the scratches times series scratches length and diameter of components than the sum of the product, shall not exceed half of the largest scratch series.
Example: a 30 mm wide 10 mm long element, element surface quality requirements for 60 -
Article 40, components on two 60 # 2 mm scratch, 40 # 3 4 mm scratches.
60 # scratches of length (
2 + 2)
;
40 # scratches of length (
4 + 4 + 4)
60 # scratch multiplied by the scratch length and the ratio of the element diameter 60 x (
2 + 2)
/ 20;
40 # scratch multiplied by the scratch length and the ratio of the element diameter 40 x (
4 + 4 + 4)
All scratch multiplied by the scratch of the series length and diameter of components than the sum of the product obtained is: [
60×（
2 + 2)
/ 20]
+ (
40×（
4 + 4 + 4)
/ 20 = 36 scratches series for maximum 60;
Half of the 60 for 60/2 = 30)
36> 30;
Components, therefore, unqualified products as shown in the figure below 0 # 20 scratches length.
7mm。
10 # scratches length 2 mm 20 - to surface quality requirement
10 the product biggest scratch length of 1 mm.
All scratch multiplied by the scratch of the series length and diameter of components than the sum of the product obtained is: 20 x 0.
7/5+10 ×2/5=2.
8 + 2 = 4.
820 # 4.
When Nick series of components does not exceed surface quality requirements of the series, and there is no scratches, the largest component of all series of scratch multiplied by the scratch length and diameter of components than the sum of the product, shall not exceed the maximum Nick series.
Ex 1: components for?
10, the surface quality indicator 60 -
40, with 50 # 2 cut 2 mm, 40 # 1 the scratch 3 mm, the other two 40 # 2 mm scratch, 2 # 20 Nick 2 mm, 10 # scratch combined 10 mm length.
All scratch multiplied by the scratch of the series length and diameter of components than the sum of the product obtained is: [
50×（
2 + 2)
/ 10]
+ (
40×(
3 + 2 + 2)
/ 10]
+ (
20×(
2 + 2)
/ 10]
+ (
10×10/10]
= 66 largest scratch is 60 # 66> 60;
Therefore, unqualified components.
The image below to see samples, images are magnified not actual size.
With 20 # cut length in figure 2.
2mm。
10 # scratches length 7 mm 20 - to surface quality requirement
10 the product is unqualified.
40 - to surface quality requirement
20 times the length of scratch is all the product series and element diameter ratio of the sum of the product obtained is: 20 x 2.
2/5+10×7/5=8.
8 + 17 = 25.
8 to 25.
August 5.
When the components quality indicators scratches of grade 20 or exceed this level, component surface are not allowed to have intensive scratch that any one in the component?
6.
35 mm area, are not allowed to have four equal to or more than four big 10 # scratches.
Example: the surface quality indicators for 20 -
10,?
20 mm, as shown in the figure below area or have long article 2 1 mm long and 2 0.
5 mm of 10 # scratches.
It is in line with 5.
1.
1;
5.
1.
4 (
Without considering 5.
1.
2;
5.
1.
3)
But it is not in conformity with the 5.
1.
Article 5.
Components is unqualified.
See sample below is, images are magnified not actual size.
In figure 5 scratches, the cell diameter of 6 mm;
20 - to surface quality requirement
10 the product is unqualified.
40 - to surface quality requirement
20 times the length of scratch is all the product series and element diameter ratio of the sum of the product obtained is: 20 x 2.
2/5+10×7/5=8.
8 + 17 = 25.
8 to 25.
8 pitting: according to the American military standard MIL -
O-
13830 b series take allow defects of pitting the actual diameter of regulations to 1/100 mm as the unit of measurement.
If the pitting irregular shape.
You should take as a maximum length and the width of the average diameter.
(
Note: the mark pitting different from scratch, pitting is quantifiable namely pitting is certain, the size of the 50 # pitting namely diameter D = 0.
5 mm of pitting)
1.
When the component is more than one surface quality requirement of pitting series, unqualified components.
Example: components for?
20 mm, quality indicators for 60 -
40, one element diameter 0.
5 mm of pitting 50> 40 components is unqualified.
2.
Every 20 mm diameter is only allowed to have 1 on the point of marijuana.
Example: components for?
20 mm, quality indicators for 60 -
40, element has two diameter 0.
Pitting components unqualified 3 of 4 mm.
Every 20 mm diameter on the sum of all hemp dot diameter shall not be more than 2 times the hemp dot: 40 mm x 40 mm components, requirements for 60 - quality indicators
40, one of them?
20 mm area 40 # ideas a, # 20 ideas a, 10 # one idea, at the same time in another?
20 mm area, 20 # 2 idea, 10 # 4, idea idea 5 # 1.

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