Third, we measured the width of part number 2a with a 2 Inch Micrometer. The results we got were such as 1.6280 Inch and 1.6298 Inch. In this case, it is very difficult to identify what the print dimension would be, however, we assumed that it would be 1~5/8 Inch with a +- 0.025 Inch tolerance. The reasons we got different results from one another would be the location we measured on the part, and could also be because the part is rusted so there is a lot of built-up particles that made the part thicker than it was made originally. Our results could be out of the tolerance. Fourth, we used the ID Micrometer and measured the diameter of a tiny hole. We used a Metric Unit System for this part because there was no ID Micrometer in English System …show more content…
The results we got were such as 1.6280 Inch and 1.6298 Inch. In this case, it is very difficult to identify what the print dimension would be, however, we assumed that it would be 1~5/8 Inch with a +- 0.025 Inch tolerance. The reasons we got different results from one another would be the location we measured on the part, and could also be because the part is rusted so there is a lot of built-up particles that made the part thicker than it was made originally. Our results could be out of the tolerance. Fourth, we used the ID Micrometer and measured the diameter of a tiny hole. We used a Metric Unit System for this part because there was no ID Micrometer in English System available in our laboratory. According to our experiment, both of us got the diameter of 10.12mm (See Table 1). Therefore, we concluded that the intended hole would be 10.12mm with a small tolerance about less than 0.01mm.
Fifth, we measured the height of part number 2 by using a Depth Micrometer. I and my partner got 0.958 Inch and 0.960, respectively. As a result, the print dimension would be around 0.958 Inch with about +-0.025 Inch tolerance. The reason we got different result could be because the location we set the measuring tool, and could also be human error. Sixth, Super