9 SI units A 12500 mm sine bar is to be used to inspect the

9. (SI units) A 125.00 mm sine bar is to be used to inspect the inclination of one of the faces in a part. While the tolerance on the part inclination is +/-0.002 degrees, its nominal value can be anything from 5 to 85 degrees. If the smallest gauge block available is 0.01 mm thick, will this sine bar be of suficient angular resolution to determine if all possible nominal part inclinations are within tolerance? Hint: Evaluate the resolution for inclination angles from 5 to 85 degrees and plot the resolution as a function of this angle. The plot will give you a visual aid to help you answer the question.

Solution

RANGE OF ANGLE = 5 TO 85 DEGREES LENGTH OF SINE BAR = 125 MM TOLERANCE ALLOWED IN INCLINATION = + OR - 0.002 DEGREES SMALLEST GAUGE BLOCK AVAILABLE = 0.01 MM THICK THA IS HEIGHT = 0.01 MM HYPOTENUSE = 125 MM SO WE HAVE THE RELATION …SIN(T) = X*0.01/125 ………………………………1 WHERE X IS THE NUMBER OF GAUGE BLOCKS THAT CAN BE USED X = 125 * SIN(T) / 0.01 = 12500*SIN(T) SINCE SINE IN THE GIVEN RANGE IS MONOTONICALLY INCREASING FUNCTION WE CAN JUST TAKE 3 POINTS TO ILLUSTRATE THE TOPIC T SIN(T) X = LOWER END 5 0.087156 1089.447 5.002 0.087191 1089.881 4.998 0.087121 1089.012 MIDDLE RANGE 45 0.707107 8838.835 45.002 0.707131 8839.143 44.998 0.707082 8838.526 TOP END 85 0.996195 12452.43 85.002 0.996198 12452.47 84.998 0.996192 12452.4 THE RESULTS SHOW THAT WE NEED A FRACTIONAL HEIGHT GAUGE BLOCK THAT IS A BLOCK OF LESS THAN 0.01 MM GAUGE TO GET THE DESIRED ACCURACY TO CLARIFY FOR EX … IF WE PUT 1089 BLOCKS , WE GET ANGLE AS SIN(T) = 4.997944 DEGREES IF WE PUT 1090 BLOCKS , WE GET ANGLE AS SIN(T) = 5.002545 DEGREES IF WE PUT 1088 BLOCKS , WE GET ANGLE AS SIN(T) = 4.993343 DEGREES SO THE DIFFERENCE IN ANGLE FROM THAT AT 1089 BLOCKS IS -0.0046      AND 0.004601 WHICH IS MORE THAN DOUBLE THE 0.002 TOLERANCE SPECIFIED SO WE NEED TO USE ATLEAST 300 MM SINE BAR FOR THE PURPOSE . OR USE GAUGE BLOCK OF THICKNESS 0.004 OR LESS ….APPROXIMATELY