00:01
The question that is given over here, this is regarding two points.
00:06
Over here, we are going to calculate the minimum and maximum distance between the two points.
00:13
Here, an image is given and we are going to include the angular tolerance in this analysis.
00:20
First, the distance between two points, it is given by d is equal to ab plus b.
00:34
Which is according to the given image and here we have been given that a is equal to two plus or minus 0 .01 b is equal to 1 plus or minus 0 .030 and c is equal to 0 .5 plus or minus 0 .005 here the theta will be 30 degree plus or minus 2 the distance bc will be equal to a by sine 30 degree this can be calculated from the given image so the bc will be equal to 2 plus or minus 0 .01 divided by sine 30 degree in order to calculate the minimum distance that is bc min, this will be equal to 2 plus minus of 0 .01, which is divided by sine of 30 degree plus 2.
01:52
So, the minimum distance of bc, that is bc min will be equal to 3 .7 .5.
02:02
And now we are going to calculate the maximum distance that is bc max.
02:07
So this will be 2 plus 0 .01 which is divided by sine of 30 degree minus 2.
02:16
So this will be 4 .2814.
02:20
So bc max will be equal to 4 .2814.
02:29
Here we have taken the tolerance in each distance...