Hooligooner
Industrial
- Jan 14, 2016
- 3
Hello
I also am trying to find the maximum and minimum sizes over pins for an external DIN 5480 spline. I am using DIN 5480:2006 parts 1 and 2. In DIN 5480-1:2006 section 10.12 of this standard it talks about using DIN 5480-15 to calculate these sizes, this is not available to us. It also suggests that it can be done using the deviation factors as described in DIN 5480-2, however no explanation of how this is achieved is given.
In DIN 5480-1:2006 section 9.1 there is given an example of a W120x3x38x8f side fit spline. I have been trying to find M1[sub]max[/sub] and M1[sub]min[/sub] for this spline using the deviation factor. The target sizes from the example are Max 126.017, min 125.956.
DIN 5480-1:2006 section 10.12 tells us the deviation of the measurements over measuring circles is A[sub]M1[/sub]=A[sub]e[/sub]*A*[sub]M1[/sub] = -0.04256
Obviously this isn't enough on it's own. So far the closest I have managed to come is to apply A[sub]M1[/sub] to the f8 limit dimensions of the 120 shaft i.e. -0.036/-0.090. This gives an answer of max 126.016, min 125.962.
I don't believe this is right, where am I going wrong?
I also am trying to find the maximum and minimum sizes over pins for an external DIN 5480 spline. I am using DIN 5480:2006 parts 1 and 2. In DIN 5480-1:2006 section 10.12 of this standard it talks about using DIN 5480-15 to calculate these sizes, this is not available to us. It also suggests that it can be done using the deviation factors as described in DIN 5480-2, however no explanation of how this is achieved is given.
In DIN 5480-1:2006 section 9.1 there is given an example of a W120x3x38x8f side fit spline. I have been trying to find M1[sub]max[/sub] and M1[sub]min[/sub] for this spline using the deviation factor. The target sizes from the example are Max 126.017, min 125.956.
DIN 5480-1:2006 section 10.12 tells us the deviation of the measurements over measuring circles is A[sub]M1[/sub]=A[sub]e[/sub]*A*[sub]M1[/sub] = -0.04256
Obviously this isn't enough on it's own. So far the closest I have managed to come is to apply A[sub]M1[/sub] to the f8 limit dimensions of the 120 shaft i.e. -0.036/-0.090. This gives an answer of max 126.016, min 125.962.
I don't believe this is right, where am I going wrong?