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Figure 5-22 Specifying a Basic Taper and a Basic Diameter (ASME Y14.5 - 2018) 1

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It might come from this: The taper ratio is 0.3 for every 1 unit of linear run. That equals 0.03 for every 0.1 linear run. And the image they're showing is at 9.9 (0.1 mm below its nominal/centered distance of 10.0). So for that 0.1 distance deviation of 0.1, the shrinkage is 0.03.

I'm pulling this explanation out of thin air, so somebody please tell me if I'm correct.
 
What this figure actually defines is:
- a conical boundary of basic diameter [23] at 9.9 distance from the surface on the left and a basic angle indirectly defined with [0.3:1] conical taper ratio, which indicates that there is 0.3 difference in the cone diameter for every 1 mm of its length,
- a similar conical boundary but this time at 10.1 distance from the surface on the left.
- a tolerance zone formed by the two conical boundaries for which at any plane perpendicular to the cone axis the radial distance between the boundaries equals to the conical taper ratio (0.3:1) divided by two (to get radial value) times the difference between 10.1 and 9.9 --> (0.3/1)/2×(10.1-9.9) = 0.15×0.2 = 0.03
 
Jassco,
You don't see the 0.03 tolerance in the orthographic view because it's not given directly. It is derived from the specified requirements. The requirement states that the basic diameter of 23 is located 9.9 to 10.1 from the base, and the basic taper is 0.3 mm in diameter per millimeter in length. This means there are two extreme conditions for the conical surface:

1. The smallest cone is when the diameter of 23 is 9.9 from the base.

2. The largest cone is when the diameter of 23 is 10.1 from the base.

If you draw these two conditions as cones that follow the basic taper, you'll find two different cones with a radial distance between their surfaces of 0.03.
 
Thanks for your replies!

But "0.3 mm in diameter per millimeter" is not a tolerance. It is equivalent to a dimension (slope).

Best regards,

Alex
 
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