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# ASTM E10 section 7.6.4 varbiage

## ASTM E10 section 7.6.4 varbiage

(OP)
Hello,

I am trying to create a verification formula based on ASTM E10 section 7.6.4:
"When indentations are made on a curved surface, the
minimum radius of curvature1 of the surface shall be two and a
half times the diameter of the ball
curved surfaces may be slightly elliptical rather than circular in
shape. The measurements of the indentation shall be taken as
the mean of the major and minor axes."

So here's the issue

1Rmin=b2/a
2a = minor axis = d0
2b = major axis = d90
=> Rmin=d902/2d0

2Rmin=2.5D

"Shall be" implies equals to. I was expecting some sort of greater than or less than situation. it seems this scenario is highly unlikely!

Therefore
d902/d0=5D

So unless I'm not thinking through this correctly.. in a case where a 10mm ball is used, the only scenario where this equation is satisfied is when the ratio of d902/d0 comes out to exactly 50mm. for example d90= 10mm and d0=2mm.

This seems like a very strict requirement...

Replies continue below

### RE: ASTM E10 section 7.6.4 varbiage

What is wrong with the word "minimum"?
That means that "shall be" is the minimum, doesn't it?

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

### RE: ASTM E10 section 7.6.4 varbiage

(OP)
No. The word minimum is referring to the minimum radius of curvature, which is the calculated value Rmin. The sentence would need to be structured as "the minimum radius of curvature of the surface shall be at a minimum two and a half times the diameter of the ball" to make sense.

Unless, again, I'm not looking at this correctly.

### RE: ASTM E10 section 7.6.4 varbiage

Rmin = 2.5D sure seems like a minimum to me.

### RE: ASTM E10 section 7.6.4 varbiage

R=b^2/a is for curvature of an ellipse. The min radius of curvature in ASTM refers to the surface of test piece. for example, flat surface has an infinite radius, the ideal situation to calculate the BHN numbers since the indentation will be symmetric if the material is isotropic.

if the radius of curvature is too small, the test surface is very non-smooth, the diameter of indentation will be too small, yielding a higher BHN than would-be for a convex surface.

in addition, avoid testing a concave surface (if not flat). No idea if ASTM specifies to use a convex surface.

### RE: ASTM E10 section 7.6.4 varbiage

(OP)
Thanks! that makes sense.

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