SME
Manufacturing Processes · Problem type
Ideal surface roughness
Use this for “A tool with 0.8 mm nose radius turns at 0.25 mm/rev. Find the ideal roughness Rₐ and the cusp height.”
Diagram
Rₐ
≈
f
2
32r
=
2.44
μm;
the
cusps
stand
about
4
×
Rₐ
high.
This is an example. Type your own values to start a new problem.
Rₐ (μm)
=
(
f (mm/rev)
)
2
32
×
r (mm)
h (μm)
=
(
f (mm/rev)
)
2
8
×
r (mm)
Formulas
Rₐ
=
1,000
×
f
2
32 ×
r
2.4414
=
1,000
×
0.25
2
32 × 0.8
h
=
1,000
×
f
2
8 ×
r
9.7656
=
1,000
×
0.25
2
8 × 0.8
Assumptions
•
The ideal surface: tool-nose arcs f apart, with no tool wear, chatter or built-up edge.
•
The feed is well under the nose radius; f and r in mm give Rₐ and h in μm.
•
Real surfaces are rougher; the ideal value is the best the feed allows.
Step-by-step
WE KNOW
f = 0.25 mm/rev, r = 0.8 mm
FIND
roughness (Rₐ), cusp height (h)
Step 1 · Find roughness (Rₐ)
Use Rₐ = 1,000 × f² ÷ (32 × r)
The
cusps
stand
f
2
8
r
high
and
Rₐ
is
a
quarter
of
that;
×
1000
turns
mm
into
μm.
Rₐ
=
1,000
×
0.25
2
32 × 0.8
Rₐ
=
1,000
×
0.25
2
25.6
Rₐ
=
1,000
×
0.0625
25.6
Rₐ
=
62.5
25.6
Rₐ = 2.4414 μm
Step 2 · Find cusp height (h)
Use h = 1,000 × f² ÷ (8 × r)
Two
nose
arcs
of
radius
r
,
f
apart,
meet
f
2
8
r
above
the
groove;
×
1000
turns
mm
into
μm.
h
=
1,000
×
0.25
2
8 × 0.8
h
=
1,000
×
0.25
2
6.4
h
=
1,000
×
0.0625
6.4
h
=
62.5
6.4
h
=
9.7656
μm
Check
Put the numbers back in. Both sides should match.
2.4414
=
1,000
×
0.25
2
32 × 0.8
✓
9.7656
=
1,000
×
0.25
2
8 × 0.8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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