4 Conversion of Tool Angles From One System to Another

8/14/2019 4 Conversion of Tool Angles From One System to Another

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Module2Mechanics of

Machining

Version 2 ME IIT, Kharagpur


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Lesson

4Conversion of tool

angles from one system

to another



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Instructional objectives

At the end of this lesson the students should be able to

(i) State the purposes of conversion of tool angles(ii) Identify the four different methods of conversion of tool angles(iii) Employ the graphical method for conversion of

Rake angles clearance angles

Cutting anglesFrom ASA to ORS and ORS to ASA systems

(iv) Convert rake angle and clearance angle from ORS to NRS(v) Demonstrate tool angles relationship in some critical conditions.

(i) Purposes of conversion of tool angles from one systemto another

To understand the actual tool geometry in any system of choice orconvenience from the geometry of a tool expressed in any othersystems

To derive the benefits of the various systems of tool designation asand when required

Communication of the same tool geometry between people followingdifferent tool designation systems.

(ii) Methods of conversion of tool angles from one systemto another

Analytical (geometrical) method: simple but tedious Graphical method Master line principle: simple, quick and popular Transformation matrix method: suitable for complex tool geometry Vector method: very easy and quick but needs concept of vectors

(iii) Conversion of tool angles by Graphical method Master Line principle.

This convenient and popular method of conversion of tool angles from ASA toORS and vice-versa is based on use of Master lines (ML) for the rake surfaceand the clearance surfaces.

Conversion of rake anglesThe concept and construction of ML for the tool rake surface is shown in Fig. 4.1.



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A

B

C

D

A'B'

C'

D'O'

O' O'

Zm

Xm

Ym

Xm

Xo

O'

Zm

Ym

x

o

y

Master line forrake surface

R

X

Y

C

T

T

T

Zo Yo

Zo

Yo

O

R

o

R

T

Xo

Fig. 4.1 Master line for rake surface (with all rake angles: positive)

In Fig. 4.1, the rake surface, when extended along X plane, meets the toolsbottom surface (which is parallel to R) at point D i.e. D in the plan view.Similarly when the same tool rake surface is extended along Y, it meets thetools bottom surface at point B i.e., at B in plan view. Therefore, the straightline obtained by joining B and D is nothing but the line of intersection of the

rake surface with the tools bottom surface which is also parallel to R. Hence,if the rake surface is extended in any direction, its meeting point with the toolsbottom plane must be situated on the line of intersection, i.e., BD. Thus the

points C and A (in Fig. 4.1) obtained by extending the rake surface along o

and C respectively upto the tools bottom surface, will be situated on that lineof intersection, BD.This line of intersection, BD between the rake surface and a plane parallel to

R is called the Master line of the rake surface.

From the diagram in Fig. 4.1,

OD = TcotXOB = TcotY

OC = TcotoOA = Tcot



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Where, T = thickness of the tool shank.The diagram in Fig. 4.1 is redrawn in simpler form in Fig. 4.2 for conversion oftool angles.

A

B

CM

E

F

Ym

Xm

XoYo

D

Masterline ofRake surface

OD = cotxOB = cotyOC = cotoOA = cot

forT=unity]

OG

Fig. 4.2 Use of Master line for conversion of rake angles.

Conversion of tool rake angles from ASA to ORS

o and (in ORS) = f (x and y of ASA system)costansintantan yxo += (4.1)

and sintancostantan yx += (4.2)

Proof of Equation 4.1:

With respect to Fig. 4.2,

Consider, OBD = OBC + OCDOr, OB.OD = OB.CE + OD.CF

Or, OB.OD = OB.OCsin + OD.OCcosDividing both sides by OB.OD.OC,

cos

OB

1sin

OD

1

OC

1+=

i.e. Proved.costansintantan yxo +=



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Similarly Equation 4.2 can be proved considering;

OAD = OAB + OBDi.e., OD.AG = OB.OG + OB.OD

where, AG = OAsinand OG = OAcos

Now dividing both sides by OA.OB.OD,

OA

1cos

OD

1sin

OB

1+=

Proved.sintancostantan yx +=The conversion equations 4.1 and 4.2 can be combined in a matrix form,

(4.3)

=

y

xo

tan

tan

sincos

cossin

tan

tan

(ORS) (ASA)

where, is the transformation matrix.

sincos

cossin

Conversion of rake angles from ORS to ASA system

x and y (in ASA) = f(o and of ORS)costansintantan ox = (4.4)

and sintancostantan oy += (4.5)The relations (4.4) and (4.5) can be arrived at indirectly using Equation 4.3.

By inversion, Equation 4.3 becomes,

(4.6)

=

tan

tan

sincos

cossin

tan

tano

y

x

from which equation 4.4 and 4.5 are obtained.

The conversion equations 4.4 and 4.5 can also be proved directly from thediagram in Fig. 4.2

HintsTo prove equation 4.4, proceed by taking (from Fig. 4.2)

OAD = OAC + OCD,[involving the concerned angles o, and x i.e., OC, OA and OD]

And to prove Equation 4.5, proceed by taking

OAC = OAB + OBC[involving the concerned angles o, and y i.e., OC, OA and OB]

Maximum rake angle (max orm)The magnitude of maximum rake angle (m) and the direction of the maximumslope of the rake surface of any single point tool can be easily derived from itsgeometry specified in both ASA or ORS by using the diagram of Fig. 4.2. The

smallest intercept OM normal to the Master line (Fig. 4.2) represents max orm as



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OM = cot mSingle point cutting tools like HSS tools after their wearing out are oftenresharpened by grinding their rake surface and the two flank surfaces.

The rake face can be easily and correctly ground by using the values of mand the orientation angle, (visualized in Fig. 4.2) of the Master line.

Determination ofm and from tool geometry specified in ASA system.

In Fig. 4.2,

OBD = OB.OD = BD.OM

or, OB.OD = OMODOB .22 + Dividing both sides by OB.OD.OM

22 OB

1

OD

1

OM

1+=

or y2

x

2

m tantantan += (4.7)

Again from OBD

OD

OBtan =

or

=

y

x

tan

tantan (4.8)

m and from tool geometry specified in ORS

Similarly from the diagram in Fig. 4.2, and taking OAC, one can prove

2o2

m tantantan += (4.9)

=

o

tan

tantan (4.10)

Conversion of clearance angles from ASA system to ORS and viceversa by Graphical method.

Like rake angles, the conversion of clearance angles also make use ofcorresponding Master lines. The Master lines of the two flank surfaces are

nothing but the dotted lines that appear in the plan view of the tool (Fig. 4.3).The dotted line are the lines of intersection of the flank surfaces concerned

with the tools bottom surface which is parallel to the Reference plane R.Thus according to the definition those two lines represent the Master lines ofthe flank surfaces.Fig. 4.4 shows the geometrical features of the Master line of the principal flankof a single point cutting tool.From Fig. 4.4,

OD = TtanxOB = TtanyOC = Ttano



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OA = Tcot where, T = thickness of the tool shank.

ML of principal flankML of auxiliary flank

Fig. 4.3 Master lines (ML) of flank surfaces.

x

y

x

y

o

x

o

yR

O

Ym

Xm

Xo

Yo

Yo

Zo

Zo

Xo

Zm

Zo Yo

A'

D'

C'

B'BC

D

A

ML forprincipalflank

T

T

T

o

Fig. 4.4 Master line of principal flank.



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The diagram in Fig. 4.4 is redrawn in simpler form in Fig. 4.5 for conversion ofclearance angles.

The inclination angle, basically represents slope of the rake surface alongthe principal cutting edge and hence is considered as a rake angle. But appears in the analysis of clearance angles also because the principal cuttingedge belong to both the rake surface and the principal flank.

A

B

MC

O

Ym YoXo

Xm

Master line for principal flank

OD = tanxOB = tanyOC = tanoOA = cot

]for

T=unity

D

Fig. 4.5 Use of Master line for conversion of clearance angles.

Conversion of clearance angles from ASA to ORS

Angles, o and in ORS = f(x and y in ASA system)Following the same way used for converting the rake angles taking suitabletriangles (in Fig. 4.2), the following expressions can be arrived at usingFig. 4.5:

cosy

cotsinx

coto

cot += (4.11)

and siny

cotcosx

cottan += (4.12)

Combining Equation 4.11 and 4.12 in matrix form



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(4.13)

=

y

xo

cot

cot

sincos

cossin

tan

cot

Conversion of clearance angles from ORS to ASA system

x and y (in ASA) = f(o and in ORS)Proceeding in the same way using Fig. 4.5, the following expressions arederived

cossino

cotx

cot tan= (4.14)

and sincoso

coty

cot tan+= (4.15)

The relations (4.14) and (4.15) are also possible to be attained frominversions of Equation 4.13 as indicated in case of rake angles.

Minimum clearance, min orm

The magnitude and direction of minimum clearance of a single point tool maybe evaluated from the line segment OM taken normal to the Master line (Fig.

4.5) as OM = tanmThe values ofm and the orientation angle, (Fig. 4.5) of the principal flankare useful for conveniently grinding the principal flank surface to sharpen theprincipal cutting edge.Proceeding in the same way and using Fig. 4.5, the following expressions

could be developed to evaluate the values ofm and o From tool geometry specified in ASA system

yxm 22 cotcotcot += (4.16)

and

=

y

x

cot

cottan

1

(4.17)

o From tool geometry specified in ORS

2

0

2tancotcot +=

x (4.18)

and

=

o

cot

tantan

1

(4.19)

Similarly the clearance angles and the grinding angles of the auxiliary flanksurface can also be derived and evaluated.

o Interrelationship amongst the cutting angles used in ASA and ORS

The relations are very simple as follows:

(in ORS) = 90o - s (in ASA) (4.20)and 1(in ORS) = e (in ASA) (4.21)



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(iv) Conversion of tool angles from ORS to NRS

The geometry of any single point tool is designated in ORS and NRSrespectively as,

, o, o, o, 1, , r (mm) ORS, n, n, n, 1, , r (mm) NRS

The two methods are almost same, the only difference lies in the fact that

o, o and o of ORS are replaced by n, n and n in NRS.The corresponding rake and clearance angles of ORS and NRS are relatedas ,

tann = tanocos (4.22)cotn = cotocos (4.23)

and cotn'= coto'cos' (4.24)The equation 4.22 can be easily proved with the help of Fig. 4.6.

o

n

o

n

n

c

ZoZn Yn

Yo

A

B

CA

B

A

C

Xo, Xn

Fig. 4.6 Relation between normal rake (n) and orthogonal rake (o)

The planes o and n are normal to Yo and Yn (principal cutting edge)respectively and their included angle is when o and n are extended belowOA (i.e. R) they intersect the rake surface along OB and OC respectively.Therefore,

AOB = o AOC = nwhere, BAC =



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Now AC = ABcosOr, OAtann= (OAtano)cosSo, tann= tanocos provedThe equation (4.23) relating n and o can be easily established with the helpof Fig. 4.7.

Fig. 4.7 Relation between normal clearance, n and orthogonal clearance, oFrom Fig. 4.7,

AC = ABcosOr AAcotn= AAcotocos

cotn= cotocos provedSimilarly it can be proved,

cotn= cotocoswhere is the inclination angle of the auxiliary cutting edge.

(v) Tool geometry under some critical conditions

Configuration of Master lines (in graphical method of tool angleconversion) for different tool geometrical conditions.

The locations of the points A, B, C, D and M along the ML will be asshown in Fig. 4.2 when all the corresponding tool angles have some

n

o

o

n

n

o

C

R

A

B

C

A

C

A

BA'

A'

Zn

Zo

Zn

Yn

Zo

Yo

C

o



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positive values. When any rake angle will be negative, the location of thecorresponding point will be on the other side of the tool.

Some typical configurations of the Master line for rake surface and thecorresponding geometrical significance are indicated in Fig. 4.8.

Configuration of ML Tool geometry

f

or ML parallel to Cx = positivey = positiveo = positive = 0m = o

for ML parallel to Yx = positivey = 0o = positive = negativem = x

for ML parallel to Xx = 0

y = negativeo = negative

= negativem= y

(c)

(b)

(a)

O

ML

ML

C B A

O

B at infinity

C

D

A

ML

A at infinity

O

DC

B



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Fig. 4.8 Tool geometry and Master line (rake face) in some typical conditions.

Tool angles relations in some critical conditions

From the equations correlating the cutting tool angles, the following criticalobservations are made:

When = 90o; x = o forX = o When = 0 ; n = o

n = o When =0 and = 90o; n=o=x pure orthogonal cutting

(N=o=X)

Exercise 4

A. Quiz test

Select the correct answer from the given four options1. The master line for the rake surface of the turning tool of geometry : -

10o, 0

o, 8

o, 6

o, 15

o, 30

o, 0.1 (inch)

(a) machine longitudinal plane(b) machine transverse plane(c) cutting plane(d) orthogonal plane

2. If the approach angle of a turning tool be 30o, the value of its principalcutting edge angle will be

(a) O deg.(b) 30

odeg.

(c) 60o

deg.(d) 90o deg.

3. The value of side rake of the turning tool of geometry : -0

o, 10

o, 8

o, 6

o, 20

o, 60

o, 0 (mm) will be

(a) 0o deg.(b) 10

odeg.

(c) 8o

deg.(d) 6

odeg.

4. The values of orthogonal clearance and normal clearance of aturning tool will be same if,

(a) =0(b) X = Y



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(c) = 0(d) none of the above

5. The angle between orthogonal plane and normal plane of a turningtool is

(a) o(b) (c) n(d)

B. Problem1. Determine the values of normal rake of the turning tool whose

geometry is designated as : 10o, - 10

o, 8

o, 6

o, 15

o, 30

o, 0 (inch)?

2. Determine the value of side clearance of the turning tool whosegeometry is specified as 0o, - 10o, 8o, 6o, 20o, 60o, 0 (mm) ?

Solutions of Exercise 4

A. Quiz test1 (a)2 (c)3 (b)4 (c)5 (d)

B. Problems

Ans. 1

Tool geometry given :10o, - 10o, 8o, 6o, 15o, 30o, 0 (inch)

y, x, y, x, e, s, r ASA

tann = tanocoswhere,tano = tanxsin + tanycos

= tan(-10o)sin(90

o- 30

o) + tan(10

o)cos(90

o- 30

o)

= - 0.065

So, o = - 3.7o

And tan= - tanxcos + tanysin= - tan(-10

o)cos(90

o- 30

o) + tan(10

o)sin(90

o- 30

o)

= 0.2408

So, = 13.54o

tann = tanocos = tan (-3.7)cos(13.54) = - 0.063So, n = - 3.6o Ans.



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Ans. 2

Tool geometry given : 0o, - 10o, 8o, 6o, 20o, 60o, 0 (mm)

, o, o, o', 1, , r (mm)

cotx = cotosin- tancos= cot 8.sin60 tan0.cos60= cot8.sin60= 6.16

So, x = 9.217o Ans.


4 Conversion of Tool Angles From One System to Another

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