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  • Friends, our subject is Manufacturing Processes II and module II is still running and subject

  • is Mechanics of Machining and today is lecture - 4. The topic today is interrelations among

  • the tool angles expressed in different systems and what are the instructional objectives

  • or content of lecture today.

  • State the purposes of conversion of tool angles from one system to another. Why do we need

  • conversion of tool angles from ASA to ORS or ORS to NSA and so on? Name the different

  • methods of the conversion of tool angles. There are different methods what are those?

  • Convert tool angles by graphical method. Rake angle from ASA to ORS and vice versa, clearance

  • angles from ASA to ORS and vice versa. Next convert tool angles from orthogonal rake system

  • to normal rake system and last ascertain tool geometry in some specific conditions. Now

  • what are the purposes of conversion of tool angles?

  • To understand geometry if expressed in a different system. Suppose you are conversant with a

  • particular system of tool designation, say orthogonal rake system but you have got a

  • book written by a foreign author who dealt the cutting tool geometry in ASA system. You

  • may not be able to understand, then what you have to do? You have to learn both ASA system

  • and orthogonal system and also convert the tool angles given in the book in ASA system

  • in to ORS system for your understanding, your visualization and study. Now next, derive

  • benefits of different system. Now there are different tool designative system ASA system,

  • orthogonal system, normal rake system, maximum rake system, work reference system, each system

  • has got certain advantage.

  • Now we have to derive the advantages. To derive for example, say ASA system is very convenient

  • for tool manufacturing tool inspection and so on. Then orthogonal system, orthogonal

  • system is very simple. It is very useful say general study, research, analysis and so on

  • but orthogonal system does not provide the exact tool geometry if lambda inclination

  • angle is not zero. Thirdly if you want to resharpen the cutting tool then ORS system

  • is not that very convenient whereas normal rake system that gives the true picture of

  • the cutting tool and it enables very easy resharpening of the tool.

  • So if you want to derive the benefits we should convert tool angle from one system to another

  • to derive the benefits. Now communicate geometry to others who follow different tool designation

  • system. Suppose you do research on a machining process with a cutting tool in ORS system.

  • Now you go to a conference and say USA and there people are more conversant in ASA system.

  • How will you communicate the tool geometry that you have dealt with to them who are knowledgeable

  • about ASA system that means, you have to convert your tool angles from ORS to ASA, and then

  • tell them this is the geometry actually in ASA system for there understanding and this

  • is mutually true vice versa also. Now what are the different methods of conversion of

  • tool angles?

  • There are basically four methods. First is analytical method it is geometrical method

  • very simple but tedious. It takes lot of time and

  • monotonous. Graphical Method based are Masterline principle, it is very simple, quick

  • and very popular. So most of the people you know learn this method first. Transformation

  • matrix method is a little complex method but very suitable for complex tool

  • geometry. Say for example, say geometry of cutting, say drill geometry for horn tools

  • then for conversion of tool angles say hob cutters or gear shaping cutters for such complex

  • shaped cutters or cutting tools the transmission matrix method is appropriate

  • for conversion and vector method is Universal, very strong powerful method as

  • well as very simple and quick but it needs concept of vectors and matrix. So our concentration

  • will be on graphical method master line principle. So conversion of tool angles

  • by graphical method using master lines.

  • Now let us start with conversion of rake angles. You know the cutting tools have got three

  • important angles. One is rake angle, next is clearance angle and third is cutting angle

  • and beside that finally the tool nose radius. So let us start with conversion of rake angles

  • from ASA system to ORS and vice versa.

  • So first of all say rake angle, we shall deal with conversion of rake angles from ASA to

  • ORS and vice versa. So let us start with this. By graphical method, this is based on master

  • line principle, then what is master line? First of all you have to understand the master

  • line concept. See this is master line, let me show you the master line for rake surface.

  • First you draw a cutting tool. This is a top view of the cutting tool. It is drawn in orthogonal

  • rake sorry pi r reference plane, then you visualize the other planes. This is machine

  • longitudinal plane pi x, this is machine transfers plane pi y, this is cutting plane pi c and

  • this is orthogonal plane pi o. Now this is xm, this is ym, this is xo orthogonal, this

  • is yo and zo and zm perpendicular to the plane.

  • So again I repeat this is pi x plane, this is pi y plane - machine transverse plane,

  • this is orthogonal plane and this is cutting plane. Now you show the rake angles in different

  • planes. Say first of all, let us take the section of the tool in machine longitudinal

  • plane and show the cutting tool section. This is section of the cutting tool and in which

  • plane the diagram has been drawn, the section is taken on pi x plane. So this is pi x, machine

  • longitudinal plane and what is this axis this is xm, what is this is axis y this is zm.

  • So this is zm and this is xm and what is this angle? Angle between the reference plane and

  • the rake surface, so this is gamma x that is side rake.

  • Now what you have to, this is the bottom surface of the tool and these are rake surface you

  • extend the rake surface until it meets the bottom surface suppose it meets at point D

  • prime. Say at certain distance, this rake surface meets the bottom surface and if you

  • project it on this diagram on the view drawn in reference plane so this is point suppose

  • D. So D is situated on the bottom plane which is parallel to the reference plane or it is

  • horizontal plane. Now you take the cutting tool section along machine transfer plane

  • in this direction. So this is the cutting tool and which plane the diagram has been

  • drawn, pi y machine transverse plane by take a section by this machine transverse plane.

  • So this is obviously zm and this is ym axis, xm is perpendicular to the plane and what

  • is this angle? this is rake angle in which plane? pi y plane, so this is gamma y that

  • is back rake. If you extend this rake surface up to the bottom plane of the tool then meets

  • at a point say this is point B prime and on this diagram if you project it this is the

  • point this is the distance suppose this is o oD this is oB. So B and D both of situated

  • on the bottom surface and if I join B and D by a straight line then what is this BD.

  • BD is situated on the bottom plane which is also parallel to the reference plane. This

  • is reference plane, this surface is reference plane, this plate now what physically what

  • is the meaning of this.

  • This is the line of intersection between the rake surface of the tool if extended and the

  • bottom surface of the tool. So this is the line of intersection between the rake surface

  • and the bottom surface if it is extended on this direction it will meet a point D and

  • B. Now you can extend it. So this line is called Master Line for rake surface for a

  • Master Line for rake surface, similarly Master Line for clearance surface will also appear.

  • Now what is this point? Say this is point C. It means if the rake surface is extended

  • along this orthogonal plane this will meet the bottom surface at point C and that point

  • C will be always situated on the line of intersection.

  • Physically it means that if you take the section draw the view of the cutting tool from here extend it. So this

  • is orthogonal plane and this is the cutting tool, this is the rake surface, this is rake

  • angle orthogonal rake and this axis is zo, this is xo and so on. Now if you extend this

  • rake surface then it will meet the bottom surface of the tool at point say C prime and

  • if it is extended it meets point C. Now what is the thickness of the tool? This is the

  • thickness of the tool say T, here this is the thickness of the tool T. Now look at this

  • tool from this side along this direction along xo and now but this view that we will get

  • of the tool will be in the cutting plane. So you draw the diagram in the cutting plane.

  • So this is yo and this is zo and what is the view?

  • Now if this is extended like this, this will also meet certain point and if it is extended

  • this will be point, this will be somewhere A prime and this will be extended here. This

  • will be A. So we get DCBA were this rake surface if extended meets the bottom surface along

  • machine longitudinal plane, machine transverse plane, orthogonal plane and cutting plane.

  • Now here you can see, that what this length OD? If we put it here OD equal to this OD

  • this intercept. This intercept is equal to this intercept how much is this? This is equal

  • T what is this angle same gamma x, so this OD. OD is equal to T multiplied tangent cotangent

  • of gamma x so T cotangent gamma x.

  • Similarly OB is how much? OB this is OB this equal to T, this is gamma y, so this is equal

  • to T cotangent of gamma y, T cot gamma y then what is OC? OC will be T what is this angle?

  • gamma O, so this will be T cotangent of gamma O and what is OA? This much this angle is

  • lambda for inclination angle of the cutting tool. So OA will be equal to T cotangent lambda.

  • So this will be very useful. Now you get the concept of master line. Now this concept will

  • be know utilized to convert the tool angles from to one system to another.

  • Now say conversion. First you start say convert rake angle from ASA to ORS. Now let us draw

  • the diagram again. This is the cutting tool top view drawn in reference plane pi and then

  • you visualize the different planes - machine longitudinal plane, machine transverse plane,

  • cutting along the cutting edge and orthogonal plane and mind that this is ninety degree

  • and this is xm, ym. This is xo, this is yo, and zo zm perpendicular to this point o.

  • Now here suppose this is the master line, suppose this is the master line for rake surface

  • and this point is D, this point is C, this point B and this point is A. What we observed

  • last time that OD is equal to T, the thickness of the tool multiplied by cotangent gamma

  • x. If T is one, say unity for T is equal to unity OD is equal to cotangent of gamma x.

  • OB is equal cotangent of gamma y. OC is equal to cotangent of gamma O and OA is equal to

  • cotangent of lambda we get it now.

  • We have to proof now we have to convert from ASA to ORS. Now what is the meaning of conversion

  • from ASA to ORS that means if the tool angle the value of the tool angles are given in

  • ASA system then what will be the value of the rake angles of the same tool that is the

  • question that means, the meaning is determine gamma O and lambda which is also a rake angle

  • from the given values of gamma x and gamma y that means, this is a function of gamma

  • O and lambda is an this is a ORS system and this is ASA system. So from ASA system to

  • ORS we have to convert now how this will be converted. Now we are discussing about conversion

  • of cutting tool angles. So first we shall show the conversion of rake angles.

  • You know the rake angles, there are side rake, back rake or orthogonal rake even lambda intersecting

  • angle is also a rake angle. So first conversion of rake angles from ASA to ORS that means,

  • let us write say tangent of gamma o is equal to tangent of gamma x sin phi plus tangent

  • of gamma y cosine phi. The other one is tangent of inclination angle lambda is equal to minus

  • tangent of gamma x cosine phi plus tangent of gamma y sin phi instead of cosine phi.

  • Say equation number one and equation number two.

  • Now actually what we are going to do now, what we are going to do that if the values

  • of gamma x, the gamma x and gamma y gamma x and gamma y are known to me then and phi

  • is also known that is the tool angles are given in ASA system we have to determine the

  • rake angles gamma o and lambda in ORS system that means to find out rake angle orthogonal

  • rake and lambda that is in ORS system as function of gamma x, gamma y and the cutting angle

  • phi. The principle cutting angle phi and we know that principle cutting angle phi is equal

  • to ninety degree minus phi s where phi s is approach angle and this approach angle is

  • given in ASA system.

  • So if phi s know from where you can determine phi and put the values of gamma x gamma y

  • and phi in to these equations number one and two we can determine the values of orthogonal

  • rake and inclination angle lambda that is from the given values of the rake angles in

  • a ASA system, we are determining the rake angles in ORS system. So now, how to proof

  • this equation? We have to proof these equations. See equation number one proof of equation

  • number one. How shall we do it? Let us take the help of this drawing. Draw this diagram

  • this is the principle cutting edge, auxiliary cutting edge. This is the tool and this is

  • machine longitudinal plane pi x, machine transverse plane pi y, cutting plane pi c and orthogonal

  • plane pi o and now you draw this master line. Suppose this is the master line for the rake

  • surface, now suppose this is point O, this is point A, this is point B, this is point

  • C and this is point D.

  • What are the intercepts I remind you that OD stands for cotangent of gamma x OB cotangent

  • of gamma y, OC cotangent of gamma O orthogonal and O is cotangent of lambda and this is phi

  • and this is phi s. Now, we have to proof with the help of this equation. The equation number

  • one now how we shall proceed? Consider triangle OBD this triangle is a right angle triangle

  • OBD. This area is equal to area of the triangle OBC plus the triangle OCD. Now all these are

  • triangles. So what is OBD area of the triangle OBD this is a right angle triangle so this

  • will be half of OB in to OD OB in to OD. What about area of the angle OBC this will be half

  • base in to altitude draw a perpendicular here see E so this will be half of OB in to CE.

  • Similarly the triangle OCD will be half of the base multiplied by the altitude this say

  • F half OC OB OD in to CF. Now what is CE? CE is equal to OC sin phi. What is half OD

  • CF is equal to OC cosine phi cosine phi OC cosine phi. Now divide both sides by half

  • OB OD OC you divide both side by half OB OD OC then what you get here. One by OC is equal

  • to one OD, OD is D OD is absent. This is sin phi plus one upon OB cosine phi or one upon

  • OC. OC is equal to cotangent of tan gamma. So this is cotangent of gamma O. So this will

  • be tangent of gamma O. Similarly one upon OD will be tangent of gamma x sin phi and

  • OB is equal to tan sin phi. So friends similarly, we can prove the equation number two.

  • Consider the triangle OAC. Lambda is equal to minus tangent of gamma x cosine phi, plus

  • tangent of gamma y sin phi. This is equation number two. This has to be proved. Now approach

  • will be same. You draw the diagram, the master line OABC and D and this is phi. Now this

  • has to be proved. Proof in the same way you choose a triangle from here, then split in

  • to two. Now in the previous one, equation we consider OBD, now we shall consider a different

  • way how. You see which angles are involved lambda that means OA has to be taken gamma

  • x which corresponds to OD. So this line has to be taken and gamma OI corresponds to OB.

  • So we have to take the angles and split in such a way that OA, OB and OD get involved.

  • Proof: Consider triangle OAD. OAD is equal to triangle OAB plus triangle OBD. Now you

  • understand how what is the trick that this triangles have to be chosen according to the

  • elements which have to be correlated. Now what is area of OAD or we can write first

  • see OBD you bring OBD first, triangle OBD or you can do it other way also not necessary.

  • Now what is the area of the triangle OAD? OAD it is the base and this is the height

  • and suppose this is k, so this is half OD multiplied by OA and what this angle phi,

  • so this is sin phi is equal to triangle OAB. So this area will be this is the base in to

  • this is the height suppose scaled and what is this angle this is phi. So this will be

  • half OB in to the height AM plus OBD that is very simple right angled triangle OB in

  • to OD.

  • Now here what is AK? This sin phi but what is say for example AM. AM is equal to OA cosine

  • phi. So this OM this is OA cosine phi. Now you divide both sides by half OA, OB and OD

  • then what happens here, we get one upon OB sin phi is equal to one upon OB OA cancels.

  • So this is OD cosine phi plus half OB OD that is one by OA. Now what is OB? OB stands for

  • cotangent gamma y OD O tangent gamma x and OA cotangent lambda. So here, we will get

  • you come to this one by OA is equal to tangent of lambda is equal to you take this one on

  • that side so that will minus one upon OD. OD is cotangent one upon OD will be tangent

  • of gamma x cosine phi plus one upon OB is equal to tangent of gamma y and sin phi. Here

  • you compare this expression with this expression, this is how it has to be proved. Now this

  • is ASA to ORS. Now you have to prove: Convert ORS to ASA.

  • What is the rake angle? We are considering rake angle, conversion of rake angle from

  • orthogonal system to ASA that means, if gamma O and lambda are given,, you determine gamma

  • x and gamma y. Suppose the tool geometry is specified in ORS system, you have to find

  • out the rake angles in ASA system that is conversion from ORS to ARS and these are function

  • of that is. Now what are the equations ultimately the equations will appear that tangent of

  • gamma x is equal to I am writing in advance, then we shall prove tangent of gamma O sin

  • phi minus tan lambda cosine phi and tangent of gamma y will be tangent of gamma O cosine

  • phi plus tan lambda sin phi. So this is equation number three and this is equation number four.

  • Now this has to be proved. How we can prove this?

  • We can go to the same method. You draw the master line for the rake surface. Suppose

  • this is the master line OA, this is B, this is C, this is D and this is phi. Now prove

  • proof of tangent of gamma x is equal to tangent of gamma O sin phi minus tan lambda cosine

  • phi. Now how to prove this? Consider you start from considering some triangles, consider

  • area of the triangle. Now it has to be done in such way that these angles under consideration

  • should be involved gamma x that means OD gamma O OC and gamma and lambda OA.

  • So we have to consider area this one and orthogonal rake this one and lambda this one. So OAD

  • area OAD which involves OA and OD that is the lambda and gamma x is equal to now you

  • involve OC triangle OAC plus triangle OCD, now you proceed you will come to this level

  • it will be proved. Now if you want to prove say similarly what you can prove that, prove tangent of

  • gamma y is equal to tangent of gamma O cosine phi plus tan lambda sin phi. Now here gamma

  • O gamma y and lambda have to be involved gamma y that is OB gamma OC and lambda OA.

  • So what we have to do? We have to start from, consider area of the triangle this one this

  • one and this one has to be involved OAC. OAC that involves OA that is lambda and OC gamma

  • O is equal to area of the triangle OAB. OAB that is OB will be involved that corresponds

  • to gamma y plus triangle OBC. Now you proceed you will be able to prove this. Now this can

  • be done, this equation number for this can be done in another way

  • What we observe that tan gamma O is equal to when we converted from ASA to ORS we got

  • tan gamma O is equal to tan gamma x sin phi plus tan gamma y cosine phi and tan lambda

  • is equal to minus tan gamma x cosine phi plus tan gamma y sin phi. Now these two equations

  • can be written say this was the equation number one and equation number two in matrix form.

  • That is tan gamma O tan lambda is equal to the matrix sin phi cosine phi minus cosine

  • phi sin phi and then gamma x gamma y tangent of gamma x and tangent of gamma y.

  • Now here this is the matrix form and this matrix is called transformation matrix, this

  • is a column matrix. From this equation, this has to be utilized to determine the value

  • of gamma O and lambda from the given value of gamma x and gamma y utilizing this transformation

  • matrix then it should be noted that the value of the determinant of this transformation

  • matrix should be always one that is unity. Now if you by inversion what do you get. If

  • you do the inversion then this will come this side, tangent of gamma x tangent of gamma

  • y will be the transformation matrix on this side we shall get tangent of gamma O and tangent

  • of lambda. So gamma x and gamma y will function of gamma and lambda and inside we shall get

  • sin phi cosine phi cosine phi sin phi and this is minus this value has to be unity.

  • Now next is conversion of clearance angles in the same system then we have to take help of this master line. Now what is the master

  • line for the flank surface or the clearance surface. You draw the tool, if you draw the

  • tool so this is the top view drawn in reference plane pi R. Now is that the complete view

  • the top view of the cutting tool. No this is not the top view, this is correctly drawn

  • because there will be two more doted lines.

  • So these doted lines will appear why because this flank surface perpendicular to this plane

  • and the auxiliary flank perpendicular to this plane are not really perpendicular. If this

  • surface the cutting plane is perpendicular to the pi R but the flank surface is inclined

  • to that to provide clearance angle. Similarly the auxiliary flank is inclined from the auxiliary

  • cutting plane to provide auxiliary plane surface. For example, say so what is this really this

  • line. This clearance surface is inclined for which it will meet the bottom surface of the

  • tool in a different line not along this line because of the angle alpha.

  • So this is nothing but the line of intersection of the principle flank with the bottom surface

  • of the tool. Now what is the master line for rake surface. The lining intersection of the

  • rake surface on and the bottom plane. Here this gives this line represents the line of

  • intersection of the principle flank and the bottom surface of the tool. So is it not master

  • line for principle flank this is the master line for principle flank and what about this

  • line this is the master line for auxiliary flank so this way you get two master lines.

  • Now if we redraw this is the cutting plane and orthogonal plane and suppose line this

  • is the master line okay say this O this D again for convenience C B and A. This is not

  • the proper we can draw a better way. Suppose this will be more convenient and then draw

  • the master line this is the master line for principle flank. Now what is the significance

  • of this. Significance of this is, if you draw this tool the sectional view of the cutting

  • tool in machine longitudinal plane then this is O, this is D, this is C, this is B and

  • this is A.

  • Now you draw this one, this D this is the clearance surface and this point D corresponds

  • to this point and this is the bottom surface and this is alpha x and this is T then what

  • is OD? OD is equal to T tangent of alpha x, if T is unity then OD for T is equal to unity

  • T is will be OD will be tangent alpha x similarly OB will be tangent of alpha y and OC along

  • the orthogonal plane tangent of alpha orthogonal and what is OA? Now OA is a typically different

  • you can see here, this is the rake these are tools draw in cutting plane and this is lambda

  • and this is T, then what is OA and this is lambda. In case of lambda, this OA will be

  • T cotangent of lambda, so this has to be noted that remains cotangent others are tangent

  • unlike rake angle it where it is cotangent. Now again we have to prove from this by the

  • same principle what we can prove, say alpha conversion of alpha from ASA to ORS.

  • By considering the same principle, the same triangle concept the master line and all this

  • things you can prove that cotangent of alpha o will be equal to cotangent of alpha x sin

  • phi plus cotangent of alpha y cosine phi. Now here you see the difference, in case of

  • rake angle these are tangent. Now these are becoming cotangent. Then lambda has to be

  • written as tan lambda because as I told you the previous frame the lambda offset is just

  • opposite.

  • So this will be cotangent of alpha x minus cosine phi plus cotangent of alpha y sin phi.

  • If you do proceed in same way alpha from say ORS to ASA then what will be the equation?

  • Cotangent of alpha x will be cotangent alpha o sin phi minus. Now when you talk about lambda

  • it should be just opposite lamb tan that has to be remembered. Tangent of lambda you can

  • prove it by the same principle as the equations have been proved from rake angles this will

  • be sin and so this will be cosine. If one is sin and the other one will be cosine and

  • cotangent of alpha y will be cotangent of alpha very easy to remember this two will

  • remain same if it is sin this will be cosine. If it is minus this will be plus others will

  • remain same it was cosine now it is sin phi so this can be proved in the same fashion.

  • Now come to rake angles normal.

  • Conversion: Convert tool angles from orthogonal system to normal rake systems. Now rake angles,

  • it means determine gamma n and also lambda from gamma o and lambda. So gamma and lambda

  • are in orthogonal plane or system and gamma n that is in normal rake system. So there

  • two basic equations as I told you importance are the rake angle and clearance angle.

  • So one equational will be tan gamma n is equal to tan gamma cos lambda that is if the values

  • of rake angle and lambda are given in ORS system you can determine gamma n.

  • Similarly from this equation if you know lambda and alpha o you can determine alpha n but

  • how to prove this this has to be proved now let us you how it can you proved?

  • Say prove tangent of gamma n is equal to tangent of gamma o cosine lambda. Now for that you have

  • to draw a diagram in a three D. Suppose this is the tool and no in a better way. Now draw

  • a velocity vector that is Zo velocity vector. This is orthogonal plane, this is Yo cutting

  • plane and this is a right angle and this is orthogonal plane pi o and this is suppose

  • this point is O and this is point A. This point A is situated on the reference plane,

  • but not on the rake surface.

  • If you extend it only then this will meet the rake surface at say point B and OB is

  • a lines of a intersection between the orthogonal plane and the rake surface. So this angle

  • is rake angle and this is nothing but orthogonal rake. Now you draw the normal plane, this

  • Zo is not perpendicular to the cutting edge. So this is the cutting edge Yn. This is Xn

  • and Xo and now you draw a line perpendicular to this one that is Zn. So this will be normal

  • plane and what is this angle lambda if you extended suppose it is meeting at point C

  • OC is situated on the rake surface and it is a line of intersection between the normal

  • plane pi n and the rake surface. This is also rake but this has been drawn in normal plane.

  • So this gamma n.

  • Now this is lambda this is also lambda here you can see that OAB, OAB this is gamma o

  • then OAC, OAC this is gamma n and ACB A this will be right angle C and B this angle is

  • lambda ah now this is lambda from here what we get that AC is equal to AB cosine lambda

  • what is AC? AC is equal to OA, OA tan gamma n and what is AB? AB OA tan gamma O then what

  • do we get out of this equation finally, tan gamma n is equal to tan gamma O cosine lambda.

  • So this is a final expression this is the proof of this one. So for this equation it

  • is evident that gamma n will be less than or equal to gamma O. Now how to prove the

  • clearance angle? That is cotangent of alpha n will be equal to cotangent of

  • alpha o cosine lambda. This can be proved by, you draw another diagram here, you extend

  • this diagram suppose this is the view, you take one orthogonal section. This is pi o

  • orthogonal section and cut it so this is point B. Suppose this is point A and this point

  • B and you cut this plane, this cutting tool what you get? This is clearance angle and

  • this is rake angle. This is point A, this point A prime and this is point B and this

  • is clearance angle in which plane orthogonal plane alpha o but this is Zo a velocity vector

  • but this is not perpendicular to the cutting edges.

  • Now you draw a plane perpendicular to the cutting edge from A. Suppose this is point

  • C and this is perpendicular to the cutting edge. Now you take the section like this.

  • Here we will see that this is C A A prime and this alpha n. Now from this triangle and

  • this triangle what we get and this lambda angle between Zzo and Zn, so this right angle

  • what is AC? AC is equal to AB cosine lambda from here. AC is equal to AB cosine lambda

  • and what is AC? AC is equal to AA prime AA prime cotangent of alpha n this is alpha n

  • and what is AB? AB is equal to AA prime cotangent of alpha o cosine lambda. So this A prime

  • cancels what remains cot alpha n is equal to cot alpha o sin lambda. So this is the

  • proof this can be proved. Now let us quickly go through some interesting features. Now

  • we got lot of equations.

  • Now when phi is equal to say ninety degree what we get when this happens, then pi x is

  • equal to pi o or gamma x is equal to gamma o when lambda is equal to zero that means

  • pi o and pi n are same that is gamma n will be equal to gamma o alpha n will be equal

  • to alpha o and when lambda is equal to zero and phi is equal to ninety than gamma x is

  • equal to gamma o is equal to gamma n and alpha x is equal to alpha o is equal to alpha n.

  • Suppose the master line is parallel to the cutting edge, then you quickly draw the planes

  • what do you understand from this? If it is parallel to the cutting plane, then where

  • is A so this is D, this is C, this is B, there is no where A so A is at the infinity that

  • is cotangent of lambda is zero so lambda is zero if this master line parallel to this

  • cutting edge which corresponds to lambda now this will be zero and this other angles are

  • gamma x gamma y and gamma o these are not zero and these are all positive on this side.

  • Now if the master line be like this parallel to this plane that means it corresponds to

  • gamma y is zero because this there is no point B where is point B. Now if it is like this

  • the master line then it is parallel to this angle so this corresponds to gamma x is zero

  • but interestingly here if you extend it the point A comes over here on the opposite side.

  • So lambda is negative and this OC if it goes in this way so C goes there in opposite direction.

  • So gamma o is also negative and D this gamma x is zero lambda. Thank You.

Friends, our subject is Manufacturing Processes II and module II is still running and subject

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