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Today, we begin our lecture with the advantages
of involute profile.
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Because of these advantages, we find involute
profiles are most commonly used in mass produced
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gears.
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So
advantages of involute gear tooth profile:
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In our last lecture, we have already seen
that a pair of involute profiles can maintain
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the conjugate action.
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That is, it satisfies the fundamental law
of gearing, that is, it maintains constant
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angular velocity ratio between the two shafts.
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So first and foremost is, maintains conjugate
action.
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However, as I said earlier, this is not so
much of importance because, if one is given
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any smooth curve as one profile, one can find
another profile which will be conjugate to
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the given profile.
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So, this is not the unique choice of involute
profiles which maintains conjugate action.
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There are other advantages of involute profile,
over and above that it maintains the most
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fundamental requirement that is, maintaining
conjugate action.
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The second advantage is that, operating pressure
angle is constant.
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If we remember in our last lecture, we denoted
this operating pressure angle by the symbol
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phi which is the angle between the line of
action and the common tangent to the pitch
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circles, maintains constant pressure angle,
operating pressure angle is constant.
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We will get back to the same figure, which
we discussed last time to show, what is the
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advantage if the pressure angle remains constant?
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what is the consequence of this pressure angle
remaining constant?
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And the third advantage is that, a pair of
involute gears maintains conjugate action,
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even if the centre distance between the gears
are changed.
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That means, we have a pair of gears, we can
mount them with a little different centre
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distance, even then this conjugate action
will be maintained.
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Even if, centre distance between the two gears
that is between a pair of, centre distance
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between the pair of gears changes.
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These two points as I said, I will explain
with reference to the figure that we discussed
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last time.
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And the last but not the least, the most important
reason why involute profiles are so common
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is that, ease of production.
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Why it is easy to produce involute gears?
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Because, gears are produced, gear teeth are
produced by a process called generation.
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That is, again we use the conjugate action
but not between a pair of gears but between
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a rack and a pinion and it is the conjugate
rack of an involute profile is a straight
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tooth rack.
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If we have an involute profile, we see this
later that the conjugate rack tooth, if this
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is a rack whose tooth profile is straight,
then it can maintain conjugate action with
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a involute gear and it is such a straight
sided rack cutter is used to produce the involute
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tooth profile on a circular gear blank.
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This is, I can say it is an involute rack
which is used as the cutter and on the cutter
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it is much easier to produce straight side
than any complicated curve and it is this
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involute rack cutter which can maintain conjugate
action with an involute tooth profile of a
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gear.
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So this is called involute rack cutter.
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Why the involute of a rack is a straight line?
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This point again will be clear in today’s
lecture a little later, that the involute
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of a straight line is another straight line.
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Like we have seen involute of a circle is
the curve.
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So, if this radius of the base circle goes
to infinity, when the gear gets converted
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into a rack, then this also becomes a straight
line.
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All these points, we will make clear in today's
lecture.
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So let me get back to that figure, where I
showed the pressure angle and now we will
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show, why what is the advantage of this pressure
angle remaining constant, in a pair of involute
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gear teeth profile?
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Let us get back to this figure, which we discussed
in our last class.
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These are the two base circles of radius rb1
and rb2.
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These are the pitch circles, of radius rp1
and rp2 and it is the common tangent to the
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base circle, which we call line of action.
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And this line of action AB makes an angle
phi with this common tangent to the pitch
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circles.
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This tt is the common tangent to this pitch
circles of radius rp1 and rp2 and this angle
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phi, what we call as operating pressure angle.
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Now as we have already noted, because the
line of action just remains AB, this pressure
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angle does not change as if the profiles are
involute.
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The pressure angles remain constant during
the entire interval.
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Now what is the advantage that the pressure
angle remains constant?
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If we neglect the friction force between the
gear teeth, which if it is well oiled and
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well lubricated, friction force is not very
large.
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The driving effort is along the common normal.
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If this gear is driving this gear, then most
of the force is acting through this, along
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this common normal.
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So if this normal force, we call say N, then
the torque that is transmitted is N times
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the base circle radius, because this line
of action N is along AB and this angle is
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90 degree.
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So, the torque acting on this gear 2 is N
into rb2.
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And if this torque is this is torque, that
is transmitted to gear 2.
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Now if this torque remains constant, then
N remains constant because rb2 is constant,
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the base circle radius.
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If N remains constant, then we see both the
magnitude and direction of this force N is
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remaining constant, which means the bearing
reactions here and here, because there is
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same N which is acting on this gear also.
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So the direction and magnitude of N both remains
constant, because phi does not change and
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if torque is steady torque, if we are transmitting
a constant torque, then this torque is also
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constant which means magnitude of N is constant,
because phi is constant means the direction
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of N is constant, which means the bearing
reactions, while the gears are transmitting
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a steady torque remains the same, which means
the bearings are not subjected to any dynamic
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reaction.
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The magnitude and direction both of the reactions
remain same.
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So long a steady torque is being transmitted.
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This enhances the life of the bearings, which
are used to mount these two shafts on the
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foundation.
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Thus, the pressure angle remaining constant
implies that under steady torque, the bearing
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reactions are not dynamic, they are static.
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That is the advantage of pressure angle remaining
constant.
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The third advantage, we discussed was that,
suppose this pair of gears, right now the
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centre distance is 1O2.
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Suppose, this gear remains where it is, only
this gear is shifted a little bit upward,
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then what we will see that it is the same
involute profiles will be useful to maintain
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the conjugate action.
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Let us see what is the centre distance?
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Centre distance O1O2 is rp1, the pitch circle
radius of the first gear plus rp2, pitch circle
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radius of the second gear.
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And we have already noted that base circle
radius rb is nothing but rp cosine of phi,
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where phi is the operating pressure angle.
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So, we know rp1 by rp2 is same as rb1 by rb2
because rb1 is rp1 cos phi, rb2 rb2 is rp2
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cos phi, cos phi cancels.
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So this ratio remains same.
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So what we see, so long the gears are same
that is the base circle radius remains same,
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rp1 by rp2 remains same.
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With the centre distance changing, it is rp1
plus rp2, this quantity is varying.
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Which means with varying center distance,
if the centre distance varies, the pitch circle
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radius rp1 and rp2 both change, rp1 and rp2
both change, but the ratio of rp1 and rp2
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that remains same.
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So it is the same angular velocity ratio omega1
to omega2 is maintained.
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Only thing what changes because rp1 and rp2
is changing means, phi is changing because
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rb1 is not changing and rb1 is rp1 cosine
phi, where phi is the pressure angle.
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rb1 is not changing because the gears are
not changing but with changing centre distance
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rp1 is changing, which means phi is changing
because this common tangent, if this circle
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is shifted a little bit upward, the common
tangent between these base circles will change,
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that means this angle phi will change.
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So pressure angle changes a little bit, but
the same angular velocity, constant angular
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velocity ratio is maintained by the pair of
gears, even when the centre distance changes
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a little bit, that was the third advantage.
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And the fourth advantage as I said, will be
explained later, when we will be able to show,
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that the involute of a straight line is a
straight line that is for a rack to maintain
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conjugate action with an involute profile,
the tooth profile on the rack will be straight,
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like trapezium, as we have shown earlier on
the board.
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Because involute profiles have all these advantages
as I said earlier, they are almost universally
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used in mass produced gears and as a consequence
of this mass production of involute gears,
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they have been standardized.
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There are various standards, but I can mention
some typical standards which are more commonly
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used, like British Standard for involute profile.
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So involute tooth profile, some standard dimensions
are: The most common value of the pressure
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angle, operating pressure angle
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phi is 20 degrees.
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Some old gears which were cast, it also had
a value phi equal to 14 and half degrees,
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but these days most gears, involute gears
will have an operating pressure angle equal
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to 20 degree.
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Similarly, the value of the addendum which
we defined earlier ‘a’ is equal to the
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module, where m is module.
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The gear teeth is described in terms of the
module of the gear teeth and the standard
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value of the addendum is equal to module.
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Similarly, dedendum which we denoted by b
and which is always more than a, the most
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common value is 1.25 times the module, m is
the module which is expressed in millimeter
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for metric gears and a is equal to the module
m and b is equal to 1.25 m.
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These three standard values we may take if
unless otherwise specified.
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As I said, involutes are most commonly used
and geometry of involute teeth is a very vast
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subject.
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We are not going to get into all the details
of involuted geometry, but I will give you
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a little glimpse or a little basic idea of
this curve involute, which we called involutometry.
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And the result that we will obtained from
this involutometry we will see, will be very
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useful to determine various proportion of
the gear teeth, which are involutes.
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So we can say application of involutometry
in gear tooth geometry.
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I repeat we will discuss of course only the
very basics of this gear tooth geometry.
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You can always refer to hand book for the
details that we may like to know.
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So let me first say, what is involutometry?
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Suppose, this is the base circle with the
centre here, and we start unwinding the string
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from this base circle from this point A and
by unwinding the string, I generate this involute.
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So at this, this is the string length which
is same as this arc length because this string
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was wound on to this cylinder and now it has
unwound up to this point and this is the involute
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from this base circle.
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So the string is tangent to the base circle,
so this radius rb, base circle radius.
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This is perpendicular to the string.
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Now at this point, the involute takes off
radially.
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So this radius is tangent to the involute
at this instant and at this configuration,
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this is the string which is perpendicular
to the involute so the tangent is perpendicular
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to the string, this is tangent, this is the
tangent at this point say B.
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This radial line is tangent to the involute
at A and this line which is perpendicular
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to the string at the point B is the tangent
at the point B. And this angle between these
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two tangents, I call the roll angle theta.
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Theta is called roll angle.
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This line doesn’t look like perpendicular,
so let me draw it correctly.
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This is O.
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So this line is perpendicular to the string,
this radius is also perpendicular to this
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string, which is tangential.
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So this is parallel to this.
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So if this angle is theta, this is also theta,
the roll angle.
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That is the string has unwound up to this
point B, where the roll angle is theta.
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This particular point on this involute, this
is the involute, this distance OB is the radius
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vector, let me call it r.
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If this circle, this represents the base circle
of the gear and the gear is rotating about
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the point O, then this particular point has
a velocity which is perpendicular to OB.
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This is the direction of the velocity and
this is the normal to the involute.
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So that is the line of action.
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So at this point, this angle between the normal
and the direction of the velocity.
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Normal really indicates the direction of the
force at this particular point.
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This angle is called psi, which we will call
involute pressure angle.
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This is the angle between the line which is
perpendicular to OB and the string at this
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00:21:29,210 --> 00:21:35,120
configuration, when I have this up to the
point B. Please note that this is not the
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00:21:35,120 --> 00:21:40,440
operating pressure angle which we discussed
in case of a pair of gears.
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This is only one gear we are talking of, one
involute we are talking of and this particular
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angle between the direction of the velocity
and the direction of the normal to the involute,
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I call it involute pressure angle.
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Now this line is perpendicular to the tangent
and this line is perpendicular to the direction
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of velocity so this angle is also psi, involute
pressure angle.
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Theta is the roll angle.
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Psi is the involute pressure angle.
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Theta is the angle between these two tangents
at A and B, which is same as these two radius
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at A and corresponding to B and psi is the
involute pressure angle.
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So let us say base circle radius is rb.
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And this is, the string length is the radius
of curvature of the involute at this point
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B.
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If this is the involute, at every instant
the radius of curvature is nothing but the
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string length, which has been unwound from
the base circle.
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So, rho is if I call this point C, then BC
is the radius of curvature rho.
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So what we see, rho is given as, square root
of r squared minus rb squared, where r is
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this radius vector of the point B measured
from the centre of the base circle and if
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I use this angle is the polar angle to define
the involute curve, what is this angle?
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That is theta minus psi.
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This angle is theta minus psi and we also
see, this rho is same as this arc length AC,
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because this is the length of the string and
this was the original length of the string.
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So, AC is nothing but rho, which is same as
rb into theta and also we can see tan psi
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is rho by rb.
216
00:24:08,490 --> 00:24:15,250
If this angle is 90 degree, this is rho, this
is rb and this is psi.
217
00:24:15,250 --> 00:24:17,880
So tan psi is rho by rb.
218
00:24:17,880 --> 00:24:28,299
We can write, tan psi is rho by rb.
219
00:24:28,299 --> 00:24:32,260
And from here, we see rho by rb is nothing
but theta.
220
00:24:32,260 --> 00:24:42,110
So I can write theta minus psi as: tan psi
minus psi.
221
00:24:42,110 --> 00:24:46,130
Theta is rho by rb and rho by rb is tan psi.
222
00:24:46,130 --> 00:24:54,520
So, this angle that the line OB makes with
this original radius OA, that angle is tan
223
00:24:54,520 --> 00:24:56,679
psi minus psi.
224
00:24:56,679 --> 00:25:05,090
And this is given a name, which is called
involute function of psi and we can also see
225
00:25:05,090 --> 00:25:08,610
rb is nothing but r of cosine psi.
226
00:25:08,610 --> 00:25:17,760
So, we can write base circle radius is related
to r as r cosine psi.
227
00:25:17,760 --> 00:25:22,220
So everything has been found in terms of this
involute pressure angle psi, which keeps on
228
00:25:22,220 --> 00:25:25,420
changing at various points.
229
00:25:25,420 --> 00:25:32,990
The distance from O, I can get as rb by cos
psi and this angle, I can get as tan psi minus
230
00:25:32,990 --> 00:25:35,960
sin psi which is called involute psi.
231
00:25:35,960 --> 00:25:41,809
This is very easy to see that if one gives
me the value of psi, I can calculate involute
232
00:25:41,809 --> 00:25:43,169
psi, very simply.
233
00:25:43,169 --> 00:25:50,970
This is called involute function just like
sin, cos, we call it involute of psi.
234
00:25:50,970 --> 00:25:55,799
Given the value of psi, it is easy to calculate
involute of psi, but not the other way around.
235
00:25:55,799 --> 00:26:00,860
So for this tables are available just like,
tables of sine, cosine, tan and log.
236
00:26:00,860 --> 00:26:07,420
I can get the value of psi if you give me
the value of involute of psi by consulting
237
00:26:07,420 --> 00:26:09,419
the table.
238
00:26:09,419 --> 00:26:14,020
Now at this stage, I will be able to show
that what happens to this radius of curvature
239
00:26:14,020 --> 00:26:17,960
of the involute profile as rb increases.
240
00:26:17,960 --> 00:26:23,010
As I said, when rb goes to infinity, the gear
gets converted into a rack.
241
00:26:23,010 --> 00:26:28,770
So under that situation, what is the radius
of curvature of the corresponding involute
242
00:26:28,770 --> 00:26:30,110
profile?
243
00:26:30,110 --> 00:26:36,770
So let me now say, what is the radius of curvature
of the involute tooth profile of a rack?
244
00:26:36,770 --> 00:26:42,919
Rack means, where rb goes to infinity.
245
00:26:42,919 --> 00:26:52,360
We have already got rho equal to square root
of r squared minus rb squared, where r is
246
00:26:52,360 --> 00:26:56,030
the instantaneous value of this distance.
247
00:26:56,030 --> 00:27:03,100
So if we start from A and go up to some point
by changing initial value of r is r, rb and
248
00:27:03,100 --> 00:27:05,580
then I go to rb plus delta.
249
00:27:05,580 --> 00:27:15,409
Lets say, as I start from A, I go a little
away from this, when r changes from rb to
250
00:27:15,409 --> 00:27:20,150
rb plus say some delta.
251
00:27:20,150 --> 00:27:23,340
If I substitute it there, what we get?
252
00:27:23,340 --> 00:27:42,059
We get twice rb into delta r plus delta r
squared to the power half.
253
00:27:42,059 --> 00:27:50,370
This is the value of rho, as we go out from
this point A. Now if rb goes to infinity,
254
00:27:50,370 --> 00:27:52,950
it is obvious that rho also goes to infinity.
255
00:27:52,950 --> 00:27:58,170
And if the radius of curvature goes to infinity,
which means this profile has become a straight
256
00:27:58,170 --> 00:27:59,170
line.
257
00:27:59,170 --> 00:28:04,830
That is what we said, that if we have an involute
rack, let the tooth profile from the base
258
00:28:04,830 --> 00:28:10,290
circle as it come out, it comes out in the
form of a straight line.
259
00:28:10,290 --> 00:28:16,090
So, the tooth profile of an involute rack
is a straight line and which can maintain
260
00:28:16,090 --> 00:28:22,289
conjugate action with an involute gear or
involute pinion and that gives the advantage
261
00:28:22,289 --> 00:28:30,590
that I can use such a rack cutter to generate
the involute tooth profile on a circular gear
262
00:28:30,590 --> 00:28:31,770
blank.
263
00:28:31,770 --> 00:28:35,900
Let me now repeat, what we have just now discussed
with reference to this figure.
264
00:28:35,900 --> 00:28:40,000
This is the base circle, this is the centre
of the base circle.
265
00:28:40,000 --> 00:28:47,539
An involute is being generated starting from
this point A. When the involute if I consider
266
00:28:47,539 --> 00:28:54,310
a point ‘I’ here, then the string at this
configuration corresponds to this line IB,
267
00:28:54,310 --> 00:28:56,700
which is tangent to the base circle.
268
00:28:56,700 --> 00:29:02,990
OB is the radius of the base circle, which
we denoted by rb.
269
00:29:02,990 --> 00:29:11,159
The instantaneous polar radius of this point
‘I’, I denote by r.
270
00:29:11,159 --> 00:29:21,490
The tangent at A to the involute is this radial
line and the tangent at ‘I’ is this line.
271
00:29:21,490 --> 00:29:27,029
The angle between this tangent at ‘I’
and the tangent at A, this angle theta, we
272
00:29:27,029 --> 00:29:38,179
call the roll angle.
273
00:29:38,179 --> 00:29:41,110
Theta was defined as the roll angle.
274
00:29:41,110 --> 00:29:46,750
If this I take as a gear and the gear is rotating
about the point O, then the velocity of this
275
00:29:46,750 --> 00:29:49,280
point ‘I’ is perpendicular to OI.
276
00:29:49,280 --> 00:29:56,070
Say this is the direction of the velocity
of the point ‘I’, if it happens to be
277
00:29:56,070 --> 00:30:02,890
a point on the gear, And the string which
is normal to this gear tooth profile or the
278
00:30:02,890 --> 00:30:11,090
involute profile, that is the normal to the
involute at the point ‘I’ and the angle
279
00:30:11,090 --> 00:30:18,940
between this velocity direction and this normal
which is nothing but BI, I call psi.
280
00:30:18,940 --> 00:30:24,360
This is normal and this is a tangent, so this
angle is 90 degree.
281
00:30:24,360 --> 00:30:29,240
This is also perpendicular to this string
BI which is tangent to the base circle.
282
00:30:29,240 --> 00:30:35,190
So this line, the tangent at ‘I’ and this
radius OB are parallel.
283
00:30:35,190 --> 00:30:40,320
So this angle is psi, then this angle is also
psi.
284
00:30:40,320 --> 00:30:47,470
Similarly, if this angle is theta, then this
angle is also theta because this line is parallel
285
00:30:47,470 --> 00:30:48,760
to this line.
286
00:30:48,760 --> 00:30:53,700
So theta is the roll angle and psi is this
angle.
287
00:30:53,700 --> 00:30:59,840
Now the polar angle of this line OI from this
vertical line, that is the radius through
288
00:30:59,840 --> 00:31:12,510
A initial point, this angle which is theta
minus psi, we defined as involute of psi,
289
00:31:12,510 --> 00:31:20,580
involute of psi is this angle theta minus
psi.
290
00:31:20,580 --> 00:31:26,950
And what we have seen, that this length AB
is nothing but rb into theta and the string
291
00:31:26,950 --> 00:31:30,120
length BI is also equal to AB.
292
00:31:30,120 --> 00:31:36,530
So this rho, which is the radius of curvature
of the involute profile at I, the string length
293
00:31:36,530 --> 00:31:38,380
is the radius of curvature rho.
294
00:31:38,380 --> 00:31:45,960
This rho is nothing but rb into theta.
295
00:31:45,960 --> 00:31:52,920
From this triangle, OBI, I can write tan of
psi because this angle is 90 degree.
296
00:31:52,920 --> 00:31:56,649
Tan of psi is also rho by rb.
297
00:31:56,649 --> 00:32:13,830
So we can write, rho is rb tan psi, which
means theta is equal to tan psi.
298
00:32:13,830 --> 00:32:26,539
So this involute psi, which is theta minus
psi, I can write it as tan psi minus psi.
299
00:32:26,539 --> 00:32:32,669
Then we also found that expression of rho
in terms of r and rb, which we wrote as rho
300
00:32:32,669 --> 00:32:42,600
equal to square root of r squared minus rb
squared.
301
00:32:42,600 --> 00:32:49,090
As we draw this involute from starting from
this this base circle, the r changes from
302
00:32:49,090 --> 00:32:58,880
rb, it increases from rb say by a value delta
r.
303
00:32:58,880 --> 00:33:01,690
Then substitute it in there, canceling rb.
304
00:33:01,690 --> 00:33:14,549
What we found the expression of rho turned
out to be twice rb into delta r plus delta
305
00:33:14,549 --> 00:33:17,450
r whole squared.
306
00:33:17,450 --> 00:33:23,250
So this clearly shows, if rb tends to infinity,
then rho also tends to infinity.
307
00:33:23,250 --> 00:33:28,630
That means, right from the point A, as soon
as rb r changes from rb to plus small delta
308
00:33:28,630 --> 00:33:34,830
r, the radius of curvature becomes infinity,
that is the involute of a straight line.
309
00:33:34,830 --> 00:33:38,850
When rb goes to infinity, this base circle
becomes a straight line.
310
00:33:38,850 --> 00:33:45,010
That is the gear is converted to a rack and
the involute profile becomes a straight line
311
00:33:45,010 --> 00:33:51,940
because radius of curvature is infinity, which
tells us a rack, involute rack tooth profile
312
00:33:51,940 --> 00:33:56,169
looks like a straight line.
313
00:33:56,169 --> 00:34:02,500
Just now, we have discussed the basic geometrical
relationship between an involute profile and
314
00:34:02,500 --> 00:34:03,520
its base circle.
315
00:34:03,520 --> 00:34:09,429
These relations are very useful for studying
gear tooth geometry as I said earlier and
316
00:34:09,429 --> 00:34:12,450
here we discuss an example.
317
00:34:12,450 --> 00:34:19,440
Suppose, this is an involute tooth profile,
which has been generated from this base circle.
318
00:34:19,440 --> 00:34:24,659
Suppose the thickness of the tooth as we see
keeps on varying, from the base circle up
319
00:34:24,659 --> 00:34:26,440
to the addendum circle.
320
00:34:26,440 --> 00:34:34,819
Suppose the thickness of the tooth at any
point X, which is defined by the polar radius
321
00:34:34,819 --> 00:34:44,500
rX, this point X is defined by this distance
OX and the thickness at this level, I denote
322
00:34:44,500 --> 00:34:46,619
by tX.
323
00:34:46,619 --> 00:34:56,060
When this polar distance changes from rX to
say rY, I get to this point Y and I want to
324
00:34:56,060 --> 00:35:00,890
find, what is the thickness of the tooth at
this level?
325
00:35:00,890 --> 00:35:01,890
That is tY.
326
00:35:01,890 --> 00:35:03,630
So tX is given.
327
00:35:03,630 --> 00:35:09,519
I would like to find tY and rX and rY are
given.
328
00:35:09,519 --> 00:35:18,940
To do this first we note, when the involute
was at X, the string is DX.
329
00:35:18,940 --> 00:35:26,829
This is the length AD, which has become DX
and when we come to the point Y, the string
330
00:35:26,829 --> 00:35:28,609
is represented by BY.
331
00:35:28,609 --> 00:35:37,940
This is the tangent, BY is same as the arc
length AB.
332
00:35:37,940 --> 00:35:42,499
So the angle between this OX and OD, we defined
as the psi.
333
00:35:42,499 --> 00:35:51,150
So I write it as psiX corresponding to X,
D is the, XD is tangent to the base circle
334
00:35:51,150 --> 00:35:59,839
and the angle between OD and OX I call the
involute pressure angle at X, which is psiX.
335
00:35:59,839 --> 00:36:05,309
Similarly, the involute pressure angle at
Y is the angle between OY and the tangent
336
00:36:05,309 --> 00:36:10,890
to the base circle from Y which is YB, that
is what I call psiY.
337
00:36:10,890 --> 00:36:16,599
We have already seen, the base circle radius
rb.
338
00:36:16,599 --> 00:36:29,770
This is also base circle radius rb and this
can be written as: rb is nothing but rX cosine
339
00:36:29,770 --> 00:36:42,259
psiX, which is also equal to rY cosine psiY.
340
00:36:42,259 --> 00:36:47,589
So if you have given the values of rX and
rY, I can get psiX and psiY.
341
00:36:47,589 --> 00:36:50,709
Now how tX and tY are related?
342
00:36:50,709 --> 00:36:58,260
For that, let me consider this angle AOC.
343
00:36:58,260 --> 00:37:09,329
OC is the mid line or the symmetric line of
this gear tooth, OC and the angle that OC
344
00:37:09,329 --> 00:37:14,579
makes with OA, I call this angle.
345
00:37:14,579 --> 00:37:34,219
So angle AOC, I can write as tX by two divided
by rX, that I’m getting this angle.
346
00:37:34,219 --> 00:37:45,539
This angle is tX by two divided by rX and
this angle is nothing but, involute of psiX.
347
00:37:45,539 --> 00:37:52,930
At the point X, the angle that OX makes with
this starting line OA, I defined as involute
348
00:37:52,930 --> 00:37:55,619
function of psiX.
349
00:37:55,619 --> 00:38:14,880
So angle AOC is nothing but tX by two divided
by OX, which is rX plus involute function
350
00:38:14,880 --> 00:38:24,249
of psiX and if we remember, involute function
is tan psiX minus psiX, involute of psiX is
351
00:38:24,249 --> 00:38:28,240
tan psiX minus psiX.
352
00:38:28,240 --> 00:38:40,989
And AOC the same angle, I can also write as:
this angle which is tY by two divided by rY.
353
00:38:40,989 --> 00:38:55,219
So tY divided by two rY plus this angle.
354
00:38:55,219 --> 00:39:02,059
This angle that is the angle between OA and
OY which is nothing but involute of psiY.
355
00:39:02,059 --> 00:39:06,559
This angle is involute of psiY.
356
00:39:06,559 --> 00:39:15,039
So this plus involute of psiY.
357
00:39:15,039 --> 00:39:19,400
If rX and rY are given, I can find psiX and
psiY from this relation.
358
00:39:19,400 --> 00:39:25,400
Once, psiX and psiY are known, I can find
involute of psiX and involute of psiY, because
359
00:39:25,400 --> 00:39:28,670
involute function is tan minus the angle.
360
00:39:28,670 --> 00:39:40,700
Involute function of any angle psiX is tan
of psiX minus psiX.
361
00:39:40,700 --> 00:39:44,109
Of course, psiX must be measured in radian.
362
00:39:44,109 --> 00:39:50,869
So from this relationship, if tX is given,
rX is given, rY is given, I can find tY which
363
00:39:50,869 --> 00:39:56,279
is the only unknown because psiX and psiY,
I have already found out from here and in
364
00:39:56,279 --> 00:40:00,380
using the involute function, I can find involute
of psiX and involute of psiY.
365
00:40:00,380 --> 00:40:06,019
So thus for the gear tooth geometry, as we
see, if you give me the thickness at any level,
366
00:40:06,019 --> 00:40:11,539
I can find the thickness at any other level
of the same involute tooth profile.
367
00:40:11,539 --> 00:40:16,779
For continuous transmission of rotation from
one gear to another, it is obvious that is
368
00:40:16,779 --> 00:40:25,109
imperative that a pair of teeth must remain
in engagement, before at least until the next
369
00:40:25,109 --> 00:40:26,719
pair of teeth comes into engagement.
370
00:40:26,719 --> 00:40:31,690
It should not happen, that one pair of teeth
has lost its engagement and the next pair
371
00:40:31,690 --> 00:40:35,630
of teeth has not come in engagement, because
then there will be no transmission.
372
00:40:35,630 --> 00:40:42,710
So to maintain continuous transmission of
rotation from one gear to another, it is imperative
373
00:40:42,710 --> 00:40:50,170
that a pair of teeth must continue to be engaged,
must continue to remain engaged, at least
374
00:40:50,170 --> 00:40:54,430
until the next pair of teeth has come into
engagement.
375
00:40:54,430 --> 00:41:00,289
This phenomena is studied in terms of a geometrical
quantity which we call contact ratio.
376
00:41:00,289 --> 00:41:07,109
Let me now define the contact ratio and try
to get the expression of this contact ratio.
377
00:41:07,109 --> 00:41:19,279
So we study contact ratio for involute tooth
profile, we use the symbol Mc, indicate the
378
00:41:19,279 --> 00:41:28,630
contact ratio.
379
00:41:28,630 --> 00:42:06,220
Lets say, this is one base circle and the
centre of this gear is at O1, and the centre
380
00:42:06,220 --> 00:42:38,219
of the other gear is at O2.
381
00:42:38,219 --> 00:42:57,799
This is the other base circle, base circle
2.
382
00:42:57,799 --> 00:43:08,869
If we remember, that it is the common tangent
to this pair of base circles, if I draw the
383
00:43:08,869 --> 00:43:21,589
common tangent to this pair of base circles,
that defines the line of action.
384
00:43:21,589 --> 00:43:41,670
This is A and this other point of tangent
is B. So we know the contact between a pair
385
00:43:41,670 --> 00:43:46,779
of gear teeth will always lie on this line
AB.
386
00:43:46,779 --> 00:43:49,999
Now let us talk of one tooth of this gear.
387
00:43:49,999 --> 00:43:56,589
Suppose this gear is rotating this way and
to the teeth action, where teeth engagement
388
00:43:56,589 --> 00:43:59,979
it is rotating this gear in this direction.
389
00:43:59,979 --> 00:44:06,519
So if we consider a tooth on this gear, the
first teeth comes into engagement at its outer
390
00:44:06,519 --> 00:44:11,700
circle or the addendum circle and because
the contact point must lie on this line, when
391
00:44:11,700 --> 00:44:27,119
the addendum circle, when the addendum circle
comes here in contact with this line AB, that
392
00:44:27,119 --> 00:44:30,599
is where the contact starts on this face of
the gear tooth.
393
00:44:30,599 --> 00:44:33,059
This is one gear tooth.
394
00:44:33,059 --> 00:44:36,760
Now, how long now this gear starts rotating
this way?
395
00:44:36,760 --> 00:44:41,930
How long this tooth will remain in contact
with the tooth on the other gear?
396
00:44:41,930 --> 00:44:43,279
Gear number 1.
397
00:44:43,279 --> 00:44:52,349
So long, the outer circle of this gear is
in contact.
398
00:44:52,349 --> 00:44:57,359
So this is the addendum circle of the gear
1.
399
00:44:57,359 --> 00:45:12,779
Let me call it radius ro1, that is the outer
radius of gear 1 and this is rb1.
400
00:45:12,779 --> 00:45:20,339
So the contact will, as this tooth rotates,
this tooth comes here and when the tooth on
401
00:45:20,339 --> 00:45:21,951
this gear which is pushing it, lets say here,
if I draw this here.
402
00:45:21,951 --> 00:45:22,951
This tooth comes and rotates, and when this
outer circle leaves this line, the contact
403
00:45:22,951 --> 00:45:23,951
is lost with this particular gear.
404
00:45:23,951 --> 00:45:24,951
So if I draw this tooth here, this is where
the contact is lost.
405
00:45:24,951 --> 00:45:25,951
So, this particular pair of teeth is in contact,
as this contact point moves from this point
406
00:45:25,951 --> 00:45:26,951
say E to this point say F. The contact between
this pair of teeth, this one and its mating
407
00:45:26,951 --> 00:45:27,951
tooth, starts the contact when the addendum
circle of this lower gear, gear 2, this is
408
00:45:27,951 --> 00:45:28,951
the addendum circle of gear 2 and this is
the addendum circle of gear 1.
409
00:45:28,951 --> 00:45:29,951
When the addendum circle intersects line AB
at the point E, the addendum circle of gear
410
00:45:29,951 --> 00:45:30,951
1 intersects the line AB at F. So, this pair
of teeth is remains in engagement from E to
411
00:45:30,951 --> 00:45:31,951
F. What we should ensure that before this
point is reach, the next tooth in this gear
412
00:45:31,951 --> 00:45:32,951
must come in engagement, must come in engagement
that is the outer circle must intersect here
413
00:45:32,951 --> 00:45:33,951
for the next tooth.
414
00:45:33,951 --> 00:45:34,951
To do this, let me define what we call base
pitch.
415
00:45:34,951 --> 00:45:35,951
We consider, this is the same tooth we are
considering, but now let me consider two consecutive
416
00:45:35,951 --> 00:45:36,951
tooth on one gear, two consecutive teeth on
one gear.
417
00:45:36,951 --> 00:45:37,951
If we remember that we define the circular
pitch on this pitch circle going from one
418
00:45:37,951 --> 00:45:38,951
point to another identical point on the adjacent
tooth, measured along the pitch circle this
419
00:45:38,951 --> 00:45:39,951
is what we defined as Pp which is 2 pi rp,
where rp is the pitch circle radius divided
420
00:45:39,951 --> 00:45:40,951
by the number of tooth.
421
00:45:40,951 --> 00:45:41,951
Similarly, you can define base pitch, that
is this point and the identical point on the
422
00:45:41,951 --> 00:45:42,951
next tooth, but on the base circle.
423
00:45:42,951 --> 00:45:43,951
This distance is called the base pitch, we
call it Pb, base pitch; that is 2 pi base
424
00:45:43,951 --> 00:45:44,951
circle radius divided by N. And if we remember
rb is nothing but rp cosine phi, where phi
425
00:45:44,951 --> 00:45:45,951
is the pressure operating pressure angle.
426
00:45:45,951 --> 00:45:46,951
Now, what is the movement along the base circle
of this particular teeth from the start to
427
00:45:46,951 --> 00:45:47,951
the end of the engagement, that is given by
this circular arc.
428
00:45:47,951 --> 00:45:48,951
Measure on the base circle, this is the rotation
that is the distance covered by the, this
429
00:45:48,951 --> 00:45:49,951
particular tooth from the start of its engagement
to the end of its engagement, the rotation
430
00:45:49,951 --> 00:45:50,951
is such that it covers so much distance on
the base circle.
431
00:45:50,951 --> 00:45:51,951
To maintain contact between two pair of teeth
that is, before this contact is lost the next
432
00:45:51,951 --> 00:45:52,951
pair must come in engagement, this distance
must be more than this base pitch.
433
00:45:52,951 --> 00:45:53,951
This distance if I defined is at Pb, then
this distance must be more than Pb such that
434
00:45:53,951 --> 00:45:54,951
the next tooth has come up to this position.
435
00:45:54,951 --> 00:45:55,951
Because the distance along the base circle
between this tooth and the adjacent tooth
436
00:45:55,951 --> 00:45:56,951
is the base pitch.
437
00:45:56,951 --> 00:45:57,951
When this point comes here, that point must
come above this point; that means, this circular
438
00:45:57,951 --> 00:45:58,951
distance must be more than the base pitch.
439
00:45:58,951 --> 00:45:59,951
Let me call this point S and T.
440
00:45:59,951 --> 00:46:00,951
So, to maintain contact all the time between
these two pair of gears, between this pair
441
00:46:00,951 --> 00:46:01,951
of gears, ST must be more than base pitch.
442
00:46:01,951 --> 00:46:02,951
And it is this ratio ST by base pitch, which
I have defined by Pb which I have defined
443
00:46:02,951 --> 00:46:03,951
just now must be more than 1.
444
00:46:03,951 --> 00:46:04,951
And we defined mc as ST by Pb.
445
00:46:04,951 --> 00:46:05,951
What is ST?
446
00:46:05,951 --> 00:46:06,951
That is the movement of this tooth along the
measured along the base circle from the start
447
00:46:06,951 --> 00:46:07,951
to the end of the engagement.
448
00:46:07,951 --> 00:46:08,951
To calculate mc we proceed as follows.
449
00:46:08,951 --> 00:46:09,951
This is what we call the pitch point P, where
the common tangent the line of action intersects
450
00:46:09,951 --> 00:46:10,951
this line of centers O1O2.
451
00:46:10,951 --> 00:46:11,951
Now, because this is the involute, I can say
and this is the tangent to the base circle
452
00:46:11,951 --> 00:46:12,951
BT is nothing but BE.
453
00:46:12,951 --> 00:46:13,951
BT, that is the distance along the base circle
is same along the length of the string; BT
454
00:46:13,951 --> 00:46:14,951
is same as BE.
455
00:46:14,951 --> 00:46:15,951
Same way I can say BS is same as BF, this
is the distance along the base circle and
456
00:46:15,951 --> 00:46:16,951
this is the length of the string.
457
00:46:16,951 --> 00:46:17,951
So, I can say BS is same as BF.
458
00:46:17,951 --> 00:46:18,951
If I subtract I get ST is same as EF.
459
00:46:18,951 --> 00:46:19,951
And the contact ratio mc which was defined
as ST by Pb, I can write this as EF by Pb.
460
00:46:19,951 --> 00:46:20,951
The expression of Pb we have already written,
Pb is base pitch 2 pi rb by N, where N is
461
00:46:20,951 --> 00:46:21,951
the number of teeth and rb is the base circle
radius, which is same as 2 pi pitch circle
462
00:46:21,951 --> 00:46:22,951
radius cos phi, where phi is the operating
pressure angle, that is this angle and also
463
00:46:22,951 --> 00:46:23,951
this angle.
464
00:46:23,951 --> 00:46:24,951
So, we have got the expression of mc in terms
of EF by Pb, Pb is given by that expression.
465
00:46:24,951 --> 00:46:25,951
Now let me try to write EF in this way as
EF plus AE plus EF plus FB.
466
00:46:25,951 --> 00:46:26,951
I have added AE, EF, FB so I subtract all
these three, so, a minus AE plus EF plus FB;
467
00:46:26,951 --> 00:46:27,951
EF is this EF, I have added AE, EF, FB I have
subtracted AE, EF, FB.
468
00:46:27,951 --> 00:46:28,951
EF plus AE as you see EF plus AE is nothing
but AF.
469
00:46:28,951 --> 00:46:29,951
Similarly, EF plus FB is nothing but BE, and
AE plus EF plus FB is nothing but AB.
470
00:46:29,951 --> 00:46:30,951
What we can see; what is AF?
471
00:46:30,951 --> 00:46:31,951
This angle is 90 degree because this is tangent
this is radius.
472
00:46:31,951 --> 00:46:32,951
So, AF I can write square root of O1F squared
minus O1A squared, and what is O1F nothing
473
00:46:32,951 --> 00:46:33,951
but ro1 this is the addendum circle that is
how we determine the point F.
474
00:46:33,951 --> 00:46:34,951
So, O1F is nothing but ro1 and O1A is rb1,
so, AF I can write square root of ro1 squared
475
00:46:34,951 --> 00:46:35,951
minus rb1 squared.
476
00:46:35,951 --> 00:46:36,951
The same way I can write BE what was E that
was the addendum circle of this gear.
477
00:46:36,951 --> 00:46:37,951
So, O2E is nothing but ro2 and this is rb2.
478
00:46:37,951 --> 00:46:38,951
So, BE I get ro2 squared minus rb2 squared
minus AB, AB is this angle is phi.
479
00:46:38,951 --> 00:46:39,951
So, O1P sin phi is AP.
480
00:46:39,951 --> 00:46:40,951
Similarly, O2P sin phi is BP.
481
00:46:40,951 --> 00:46:41,951
So, O1O2 sin phi will give me AB.
482
00:46:41,951 --> 00:46:42,951
O1P plus O2P is O1O2, that is the centre distance
and AP plus BP which is AB if you take sin
483
00:46:42,951 --> 00:46:43,951
phi O1P sin phi you get AP, O2P sin phi you
get BP.
484
00:46:43,951 --> 00:46:44,951
So, this is the centre distance which I write
as C is sin phi, where C is O1O2.
485
00:46:44,951 --> 00:46:45,951
So, I get EF in terms of base circle radius
and the outer circle radius, addendum circle
486
00:46:45,951 --> 00:46:46,951
radius, similarly base circular radius, addendum
circle radius of both the gears, center distance
487
00:46:46,951 --> 00:46:47,951
between the gears and the operating pressure
angle.
488
00:46:47,951 --> 00:46:48,951
And this divided by Pb which again I expressed
in terms the operating pressure angle, number
489
00:46:48,951 --> 00:46:49,951
of teeth and the pitch circle radius.
490
00:46:49,951 --> 00:46:50,951
This I should have taken second gear.
491
00:46:50,951 --> 00:46:51,951
So, 2 pi rb2 N2.
492
00:46:51,951 --> 00:46:52,951
So, 2 pi rp2 N2.
493
00:46:52,951 --> 00:46:53,951
And rp1 and N1 this is the same.
494
00:46:53,951 --> 00:46:54,951
So, rp2 by N2 is same as rp1 by N1.
495
00:46:54,951 --> 00:46:55,951
So, it is the same expression we have to take
the pitch circle radius and put the number
496
00:46:55,951 --> 00:46:56,951
of teeth of the same gear.
497
00:46:56,951 --> 00:46:57,951
So, that gives me the expression for the contact
ratio.
498
00:46:57,951 --> 00:46:58,951
The recommended value of the contact ratio
for a pair of involute gears is 1.4.
499
00:46:58,951 --> 00:46:59,951
What does that mean?
500
00:46:59,951 --> 00:47:00,951
This is a not an integer, what does it mean
that, during some phase of the contact only
501
00:47:00,951 --> 00:47:01,951
one pair of teeth is in contact, for certain
other phase of contact at least two teeth
502
00:47:01,951 --> 00:47:02,951
is in contact.
503
00:47:02,951 --> 00:47:03,951
Because there must be a region when two teeth
must be in contact so, that continuous transmission
504
00:47:03,951 --> 00:47:04,951
is ensured and on an average 1.4 pair of teeth
remain in contact if I consider the entire
505
00:47:04,951 --> 00:47:05,951
cycle.
506
00:47:05,951 --> 00:47:06,951
That is the physical meaning of this contact
ratio as 1.4.
507
00:47:06,951 --> 00:47:07,951
That on an average within the entire cycle
1.4 pair of teeth remain in contact; that
508
00:47:07,951 --> 00:47:08,951
means, sometimes there is only one pair of
teeth in contact, and just when the contact
509
00:47:08,951 --> 00:47:09,951
is being lost within one pair, other pair
comes into contact before that.
510
00:47:09,951 --> 00:47:10,951
So, for some time at least two pair of teeth
remain in contact.
511
00:47:10,951 --> 00:47:11,951
So this contact ratio gives you the average
number of pair of teeth that remain in contact
512
00:47:11,951 --> 00:47:12,951
during the entire cycle.
513
00:47:12,951 --> 00:47:12,952
3