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We finished our last lecture with the contact
ratio for a pair of involute gears.
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What is the expression of the contact ratio?
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Today, we start our discussion with the expression
of the contact ratio for a rack and pinion
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drive.
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So
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contact ratio we use the symbol mc, for a
rack and pinion.
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Lets say, this is the pitch circle of the
pinion.
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For the rack, as we know the pitch circle
radius is infinity, that is the pitch circle
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gets converted into a straight line.
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As in the case of a pair of gears, two pitch
circles was touching each other, hear again
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the pitch line of the rack is in as a tangent
to the pitch circle of the pinion.
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So this is what we call pitch line of the
rack, and this is the pitch circle of the
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pinion.
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If we draw the base circle of the pinion,
this is the base circle, this point of contact
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between the pitch circle and the pitch line
is the pitch point, which we denote it by
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P. From this pitch point, if I draw a tangent
to the base circle from this pitch point let
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me draw a tangent to the base circle.
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This line represents the line of action.
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Let this point of tangency be I denote by
A, this is the centre of the pinion, then
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this angle is the operating pressure angle
phi which is same as this operating pressure
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angle phi.
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The point of contact between the teeth of
the pinion and the rack always moves along
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this line of action.
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Let me talk of a particular tooth on this
pinion, which is just beginning the engagement
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that means the contact with that particular
tooth is just starting at this configuration.
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Let this be the tooth, which is coming in
contact with the tooth of the pinion at this
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instant.
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Now where does the contact start?
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Contact obviously starts at the addendum line
of the rack that is the highest point of the
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rack tooth.
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So the rack tooth if I draw, this is the addendum
line of the rack.
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So the rack tooth is perpendicular to this
line of action; so let this be the rack tooth.
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The contact starts at this point between the
pinion tooth and the rack tooth on the line
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of contact; this is the addendum line of the
rack.
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If I complete the rack tooth, it will look
something like this.
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This is the rack.
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At this beginning of the contact, this point
let me call E. Now this gear, this pinion
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rotates in the clockwise direction, this pinion
rotates in the clockwise direction drives
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the rack to the left.
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Now when, when does the contact between this
pair of teeth is lost?
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It is lost when the contact comes at the addendum
circle of this pinion.
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Let this be the addendum circle of this pinion
tooth; if I complete the pinion tooth it looks
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like this.
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When this addendum circle intersects this
line of action, lets say this point I call
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F.
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This tooth has rotated and lets say this is
the this is the addendum circle and this point
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is F. Now how much is the movement of this
tooth along measured along the base circle,
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let me call this point S, this point on the
base circle and the tooth has rotated, and
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the movement along the base circle is from
S to T. This is due to the rotation of the
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gear, this point S of this tooth has moved
up to this point T.
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Now to ensure that the teeth remain in contact
the next tooth that is the adjacent tooth
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on this gear, this point must get into beyond
this point.
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This point must move beyond S such that this
tooth comes in contact with the next tooth
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and this distance measured from this point
to this point, let me mark it, the distance
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between these two points on adjacent tooth
along the base circle, that is what we denoted
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as base pitch or pb.
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The distance between two adjacent teeth along
the base circle that means from this point
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to this point, that is what we call pb.
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An expression for pb we have used last time,
which is 2 pi rb divided by N; where rb is
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this base circle radius, OA this is rb, and
N is the number of teeth on the pinion, rb
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is this base circle radius of the pinion.
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Contact ratio mc was defined as: the movement
of one tooth along the base circle, which
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is ST divided by the base pitch.
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And as we said, this must be more than one
so that continuous transmission of motion
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from between these two rack and opinion is
ensured.
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To get the expression of this mc, let us see
the contact started at E and finished at F
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and these are involute profiles.
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So, AF is nothing but AE, AE is the string
length which was wound on to this base circle
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and by unwinding I have generated this involute
profile.
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Because it is an involute profile, the distance
AE is same as the distance measured along
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the arc of the base circle AS.
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So, AE is same as AS.
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Similarly, this is also an involute from the
same base circle, this is the same tooth.
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So AT this measured along the base circle
will be AF, that is the length of the string
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which has been unwound from this base circle.
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So AT is AF, length of the string AF is the
length of the circular arc from A to T. So
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if I subtract AF minus AE, I get EF and if
I subtract AT, AS from AT, I get TS ST, so
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ST is same as EF.
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So that I can write again contact ratio is
nothing but EF by pb.
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EF, I can write as, EF this is EF this point
is E this point is F, so I add AE
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and I also subtract AE, so that we I get,
okay let me now try this way, let me try write
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it little differently I write is as EP, EF
I write EP plus PF.
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EP is this distance, PF is this distance divided
by pb.
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Now I add PF, AE plus EP.
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EP I have written, EP I add AE to PF, and
I again add EP and subtract AE plus EP that
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is, AP.
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EP I have written here, PF I have written
here, AE plus EP I have added so I subtracted
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AE plus EP which is nothing but AP, AE plus
EP which is nothing but AP divided by pb.
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So finally I get mc, the expression of mc
contact ratio is EP.
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What is EP?
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EP is this height, which I can see, if this
angle is phi and this vertical height is addendum
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of the rack, which I denote by ar, ar is the
addendum of the rack.
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So this EP is nothing but ar by sine phi,
where phi is the operating pressure angle,
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this is EP and PF plus EP plus AE is nothing
but AF, whole thing is divided by pb, this
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is AF and minus AP.
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A is this contact point between the tangent
from the pitch point to the base circle, so
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this angle is 90 degree.
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This is radius, this is tangent, so this angle
is 90 degree and O1P is nothing but the outer
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radius of the pinion.
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This is the outer circle, this addendum circle,
so O1 to F is nothing but the outer circle
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of the, outer radius of the pinion.
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So AF is square root of outer radius squared
minus rb squared.
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So this I get, ar by sine phi plus, this is
ro1 squared, that is the outer radius or the
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addendum circle radius, this is what I call
ro r outer 1, minus rb1 squared minus AP.
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AP is nothing but, this is phi and this is
rb1, so AP is rb1 tan phi, tan phi is AP by
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rb1, so AP is rb tan phi minus let me call
it rb1, I’m writing rb1 here, rb1 tan phi
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divided by pb, for which I already wrote the
expression.
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So finally we have got the expression for
the contact ratio for a rack and pinion drive
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involving the addendum of the rack, operating
pressure angle, outer radius of the pinion,
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base circle radius of the pinion and the base
pitch, which is pb, is nothing but two pi
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rb divided by N, rb1 N1, rb1 is the base circle
radius, N1 is the number of teeth of the pinion.
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This completes our discussion on the contact
ratio for a rack and pinion drive.
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The methodology is just similar to what we
used for a pair of involute gear teeth profile.
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Now that we have finished discussion on contact
ratio, let us start with a very important
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parameter for gear tooth geometry, that is
known as interference and undercutting.
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First, let me explain what do we mean by interference
and what is undercutting.
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If we remember from the very definition of
involute profile, it is obvious, that the
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involute does not exist inside the base circle,
it is from unwinding from this base circle
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outward, we created the involute profile.
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Lets say, it starts from this point A it starts
radially, and as the string unwinds, we generate
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this involute profile.
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So the involute profile exists only outside
the base circle, does not exist inside the
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base circle.
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Suppose, we use very few numbers of teeth
on a gear and the teeth size is very big,
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then it may so happen that the addendum circle
of the mating gear gets into the base circle
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of the other gear.
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If the mating profile, which is another involute
for another base circle gets into the, inside
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the base circle.
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As we see, there is no involute of this gear
say, this is gear number 1 and this is 2,
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the gear number 1 does not have any involute
inside this base circle.
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Whereas, the involute of the other mating
gear is getting into the base circle so there
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is no conjugate profile.
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So what one can do?
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One can extend this involute of gear 1 may
be radially, even then there is interference.
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As we see, this tooth will interfere with
this tooth because this tooth at once to occupy
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this position, whereas this tooth is lying
there, this is called interference.
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One solution to this interference is to remove
some portion of gear 1 - we do not have the
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involute profile of gear 1 anyway - so let
me cut this portion from gear 1.
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Then as we see, this gear tooth is not interfering
with gear 1, the tooth of gear 2 is having
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a space inside gear 1, though there is no
conjugate profile to maintain contact.
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However, the conjugate action maybe taken
care of by the other pair of teeth in engagement,
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so the continuous motion is transmitted maintaining
the constant angular velocity ratio, but this
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tooth is not transmitting any power because
this is getting into this portion and this
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is what is called undercutting.
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We have avoided the interference by undercutting
the tooth of gear 1, inside the base circle.
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But this is not a very satisfactory solution,
because this undercutting weakens the tooth
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of gear 1, undercutting which weakens the
undercut tooth and it is here at the root
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of the tooth where failure takes place so
if we have weakened the tooth near its root,
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that is not a very desirable solution, only
if it is a must, then only we should undercut
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the tooth.
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Next, we shall discuss how to ensure that
that there is no interference, how to avoid
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interference by ensuring some minimum number
of teeth on a gear.
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We cannot have number of teeth on a gear less
than a minimum number to avoid interference
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– that will be our next topic of discussion.
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As we have just now shown the possibility
of interference between a pair of gear teeth,
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because involute does not exist inside the
base circle, one not so satisfactory solution
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was to undercut the tooth.
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But a better solution is to have a minimum
number of teeth on a gear to avoid interference.
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We have already seen how does the interference
took take place first of all, because the
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addendum circle gets into the base circle.
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So now, we talk of a minimum number of teeth
to avoid interference to transmit a particular
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angular velocity ratio, to transmit angular
velocity ratio – say we call it lambda,
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which is defined as omega2 by omega1 which
is less than equal to 1, that means omega1
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is more than omega2.
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When we transmit continuous rotation from
one shaft to another at the constant angular
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velocity ratio, let us forget about the sign
for the time being, I defined angular velocity
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ratio only by the positive number and I define
it in such a way, that it is less than equal
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to one, that means the higher speed is at
the denominator.
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For example, suppose lambda is 8 upon 9.
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To transmit a constant angular velocity 8
by 9, I can have 16 and 18 teeth or 24 and
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27 teeth or 32 and 36 teeth.
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The question is, what must be the minimum
number of teeth to transmit angular velocity
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8 by 9, can I use 8 and 9 teeth to transmit
constant angular velocity 8 by 9.
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The question we are trying to answer – what
is the minimum number of teeth such that interference
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is avoided while transmitting a particular
angular velocity ratio which is given to us,
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the value of lambda?
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Towards this end, let me first draw the two
pitch circles.
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Suppose these are the two pitch circles, omega1
is the higher speed, so this is gear1 which
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rotates at a fastest speed and this is gear2
which is rotating at a slowest speed.
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Suppose, this gear is rotating this way and
this gear is rotating this way, these are
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the pitch circles.
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Now let me draw the two base circles, these
are the two base circles.
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This point is O1, this point is O2, this is
the pitch point which we call P and if I draw
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the common tangent to these two base circles,
that is the line of action AB.
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Now as this is very clear, because this is
a smaller gear AP is smaller than BP this
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is a larger gear, which is rotating at a slower
speed.
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So lambda is N1 by N2, where N1 is the number
of teeth on this gear and N2 is the number
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of teeth on this gear, which is also same
as O1P divided by O2P, which is same as.
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Now the question is, that addendum circle
of either gear should not get into the base
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circle of the other gear and as we know of
the contact point is restricted to move on
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this line of action that is line AB, so which
addendum circle will first get into the base
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circle.
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Addendum circles are of same size – both
the gears teeth are of same size, they have
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the same addendum.
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So it is obvious, that the bigger gear addendum
will first get into the base circle.
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If this is the addendum circle, if this touches
A the addendum circle of this gear, which
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has exactly same addendum will still pass
through the below the point B, it will not
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get into the base circle.
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The point of contact always lies on this line,
so I am only interested in seeing whether
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the addendum circle of this line intersecting
AB beyond A or beyond B. So interference obviously
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occurs first with the pinion when the addendum
circle of the gear just passes through the
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point A. Because, even at that time the addendum
circle of the pinion does not pass through
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B, it is still to interfere with this gear.
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So first point to note interference occurs,
first at the pinion, that is the smaller gear,
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and we are considering as if the interferences
has just started.
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From this diagram, we should be able to find
out, what is the minimum number of teeth that
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is N1 to avoid interference, that is the task.
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And this common tangent to the pitch circles,
this is the point P, and this angle is the
209
00:29:12,379 --> 00:29:17,669
operating pressure angle phi.
210
00:29:17,669 --> 00:29:32,119
Now this distance O2A which is nothing but
the circle of the, radius of the addendum
211
00:29:32,119 --> 00:29:33,850
circle of gear2.
212
00:29:33,850 --> 00:29:42,710
Let me first write an expression for this
O2A.
213
00:29:42,710 --> 00:29:54,119
We consider the triangle O2PA, this point
is P. Let me write it here again, O2PA let
214
00:29:54,119 --> 00:29:56,519
us consider this triangle.
215
00:29:56,519 --> 00:30:13,320
So O2A I can write, O2A squared is O2P squared
plus AP squared minus, just using the triangle
216
00:30:13,320 --> 00:30:22,570
law twice O2P into AP into cosine of the angle
between O2P and AP, which is pi by two plus
217
00:30:22,570 --> 00:30:29,850
phi.
218
00:30:29,850 --> 00:30:39,039
We consider this triangle O2A sorry O2AP,
O2A squared is O2P squared plus AP squared
219
00:30:39,039 --> 00:30:46,289
minus twice O2P into AP into cosine of the
angle between them, which is pi by two plus
220
00:30:46,289 --> 00:30:47,409
phi.
221
00:30:47,409 --> 00:31:03,889
So this I write O2P squared plus AP squared
minus twice cosine pi by two plus phi is minus
222
00:31:03,889 --> 00:31:20,179
sine phi so I get plus two O2P into AP into
sine phi and what is AP?
223
00:31:20,179 --> 00:31:27,440
This angle if we remember, we have used it
number of times this angle is also phi.
224
00:31:27,440 --> 00:31:30,409
So what is AP? that is O1P sine phi.
225
00:31:30,409 --> 00:31:45,999
So this I can write O2P squared plus O1P squared
sine squared phi plus twice O2P and for AP
226
00:31:45,999 --> 00:31:53,740
I write O1P into sine phi and another sine
phi, so I get sine squared phi.
227
00:31:53,740 --> 00:32:10,370
So now if I take, O2P common from this expression
and take the square root, I get O2A is O2P
228
00:32:10,370 --> 00:32:22,220
into from here I get 1, here I get O1P squared
divided by O2P squared and if we remember,
229
00:32:22,220 --> 00:32:29,769
lambda this was equal to O1P by O2P.
230
00:32:29,769 --> 00:32:34,330
The speed ratio is nothing but the pitch circle
radius ratio.
231
00:32:34,330 --> 00:32:36,450
So lambda is that.
232
00:32:36,450 --> 00:32:45,649
So if I take O2P squared, I get here lambda
squared sine squared phi and from here I get
233
00:32:45,649 --> 00:32:55,720
one lambda, so two lambda sine squared phi
to the power half.
234
00:32:55,720 --> 00:33:03,909
I have taken O2P common from these expressions,
so I get O1P by O2P, which gives me this lambda,
235
00:33:03,909 --> 00:33:11,299
here I get O1P squared by O2P squared, which
I get lambda squared and then I have taken
236
00:33:11,299 --> 00:33:13,460
the square root.
237
00:33:13,460 --> 00:33:20,369
O2A minus O2P, O2A is the addendum radius
O2P is the pitch circle radius.
238
00:33:20,369 --> 00:33:30,220
So the difference of this is nothing but m,
sorry addendum, O2A O2P O2A minus O2P, that
239
00:33:30,220 --> 00:33:38,070
is the addendum, O2A minus O2P.
240
00:33:38,070 --> 00:33:58,599
So if I use this expression of O2A, I get
this as, O2P one plus lambda into lambda plus
241
00:33:58,599 --> 00:34:08,960
two sine squared phi to the power half minus
one.
242
00:34:08,960 --> 00:34:21,500
This is the expression for
243
00:34:21,500 --> 00:34:22,500
the addendum.
244
00:34:22,500 --> 00:34:29,270
Addendum is always expressed as a fraction
of the module, addendum is always expressed
245
00:34:29,270 --> 00:34:32,500
as a some fraction f of the module.
246
00:34:32,500 --> 00:34:38,720
If we remember, for the British Standard for
20 degree pressure angle, we said f is normally
247
00:34:38,720 --> 00:34:43,580
one, but I keep it as f, which is of the order
of one.
248
00:34:43,580 --> 00:34:58,980
So ‘a’ which is f into m is O2P into this
expression, which is one plus lambda into
249
00:34:58,980 --> 00:35:12,780
lambda plus two sine squared phi to the power
half minus one and if we remember what was
250
00:35:12,780 --> 00:35:15,400
the module?
251
00:35:15,400 --> 00:35:25,260
Module was 2 into rP divided by number of
tooth.
252
00:35:25,260 --> 00:35:31,940
Module by definition is pitch circle diameter
by the number of teeth, number of teeth of
253
00:35:31,940 --> 00:35:35,220
the second gear, pitch circle diameter of
the second gear.
254
00:35:35,220 --> 00:35:49,330
So this O2P, I can write as mN2 by 2, if I
write mN2 by 2, m, m cancels, so I get N2
255
00:35:49,330 --> 00:36:34,780
is O2P, I am writing mN2 by 2, so 2f divided
by this expression.
256
00:36:34,780 --> 00:36:41,420
We have considered the situation where interference
just started with the pinion and it will get
257
00:36:41,420 --> 00:36:50,710
this expression for N2 which means if I multiply
by N1 by N2, which is nothing but lambda.
258
00:36:50,710 --> 00:37:04,840
N1 minimum tooth on the pinion, we get 2f
lambda divided by this expression, the denominator
259
00:37:04,840 --> 00:37:13,750
remains the same.
260
00:37:13,750 --> 00:37:25,859
2f lambda divided by 1 plus lambda into lambda
plus 2 sine squared phi, this whole thing
261
00:37:25,859 --> 00:37:28,360
to the power half minus 1.
262
00:37:28,360 --> 00:37:34,150
This is the minimum number of teeth required
to avoid interference while transmitting an
263
00:37:34,150 --> 00:37:40,370
angular velocity ratio lambda and lambda as
we have defined, is always less than 1.
264
00:37:40,370 --> 00:37:44,880
So this is how we get the minimum number of
teeth to just avoid interference.
265
00:37:44,880 --> 00:37:50,090
N1 at least should be this much, if N1 is
more, then there will be teeth teeth will
266
00:37:50,090 --> 00:37:52,330
be smaller, so there will be no problem.
267
00:37:52,330 --> 00:38:00,400
This is just when the interference starts,
so N1 should be more than this N1 minimum.
268
00:38:00,400 --> 00:38:05,531
Just now, we derived the expression for the
minimum number of teeth that must be on a
269
00:38:05,531 --> 00:38:12,000
pinion so that interference is avoided while
transmitting a constant angular velocity ratio
270
00:38:12,000 --> 00:38:14,150
lambda.
271
00:38:14,150 --> 00:38:30,200
Number of teeth on a pinion must be greater
than (N1) minimum, where (N1) minimum was
272
00:38:30,200 --> 00:38:49,070
given by 2f lambda divided by 1 plus lambda
into lambda plus 2 sine squared phi to the
273
00:38:49,070 --> 00:39:04,260
power half minus 1; where lambda by definition
is between 0 and 1.
274
00:39:04,260 --> 00:39:10,700
We always define lambda as less than 1, that
is the slower angular velocity divided by
275
00:39:10,700 --> 00:39:14,740
the faster angular velocity and phi is the
operating pressure angle.
276
00:39:14,740 --> 00:39:15,740
What is f?
277
00:39:15,740 --> 00:39:22,630
f is addendum is expressed as a constant fraction
of the module, it is this constant fraction
278
00:39:22,630 --> 00:39:29,010
is f; f is normally 1 unless otherwise stated.
279
00:39:29,010 --> 00:39:35,480
From here let me try to drive the expression
for the minimum number of teeth that must
280
00:39:35,480 --> 00:39:43,810
be there on the pinion while it is in contact
with a rack, for rack and pinion.
281
00:39:43,810 --> 00:39:48,650
If it is a rack and pinion, then what must
be the minimum number of teeth on the pinion
282
00:39:48,650 --> 00:39:49,880
to avoid interference.
283
00:39:49,880 --> 00:39:58,210
For rack and pinion, as we see lambda is 0
because the angular velocity of the rack is
284
00:39:58,210 --> 00:40:01,520
0, rack has only translational velocity.
285
00:40:01,520 --> 00:40:05,620
So lambda is 0, omega2 by omega1 is 0.
286
00:40:05,620 --> 00:40:13,340
Now if I put lambda equal to 0 here, then
I get 0 and this is 0, so 1 this is 0 by 0.
287
00:40:13,340 --> 00:40:25,800
For this (N1) minimum from this expression
is standing out to be of the form 0 by 0.
288
00:40:25,800 --> 00:40:29,920
That is now nothing problematic as we know,
we can always take the limit.
289
00:40:29,920 --> 00:40:35,730
Let us take the limit of this as lambda tends
to 0.
290
00:40:35,730 --> 00:40:45,010
So for rack and pinion, I can get (N1) minimum,
by taking the limit of this expression as
291
00:40:45,010 --> 00:40:46,830
lambda tends to 0.
292
00:40:46,830 --> 00:40:51,510
Because it is of the 0 by 0 form, I can use
L'Hospitals rule, that means differentiate
293
00:40:51,510 --> 00:40:56,380
the numerator with respect to lambda, denominator
with respect to lambda, then put lambda equal
294
00:40:56,380 --> 00:40:59,050
to 0 and see what is the limiting value.
295
00:40:59,050 --> 00:41:04,960
So that way, if I take derivative of this
with respect to a lambda, on the numerator
296
00:41:04,960 --> 00:41:11,550
I am getting 2 f and the denominator if I
differentiate with respect to lambda, I get
297
00:41:11,550 --> 00:41:32,190
half then this goes at the top, 1 plus lambda
into lambda plus 2 sine squared phi to the
298
00:41:32,190 --> 00:41:45,500
power half and here I am getting lambda squared
is 2 lambda plus 2.
299
00:41:45,500 --> 00:41:49,930
If I differentiate, derivative of the numerator
and derivative of the denominator, this is
300
00:41:49,930 --> 00:41:55,280
what I get: Half into this to the power minus
half, then while differentiating this, I am
301
00:41:55,280 --> 00:42:09,180
getting with respective lambda if I, I’m
getting sine squared phi.
302
00:42:09,180 --> 00:42:14,560
Sine squared phi lambda squared gives me 2
lambda and 2 lambda gives me 2 and now I put
303
00:42:14,560 --> 00:42:15,760
lambda equal to 0.
304
00:42:15,760 --> 00:42:25,920
If I put lambda equal to 0, this goes to 0,
so I get 2f and here if I put lambda equal
305
00:42:25,920 --> 00:42:33,630
to 0, then 2 cancels, we get sine squared
phi.
306
00:42:33,630 --> 00:42:44,680
So this is the minimum number of teeth on
the pinion which engages with a rack and without
307
00:42:44,680 --> 00:42:46,820
interference, there will be no interference.
308
00:42:46,820 --> 00:42:51,410
This expression of course we could have gone,
got geometrically very simply by drawing the
309
00:42:51,410 --> 00:42:54,920
rack, just undergoing interference with the
pinion.
310
00:42:54,920 --> 00:42:58,020
That is what we are going to do next.
311
00:42:58,020 --> 00:43:03,030
Just now we derived the minimum number of
teeth on a pinion that is required to avoid
312
00:43:03,030 --> 00:43:08,140
interference with a rack, but that we did
analytically from the expression that we derived
313
00:43:08,140 --> 00:43:12,690
earlier for a pair of involute gears, a gear
and a pinion.
314
00:43:12,690 --> 00:43:18,740
Let me now get the same expression of the
minimum number of teeth geometrically.
315
00:43:18,740 --> 00:43:34,450
So (N1) minimum on the pinion for a rack and
pinion to avoid interference.
316
00:43:34,450 --> 00:43:54,070
Towards this end, let me say this is the pitch
circle of the pinion and, this is the pitch
317
00:43:54,070 --> 00:44:21,800
line of the rack; this is the base circle
of the pinion.
318
00:44:21,800 --> 00:44:33,380
If I draw from this pitch point P, if I draw
a tangent to the base circle, this angle is
319
00:44:33,380 --> 00:44:48,420
phi, this is O1, this is A. Now if the addendum
line passes through the point A, if this is
320
00:44:48,420 --> 00:44:54,790
the addendum line of the pinion, of the rack
that is the rack tooth looks something like
321
00:44:54,790 --> 00:44:57,860
this.
322
00:44:57,860 --> 00:45:07,220
If the addendum line intersects this base
circle just at A, that is when the interference
323
00:45:07,220 --> 00:45:12,940
starts and we get we can use this diagram
for getting the minimum number of teeth on
324
00:45:12,940 --> 00:45:16,320
the pinion.
325
00:45:16,320 --> 00:45:34,490
This is the rack of the addendum ar which
we write as f into m and this angle is the
326
00:45:34,490 --> 00:45:36,770
operating pressure angle phi.
327
00:45:36,770 --> 00:45:53,510
So AP is ar sine phi just when the interference
starts, AP is ar divided by sine phi.
328
00:45:53,510 --> 00:46:05,280
And AP is also O1P sine phi, this is O1P sine
phi, this angle is 90 degree.
329
00:46:05,280 --> 00:46:10,690
This is the base circle radius and this is
the tangent, this angle is phi, so pitch circle
330
00:46:10,690 --> 00:46:26,610
radius is O1P, and AP is O1P sine phi, which
tells me ar which is f into m is O1P sine
331
00:46:26,610 --> 00:46:30,870
squared phi and what was m?
332
00:46:30,870 --> 00:46:40,530
m if we remember is nothing but 2 into O1P
by N1, pitch circle diameter by the number
333
00:46:40,530 --> 00:46:43,830
of teeth, that is the definition of the module.
334
00:46:43,830 --> 00:47:01,930
So this I if I place that, I get N1, m I am
writing twice O1P by N1, O1P cancels, N1 is
335
00:47:01,930 --> 00:47:05,700
2f by sine squared phi.
336
00:47:05,700 --> 00:47:11,430
This is the same expression, that we got earlier
by taking the limiting value from the expression
337
00:47:11,430 --> 00:47:16,550
of (N1) minimum for a pair of involute gears.
338
00:47:16,550 --> 00:47:21,660
Next we shall solve an example to show the
application of this expression that we got
339
00:47:21,660 --> 00:47:22,750
for (N1) minimum.
340
00:47:22,750 --> 00:47:27,900
How do we determine the minimum number of
teeth or the number of teeth that must be
341
00:47:27,900 --> 00:47:33,860
used to transmit a particular given angular
velocity ratio lambda?
342
00:47:33,860 --> 00:47:40,800
Let me now solve an example, for a pair of
involute gears to transmit an angular velocity
343
00:47:40,800 --> 00:47:44,900
ratio lambda as 8 by 9.
344
00:47:44,900 --> 00:47:51,030
Lambda is given to be 8 by 9 and the operating
pressure angle phi just as an example, let
345
00:47:51,030 --> 00:48:03,710
me take 14 and a half degree which means sine
phi will be 0.25.
346
00:48:03,710 --> 00:48:20,340
If we remember the expression for (N1) minimum
was given as 2f lambda divided by 1 plus lambda
347
00:48:20,340 --> 00:48:31,390
into lambda plus 2 sine squared phi square
root minus 1.
348
00:48:31,390 --> 00:48:37,670
Now ‘f’ the most usual value is 1, so
let me say f is taken to be 1.
349
00:48:37,670 --> 00:48:44,400
If I put f equal to 1, lambda equal to 8 by
9 and sine phi equal to 0.25, (N1) minimum
350
00:48:44,400 --> 00:48:47,910
from this expression turns out to be 22.94.
351
00:48:47,910 --> 00:49:02,050
But, number of teeth cannot be 22.94, it has
to be an integer, so (N1) minimum
352
00:49:02,050 --> 00:49:03,480
looks like it should be 23.
353
00:49:03,480 --> 00:49:10,550
But if N1 is 23 and to transmit an 8 by 9
angular velocity ratio, N2 will turn out to
354
00:49:10,550 --> 00:49:20,460
be non-integer which means, this will imply
N1 has to be a multiple of 8, such that N2
355
00:49:20,460 --> 00:49:23,660
can be a multiple of 9 and they will be both
integers.
356
00:49:23,660 --> 00:49:34,930
So (N1) minimum is actually 24 and corresponding
N2 is 27.
357
00:49:34,930 --> 00:49:43,550
So if I use 24 and 27 teeth, I can maintain
a constant angular velocity 8 by 9 and without
358
00:49:43,550 --> 00:49:49,090
interference.
359
00:49:49,090 --> 00:50:11,190
So this is what we mean by minimum number
of teeth to avoid interference.
360
00:50:11,190 --> 00:50:29,569
3