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In the first two lecturers of this module, we discussed some special mechanism namely,
approximate straight line mechanism and exact straight line mechanism. In today's lecture,
we will discuss another special mechanism, which are used for steering the automobiles
and are called 'steering mechanisms'. As we know, when an automobile wants to take
a turn, the front two wheels have to rotate about the vertical axis.
Let us say, this is the front view of the one of the wheels, front wheel's front view.
As we see, the front wheel is normally mounted on a short axle
which is pivoted to chassis and the pivot axis can be vertical, this is the body of
the automobile and this is the pivot axis. When car wants to take a turn, it is this
front wheel, rotates about this pivot axis. There are other types of steering mechanisms,
where this pivot axis may not be vertical, it is inclined from the vertical. Again, this
is the front view of another front wheel. This is the road surface and this is the short
axis about which the front wheel can rotate and during the steering action, when the car
wants to take a turn, this wheel will rotate about this pivot axis and this is the body
or chassis of the automobile. What you see in this arrangement, while taking a turn,
the body of the automobile is lifted. Consequently, when the steering wheel is released, the car
automatically comes to its original position and the wheel becomes straight. We shall discuss
this type of steering mechanism in our subsequent discussion. First, we should discuss, what
is the steering requirement? When the automobile takes a turn, either a
right hand turn or a left hand turn, it is imperative that all the wheels that is the
two front wheels and the two back wheels continue to roll. What is the rolling condition? That
are the contact point velocity must be zero. It is easy to see that, if I consider this
contact point C and it is center of the wheel, if I call 'O', we know velocity of C is velocity
of the center of the wheel O plus omega of the wheel cross OC. Now, for Vc to be zero,
it is this vector must cancel this vector, that means, these two vectors must lie in
the one plane. Now, omega cross OC lies in the plane of the wheel. So, the Vo must lie
in the plane of the wheel. This is the basic steering requirement that the velocity of
the center of the wheel must lie in the plane of the wheel, such that, when the wheel rotates,
the contact point velocity can be zero that is the perfect rolling condition can be maintained.
Steering requirement is rolling condition must be maintained. I hope this is clear that
when the car takes a turn, the velocity of the center of the wheel, if it does not lie
in the plane of the wheel, then there will be velocity of the contact point, because
this lie in the plane of the wheel. Omega is this way, OC is this way, so omega cross
OC is in this direction, that is in the plane of the wheel. So, Vo must lie in the plane
of the wheel such that Vc can be zero, which is required for perfect rolling conditions.
So the front wheels must turn appropriately, such that the center velocity must lie in
the plane of the wheel. To maintain this steering requirement, let us see what is the condition
needed. Let us now see the condition that needs to
be satisfied, such that the steering requirement is fulfilled.
Let us consider the two back wheels. These are the gear wheels. We know these two gear
wheels are connected to the differential, which is fixed to the body of the automobile.
This differential allows the two gear wheels to rotate at different speed when the car
is taking a turn. For example, if the car is taking a turn like this, then this becomes
the outer wheel and this becomes the inner wheel and we need that the outer wheel should
rotate faster as compared to the inner wheel. This differential speed can be created by
this differential, as we shall discuss later on when we discuss gear box.
The distance between these two wheels is called wheel track. Let us say wheel tracking (a).
This is the top view of the car and the distance between the gear axle and front axle, this
distance let me call b, which is called the wheel base. Similarly, there is another front
wheel. So, these are the two gear wheels and these are two front wheels. As we see, the
front wheel axles are mounted on these pivots on the body of the automobile. Now, when the
car takes a turn in this direction, these two front wheels must rotate. Let us say,
this axle rotates about this pivot and this axle rotates about this pivot. These two rotations
must be such that, all these axles must intersect at one point. The gear axle and the two front
axles, when the car is taking a turn must intersect at one point, let me call it 'I',
that is, I is the instantaneous axle of rotation for this entire car body as the rigid body.
So, this is instantaneous axis of rotation for the car body.
As we see, in this situation, the wheel which is now rotated like this, the velocity of
center of the wheel is in this direction that is contained in the plane of the wheel and
this wheel which is rotated like this. The velocity of the center of the wheel is perpendicular
to this line and again contained in the plane of the wheel, which is the requirement for
perfect steering. These two wheels must rotate say, this is phi and this angle is psi. So
for perfect steering, this phi and psi should be such that these two front axles and the
gear axles pass through one point. This is the steering requirement. So, let me find
out what is the condition? Let us see, if we say that the distance between
these two pivots in this direction is p. If I say that distance of this instantaneous
axis of rotation from this pivot, say x. So, what we see that this distance is this is
the wheel base which I call b. What we can write that, cot phi is x divided by b. Similarly
cot psi is x plus p divided by b. If I subtract, we are getting cot psi minus cot phi is p
by b. So, for perfect steering requirement the rotation of the two front wheels - the
inner wheel by phi which has to rotate more and outer wheel by psi, has to rotate less.
This phi and psi should be related by this equation cot psi minus cot phi is p by b,
but p is the distance between the two pivots on to which is front short axles are mounted
and b is the wheel base. Now, we will see how we can achieve this condition
by using different types of steering mechanism. Now, that we have obtained the steering requirement,
that is the condition that must be fulfilled by steering mechanism. Let us discuss a particular
steering mechanism known as 'Davis steering gear'.
In this Davis steering mechanism, we use a six link mechanism with seven lower pairs.
Let me first draw the mechanism. This is the Davis steering mechanism, as we see these
are the two front wheels mounted on short axles and the axles are pivoted on the body
of the automobile. Let us consider that the car is taking a turn in this direction. As
we see, let me call this point O2, this point A, this point B, and this point O6. This is
the six link mechanism, link number one refers to the body of the automobile. This is link
number two and this block in which this link number 2 can slide, this is link number 3,
link number 3 is hinged to this link which is link number 4 and this block which is hinged
to link number 4 at B is link number 5 and this is link number 6, which is nothing but
the axle of this outer link. Similarly, link number 2 is nothing but the
axle of the inner link. These are the axles which can slide in this link. This is another
axle which can slide in this link and these two links are connected by link number 4 to
revolute pairs. So, over all, we get 6 link mechanism, 1, 2 is the axle of the inner link,
3 is this block in which this link 2 can slide. Link 4 is this rod this link which can slide
in the body of the automobile parallel to the direction of O2O6. Link 4 is in sliding
motion with respect to link number 1, that is, body of the automobile, there is a prismatic
pair here. Link number 4 has a revolute pair of the link number 5 and the link number 5
has the prismatic pair in this direction with link number 6, which is the nothing but the
axle of the outer link. Obviously, the mechanism has to be symmetric,
such that similar turn can take place either in this direction or in this direction. The
car can turn in this way or the car can turn that way. As we see, this mechanism is symmetric,
that means this angle is same as this angle. Now, our only requirement is to find the value
of alpha, such that the steering requirement is satisfied. Let us say, the rotation of
this link 2, how it is steered? It is this rod is moved in this direction. That means,
suppose the A comes to this point equal A prime and B comes to exactly by the same distance,
here let me call this point B. This rod AB moves horizontally parallel to O2O6 from AB
to A prime B prime. Consequently, this link 2, rotate by theta2 and if I join O6 with
B prime, link 6 as we rotate by this angle, which is theta6 that is this angle is fixed.
So, theta2 is rotation of this axle, theta6 is the rotation of this axle.
As AB goes from AB to A prime B prime, you have to find the value of alpha such that
the steering requirement is fulfilled. If we remember, this distance between the pivots
is denoted by p and steering requirement was cot of theta6 minus cot of theta2 was p by
b. In our previous diagram, theta6 is denoted as the rotation of the outer wheel, which
is psi and theta2 as the rotation of the inner wheel, which is denoted by phi and we said
cot psi minus cot psi should be p by b. The p is the distance between the pivot and b
is the wheel base. Let us say, in the state configuration, these two lines meet here and
in this rotated configuration, these two lines meet there. We call this point R and this
point S. Extension of link 2 and extension of link 6. In the state configuration, the
wheel was straight, meet at R and when the wheel turns, they meet at S. It is very easy
to show that this line RS, let me use a different color, this line RS is parallel to O2O6. We
can easily prove that RS is parallel to O2O6. To prove that, we have to remember that this
link length four between the two revolute pairs is same, that is A prime B prime is
same as AB. BB prime is same as AA prime and AB, that is the length of the rigid link 4
is A prime B prime. We already stated that this prismatic pair was parallel to O2O6 that
is AB is parallel O2O6. What we can write? That RA by O2R is same
as RB by O6R is Ab by O2O6 because this line is parallel to this, these two triangles are
similar triangles, RAb and RO2O6 are similar triangles, because AB is parallel to O2O6.
So, we can write RA by O2R is same as RB by O6R is AB by O2O6. Now, A prime B prime is
again parallel to O2O6. So, for these two triangles, S A prime B prime and SO2O6, we
can have similar relationship, that is SA prime divided by O2S is SB prime by O6S is
A prime B prime by O2O6. These two white triangles are similar and this triangle and this blue
triangle are similar, because this line is parallel to this line. Now, AB is same as
A prime B prime. So, these are also same. So, we can write RA by O2R is same as SRA
by O2R is same as SA prime by O2S, RA by O2R is AB by O2O6 and SA prime by O2 S is A prime
B prime by O2O6. AB is same as A prime B prime, so this is equal to this. Now, this RA by
O2R is same as S A prime by O2S, then, I can say this line A prime is parallel to RS which
means RA is also parallel to O2O6. So we have just proved that this line RS is parallel
to O2O6. To determine the required value of alpha, such that the steering condition is
fulfilled, let us note that because RS is parallel to O2 O6, this angle is also alpha.
Similarly, this angle is also alpha. Now, this is the angle theta6, this angle is alpha.
So, this angle is alpha minus theta6. This angle is alpha and this angle is theta2 so
this angle is pi minus alpha plus theta2. Now, we use the triangle sin law for this
triangle namely, O6RS, we can write RS by sin of this angle which is theta6. RS by sin
of theta6 is same as O6R divided by sin of this angle, which is alpha minus theta6. So,
this is O6R sin of alpha minus theta6. This angle is alpha and this angle is theta6, so
this angle is alpha minus theta6. Now, in this triangle, we apply the sin law RS by
sin of theta6 is O6R divided by sin of this angle, which is alpha minus theta. Again,
we apply the triangle sin law for O2RS, this triangle. So, O2R divided by sin of this angle,
which is O2R by sin of pi minus alpha plus theta2 is same as alpha plus theta2 is same
as RS divided by sin of this angle, which is theta2. So, we apply the sin law for triangle
O6RS and O2RS to get these two relations.
Now, from here, if I divide this equation by this equation, we get sin theta2 divided
by sin theta6 is same as sin alpha plus theta2 divided by sin alpha minus theta6. We just
divide this equation by this and this equation by this to get these relations. If we expand,
we get sin theta2 sin alpha cos theta6 minus sin theta2 six cos alpha sin theta6 is same
as sin theta6 sin alpha cos theta2 plus sin theta6 cos alpha sin theta2. Just multiplying
this with this, this with this, we get this relation.
Now, we can divide all these four terms by sin theta2 sin alpha sin theta6, so that I
get, Cot of theta6, sin theta2 sin alpha cancels, divided by sin theta6 I get cot. Again, from
here, I get minus cot alpha. This sin cancels and this gives me cot alpha. Here, I get equal
to cot theta2, then here I get plus cot alpha. So that gives me, cot theta6 minus cot theta2
is two cot alpha. For perfect steering condition to be maintained, I know cot theta6 minus
cot theta2 should be p by b, while p is the distance O2 O6 between these two pivots and
b is the wheel base. So, this is equal to p by b. So, alpha is half of cot inverse p
by b. Let me explain this Davis steering gear once
more. This is the six link mechanism, where link 1 refers to the body or chassis of the
automobile. Link 2 is inner wheel axle, we consider a turn like this and with this front
wheel becomes the inner wheel and link 6 is the short axle of the outer wheel, that is
link number 6. Link 3 has a prismatic pair between 2 and 3. Link 5 and 6 has a prismatic
pair between them. Link 4 has two revolute pair at its end and a prismatic pair with
link 1 and the direction of this prismatic pair is parallel to O2O6.
The mechanism is symmetric, because we have to take identical left turn and right turn.
Only, design parameter is to find out this angle alpha and we found out that alpha should
be half cot inverse p by b, where p is distance between O2, O6 and b is the wheel base. Then,
this steering action will be maintained.
So this is a six link mechanism, where you have, R and a P pair, then again an R pair,
then again a P pair then again an R pair, then again P pair and an R pair, so it has
seven lower pairs, alternately R-P-R-P-R-P-R. Though this Davis steering gear can maintain
the perfect steering condition for all radius of turning, it is cumbersome to make because
there are so many prismatic pairs and it is difficult to maintain.
So, Davis steering gear is a six link mechanism with one, two, three prismatic pair and four
revolute pairs. Due to difficultly of this prismatic pairs, it is decided to forego this
advantage of maintaining the steering condition for all turning radius. We can go for a simple
mechanism, which will maintain the steering conditions for certain values of turning radius
and will not be exactly satisfied for all other turning radius. There will be a little
tendency of slight slip or the transverse motion of the wheel at the point of contact.
Let me now summarize, what we have covered today. We started discussion of steering mechanisms
which are used in automobile. The front wheels of the automobile are mounted are short axles,
which are pivoted to the body of the automobile. For perfect steering condition, that is the
wheel continued to roll even when the car is taking a turn, we need that the velocity
of the wheel center must lie in the plane of the wheel, so that the contact point of
the wheel has zero velocity maintaining pure rolling. This requires that the two front
wheels must be rotated by different angles, depending on the turning radius. The condition
we obtained was in terms of the distance between the pivots of these two short axles and the
wheel base. We showed that is six link mechanism, using alternately R and P pair, seven of these
4- R, 3 - P pairs; which we call Davis steering gear can maintain this perfect steering condition
for all turning radius, but there is difficultly of this complicated mechanism, both for manufacturing,
it is costly and also for maintenance because of three prismatic pairs.