Math%/%CT%Exemplar%Task%
CT#Course:#Automotive#Technology#
CT#Skillset:#Engines/Valve#Train####
CT#Concept:#Camshaft#
Contributor(s):#Adrienne#Brellahan#
Task#Name:#Cam#&#Follower#
Description#of#Math#Content#
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WV#Math#Objective(s):#
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Evidence#of#Standards#for#
Mathematical#Practice#
M.TM.REI.6,#M.TM.REI.7,#M.TM.GPE.1#
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1) #Make#sense#of#problems#&#persevere#in#solving#them#
2) Reason#abstractly#and#quantitatively#
3) Construct#viable#arguments#&#critique#the#reasoning#of#others#
4) Model#with#mathematics#
5) Use#appropriate#tools#strategically#
6) Attend#to#precision#
7) Look#for#and#make#use#of#structure#
8) Look#for#and#express#regularity#in#repeated#reasoning#
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Description#/#Framing#of#the#Task:#
Can#be#implemented#when#performing#NATEF#tasks#associated#with#the#knowledge#objectives#1625.22#and#1625.30#
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Cam & Follower
In engines the camshaft is responsible
for providing the motion to open and
close each of the intake/exhaust valves.
The camshaft is a shaft with offset
eccentrics called cam lobes.
The cam lobe determines when each
valve opens, how far it opens, and how
long it stays open. The profile of the
cam lobe determines these functions.
A cam and follower are imposed on an
x-y axis as shown below. The follower is
in position on the positive x axis. We will
consider this position 0°. The center of
rotation of the cam is at the origin (0,0).
In overhead camshaft (OHC) engines,
camshaft followers (instead of rocker
arms) open the valves. The follower has
a pivot on one end and a slider or roller
near the center that directly contacts the
cam lobe.
CAM$
FOLLOWER$
Point$of$Tangency$
In order for the cam and follower to
function smoothly there must be no
bumps or drags on the cam, and the
follower must always be touching the
cam in exactly one point (this is called
tangent)
The cam lobe pictured has a profile that
can be written as two mathematical
equations. The cam profile and follower
outline has been imposed onto a
coordinate plane.
$
The cam rotates clockwise about the
origin (0,0).
For the purpose of this lesson the
position as shown at the right will be
considered 0° rotation.
$
With the help of the unit circle, we will
determine the point at which the follower
touches the cam for a given angle of
rotation.
! ! + ! ! = 225$
! = −0.1! + 25$
To simplify the mathematics involved here, consider the following,
suppose the camshaft rotated 45°. The resulting image is below.
In order for the cam lobe to satisfy the mathematical equations it must remain in it’s
original position, therefore we can move the follower as it would be positioned along the
cam lobe through the rotation. On the coordinate plane, the 45° rotation of the cam
would position the follower as shown below.
Represents$45°$angle$
of$rotation$in$standard$
position$
In terms of the 45° line and the cam profile what geometric term can be used to
represent the point at which the follower touches the cam.
Which equation that represents the cam profile pertains to this situation? What type of
equation is it?
Using the unit circle, you can determine that a point that lies on the 45° line is
!
(! ! !,
!
!
!!). Since the line passes through the origin, you also know that it contains the
point ( 0 , 0 ). Find the slope and the equation of the line in slope intercept form.
Now that you have the equation of the line and the equation of the cam lobe, you can
find the point of intersection. How would you set up the equations to solve?
What method can you use to solve a quadratic equation when it is not easily factorable?
Use this method to find the values of x. You will end up with 2 values of x. Looking at
the location of your graph, which answer is the logical choice?
What are the coordinates of the point at which the follower touches the cam after 45° of
rotation (point of intersection of the parabola and line)?
Using this same method, suppose the camshaft rotates 15° more, what are the
coordinates of the point at which the follower touches the cam? Justify your answer.
If the camshaft rotated a total of 225°, at what point would the follower touch the cam?
(HINT: pay attention to where 225° lies on the unit circle in relation to other angles you
may know) Sketch a diagram and show your work.
In terms of the 45° line and the cam profile what geometric term can be used to
represent the point at which the follower touches the cam.
The tangent point is
represented by the
intersection of the line
(terminal side of a 45° angle
in standard position) and
the cam lobe equation.
!
Which equation that represents the cam profile pertains to this situation? What type of
equation is it?
! = −0.1! + 25!
quadratic equation
Using the unit circle, you can determine that a point that lies on the 45° line is
!
(! ! !,
!
!
!!). Since the line passes through the origin, you also know that it contains the
point ( 0 , 0 ). Find the slope and the equation of the line in slope intercept form.
!=
!
!!
!
!
!!
!
=1
!−0=1 !−0
!=!
Now that you have the equation of the line and the equation of the cam lobe, you can
find the point of intersection. How would you set up the equations to solve?
To find the point of intersection you set the two equations equal
! = −0.1! ! + 25
What method can you use to solve a quadratic equation when it is not easily factorable?
Quadratic Formula: ! =
!!± ! ! !!!"
!!
Use this method to find the values of x. You will end up with 2 values of x. Looking at
the location of your graph, which answer is the logical choice?
Set the equation equal to 0. −0.1! ! − ! + 25 = 0
!=
1 ± −1! − 4(−0.1)(25)
2(−0.1)
!=
1 ± 11
−0.2
! = −21.58!!"#!11.58
by referring to the graph it is obvious that in this case -21.58 is not a possible
solution, therefore 11.58 is the x-coordinate of the point of tangency of the
follower.
What are the coordinates of the point at which the follower touches the cam after 45° of
rotation (point of intersection of the parabola and line)?
Substitute the x-coordinate into either of the original equations to find the ycoordinate. Sine the line has the equation ! = !, the y-coordinate and xcoordinate are the same value at the point of tangency, (11.58, 11.58)
Using this same method, suppose the camshaft rotates 15° more, what are the
coordinates of the point at which the follower touches the cam? Justify your answer.
An additional 15° rotation would result in a total rotation of 60°. A point on that
!
!
line (from the unit circle) is !! !, ! ! . Find the slope and equation of the line that passes
through that point and ( 0 , 0 ). Then since the follower is still touching the parabolic part
of the cam, you follow the same steps as above to find the coordinates of the point.
!=
!
!!
!
!
!!
!
!−0= 3 !−0
! = 3!! ≈ !1.73!
3! = −0.1! ! + 25
−0.1! ! − 3! + 25 = 0
!=
!=
!±
!
! !!(!!.!)(!"
!(!!.!)
!± !"
!!.!
! = −26.68!!"#!9.38 the reasonable answer is 9.38
If ! = 3! then ! = 3 9.38 = 16.25
The coordinates of the point of tangency of the follower are ( 9.38 , 16.25 )
If the camshaft rotated a total of 225°, at what point would the follower touch the cam?
(HINT: pay attention to where 225° lies on the unit circle in relation to other angles you
may know) Sketch a diagram and show your work.
225° is a 180° rotation from 45°,
therefore, it lies on the same line
so the same equation can be
used, however, this time the
follower is touching the cam lobe
at a point represented by the
equation ! ! + ! ! = 225. This is
an equation for a circle. By using
the same method of substitution
(setting the equations equal) the
coordinates of the point of
tangency can be found.
If ! = !, then we can replace any ! with an ! in the equation ! ! + ! ! = 225
! ! + ! ! = 225
2! ! = 225
! ! = 112.5
! = ± 112.5
! = 10.61!!"# − 10.61
because the follower is now in quadrant III, the x value of the coordinate must be
negative. If ! = ! then the x and y coordinates are the same, so the point of tangency is
at ( -10.61 , -10.61)