Unit 23: Applied mathematics for engineering
LO1 and 2
Analysis of a cam
Instructions and answers for teachers
These instructions should accompany the OCR resource ‘Analysis of a cam’ activity which supports
Cambridge Technicals in Engineering Level 3.
Version 1
The Activity:
This task involves the analysis of a rotating cam and requires knowledge from LO1 and LO2.
Learners must develop the required formulae and evaluate these with various parameters. This has been designed to
stretch the learners understanding and application of 2-D geometry.
The spreadsheet Anal_Cam_Data is provided that will allow cam dimensions to be specified; the spreadsheet will
then perform the necessary calculations and plot a graph of the cam lift for corresponding angles of rotation.
This activity offers an
This activity offers an
opportunity for English
opportunity for maths
skills development.
skills development.
!
Supporting
documents:
!
Formulae Booklet for Level 3 Cambridge Technicals in Engineering
(available from: http://www.ocr.org.uk/qualifications/cambridge-technicals-engineering-level-3-certificate-extendedcertificate-foundation-diploma-diploma-05822-05825/)
Spreadsheet Analy_Cam_Data
Suggested timings:
Task 1: 1 hour
Task 2: 1 hour
Task 3: 1–2 hours
Activity 1
Solution
We start this solution by generalising the geometry of the cam by replacing the radius of the heel with
variable r1, the radius of the nose with variable r2 and the distance between the centres of the heel and
the nose with variable d. If we can find solutions to the problem in terms of these variables then we could
insert the values 16, 12 and 14. In this way we could easily change the values to represent any cam with
the same configuration but with different dimensions.
The diagram below shows the generalised cam including the angle α between the horizontal centre line
and the line OA.
Version 1
A
r1
B
α
α
O
r2
d
In order to determine the length of the flank, D, and the angle α, a new (rotated) partial diagram showing
the main geometry may be helpful.
A
r1
r1 - r2
D
α
O
B
r2
d
The following should now be obvious from this diagram.
cos  
r r
r1 r 2
and so   cos1 1 2
d
d
sin  
D
and so D  d sin
d
The total perimeter is the length of AB (i.e. D) plus the length of arc BD plus the length of CD (this is the
same as the length of AB) plus the length of arc AC.
Perimeter = D  2r2  D  (2  2 )r1  2( D  r2  (   )r1 )
 2(d sin  r2  (   )r1)
With r1 = 16, r2 = 12 and d = 14
α = 1.2810 (73.3985 o), D = 13.4164, perimeter = 117.1154 mm.
Version 1
Learners should be encouraged to enter these equations and values into an Excel spreadsheet and
confirm the results. Values of r1, r2 and d could then be changed for a cam with different dimensions.
Version 1
Activity 2
The total area of the cam is the sum of the heel sector, the nose sector and the central section, half of
which is shown in the rotated diagram above.
Area of heel sector = (   )r1
2
Area of nose sector = r2
2


Area of central section = 2 Dr2 
D(r1  r2 ) 
 = d (r1  r2 ) sin
2

You should confirm that an alternative expression for this final area is tan  (r1  r2 ) although this will not
2
2
be valid for r1 = r2
With r1 = 16, r2 = 12 and d = 14: Total Area = 1036.43014 mm2.
Again it would be useful to enter these formulae into an Excel spreadsheet with various parameters and
check that the values obtained are reasonable.
Activity 3
The answers to these parts will depend upon which section of the cam is directly above the point O.
There are four sections: line AB, arc BD (the nose), line CD and arc AC (the heel). We need to discover
the angles, θ, for which these sections apply.
It should be clear that for any angle θ from 0 to


  and for any angle from   to 2 the heel is
2
2
directly above point O and therefore the lift will simply be zero.
For this particular cam these angles correspond to 0 to 16.6016o and 163.3985o to 360o
(i)
Since 250o is within the range 165.5225o to 360o lift = 0
Version 1
For angles greater than

  but less than the angle at which the section above point O changes from
2
line AB to arc BD, the lift will vary depending upon angle θ. We need to discover the angle β in the
following diagram.
When heated and subject to other processing (controlled oxidation and carbonization) these honeycomb
shaped crystals organize themselves in long flattened ribbons.
A
B
r1
β
O
α
r2
d
Using this diagram you should confirm that:
OB2  D2  r12
and
 OB 2  d 2  r22 

2OBd


  cos 1 
So, for

2
   

2
  the line AB will be directly above point O
With r1 = 16, r2 = 12, d = 14 and β = 33.4178o these angles correspond to 16.6016o and 56.5822o
You should now confirm that within this interval the distance between O and the point on line AB directly
above is
r1
cos(   
Version 1

2
or
)
r1
and therefore:
sin(   )


1
 1
 sin(   ) 
lift = r1 
(ii)
Since 45o is within the range 16.6016o to 56.5822o, using the formula above lift = 2.1888 mm.
For angles between 56.5822o and 90o, the nose section is directly above point O.
Analysis will show that the distance L between the point O and the point on the nose directly above O
satisfies:
r22  L2  d 2  2 Ld cos(

2
  ) or r22  L2  d 2  2Ld sin( )
Rearranging this as a quadratic in L gives:
L2  bL  c  0
where
b  2d sin( )
c  d 2  r2
2
This leads to:
L
 b  b2  4c
2
Using only the positive square root expression:
Lift =
 b  b 2  4c
 r1
2
(iii)
80o is within the range 56.5822o to 90o, and so using the formula above:
lift = 9.5385 mm.
Since the cam is symmetrical the lift for any angle 90o + xo ≤ 180 will be the same as the lift for angle 90o
– xo
(iv)
130 = 90 + 40 and 90 – 40 =50
Since 50o is within the range 16.6016o to 56.5822o, using the appropriate formula:
Version 1
lift = 3.1648 mm.
Note that for 180 ≤ θ ≤ 360 the heel is directly above point O where lift = 0
(v)
This graph has been produced using the spreadsheet ‘Cam_analysis’.
Lift v Angle of rotation
12
Lift (mm)
10
8
6
4
2
0
0 20 40 60 80 100120140160180200220240260280300320340360
Angle of rotation θo
We’d like to know your view on the resources we produce. By clicking on ‘Like’ or ‘Dislike’ you can help us to ensure that
our resources work for you. When the email template pops up please add additional comments if you wish and then just
click ‘Send’. Thank you.
If you do not currently offer this OCR qualification but would like to do so, please complete the Expression of Interest Form
which can be found here: www.ocr.org.uk/expression-of-interest
Version 1
OCR Resources: the small print
OCR’s resources are provided to support the teaching of OCR specifications, but in no way constitute an endorsed teaching method that is required by the Board,
and the decision to use them lies with the individual teacher. Whilst every effort is made to ensure the accuracy of the content, OCR cannot be held responsible
for any errors or omissions within these resources.
© OCR 2015 - This resource may be freely copied and distributed, as long as the OCR logo and this message remain intact and OCR is acknowledged as the
originator of this work.
OCR acknowledges the use of the following content: English and Maths icon: Air0ne/Shutterstock.com; OS Map © Ordnance Survey 2015.
Please get in touch if you want to discuss the accessibility of resources we offer to support delivery of our qualifications: resources.feedback@ocr.org.uk
Version 1