Welcome to our community

Be a part of something great, join today!

Trigonometry Is there an alternative method to solve this problem

Casio

Member
Feb 11, 2012
86
With reference to the Force Diagram, which represents a single piston cylinder rotating and producing 26NM of torque.

The technical data given is;

I = 70mm
r = 18mm
T = 26Nm
Theta = 17.13 degress
Phi = 4.34 degress

Using the diagram calculate the force on the piston Fp at 1mm before TDC.

Interestingly I can't solve this using the above data, but using trig in conjunction with the above I calculated the following angles;

Triangle OMC

M = 85.66 degress
O = 72.87 degrees
C = 21.47 degress

I worked it out like this;

PC = 18(sin 17.13) / (sin 4.34) = PM = I = 70.06 mm

The length (I) has now been proven.

Using the sine rule;

OM = 18(sin 21.47) / (sin 85.66) = 6.61

Force on piston Fp = 26 / 6.61 x 1000 = 3933 N

Once you follow it through you will see that I have not used much of the original data to solve the problem, but I am thinking there is an alternative method because I am to believe that the actual solution is

Fp = 3935 N

Although it is only two newtons difference I could be using the wrong techniques?
 

Attachments

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi Casio, :)

Casio said:
PC = 18(sin 17.13) / (sin 4.34) = PM = I = 70.06 mm
How did you equate PC and PM?

Kind Regards,
Sudharaka.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hi Casio, :)



How did you equate PC and PM?

Kind Regards,
Sudharaka.
Building on that, did you have your calculator set to use degrees and not radians?
 

Casio

Member
Feb 11, 2012
86
Hi Casio, :)



How did you equate PC and PM?

Kind Regards,
Sudharaka.
Hi Sudharaka:)

Thanks for replying to the thread, I have worked out a full solution now to the problem, so will post it soon.

Kind regards

Casio
 

Casio

Member
Feb 11, 2012
86
Building on that, did you have your calculator set to use degrees and not radians?
Hi Ackbach,

I didn't adjust the calculator to radians as the unit measurements were given in degress, but I have solved it and will post full solution soon.

Kind regards

Casio
 

Casio

Member
Feb 11, 2012
86
Sorry for not replying earlier:D

I have worked on the problem and this is what I have now calculated.

With reference to the force diagram, which should be drawn as a rectangle to find the angles required.

The length of OM is;

OM = r 18(sin21.47) / (sin 85.66) = 6.6072mm

The length of "I" is calculated as;

I = PC = 18(sin 17.13) / (sin 4.34) = 70.0594 mm

The length OP is calculated as;

OP = Sqrt PC^2 - OM^2 = 70.0594^2 - 6.6072^2 = Sqrt 4856 = 69.69mm

Length of OP = OC + OP = 18 + 69.69 = 87.69mm

Force Fp = (r x F x sin(158.53))

Fp = T / r (sin 158.53) = 26 / 6.5883 = 26K / 6.5883 = 3946.4N

The compoent force acting on the piston is calculated from;

Fc = cos 4.34 = (3946.4 x cos (4.34)) = 3935.1N

I have moved on from this now and also calculated other areas like;

Diameter of the cylinder, area of the cylinder, swept volume of the cylinder, clearance colume of the cylinder, compression ratio of the cylinder, and the compression pressure:)

Casio:cool: