MHB Is there an alternative method to solve this problem

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The discussion revolves around calculating the force on a piston in a rotating cylinder setup, using a force diagram and trigonometric methods. Initial calculations yielded a force of approximately 3933 N, but the final refined calculation determined the force to be around 3935.1 N. Participants emphasized the importance of ensuring calculators are set to degrees rather than radians for accurate results. Additional calculations related to the cylinder's dimensions, volume, and pressure were also mentioned. The conversation highlights the complexity of the problem and the potential for alternative methods to arrive at the solution.
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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?
 

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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.
 
Sudharaka said:
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?
 
Sudharaka said:
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
 
Ackbach said:
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
 
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:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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