MHB Is there an alternative method to solve this problem

  • Thread starter Thread starter Casio1
  • Start date Start date
  • Tags Tags
    Method
AI Thread Summary
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.
Casio1
Messages
86
Reaction score
0
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

  • Crank Angle Torque Diagram.png
    Crank Angle Torque Diagram.png
    2.7 KB · Views: 139
Mathematics news on Phys.org
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:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top