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• # Crank Torque/Piston Force Question

Discussion in 'Calculations' started by Idiotic, Jun 29, 2012.

1. ### IdioticNew Member

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Hi all,

Before I throw myself out of the office window could someone please help me out a bit here...I'm sure I used to handle these sort of calcs years ago but now I'm not sure of anything!

Got a pneumatic press module here and I'd like to calc the force on the piston at 1mm from TDC, I've drawn it out here as clear as possible and it's like this: A formula I came across in many of the books I've been looking through said Fp = T/OM
I worked out OM as OM=OC(Sin(17.13+4.34)/Sin(90-4.34); then transposed and came up with Fp=3935N

Could some one please let me know if I'm along the right lines or am I going mad?

Many thanks

I've just previewed the post and it looks a lot like a school/college homework question...I can assure you it's not, although that doesn't make me feel any better about my maths!

2.
3. ### ErichWell-Known Member

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I approached the problem a different way, but I got the same answer. Go with it.

4. ### IdioticNew Member

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Phew, that's good to know. Thanks Erich

I'd be interested to know the way you approached it - I originally did it a different way, got a crazy answer which completely threw me and then it went downhill from there!

5. ### ErichWell-Known Member

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I just broke the problem down and drew Free body diagrams.

Calculated the Force in the conrod as a function of the piston force.

The I went up to the crank and applied that force to the crank. I broke that force into a component normal to the crank and a component in line with the crank.

Came up with equation relating Torque to Conrod force.

substituted and solve for Fp vs T

6. ### IdioticNew Member

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Got it...finally! Thanks for explaining your methods and also thanks very much for taking the time and trouble to help me out, it is very much appreciated.

7. ### Auto EngineerActive Member

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Looking at the diagram but applogies in advance if anyone thinks I am slightly off the mark, but the obtuse triangle appears to me that the length I would not equate to 70mm?

8. ### Auto EngineerActive Member

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Sorry my mistake should have worked it out first

9. ### Auto EngineerActive Member

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Yep I was thinking about something should be done with the value of I, but I found this method works as well;

Working with trig, POC triangle OC = 18mm, theta = 17.13 degress, Phi = 4.34 degress.

Using the sine rule we have; PC = 18(sin 17.13/sin4.34) = 70.06 which is I.

OCM triangle = angle C = Phi + theta = 21.47 degress
POM = 90 degress

Angle m = 85.66 degress

Using the sine rule we get;

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

So Fp = 26 / 6.61 x 10^3 = 3.94 x 10^3 N

10. ### BobInOHNew Member

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A simpler approach to solve is: the applied moment is constructed of vectors, which are the applied force, Fp, and the moment arm, r, or, T = r X F = r * F * sin(angl OCP).
Since the sum of angles in a triangle is 180 degrees, angle OCP = 180 - 4.34 - 17.13 = 158.53 degrees.
So Fp = T / r sin(158.53) = 26 Nm / [18mm * sin(158.53)] = 26000 Nmm / (18mm * 0.3660) = 3946.42 N. QED!

11. ### Auto EngineerActive Member

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