- #1
Ole SeaBee
- 14
- 0
Hello,
This is for sure not a 'Homework' problem. I have no clue! I understand that a crankshaft looses power from a piston because the force is pressing the Crank outward verses around it's shaft. I have no clue as to how to figure this out. I have looked all over the web and could find nothing that I could understand well enough to come up with the figure I need. When you start talking sin cosine I'm totally lost.
I can give the positions of the Crankshaft and the Rod Arm, I know the abgles and lenghts I'm working with. The Crankshaft has a R of 5.25" It is at 22.5 degrees above horizontal. The Rod length is 8.6875" and is at 13.3 degrees above the horizonal. If my 'piston is pushing on the Crank with 1000 lbs of force, How much of this force is going into turning the crank or not into tearing the crankshaft apart?
Thanks,
ed aka Ole SeaBee.
This is for sure not a 'Homework' problem. I have no clue! I understand that a crankshaft looses power from a piston because the force is pressing the Crank outward verses around it's shaft. I have no clue as to how to figure this out. I have looked all over the web and could find nothing that I could understand well enough to come up with the figure I need. When you start talking sin cosine I'm totally lost.
I can give the positions of the Crankshaft and the Rod Arm, I know the abgles and lenghts I'm working with. The Crankshaft has a R of 5.25" It is at 22.5 degrees above horizontal. The Rod length is 8.6875" and is at 13.3 degrees above the horizonal. If my 'piston is pushing on the Crank with 1000 lbs of force, How much of this force is going into turning the crank or not into tearing the crankshaft apart?
Thanks,
ed aka Ole SeaBee.