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12 Mar 2004, 10:42 (Ref:903039) | #1 | ||
Racer
Join Date: Feb 2004
Posts: 146
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Crankshaft angle-stroke
Is there a formula for the calculation of crankshaft angle relating to piston movement.
4" Stroke - bottom dead centre 180 deg . tdc 360 deg, piston halfway 2"- 90 or 180 degree But what about the angles in between. eg if piston up 3.5", what is crank angle? Lister |
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Rattie,on the road with Mole.Beep,beep! |
12 Mar 2004, 15:44 (Ref:903381) | #2 | ||
Racer
Join Date: May 2003
Posts: 233
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Lister
You might regret asking this! The position of the piston also depends on the length of the conrod and the height of the piston crown from the little end. If the length of the conrod is "l", the stroke "s", the height of the piston crown from the little end "c" and the angle of the crank "a" (I can't do theta on this forum!), then the position of the top of the piston crown from the centreline of the crankshaft is: distance = c + s cos(a) + squareroot ( l squared - (s sin(a))squared) As my brain is addled with pupils hyped up with the snow, I can't see how to simplify the equation any further at the momemt. HTH Duncan PS On a more design orientated note, if you look at the equation you can see how having a long rod increases the time available for combustion by accelerating the piston slower past TDC. This is always traded off against the increase in reciprocating weight and the overall height of the engine in practice. PPS Good job you didn't ask for the version for a desaxe engine! |
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12 Mar 2004, 16:22 (Ref:903418) | #3 | ||
Racer
Join Date: Feb 2004
Posts: 146
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Thamks Duncan,
I've received the same formula from my Lotus Engineering pal. He said he'd work it all out for me in the garage here on Sunday. What a pal!!!!!! Lister |
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Rattie,on the road with Mole.Beep,beep! |
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