copemech Posted April 25, 2007 Author Report Share Posted April 25, 2007 (edited) ScottT, If you will please notice that the factory "tic" mark(noted as 4) has rotated approx 4mm counterclockwise from the(STD) position alignment with the right edge of the casting boss. To plot the exact amount of piston travel vs crankshaft rotation you would need to plot on a parabolic curve. At exactly top or bottom center, there is NO movement of the piston in relation to the crankshaft. Piston travel is maximum in relation to degrees of rotation at 90 and 270 degrees from TDC and declines in either direction from that point. (almost) Edited April 26, 2007 by copemech Quote Link to comment Share on other sites More sharing options...
john.b Posted April 26, 2007 Report Share Posted April 26, 2007 (edited) Agree with Scott T. To retard the ignition loosen the bolts and rotate the backplate counter clockwise. Page 12 of 16. sherco ignition guide. Edited April 26, 2007 by scudo Quote Link to comment Share on other sites More sharing options...
copemech Posted April 26, 2007 Author Report Share Posted April 26, 2007 At this point, I no longer care which way you rotate! Quote Link to comment Share on other sites More sharing options...
ikb Posted April 26, 2007 Report Share Posted April 26, 2007 I've got it! How about a lever on the bars with a cable that goes to the stator which can advance or retard the timing as desired But I do think its been done before (1922!)! Quote Link to comment Share on other sites More sharing options...
scottt Posted April 26, 2007 Report Share Posted April 26, 2007 ScottT, If you will please notice that the factory "tic" mark(noted as 4) has rotated approx 4mm counterclockwise from the(STD) position alignment with the right edge of the casting boss. Ok thanks, its just your diagram was just a little bit confusing as to which way you had turned it, perhaps arrows might have been better. To plot the exact amount of piston travel vs crankshaft rotation you would need to plot on a parabolic curve. No, the piston moves 60mm that is its stroke, as it moves fully up and down for every revolution of the crank. 1 revolution equals 120mm of piston movement 60mm up 60mm down, thats how 2 strokes work, so if its moving 120mm for 360deg of revolution then every 3deg of revolution the piston must move 1mm Quote Link to comment Share on other sites More sharing options...
marlintec Posted April 26, 2007 Report Share Posted April 26, 2007 ...so if its moving 120mm for 360deg of revolution then every 3deg of revolution the piston must move 1mm Certainly not. Here you have the exact relationship and an approximate equation very useful for trials. http://www.terra.es/personal9/marlintec/ta...ientopiston.pdf Cheers, JM. P.D. Great job, Cope! Quote Link to comment Share on other sites More sharing options...
scottt Posted April 26, 2007 Report Share Posted April 26, 2007 Sorry I cant understand what that site is saying but now i see it i realise it isnt as straight forward as i thought. Quote Link to comment Share on other sites More sharing options...
copemech Posted April 26, 2007 Author Report Share Posted April 26, 2007 (edited) Unfortunately, the simple theory of height vs rotation does not work, mostly because the piston accellerates and decellerates twice per revolution at a given crankshaft speed. Here is a short study to relate this, makes my brane hurt! piston speed Gee, I should not have slammed this up in a hurry, as Marlintec already had a better timing diagram and formula, you do not have to read Spanish, just math! Spanish is easier! Edited April 27, 2007 by copemech Quote Link to comment Share on other sites More sharing options...
brian r Posted April 27, 2007 Report Share Posted April 27, 2007 Quote Link to comment Share on other sites More sharing options...
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