To say it with Bodo Bach: “I’d like to have a problem there:” I have now noticed several times how much the cross-section (landwise called “voltage”) of a drive chain changes when the motorcycle springs in the rear, i.e. a chunk of my caliber sits on it. That this is due to the fact that the pivot point of the swing and pivot point of the chain sniper, i.e. the drive and the swing axis, lie apart, is clear to me. I have just finished a 650 Enduro and their If this enduro stands unloaded on the side stand and the chain cross-section is about 40-50 mm (difference of top and bottom point) the chain is tight when I load the bench with more than 100 kilos. I therefore ask the math geniuses here, how to explain it or calculate it. I was never a math genius, but somehow I think that the distance of the drive-back axle is greatest when the drive-wing and rear axle are defined on a line. This is the case most likely in the springed state. If the planer is springed out again, the rear axle wanders in a circular arc-shaped movement from this escape downwards, whereby the distance from the output axis to the rear axle decreases, or am I wrong? Does it have something to do with the size of the pinion and the chain wheel? But they remain constant in all positions… I think that these are minimal values, but that would have to be calculated if you could calculate the Distan The length of the tangent of the circular arc motion of the HA is almost 10 cm. On the other hand, you also need only a few mm at the chain tensioner to get from 40 mm game to 70-80. If the chain of the 650 has a play of 30-40 mm at load, the cross-hang in spring-out condition is already almost dramatic. Can someone explain this to me, better click on how to calculate this ?