Imagining that 12:00 is TDC, 1 revolution of the crank from TDC to TDC is a symmetric operation. With the piston at BDC (6:00) and the piston stop in place, the piston will encounter the stop at the same height in the bore (distance from TDC) whether the crank is turned CW or CCW. For example, say the stop is installed in such a way that starting at BDC and going CW, the piston contacts it at 10:00. Conversely, going CCW the piston will contact it at 2:00. The midway point, or 12:00, would then be TDC. Basically the piston stop prevents you from completing the circle. The midpoint we're interested in is the midpoint of the uncompleted arc (TDC), not the midpoint of the completed stop-to-stop arc (BDC).
Note that you don't have to start with the piston at BDC, but if done this way the amount of CW crank rotation will be different from the amount of CCW rotation. However, that doesn't matter because the piston stop will still contact the piston with the piston at the same distance from TDC regardless of rotational direction. Brad > I think I need the hand puppets and/or flash cards, because I'm still not > getting how TDC is between the tops of the two upstrokes. Wouldn't that be > the bottom? You're drawing random marks and saying TDC is between them? I'm > obviously just not getting this. Well, the cork and straw method never lies. > You also never mentioned how you know which stroke you are on when the > piston hits the stop, so it sounds like a good way to end up with your > distributor (the reason this thread started) 180 degrees out roughly 50% of > the time. > > John "lost cause" Nasta ------------------------------------------------ Get the award winning ISP, AT&T WorldNet Service http://download.att.net/webtag ----------------------------------------------------- To Unsubscribe please visit www.chevelles.net/list.html To start a new topic, send mail to: [EMAIL PROTECTED]
