Editing Camshaft timing events, 0.006" to 0.050"
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*This poster asked me how to find the 0.050" camshaft timing events when you have the 0.006" timing events. *Originally Posted by CJ_1080 *Tech, I have to ask....how do you figure out the 0.050 value from the 0.006 value? Just wondering if there was a formula, or if you could *explain how you arrived at 30 I would apreciate the knowledge. *It's magic. No, here I'll explain it. *Draw two circles on a piece of paper, maybe 2-3" in diameter. The circle on the left will be intake and the circle on the right will be *exhaust. Inside the left circle, near the top, write in the 0.050" intake duration. We'll use 202 in this case. Draw a line under it and *write in half the intake duration (101). Inside the circle, near the bottom, write in the intake valve lift (0.550). *Using the face of an analog clock for reference, make a mark on the perimeter of the intake circle at 11 O'Clock and label it Intake Opens. *Make a mark at 7 O'Clock and label it Intake Closes. Make a mark at 4 O'Clock and label it Intake Centerline (ICL). *Inside the exhaust circle, near the top, write in the exhaust duration @0.050". We'll use 212. Draw a line under it and write in half the *exhaust duration (106). Inside the circle, near the bottom, write in the exhaust valve lift (0.546"). *Make a mark on the perimeter of the exhaust circle at 1 O'Clock and label it Exhaust Closes. Make a mark on the circle at 4 O'Clock and *label it Exhaust Opens. Make a mark at 8 O'Clock and label it Exhaust Centerline (ECL). *You need at least two values to begin, intake centerline and exhaust centerline. We'll use the timing card referred to in this thread *because we already have a link to it. You'll see that they give you the intake centerline (109), but not the exhaust centerline. They do, *however, give you the lobe separation angle (113). Since we know that the lobe separation angle is reached by adding the intake centerline *and exhaust centerline together and dividing by 2, we can do a little quick figuring in our head (add 4 to 109 to reach 113, so add 4 to 113 *to reach the exhaust centerline of 117) (109 + 117 divided by 2 = 113). *So, on your intake circle, write in the intake centerline 109 at 4 O'Clock. This is the mid-point of duration of the cam. Put another way, *this is halfway through the intake duration period and the point of highest lobe lift. Earlier, you wrote in the intake duration and half *the intake duration at the top of your intake circle. We will now travel from the intake centerline point, counter-clockwise back toward top *dead center, the number of half the intake duration that you wrote down. In this case, that would be 101 degrees. Now, because the intake *centerline is at 109 degrees after top dead center and a point 101 degrees counter-clockwise back toward top dead center from the 109 point *would be at 8 degrees after top dead center. So, at your 11 O'Clock mark, you will write in (-8), showing a dash for negative, that the *intake is opening after top dead center in this case instead of before top dead center. Looking back at the circles shown in the Comp timing *card on the link, you will see that if we were plotting the timing points at 0.006" tappet rise, the intake would be opening at 17 degrees *before top dead center. But, since we are checking at 0.050" tappet lift, the valve opens later, on the other side of TDC. *OK, now we have the intake opening point at 8 degrees after top dead center. From that point, clockwise to bottom dead center, will be 172 *degrees. Referring to the numbers you wrote inside the circle, the total intake duration is 202 degrees. Now, if we subtract 172 from 202, *we find that we have to place the number 30 at the 7 O'Clock mark on the circle, because we now know that the intake closes at 30 degrees *after bottom dead center. Now, prove your figures. 30+172=202. *There, we have all the info we need on the intake circle. Now, let's move to the exhaust circle. *We have already figured that the exhaust centerline is at 117 degrees before top dead center, so write 117 in at your 8 O'Clock mark on the *exhaust circle. Inside the circle, close to the top, write in the 0.050" exhaust duration of 212. Draw a line under it and write in half the *duration, 106. *Now, if we begin at 117 degrees BTDC and move clockwise up toward TDC by half the duration (106), then we will stop at 11 degrees this side *of TDC. That would be the exhaust closing point (-11), so write -11 in at your 1 O'Clock mark. From that point of 11 degrees BTDC, counter-*clockwise around to bottom dead center would be 179 degrees (180 less 11). If we subtract 179 from the known exhaust duration of 212, we get *33, so write 33 in at your 4 O'Clock mark for the exhaust opens event. To prove our figures, 33+179=212. *This is a very simple operation and you will do it almost by second nature after you have done a few of them. I still have to write *everything down, but you guys who are sharper than I am will be able to do it in your head.
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