Qverlap question
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Qverlap question
A search really didn't get me what I am looking for. I think Im about 80% there learning about cams and cam theory. I get the straight forward stuff, now its onto the tougher questions and just for my general knowledge and curiosity I have a question about Overlap in a Camshaft. I see it referred to as what gives the chop/lope I.E more overlap means a rougher idle. I also understand overlap refers to the time in degrees of rotation that the intake and exhaust valve are both open. My Question is what is the effect of a negative overlap?
I recently did a cam swap from a -1 degree overlap cam 224/230 114lsa
to a +15 degree overlap cam 238/240 112lsa
The reason I ask is because however my idle is more radical is wasn't what I expected going from the first cam to the second. They're fairly close sounding.
I have a bunch of questions to add to this main one but maybe if someone could explain what exactly overlap will theoretically change I might can gather from that what I am trying to figure out.
Thanks in advance. If I need to stop taking up space with my theory questions just let me know and I will delete this.
I recently did a cam swap from a -1 degree overlap cam 224/230 114lsa
to a +15 degree overlap cam 238/240 112lsa
The reason I ask is because however my idle is more radical is wasn't what I expected going from the first cam to the second. They're fairly close sounding.
I have a bunch of questions to add to this main one but maybe if someone could explain what exactly overlap will theoretically change I might can gather from that what I am trying to figure out.
Thanks in advance. If I need to stop taking up space with my theory questions just let me know and I will delete this.
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I understand the LSA and all. Im just not understanding a negative overlap. or what I would interpret as one degree short of actually overlapping. I actually asked him to make it choppy for me. And it is much choppier than it was just not right off that bat. You have to really listen to it and on video it sounds very close to what it did before. Thats not a problem just curious why since I added 16 degrees of overlap
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I understand the LSA and all. Im just not understanding a negative overlap. or what I would interpret as one degree short of actually overlapping. I actually asked him to make it choppy for me. And it is much choppier than it was just not right off that bat. You have to really listen to it and on video it sounds very close to what it did before. Thats not a problem just curious why since I added 16 degrees of overlap
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I agree. You can tune a 15 deg overlap cam to sound like a 0 overlap cam. You can also tune the 0 overlap cam to be very smooth, yours just wasn't, apparently.
there are lots of timing and airflow tables that are used to make a car idle smoothly. bottom line is, All things equal, the bigger cam will have more trouble holding a steady idle RPM, and that's all lope/chop is.
there are lots of timing and airflow tables that are used to make a car idle smoothly. bottom line is, All things equal, the bigger cam will have more trouble holding a steady idle RPM, and that's all lope/chop is.
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I think its a case here of a better tuner (my guess tune with the bigass cam drove better than the final with the small one) a few more RPM at idle, (i think its roughly about 100 RPM higher), and maybe I'm just crazy and want the thing to hit like my life long goal of building a motor to sound like a Nascar car. Which I'm in the process of doing btw. It definitely pulls harder and longer. It definitely drives a bit bigger. and revs quicker also. Not sure it thats in the tune also. Could be due to coming off of a higher idle speed im guessing. Im not complaining at all by any means more just extremely curious as to just exactly how overlap works
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The only thing I wanted to add was to make sure you are aware that the -1* overlap is measured at 0.050" lift. At 0.006" lift the overlap on that same cam will be much higher. In fact, I think even the tamest of cams will have a pretty significant positive overlap at 0.006.