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Power Strokes
7.3 Aftermarket
Peak Torque High vs Low RPM
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[QUOTE="TARM, post: 317185, member: 578"] Sorry that is what I meant. I was in a rush and was going to quickly. I meant distance not work. Since this has not really gone the direction I thought it would I will try to post my thoughts on it At first I was looking at it that you have same compression pressure curve to produce the same amount of force ( tq) say 1000ftlb whether it be at a low rpm, say 2000, as you do at a faster RPM of 4000. These are only representational rpm figures. So it took 3500psi of CP at each cylinder to create the 1000ftlb TQ Nothing there is going to change from a low rpm to high. Then if you add in all the strain of higher rpms it looked to me that actually the same tq @ higher rpms created more stress. As I started kicking it around a bit more I began to consider the CP and the actual crank revolutions and angles along with considering the actual instantaneous tq effects. Then I started to look at the actual combustion curve and time and how its spread out over the crank degrees of revolution, angles etc.. That is when things started to come together better or get more confusing. Considering that no matter 2000rpm or 4000 rpm the speed of the actual rate of combustion is relatively happening at the same speed. Yet at the same time the crank and rods are moving in this example either twice as fast (4000rpm)or half as fast(2000rpm). Consider what that means for the force applied to that piston, rod and crank at the two speeds. The slower the rotating assembly is moving (low RPMs)the greater the TQ spike each part in the assembly is having to deal with as its all happening in fewer crank degrees of movement. When its moving faster (higher rpms)that pressure curve while still taking close to the same amount of time is now being spread out over more crank degrees as the assembly is moving that much faster. Just throwing some numbers out as examples only. Let say the pressure curve @ 2K is barely taking say 20 degrees ATDC to complete but then at 4-5K that same pressure curve is now spread out over say 40 degrees of rotation. Those are just examples to help illustrate the explanation and are not meant to be correct in ratio of degrees to rpms. This lead me then to start thinking about how the 4 stroke cycles played into all of this as you have one power stroke per 720 degrees at each cylinder. So, even though TQ is not suppose to be a factor of time a 8 cyl 4 stroke engine has to go thru 2 rev or 720 degrees to complete a full power stroke cycle. So TQ as we measure it for engine performance is really an average created over that cycle. With CP @ crank angle I would think instant tq could be calculated. I would have to research to figure out the formula but figure it is doable and likely not that complex. Regardless the conclusion I came to is the lower the rpms the less combustion cycles you have for a given period of time. You also have that much more time between each combustion event and cycle. Yet you still are creating the same average torque @ that rpm. That would mean that the TQ spikes absorbed by the rotating assembly would be that much greater as they have to create the same average tq. As rpms increase the number of cycles increases and the time between them decreases there for it would create lower tq psikes to create the same average tq. I guess the best way to frame this is that the stresses per cycle of revolution is greater at lower RPMS than at higher for the same mean tq level. Now I have gone back and forth on this as I look at it from various points. The focus for me is from a stand point of stress on the individual parts of the rotating assemble per combustion cycle. For our engines rods and their fasteners seem to be the weak link so I focus on that and try to visualize the effects there. I am not sure if I am looking at this correctly or if I have over thought it too much to where I have got it all twisted. I may have not explained in the correct manner but I am hoping that those with more expertise in these areas can understand what I am trying to convey. Dave, Jason, Charles and many others of you that I am sure have far more knowledge about this kind of stuff than I. [/QUOTE]
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