Quote:
Originally Posted by GeorgeWiseman
No, the 'lie' math has absolutely nothing to do with wide open throttle. The math is for normal driving with normal throttle use. The math simply takes the weight of air that ACTUALLY goes through the engine and compares it to the weight of fuel that ACTUALLY goes through the engine during NORMAL vehicle operation.
You can confirm this by using your own scan gauge on your own vehicle just like I show you. I never mention anything about wide open throttle.
Can you tell me why you think I'm talking about wide open throttle?
|
I mention wide open throttle because your VE is not going to be 80% at part throttle cruise. It will be much less. From this blog, "At part throttle conditions, like cruising down the road with the throttle plate nearly closed, the volumetric efficiency can be as low as 20%."
Performance Trends Blog » Better than Perfect
Using this referenced website,
Volumetric Efficiency (and the REAL factor: MASS AIRFLOW), by EPI Inc. I calculated the VE requirement for some assumed numbers, like HP needed cruise at 60 mph and BSFC.
Then, using your math and fuel flow rate, I came up with a much different air/fuel ratio. Also, at -2* F, which your video showed, the weight of air, according to Wiki
Specific weight - Wikipedia, the free encyclopedia is different than what you used in the video. You can see my spreadsheet:
Granted, there are a few assumptions, but you can see that with your math, with the exception of VE, I got an air/fuel ratio that the rest of the world would consider "normal".
Here is a snapshot of my air/fuel ratio from my drive to work this morning. This isn't a wideband, but even taking into account there could be some error in accuracy, the air/fuel ratio is right where everyone but yourself thinks it should be.
Can you explain how my data logged A/F ratio is false?
I took two semesters of calc and two semesters of calc based physics in college. I am hardly an expert on any of this and would never claim that I know a lot on the subjects at hand (I have no idea if your math is correct, either...). But, using your references and assuming your A/F calculation method is correct, I was able to come up with a much different A/F ratio. Please explain.