| Vegas69 |
03-26-2009 02:17 PM |
Nice...we tried to look it up but he couldn't find it. I think you are onto something with the humidity. I went back and looked at the weather Tuesday at around 2 PM when the car was dynoed. I used a calculator that uses temp vs dew point and the humidity should have been 15%. When I got home that day I checked and it was 15%. The lowest it got was 11% on Tuesday. http://www.wunderground.com/weathers...y=24&year=2009 I thought the HG was off but it's a calculation based on absolute pressure not barometric. I looked at my motorcycle dyon from two years ago and it's right on the money. I also found that the correction factor is based on 0% humidity. So if the humidity was really 15% and the computer used 5% the correction would be smaller than it should have been. I took it one step further and found a calculation. http://wahiduddin.net/calc/calc_hp.htm With the correction factor at 5% humidity, my engine would have made 93.1% of the power it would have created with perfect conditions or the standard. With 15% it would have made 92.8% of perfect conditions or the standard. So only .3% difference. Still shy but right at about 499 ftlbs. One more degree of timing and I'm there easy.:D
Just for fun...in Huntington Beach today I would be making 102.2% of the standard. 66 degrees/58% humidity/3 feet of elevation They would actually have to decrease the numbers. So I would have 9% more power today over there than here in the seat of the pants wich would be the equivelant of 44 ftlbs and 44 hp. California is good for something. LOL
Now if I was in Denver, CO today, I would only be making 87% of the standard thus losing 6%. So I gain more power driving to the coast than I lose going to Denver! It's 19 degrees, 90% humidity, and a mile high. I would lose approx 30 horse and ft lbs driving to Denver today. Little to cold for my taste. |
