Of course that said, it is a rough estimate to be used for the sake of comparison and not in any way meant to be used to obtain exact figures.
Thanks! Some people will take things too literally. He took your ballpark estimate of double/half as 2.0/0.50.
Appears also to be 2015, published in the 4th month according to the link.
Nope. If you look at the lubes there, they are the same lubes as in D5293-08, not D5293-15. Therefore, they didn't get a chance to update to the latest standard at the time of their publication. The lubes in D5293-15 are entirely different. The problem with the numbers in their catalog is that they are all over the place, between 1.7 and 2.3. Taking the average with such random numbers doesn't help. However, if you look at the numbers in D5293-15, DV-35/DV-30 = 1.677, 1.711, 1.724, 1.742, 1.760, 1.791, and 1.813 for CL130 - CL200, respectively, and they are also uniformly increasing. Clearly these lubes are systematic and they have fixed a problem in D5293-08.
Moreover, as I stated before, using 2.00 instead of 1.75 would favor thinner oils over thicker oils when the base-oil-quality index (BOQI) is calculated. Therefore, I'm not trying to bias the calculations in the direction of thinner oils as you may be thinking. What's the point of doing all this analysis if you're not even going to be honest to yourself? It would be a great waste of time doing so and where is the fun of doing that when you aren't learning anything from your results because you doctored them and can't trust them?
When I was at school, 1.96 rounded up, but I guess things have changed, yet again.
Ha? I wasn't even using a calculator and I just estimated the average to be around 1.9 when looking at your data. I don't think we were even looking at the same numbers. Why bring this up in this manner?
LOL, you and the rounding errors...14% is clearly 20%
Re NOACK...
If the repeatability is 0.81%, and the reproducability is 1.62% (both in terms of actual total evaporation), then 9% could be either 8%, or 10%...there's no "two significant figure" accuracy in that band.
Are you kidding?
Let's use the scientific notation:
14 = 1.4 x 10^1 (two significant figures) = 1 x 10^1 (one significant figure) = 10 (one significant figure)
So, you need at least two significant figures in this case. "Significant figure" does not mean that the figure has 100% accuracy. You can have a +/- 7% error in 14, which makes it in the range 13 - 15. This doesn't mean that you should reduce it to one significant figure and express it as 10. That would make no sense and represent a +/- 50% error instead of the actual +/- 7% error. (Also note that when you do calculations, you never round off until the final answer.)