denton
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What's your R^2 with just an X term?
Velocity as pressure sign
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denton
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Ah, my friend, please take no offense, but it is extremely linear. An R^2 of .985 says you have an excellent fit to a straight line.
In Multiple Regression, you add new variables cautiously because the more variables you put in the model, the more shakey it becomes. With only 1.5% of the variation unexplained by a linear model, adding higher order terms to the model adds risk, but practically no new information. I usually won't add another term unless it lifts the R^2 by at least 5 or so, and there isn't room for higher order terms to do that.
With X^2 and X^3 terms in the mix, you introduce multicolinearity, soundly violating the fundamental assumptions of the tool. You may get an R^2 of 1, but you have a busted model.
So.... we had fun. I don't often get conversations like this, and I do very much respect what you post.
Well.... by eyeball inspection of the velocity points themselves alone, the curve is rolling over.
(One does not need a regression analysis drawn to see that. Such analysis only
describes the shape of the best-fit curve as it conforms in mathematical terms.)
Note that the positive ^3 term still causes velocity to rise with increasing pressure.
All the negative ^2 term does is control that rise with slight rollover within the interval.
If there's a problem, it's not with the curve fit/description, but rather the internal ballistics
algorithms at work in QuickLoad -- again within that interval. The key there is there realization
that final velocity is dependent upon total area under the pressure curve as the bullet proceeds
down the barrel, not the pressure peak alone, and....
the fact that Burn rate/pressure is neither constant nor linear, but rather rises then falls during
burn progression with modern powders. (higher pressure --> faster burn --> sooner roll-off)
Much like a helicopter... `lotta moving parts going in totally different directions -- but all trying to tear things apart.
~~~~~~ Again, UncleNick weigh in on reality here
~~~~~~~~
(One does not need a regression analysis drawn to see that. Such analysis only
describes the shape of the best-fit curve as it conforms in mathematical terms.)
Note that the positive ^3 term still causes velocity to rise with increasing pressure.
All the negative ^2 term does is control that rise with slight rollover within the interval.
If there's a problem, it's not with the curve fit/description, but rather the internal ballistics
algorithms at work in QuickLoad -- again within that interval. The key there is there realization
that final velocity is dependent upon total area under the pressure curve as the bullet proceeds
down the barrel, not the pressure peak alone, and....
the fact that Burn rate/pressure is neither constant nor linear, but rather rises then falls during
burn progression with modern powders. (higher pressure --> faster burn --> sooner roll-off)
Much like a helicopter... `lotta moving parts going in totally different directions -- but all trying to tear things apart.
~~~~~~ Again, UncleNick weigh in on reality here
~~~~~~~~
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denton
Member
You are absolutely correct that MV is driven by the area under the pressure curve. However, the area under the curve, in turn, is driven by the peak pressure as I explained earlier. I've run data for several firearms, always with the same result.
Bear in mind that you have built a regression model of QuickLoad's model of ballistics... a model of a model. And you have soundly violated the assumptions of the regression model in so doing.
There might be some curvature under the statistical noise in real-world data, but if it's there, it's so tiny that it would take quite a bit of data to establish its presence.
I had a lovely data set from Ken Oehler that covered a broader range than any I have done personally, and that popped an R^2 of 99.9%, leaving only .1% of the variation to be explained by the sum of curvature, measurement error, and all other sources of variation.
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redcon1
member
Hodgdon reloading tells me that 42 gr of Varget will launch a 168 gr sierra HPBT 2,386 fps at 41,200 CUP.
My AR10T will launch a 168 gr AMAX at 2,567 fps with 40.8 gr of Varget.
Hodgdon, annoyingly, is using a 15" barrel as their test gun. I'm using a 24" barrel.
So, because I have 9 more inches of barrel, I exceeded their velocity by 181 fps with 1.2 gr less powder. Did I also exceed 41,200 CUP? I'd have to say, no way. In fact, I'd guess that my peak pressure is actually less than 41,200 CUP. The longer barrel creates more "area under the pressure/time curve". The force of the powder is acting on the bullet for a longer period of time and thereby accelerating it more. I have exceeded their velocity at a lower peak pressure.
Not hardly.
What is revealed is the inner characteristics of the QuickLoad Model -- which is 3(+)-order polynomial at its roots.
The least squares methodology shows that empirically... without ever having access to the multifactor equations themselves.
denton
Member
No, not not hardly. Heartily!
You're one of the people here who has the background and skill to understand what is going on. So I'll take one more swing at explaining why your model is severely busted.
In normal algebra, we're trying to solve for the values of the Xs. In regression (curve fitting) we are trying to solve for the coefficients of an expression of the form Y = a0 + a1X1 + a2X2 + a3X3.... In your case X1 is your X^3 term, X2 is your X^2 term, and X3 is your X term.
The required assumption is that X1, X2, and X3 are independent, or, if you prefer, orthogonal like the X Y and Z axes. The fundamental math of regression runs on variances, and if this condition is not met, then X1 contributes to the X2 and X3 variances, X2 to X1 and X3, etc. One of the ways to test for this is to calculate the Variance Inflation Factor. For small data sets, a VIF of 5 or more signals trouble. If your software calculates VIF, you might want to see what it is telling you. If it doesn't calculate it, then it is leaving you adrift.
With your data set, the contributions of the exponential terms are small. What that means is that your exponential terms practically overlay your linear term. In other words, they are far from independent but are highly correlated, or colinear. That busts the tool you are trying to use, causing non-credible results, and it frequently produces negative coefficients, such as X^2, where there should be none.
For a plethora of reasons, most reputable curve fitting software doesn't even use terms above X^2. And if you already have an R^2 of .985, it is unwise to continue adding terms. More terms = more wobbly model, and virtually no additional information. There are tools where that is not the case, but it is very much the case with regression.
Trust me on this. I co-wrote a major statistical application, and wrote a couple of books on the topic.
In the real world, I have run many data sets like the 8mm data I showed earlier. I have never found curvature. The hardest thing is to find a black cat in a dark room especially if there is no cat.
So is there curvature or not? There might be. We don't know. Under the circumstances, it is improper to assert that it is there. If it is there, it makes a very tiny contribution and a simple linear model gives a more credible result that has practically all the information that can be extracted from the data, and it is proper to say that the data are linear.
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denton
Member
LOL.... Well, we've had about as much fun with this as we are going to have.
A straight line fits the data exceptionally well. Any time you get R^2 = .985, that is true. Adding more terms is bringing coals to Newcastle, and, under the circumstances, creates a worrisome multicolinear model. Perhaps a polynomial does fit the data better, but if it does, you have no idea which polynomial, because you grossly violated the assumptions of the tool.
With real world data, I typically get R^2 of .98-.99 for a straight line fit, unless I load into the plateau region. Other than that, I have never detected curvature in my MV vs. pressure data. If it were there, I suspect that I would have detected it, but then again, maybe not. I can't prove that it doesn't exist, but as far as I know, no one has proved that it does. Without evidence, I think it is improper to assert that the curve is cubic.
Keep well..... Another day. It was a good wrestle.
A straight line fits the data exceptionally well. Any time you get R^2 = .985, that is true. Adding more terms is bringing coals to Newcastle, and, under the circumstances, creates a worrisome multicolinear model. Perhaps a polynomial does fit the data better, but if it does, you have no idea which polynomial, because you grossly violated the assumptions of the tool.
With real world data, I typically get R^2 of .98-.99 for a straight line fit, unless I load into the plateau region. Other than that, I have never detected curvature in my MV vs. pressure data. If it were there, I suspect that I would have detected it, but then again, maybe not. I can't prove that it doesn't exist, but as far as I know, no one has proved that it does. Without evidence, I think it is improper to assert that the curve is cubic.
Keep well..... Another day. It was a good wrestle.
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