MV AND POWDER CHARGE, THE SLOPE AND EQUATIONS

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  • Last Post 30 July 2018
joeb33050 posted this 14 July 2018

Within the range 5.5-8.5 grains of Titegroup, the AVERAGE slope for 22-250 is 179 fps/gr; and for 223 is 205 fps/grain.

Individual slopes for cartridge/bullet weight/barrel length vary; more data is needed. 

A sample equation is shown after the tables.

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joeb33050 posted this 14 July 2018

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joeb33050 posted this 14 July 2018

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joeb33050 posted this 14 July 2018

Example:

?Estimated mv of 223, M12 OLD bbl, 7.5 gr Titegroup?

5.5 gr mv = 2058 fps

7,5-5.5 = 2 gr = delta gr Titergroup

2 gr * 203 fps/gr = 406 fps

2058 + 406 = 2464 fps = estimated mv

See test mv = 2503 fps @ 7.5 gr

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JimmyDee posted this 27 July 2018

JoeB,

 

There’s more than one way to do it.

 

You’re spending lots of time collecting and analyzing data; I’d like to suggest a couple tools you might find helpful.

 

Take a look at non-linear approximations of muzzle velocity v. charge.  Using a spreadsheet power model of the form y = mx^b applied to your data, I came up with a coefficient m value of 565.272 and exponent b value of 0.644655.  I plugged that into a spreadsheet and compared the values from the approximation (“Estimate&rdquo v. your data; the difference was within +/- 1% at your data points.  (In case you’re wondering, I also got an r*r value > .998.)

 

You can take that approximation to Google: type in “plot y = 565.272 * x ^ 0.644655” and you’ll get an interactive plot.  In the screen shot I’ve attached, I simply dragged the dot to 7.5 (i.e., a charge of 7.5gr) and the plot gives a y value (velocity) of ~ 2072.

 

 

If you want linear approximations around various charges, you can use the web to compute the first derivative of the approximation formula and evaluate that at various points.  (I’ve done that in the last two columns of the spreadsheet in the screen shot.)

 

Again, the web is your friend.  Go to Google, search for “compute derivatives,” and you’ll get a few results.  You’ll have to select one you like.

 

You may have to use their formula writer to plug in your formula and the site will probably “rationalize” your co-efficient and exponent.  (“Rationalize” simply means “give me a ratio of two integers which approximate a value with digits to the right of the decimal point.&rdquo

 

What you get back may look weird and have lots of zeros in it but your spreadsheet will understand it.

 

This derivative may be reported as [d/dx] but, for this purpose, it is “the derivative with respect to charge” and will give you the slope of the line tangent to the approximation when evaluated at various charges.  I’ve gone ahead and computed the y-intercept so line is described in slope-intercept form.

 

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  • Ross Smith
joeb33050 posted this 27 July 2018

 

Jim;

Thanks for the reply.

The data, “powder charge and mv”, looks pretty linear to me.

Excel, when asked, will apply a trend line to the graph, calculate a formula-based on the equation-type chosen, and give us R^2.

For the data in question, the linear equation is: y = 202.07 x + 967.29; R^2 =  .9941.

The exponential equation is:  y = 1304e^.0856x; R^2 = .9881.

The logarithmic equation is: 1397.3 ln (x)-322.81; R^2 = .9979.

The power equation is: y=752.56x^(0.5933); R^2 = .9966.

(Brings to mind carrying 100 pounds of Hollerith cards, at 3AM, to THE COMPUTER for a regression analysis, that might be complete ~ 5PM. Now its as fast as I can click.)

I suspect that the similar R^2 values have to do with a pretty linear set of data-non-linear data might/would have varying values of R^2 amongst the equations.

I had reviewed the linear equations, tested some, and found that subtracting 5.5 GR mv from 8.5 GR mv and dividing by 3 (GR) was close enough for me.

The data, “powder charge and mv”, looks pretty linear to me. However, I could be, and often am, wrong.

 

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Ross Smith posted this 29 July 2018

Very interesting work Joe. There are so many variables that are all inter-related that it is confusing. Is bullet stability more influenced by bullet length and rpm's?

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JimmyDee posted this 29 July 2018

The data may look linear but very few things in "the real world" are linear.

I chose the power model because, over a relatively small operating area, one can use the same model to examine pressure, too, which, for our purposes, is the constraint which must not be ignored.

It's difficult to get enough pressure data for fast powders under cast bullets to come up with reliable approximations (maybe QuickLoad can help -- but if you have that you don't need all this number putzing) but the maximum pressure v. charge curve probably looks like this

and we see that, as charge increases and we're making more and more heat, pressure increases at an increasing rate while velocity increases at a decreasing rate.

So I'm wondering: why are you messing around with linear models?

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joeb33050 posted this 29 July 2018

I've ever understood the LG rpm business, and I've tried.

RPM = mv (fps) X 720)/twist (") = (mv/twist) X 720. Since stability is a function of twist and mv, as is rpm; then stability is a function of rpm-all for a given bullet. But this is a tautology, nothing has been brought to the table.

Here's where I am on twist and stability:

 

 

TWIST, MV, STABILITY AND ACCURACY SUMMARY

 

As mv falls; bullets begin to tip and accuracy remains reasonable; then, as mv continues to fall, bullets print sideways and accuracy may remain reasonable

 

A 68 gr Hornady BTHP .22, .983”, Greenhill min. twist = 7.7”, shoots accurately in a Stevens 223 M200, 9” twist bbl., down to 1920 fps, where it begins to tip. Reasonable accuracy continues to 1490 fps.

 

Same bullet, Shilen 22-250, 9” twist, begins to tip at 1713 fps, reasonable accuracy continues 1340 fps, where bullets print sideways and accuracy is still reasonable.

 

(More examples in the EXCEL workbook)

 

So, the definition of “stability” looks to be sorta fuzzy, tipping bullets and sideways bullets can be accurate, and reasonable accuracy extends over a ~400 fps window.

 

Greenhill is about short, fat bullets. As bullets lengthen, get pointy, and acquire boattails, the Greenhill solution increases, twist # rises.

 

Greenhill is about bullet density. As alloy density decreases, as grease grooves increase effective size, the Greenhill solution decreases, twist # falls.

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Still; working on it, shooting 2-3 times a week.

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joeb33050 posted this 29 July 2018

So I'm wondering: why are you messing around with linear models?

Because most of the data so far looks pretty linear. I suspect that the energy/charge curve becomes non-linear somewhere < 5,5 and > 8.5 gr of Titegroup, but don't know. If so, I would expect mv to get non linear.

BUT, in the region 5.5-8.5, energy is pretty linear.

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joeb33050 posted this 29 July 2018

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joeb33050 posted this 29 July 2018

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joeb33050 posted this 29 July 2018

All the data I have is from uncomfortably small samples; 2 guns, 2 calibers, 3 bullets, 7 powder charges, lotsa variation.

d anything/d anything else is dead linear over some range. Maybe 5.5-8.5 is that range.

I'd be happy to send you the data,just pm your email address. Delighted to9 have some interest.

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Ross Smith posted this 29 July 2018

Thanks for the answer Joe.

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Ken Campbell Iowa posted this 29 July 2018

i have been confused that both velocity and energy out slopes can seem flat compared to energy ( charge ) input ....  since energy increases as a factor of the square of the velocity ..... ... usually ...

ken

 

 

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JimmyDee posted this 30 July 2018

"...most of the data so far looks pretty linear" because you're using appropriate charges of suitable canister powders.

The domain of charges and range of velocities denote an operating area.  As a consequence of design and application, part of that area is considered safe and, oftentimes, the operating curve in the safe area is rather linear.  Linearity gives us some slack: small differences in things we didn't consider -- in this case, ambient temperature or case volume or projectile mass -- are less likely to have a drastic effect on our results. The approximations I've plotted above are decidedly non-linear outside the safe operating area.

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joeb33050 posted this 30 July 2018

i have been confused that both velocity and energy out slopes can seem flat compared to energy ( charge ) input ....  since energy increases as a factor of the square of the velocity ..... ... usually ...

ken

I use (mv^2 fps * bullet weight grains)/450,240 = energy. Checked workbook against online energy calculator, no difference. I think the workbook is correct. We know that energy varies as the square of velocity, unless something changed recently, and that the slope curves up. However, over the mv with charges 5.5-8.5, energy is pretty linear. I'd like someone to check, mv  table is below.  

 

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joeb33050 posted this 30 July 2018

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JimmyDee posted this 30 July 2018

"[ d anything/d anything else is dead linear ]"  Of course: that's what a derivative is.  And evaluating a derivative at any point in a curve's domain gives a line which is tangent to the curve at that point -- unless the curve is linear; then, you get the slope of the line.

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