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The Troll Whisperer (Moderator)
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15" bbl = 2500 fps


25" bbl - 2850 fps


Guess the old rule of thumb of 30 fps +_ per inch of bbl gain/loss of velocity holds true.
 

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It works for that example. It is more fps/in with a short barrel and less with a long barrel. Cut an inch off a 32" Palma barrel and you get only around 15 fps velocity loss. It's more fps/in with a small expansion ratio, less with a large expansion ratio. Cut an inch off a 24" barrel chambered for .22 Long Rifle and you may see no change in velocity at all. Indeed, velocity may even increase slightly as the powder has burned, the gas has expanded and started to cool by expansion and the bullet has often begun to be slowed by long barrel friction (long in terms of expansion multiples rather than inches or calibers)).

One of the things I like about QuickLOAD is it gives you the option to see pressure versus time or displacement.
 
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OK, so I am a little over two years late on this one, but the Leduc equation is
V = ax/(b + x) where V is the velocity, x is the distance travelled by the projectile and 'a' and 'b' are constants

so, how did you get the constants a and b from the information above?
 

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Ah yes, but the value of C was not mentioned in the OP.

As for 'k', I did a similar run-around many years ago in my "Ballistics Handbook" to get 'a' and 'b' in terms of loading values.

Actually, in the original paper by Challeat [1] where the Leduc system was described, some forms were given whereby 'a' and 'b' could be derived based on propellant properties and loading conditions (no need for the chamber pressure and muzzle velocity). But they were empirical forms based on French powders and they were derived with little support from thermodynamic theory. They were described as “fallacious” by Hunt [2]. They were modified by Alger [3] for use with US Navy propellants, for which they were used until 1942. Serabryakov [4] modified them for Soviet propellants. But they have never proved to be of much use in a general sense.

But it remains true that the Leduc equation can be made to agree with the experimental values very accurately for a given gun, once the two constants have been found, as has been shown for large guns [5] and for small arms (see, for example [6]). As a result, the Leduc formula for velocity as a function of distance travelled was used in the determination of some internal ballistic properties in the United States right up until the 1970s [7]. These days, the Leduc equation is probably the best known internal ballistics equation and is often cited in the popular media.

It is, however, possible to derive the two constants of the Leduc equation in terms of the loading conditions and propellant properties, based on sound thermodynamic principles, (provided certain assumptions are made), and create a very neat, simple - and surprisingly accurate - internal ballistics system....

[1] CHALLEAT, J. Theorie Des Affutes à Deformation. (in French) Rev. d’Art, Vol. LXV. 184-186, 1904/5.
[2] HUNT, F. R. W. Internal Ballistics The Philosphical Library, 1951, p. 141-143
[3] ALGER, P. R. The Le Duc Velocity Formula. US. Nav. Inst. Proc. Vol. 37, No. 138, June 1911, p. 535-540
[4] SERABRYAKOV, M. E. Interior Ballistics, Air Technical Intelligence Translation, Wright-Patterson Air Fore Base, Ohio, 1968, p. 712
[5] PATTERSON, G. W. The Le Duc Ballistic Formulae. US. Nav. Inst. Proc, Vol. 38, No. 143, September 1912, p. 885-892
[6] WEBSTER, A, G, On the Springfield Rifle and the Leduc Formula Proc. Nat. Ac. 345 Sci, Vol. 6, 1920, p. 289
[7] KRIER, H. Interior Ballistics of Guns. Progress in Astronautics and Aeronautics. Vol. 66, 1979, P. 43
 

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Discussion Starter · #10 · (Edited)
You are the first to notice in two years, including me. I have several of those references.
I vaguely recall that the board had a prohibition against listing specific loads.
 

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Solve for k
I would if I could.

But I don't see how even when given all the variables. Charge Weight (C) was not specified, but let's say 50 gr.
So we can now get a value for the scary part of the equation.
But in the end we get (2 + k)^2 / k = "some value"
I don't see how we can extract "k" from that. Even accounting for the other relations that include k.
There is always another unknown variable, in this case the "travel to peak pressure".
Note: I'm guessing 336 is a constant?

I'm going to reference an older thread of yours here:
It seems there was an attachment (in your first post) that's no longer available.
In your second post you're detailing the F2 formula, and the term (k) pops up but there's no formula provided for the "bullet travel to peak pressure". I'm guessing/hoping it was shown in the attached picture.

So basically is there an equation for "k" and thus a way to obtain Powley's F2 value?
Because I can't see it from the info provided in these two threads.

Thanks.
 

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Discussion Starter · #12 ·
Solve for k:

101488
 
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@F2G1D

Oh boy. It was a quadratic equation, but somehow I missed it the first time around.
Talk about brain throttling... and rusty synapses.

OK so using the values in the OP and a 50 gr charge weight, I got k = 10.55.
From that, Xp = L/k = 21.8/10.55 = 2.066 in
And then F2 using the formula given in your other thread.

The problem with Powley's F2 term is that his formula for pressure involved F2 as an input.
But to get F2 via k you need pressure as an input. And you get a circular reference.
So... what now? Oh right, back to interpolating the F2 table. ROFL
Seriously is there a way out of this maze?

I'm asking because I'm sure that by interpolating the F2 table I get really close to the actual number that Powley built into the slide rule, so to speak, although from what I've gathered that table, given by Davis, is sort of a reverse engineered "Powley-Leduc formula" because it was never present in the slide rule. The slide rule's ER range is 4 - 14, whereas the ER range for the table is 5 - 13. So these numbers in the table are obviously rounded off which creates a minuscule degree of error. The back of the slide rule (the insert) has velocity figures. Which themselves look rounded off in order to make them appear more aesthetic (which is understandable). I suppose that from those velocity figures the F2 table was "reverse engineered".
Getting to the source would be interesting, the results would be practically the same, +/- a few fps most of the time, but it would look cleaner and it would bypass all the shortcuts and rounded off figures.
 

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Discussion Starter · #14 ·
This solves that problem:

F2 = (0.01558)(11.65 - C/W)(2.340 + E)
 
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@F2G1D

Thank you!
I have two other equations that both use MR and ER, I'll check them all and see which gives the tightest groups, so to speak.
 

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Discussion Starter · #16 ·
I created F2 using first principles. There is another one due to Miller based on the table in Davis.
 

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@F2G1D

Do you have Miller's formula? If so could you share it, please?

I might have it without knowing it's his because regarding the two formulas I have I don't know the authors, I only know the sources.
The first formula is from an Excel spreadsheet on an ammo site and the second formula is from an old post on a different forum.
Right so here they are:

F2 = 1.85 + 0.225 * (1 - A) + (R - 9) * [0.19 + 0.025 * (0.2 - A)]

F2 = (1 - A / 9.5) * [0.25 + (0.2213 * R) - 0.00206 * (R^2)]

A = Mass Ratio
R = Expansion Ratio
 

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So, here is my take on this:

Leduc's equation is

101501


where V is the velocity, x is the distance travelled up the barrel, a and b are constants. At large values of x, V will tend to a, so a is the maximum velocity in an infinitely long barrel. Since we have the muzzle velocity, we can write,

101502


where x here is the distance the bullet travels to the end of the barrel.

The acceleration up the barrel can be written as,

101503


The maximum pressure will occur at the point of maximum acceleration. So, we can differentiate again to find the maximum acceleration

101504


It follows that at the point of maximum acceleration,

101505


The acceleration of the moving masses (projectile plus half the charge weight) will be directly related to P, the maximum pressure, where,

101506


A is the area of the bore, g is the acceleration due to gravity, m is the projectile mass and C is the charge weight.

Re-arranging in terms of b,

101516


Substituting for a,

101515


x here is the distance the bullet travels to the end of the barrel.

Define a constant k, where

101517


To be continued in the next post:
 

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then

101518


or

101512


b can now be solved directly by taking roots, so that

101513


Now, plugging in the numbers, assuming a charge weight of 50 grains, (x here is the distance the bullet travels to the end of the barrel)

b is .5496 feet and a is 3582 ft/sec.
 

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Discussion Starter · #20 ·
@F2G1D

Do you have Miller's formula? If so could you share it, please?
F2 = (0.024075)(9.3 - C/W)(1.071 + E - 0.009736E^2), Don Miller
 
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