Weapon Accuracy for Devices

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GaryChildress
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Weapon Accuracy for Devices

Post by GaryChildress »

I swear I asked this once before, maybe not. I can't find it when I do a search of all threads I've contributed to, but what determines accuracy for devices? So for instance the 16" Mk7 gun has an accuracy of 20 in the editor. How was this number arrived at? What formula is used or whatever? I'm trying to add some German guns. I was thinking of just mimicking their accuracy with their closest Allied equivalents but I don't know if that is a good idea or not.

Thanks.
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Symon
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RE: Weapon Accuracy for Devices

Post by Symon »

ORIGINAL: Gary Childress
I swear I asked this once before, maybe not. I can't find it when I do a search of all threads I've contributed to, but what determines accuracy for devices?
The code.
So for instance the 16" Mk7 gun has an accuracy of 20 in the editor. How was this number arrived at? What formula is used or whatever??
You start by graphing what already exists. Then you do a curve fit on the data (it will take you 3 iterations). Once you get it down, the rest is brainless, just plop in your device specs and viola, ...
I'm trying to add some German guns. I was thinking of just mimicking their accuracy with their closest Allied equivalents but I don't know if that is a good idea or not.
Not
Thanks.
Ciao. John
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inqistor
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RE: Weapon Accuracy for Devices

Post by inqistor »

ORIGINAL: Gary Childress

I swear I asked this once before, maybe not. I can't find it when I do a search of all threads I've contributed to, but what determines accuracy for devices? So for instance the 16" Mk7 gun has an accuracy of 20 in the editor. How was this number arrived at? What formula is used or whatever?
For most Naval Guns it is Rate of Fire per Minute times 10.
So, 16" Mk7 have 20, because it had Rate of Fire equal to 2 per minute.
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Symon
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RE: Weapon Accuracy for Devices

Post by Symon »

ORIGINAL: How was this number arrived at? What formula is used or whatever?
Not a good idea to use 10 x RoF. That was back in UV, WiTP and the AE initial release. It caused so many problems, that we decided enough of that nonsense.

Michaelm adjusted the accuracy code and we changed everything back in 2010. Naval and AA accuracy are both changed to take advantage of the new algorithm. I calculated a curve that worked within the algorithm’s sweet spot. Did a spread of 3 iterations – high, low and center, with all three pegged to the same spot, at 16”. Data was very simply extracted from one or another of the curves. This is what exists today.

There is a newer system that performs individual calculations on each gun and it seems to represent historical capabilities better, as well as give smoother results. The resulting data fits into the spread of the graphical curve more reasonably and actually models “better” guns as having better characteristics.

This is how it’s done: Calculate “nominal” RoF (very often less than maximum). Then calculate “factor” RoF = 2x(Sqrt)Nominal. Add Nominal and Factor. Multiply by shell sectional density. Multiply by shell diameter ratio (using 16.1” basis, i.e., 16.1/(shell diameter)). Divide by 2 = Value. If you want to get really fancy, you can factor in crh and L ratio (ratio of ballistic length as a function of tube length compared to 16.1” basis); this is especially useful for representing the Japanese “short” guns.

A very simplified version of this is what was used to generate the graphical curves. You just need to do it backwards; do the curves and pick off your gun stats. [:)] Believe me, they are way different from WiTP and the initial release of AE.[;)]

Listening to little clones using 2006 RHS data is noooot the greatest idea I ever heard.[8D]
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Symon
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RE: Weapon Accuracy for Devices

Post by Symon »

Btw, this is all coming to you at a BigBabes theater in your neighborhood, soon. JuanG, I got the spreadsheets.

Ciao, John
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GaryChildress
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RE: Weapon Accuracy for Devices

Post by GaryChildress »

ORIGINAL: Symon
ORIGINAL: How was this number arrived at? What formula is used or whatever?
Not a good idea to use 10 x RoF. That was back in UV, WiTP and the AE initial release. It caused so many problems, that we decided enough of that nonsense.

Michaelm adjusted the accuracy code and we changed everything back in 2010. Naval and AA accuracy are both changed to take advantage of the new algorithm. I calculated a curve that worked within the algorithm’s sweet spot. Did a spread of 3 iterations – high, low and center, with all three pegged to the same spot, at 16”. Data was very simply extracted from one or another of the curves. This is what exists today.

There is a newer system that performs individual calculations on each gun and it seems to represent historical capabilities better, as well as give smoother results. The resulting data fits into the spread of the graphical curve more reasonably and actually models “better” guns as having better characteristics.

This is how it’s done: Calculate “nominal” RoF (very often less than maximum). Then calculate “factor” RoF = 2x(Sqrt)Nominal. Add Nominal and Factor. Multiply by shell sectional density. Multiply by shell diameter ratio (using 16.1” basis, i.e., 16.1/(shell diameter)). Divide by 2 = Value. If you want to get really fancy, you can factor in crh and L ratio (ratio of ballistic length as a function of tube length compared to 16.1” basis); this is especially useful for representing the Japanese “short” guns.

A very simplified version of this is what was used to generate the graphical curves. You just need to do it backwards; do the curves and pick off your gun stats. [:)] Believe me, they are way different from WiTP and the initial release of AE.[;)]

Listening to little clones using 2006 RHS data is noooot the greatest idea I ever heard.[8D]

Wish I had the math competency I had coming out of highschool but my math is about as rusty as my Spanish, "no comprende" is about all I know now. lol

I guess it's pretty much the kiddie pool for me and this mod. I'll probably have to stick with something simpleminded like RoF x 10 (which means I'll probably have to recalibrate the existing data to match the new). Though, I do appreciate the attempt to educate me, JWE. I apologize to waste your time. I didn't realize the calculation is as complex as it is. Its reassuring to know that real thought has been put into stock AE and into DaBabes. Kudos! [&o]
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JuanG
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RE: Weapon Accuracy for Devices

Post by JuanG »

ORIGINAL: Symon
This is how it’s done: Calculate “nominal” RoF (very often less than maximum). Then calculate “factor” RoF = 2x(Sqrt)Nominal. Add Nominal and Factor. Multiply by shell sectional density. Multiply by shell diameter ratio (using 16.1” basis, i.e., 16.1/(shell diameter)). Divide by 2 = Value. If you want to get really fancy, you can factor in crh and L ratio (ratio of ballistic length as a function of tube length compared to 16.1” basis); this is especially useful for representing the Japanese “short” guns.

Hey John, very interesting.

Any chance you could just clarify a couple of things regarding this formula;
Sectional density - I presume this is in lbs / sqin?
L ratio - is this in caliber lengths or physical length?

I'm curious about the choice of some of these variables - a lot of them are more directly related to ballistic accuracy rather than firecontrol accuracy, so I'm left wondering if this is deliberate.

Also - just from some back of the envelope math (waiting on answers to the above units before I mess with it too seriously), it seems like some weapons might be a little undervalued by this approach. In particular, the British 15in/42 which had some of the best dispersion characteristics of any large caliber gun in the war seems to be victim.

Would really appreciate hearing your thoughts (and if you're willing to share that spreadsheet).
Juan
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Symon
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RE: Weapon Accuracy for Devices

Post by Symon »

ORIGINAL: JuanG
Hey John, very interesting.

Any chance you could just clarify a couple of things regarding this formula;
Sectional density - I presume this is in lbs / sqin?
L ratio - is this in caliber lengths or physical length?
Hi Juan. Thanks.
Yep, secden is shell weight/diam^2. L ratio is in calibers, so you have to convert your shell ballistic length into its caliber basis.
I'm curious about the choice of some of these variables - a lot of them are more directly related to ballistic accuracy rather than firecontrol accuracy, so I'm left wondering if this is deliberate.
Yes it is. One must start somewhere and I thought the tube/round combo was a good place to begin. If ya put a stake in the ground it's good that everything bats from the same box. These thingies are 'devices' and I thought it appropriate to treat them as such. So made an algorithm to treat the 'devices'. The game code doesn't have any way of representing fire control, except as a "chunky" update to Allied capability in certain "years". Very unfortunate, but that's how the stock executable works.
Also - just from some back of the envelope math (waiting on answers to the above units before I mess with it too seriously), it seems like some weapons might be a little undervalued by this approach. In particular, the British 15in/42 which had some of the best dispersion characteristics of any large caliber gun in the war seems to be victim.
Well, yes, perhaps they are. The algorithm is just a basis to get everyone in the same box. It doesn't (can't) allow for the skill of the batter. As you know very well, there's tubes and then there's tubes; and every once in a while somebody gets a tube and a shell design together that hits the sweet spot (and sometimes, it's an utter roach). The algorithm simply provides a consistent and reproducable starting point for individual weapon adaptations. I tweak the algorithm, myself, for those weapons I am familiar with. Folks like yourself, that know other weapons can (and should) do the same. The algorithm is not cast in stone, it's a self-consistent starting point from where folks can depart without wondering "what do I do with this crap, Batman."
Would really appreciate hearing your thoughts (and if you're willing to share that spreadsheet).
Juan
All I have is yours, my friend. I'm finalizing a PI scenario for porting into AE but soon as that's done, I'll complete the spreadsheet (right now, only Japan and a third of US is done). Btw, it has the complete algorithm as column math and has not a few individual devices manually adjusted, in red.

Regards. John
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JuanG
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RE: Weapon Accuracy for Devices

Post by JuanG »

Gary, if its of any help, the simplest form of the equation John posted can be condensed to;

Accuracy = Image

where
W is shell weight in pounds,
D is bore diameter in inches,
T is nominal rate of fire in rounds per minute.

This excludes the barrel L ratio and the shellform crh ratio.

So for example, the US 16in/50 Mk7 would work out as;
W = 2700
D = 16
T = 1.8
Therefore A = 30.3

The German 38cm/52 C/34 (Bismarck main battery) would work out as;
W = 1764
D = 14.96
T = 2.3
Therefore A = 28.8

If you factor in the L ratio, just multiply the 38cm/52 by 1.04 (52/50) for an A of 30.
With the shellform crh ratio, the 38cm/52 is multiplied by 0.98 (4.4/4.5) for an A of 29.4.

John,
I find the choice of T + 2sqrt(T) for the RoF factor interesting - I use T^0.34 as part of my own formula, and our final numbers are not massively different (~30 vs ~39 for the 16in/50 Mk7) - the main factor being that my factor tends to favour slower firing (ie battleship) guns slightly more than your formula does.

Its also very intriguing that the W/D^3 factor pops up if one simplifies your equation - this is nice because its both a key ratio for shell design as well as being easy to calculate. Its also fairly constant for shells of the same design generation and nation.
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MateDow
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RE: Weapon Accuracy for Devices

Post by MateDow »

ORIGINAL: JuanG

Gary, if its of any help, the simplest form of the equation John posted can be condensed to;

Accuracy = Image

where
W is shell weight in pounds,
D is bore diameter in inches,
T is nominal rate of fire in rounds per minute.

This excludes the barrel L ratio and the shellform crh ratio.

Where did the constant of 10.25 come from? Did I miss that in the explanation earlier?
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JuanG
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RE: Weapon Accuracy for Devices

Post by JuanG »

ORIGINAL: MateDow

ORIGINAL: JuanG

Gary, if its of any help, the simplest form of the equation John posted can be condensed to;

Accuracy = Image

where
W is shell weight in pounds,
D is bore diameter in inches,
T is nominal rate of fire in rounds per minute.

This excludes the barrel L ratio and the shellform crh ratio.

Where did the constant of 10.25 come from? Did I miss that in the explanation earlier?

It comes from simplifying the constants in the equation. These are 16.1 (from the caliber basis), Pi (from the sectional density calculation), 4 (also from the sectional density calculation, a consequence of converting R^2 -> (D/2)^2 -> (R^2)/4 ) and 1/2 (from the 'divide by 2' final operation).

Equation wise;
Initial equation;
Image

Convert R^2 -> (D^2)/4, 4 moves to top;
Image

Rearrange constants;
Image

Resolve constants = 10.2495 = 10.25.
Image
Buck Beach
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RE: Weapon Accuracy for Devices

Post by Buck Beach »

ORIGINAL: JuanG

ORIGINAL: MateDow

ORIGINAL: JuanG

Gary, if its of any help, the simplest form of the equation John posted can be condensed to;

Accuracy = Image

where
W is shell weight in pounds,
D is bore diameter in inches,
T is nominal rate of fire in rounds per minute.

This excludes the barrel L ratio and the shellform crh ratio.

Where did the constant of 10.25 come from? Did I miss that in the explanation earlier?

It comes from simplifying the constants in the equation. These are 16.1 (from the caliber basis), Pi (from the sectional density calculation), 4 (also from the sectional density calculation, a consequence of converting R^2 -> (D/2)^2 -> (R^2)/4 ) and 1/2 (from the 'divide by 2' final operation).

Equation wise;
Initial equation;
Image

Convert R^2 -> (D^2)/4, 4 moves to top;
Image

Rearrange constants;
Image

Resolve constants = 10.2495 = 10.25.
Image


Uhhh, what ever happened to "you push the little value down and the music goes round and round and it comes out here"[:D][:D]
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Symon
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RE: Weapon Accuracy for Devices

Post by Symon »

ORIGINAL: JuanG
John,
I find the choice of T + 2sqrt(T) for the RoF factor interesting - I use T^0.34 as part of my own formula, and our final numbers are not massively different (~30 vs ~39 for the 16in/50 Mk7) - the main factor being that my factor tends to favour slower firing (ie battleship) guns slightly more than your formula does.
Juan, you are a man after my own heart. Oh, yeah, I wish to gosh we could have done it right and then tweaked the code to match. But we had to work with the legacy algorithm, that required some 'squeezing' of the data to fit the game curve sweet spot.
Its also very intriguing that the W/D^3 factor pops up if one simplifies your equation - this is nice because its both a key ratio for shell design as well as being easy to calculate. Its also fairly constant for shells of the same design generation and nation.
I may not be the brightest Nav Gun guy in the box (in fact, I think you are far more knowledgeable) but I was trained at Sill and I do know my ballistics. [;)]

Thanks for all your help and comments. John
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Dili
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RE: Weapon Accuracy for Devices

Post by Dili »

Rate of fire wasn't relevant in BB and in practice was lower since some guns missed the salvoes.
GaryChildress
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RE: Weapon Accuracy for Devices

Post by GaryChildress »

Thanks for the explanation JuanG! [&o]
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MateDow
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RE: Weapon Accuracy for Devices

Post by MateDow »

Are there empirical methods that are used for calculating the ammunition for guns as they are inserted?

It looks like Vanilla uses the following...

Large Caliber - 1 point per 10 rounds
Medium Caliber - 1 point per 19 rounds

I know that there has to be a better method to calculate this.

Any suggestions?
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MateDow
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RE: Weapon Accuracy for Devices

Post by MateDow »

ORIGINAL: Symon
This is how it’s done: Calculate “nominal” RoF (very often less than maximum). Then calculate “factor” RoF = 2x(Sqrt)Nominal. Add Nominal and Factor. Multiply by shell sectional density. Multiply by shell diameter ratio (using 16.1” basis, i.e., 16.1/(shell diameter)). Divide by 2 = Value. If you want to get really fancy, you can factor in crh and L ratio (ratio of ballistic length as a function of tube length compared to 16.1” basis); this is especially useful for representing the Japanese “short” guns.

ORIGINAL: JuanG
If you factor in the L ratio, just multiply the 38cm/52 by 1.04 (52/50) for an A of 30.
With the shellform crh ratio, the 38cm/52 is multiplied by 0.98 (4.4/4.5) for an A of 29.4.

How is crh ratio used? Do I compare the crh of the shell I'm trying to model vs the Japanese 16.1"?
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JuanG
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RE: Weapon Accuracy for Devices

Post by JuanG »

ORIGINAL: MateDow

ORIGINAL: Symon
This is how it’s done: Calculate “nominal” RoF (very often less than maximum). Then calculate “factor” RoF = 2x(Sqrt)Nominal. Add Nominal and Factor. Multiply by shell sectional density. Multiply by shell diameter ratio (using 16.1” basis, i.e., 16.1/(shell diameter)). Divide by 2 = Value. If you want to get really fancy, you can factor in crh and L ratio (ratio of ballistic length as a function of tube length compared to 16.1” basis); this is especially useful for representing the Japanese “short” guns.

ORIGINAL: JuanG
If you factor in the L ratio, just multiply the 38cm/52 by 1.04 (52/50) for an A of 30.
With the shellform crh ratio, the 38cm/52 is multiplied by 0.98 (4.4/4.5) for an A of 29.4.

How is crh ratio used? Do I compare the crh of the shell I'm trying to model vs the Japanese 16.1"?

I actually used the USN 16in 2700lber shellform ratio (4.5) for the Bismarck calculation, since it meant I didn't have to apply it to both. In retrospect the choice of the IJN 16in 2249lbs with crh of 4.24 is probably the original intent.

John, any chance you could clarify which you intended as the basis? Also, should the L and crh basis be applied before or after the divide by 2 bit? I applied them afterwards, but that was in the interest of simplification for the example.
Mac Linehan
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RE: Weapon Accuracy for Devices

Post by Mac Linehan »

Original:

Equation wise;
Initial equation;


Convert R^2 -> (D^2)/4, 4 moves to top;


Rearrange constants;


Resolve constants = 10.2495 = 10.25.

End

I think my head hurts...<grin>

Good Stuff; thanks for the under the hood look, Gents!

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MateDow
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RE: Weapon Accuracy for Devices

Post by MateDow »

ORIGINAL: JuanG

I actually used the USN 16in 2700lber shellform ratio (4.5) for the Bismarck calculation, since it meant I didn't have to apply it to both. In retrospect the choice of the IJN 16in 2249lbs with crh of 4.24 is probably the original intent.

Have you found a good site that lists crh and information like ballistic coefficient for naval shells? I looked on NavWeaps, and they have some references, but it isn't consistent. I love NaAB for calculating penetration, but not all of the shells have that data, so I have been trying to interpolate based on era and such.


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