Dispersion Angle for rapid fire WW2 aircraft guns

[Scherf]

Originally posted by hitech
You are totaly miss informed and full of dog dodo.

HiTech

sig material at last!

[HoHun]

Hi Crumpp,

>I have some ballistic diagrams for the Mk 108 I could dig out and send if you would like.

That would be great! :-)

>A surprisingly flat shooting weapon in comparision.

You're quite right - it's often forgotten that depressing the sight line (or elevating the bore axis) will help to make the weapon hit close to the aiming point  - which is just what the shooter wants.

When aiming right at the centre of the target without considering bullet drop makes the bullet strike the target, this is considered "point blanc range" I believe. Using your Ta 152 example, the MK108 really has a point blanc range of more than 400 m, which is not what you'd expect from the comments on trajectory curvature.

The point is that trajectory curvature begins to have a serious impact only at ranges far beyond what was normal engagement range in WW2.

Regards,

Henning (HoHun)

[gripen]

Originally posted by HoHun

You're quite right - it's often forgotten that depressing the sight line (or elevating the bore axis) will help to make the weapon hit close to the aiming point  - which is just what the shooter wants.

In the deflection shooting, the shooter does not know the aiming point, it just had to be quessed if there is no gyroscopic sight or other similar device.  

gripen

[Crumpp]

The point is that trajectory curvature begins to have a serious impact only at ranges far beyond what was normal engagement range in WW2.

Exactly!

I will scan those documents and email them to you soon.

All the best,

Crumpp

[Charge]

"The total energy of mine shells is preserved pretty well downrange because it's primarily a function of its explosive content."

I'd say it fully retains its energy as upon impact the velocity it hits the target has no significance at all, its the chemical energy that counts. Even the shrapnel does not get any added energy of the impact speed as its only a very thin steel core with negligible mass.

"the MK108 really has a point blanc range of more than 400 m, which is not what you'd expect from the comments on trajectory curvature."

This is really surprising. Then again the 30mm HE grenade is probably rather light, so after that distance (400m) it is probably mostly aerodynamics which causes it to slow down quite rapidly. I wonder what is the trajectory like for MK103 which probably used the same grenade?

-C+

[hitech]

Gripen:

When considering the probabilty of a hit one must ask a slightly different question as regards to dispersion.

What is the probabilty of a hit or kill, in a given time frame.

This case represents large dispersion.

As an extreem case of what I am talking about, Assume have one gun that fires 100 rounds a sec. And has a 1 foot dispersion at 100 yards. And each round has a hit probablity of 10%.

This case represents 0 dispersion, but aiming errors.
In the second case assume you have 100 guns each firing 1 round a sec all at exatly the same point. But with each shot all guns disperse together to exatly the same point.

And just as before they have a hit probabilty of 10%.

Now do not look at the probabilty from each bullets perspective, because those chances are the same.

But wrather look at the chances of scoring a single bullet in in 1 sec.

This is more representive of how dispersion effects fire power.

Basicly the Bigger the round, (I.E. more damge each can inflict) will lower the average time required for a kill under the first example.

In the 2nd case bigger rounds would not improve your average time required for a kill, assuming 100 rounds is enof for a kill.

[gripen]

Hitec,
I'm trying to modify your example to the form of two planes with different armament to make sure we talk about same issue.

Plane A has 2 guns being able to shoot 20 rounds/second/gun ie 40 rounds/s with 1 m dispersion at 100 m causing each round having hit probability 10%.

Plane B has 8 guns being able to shoot 5 rounds/s/gun ie 40 rounds/s to the exactly same point with no dispersion but aiming error causing 10% probability of the hit.

So every single round in both planes has a 10% probability of hit but in the plane A it takes in average 0,25 s to get one hit. In the case of the plane B it takes in average 2 s to get 8 hits.

If we assume that it takes 8 hits to down the target with a given round and 4 hits with bigger round. Then increasing the size of the round would improve the effectivity of the plane A reducing the average time to for a kill to 1 s. The plane B does not benefit at all from the bigger round because when it hits, it got a kill regardless the size of the round.

I think that this kind of example can be extended further but notify if I got it right.

gripen

[HoHun]

Hi Charge,

>I'd say it fully retains its energy as upon impact the velocity it hits the target has no significance at all, its the chemical energy that counts.

Well, naturally the kinetic energy of the projectile is present in the explosion gases too, increasing the blast damage. However, it's about 90% chemical 10% kinetic energy at the muzzle, and the kinetic energy decreases downrange, so you got the relative significance right :-)

>>"the MK108 really has a point blanc range of more than 400 m, which is not what you'd expect from the comments on trajectory curvature."

>I wonder what is the trajectory like for MK103 which probably used the same grenade?

Point blanc range about 650 m. (The Browning M2 is close to 800 m, for comparison.)

Regards,

Henning (HoHun)

[Tony Williams]

Originally posted by HoHun
You're quite right - it's often forgotten that depressing the sight line (or elevating the bore axis) will help to make the weapon hit close to the aiming point  - which is just what the shooter wants.

An important issue here is the height of the sights above the gun: the greater the vertical separation, the flatter the gun seems to shoot.

This is because the projectile trajectory is curved but the sightline is laser-straight. If the sights are at the same level as the gun, then to zero-in the gun for 400m (say) means that all of the trajectory will be above the sight-line, which in a low-velocity gun like the MK 108 would mean that at 200-250m the gun would be shooting very high. When the sights are mounted significantly higher than the gun (as they are in most aircraft installations) then the first part of the trajectory is below the sight-line, which means that the gun has two 'zero points'; one when the shells cross the sight-line on the way up (perhaps at 100m or so) and one on the way down, at 400m. The highest point of the trajectory is not as high above the sight line as in the first example, so the gun appears to shoot 'flatter'.

You get a similar effect in rifles with a scope sight high above the bore, as opposed to iron sights mounted on the barrel.

Tony Williams: Military gun and ammunition website and discussion forum

[Charge]

"Well, naturally the kinetic energy of the projectile is present in the explosion gases too, increasing the blast damage. However, it's about 90% chemical 10% kinetic energy at the muzzle, and the kinetic energy decreases downrange, so you got the relative significance right :-)"

If we consider the velocity of the explosion (gases) and the momentum component of the projectile I' day the significance of the kinetic energy upon impact is very very small, so I'd say, without calculations, that 10% is very much exaggarated. I'd be surprised if it is even as much as 1% of the full energy of the blast on target.



-C+

[HoHun]

Hi Charge,

>I'd be surprised if it is even as much as 1% of the full energy of the blast on target.

Time to be suprised then ;-)

MK108 330 g mine shell @ 505 m/s:

Kinetic Energy: 42.1 kJ
Chemical Energy: 461 kJ

Regards,

Henning (HoHun)

[Charge]

Thx HoHun.

But my point is, what does that kinetic part consist of? If half of it consists of the weight of the chemical part so upon explosion that part of mass is actually reduced of the damage as it is transformed into chemical energy? OR That mass is added to the forward velocity component of the explosion and the other half is reduced as it works against the explosive force, so what remains?

Can you just add the kinetic energy to chemical energy like that?

That is the energy at muzzle?

-C+

[Charge]

Then again, the energy does not disappear...

-C+

[HoHun]

Hi Charge,

>That mass is added to the forward velocity component of the explosion and the other half is reduced as it works against the explosive force, so what remains?

It's tricky to visualize, so here is a simplified example: Let's assume we have not an explosion, but rather a recoilless rifle firing a slug of equal weight out of both ends of the barrel.

Let's say it fires 200 g each way at 500 m/s, so we get Ekin0 = 2 * 1/2 * 0.2 kg * (500 m/s)^2 = 50 kJ.

As the kinetic energy of both projectiles was zero before we exploded fired the rifle, these 50 kJ obviously were stored in the driving charge as chemical energy (Ekin0 = Echemical).

Now fire the entire recoilless contraption as a single projectile. Let's assume we fire it at 200 m/s. Barrel mass and explosive are assumed to be 0 kg.

We get a kinetic energy of Ekin = 1/2 * 0.4 kg * (100 m/s)^2 = 2 kJ.

On firing the rifle, both projectiles now travel at different speeds:

Foward fired projectile: EkinForward = 1/2 * 0.2 kg * (500 m/s + 100 m/s)^2 = 36 kJ

Rearward fired projectile: EkinRearward = 1/2 * 0.2 kg * (500 m/s - 100 m/s)^2 = 16 kJ

So the balance:

Echemical + Ekin = EkinForward + EkinRearward

or 50 kJ + 2 kJ = 36 kJ + 16 kJ

In short: The forward-fired projectile gains more destructive energy than the rearward-fired projectile loses.

You could run a similar sum for all molecules of the blast cloud of a mine shell, and would of course find a perfect analogy there. The forward traveling front of the blast wave would gain more destructiveness than the rearward traveling front would lose.

Regards,

Henning (HoHun)