Stupid German 30mm Mk108 cannons!

[Serenity]

Yer smert.

lol. I am BARELY passing high-school at the moment. I just FINALLY found a practical application for my AP Physics, lol.

[Saurdaukar]

Well your understanding of physics is well beyond my own.  "As we all know..." does not apply, methinks.  

I'm still struggling with my conceptual vision, though.  Let me try to explain it differently and you tell me what you think.

As most tater dweebs will confirm; the shot with the highest probability for success is a near 90 degree snapshot with wings perpendicular to the flight path of the EAC.

For this reason, I will use the highway example again - but with a twist.

Here are the facts:

1.)  The subject area consists of a flat, grassy field to the west of center and a two-lane, single-direction highway to the east of center.

2.)  Each lane has an on-ramp to the north of center and cars travel south in both lanes.

3.)  Lane A of the highway is the slow lane.  Lane B of the highway is the fast lane.  Each lane is 10m wide.

4.)  Cars enter the on-ramp at the same rate: One car every second.  Spacing remains the same between cars in the same lane: 20m apart.

5.)  Cars in Lane A travel at 20m/s.  Cars in Lane B travel at 40m/s.

6.)  You are driving across the field at 20m/s, towards the highway, at a perfect right angle, with the intention to cross the highway.


Now...

This makes the point (time and space) at which you cross the highway the "target area" and it makes you the target.

The target area is, for the sake of argument, a perfect square of 400m^2 (20m*20m).

It will take you 1 second to cross the target area.

Along the perpendicular path, it takes the cars in Lane A 1.0 second to pass through the target area and it takes the cars in Lane B 0.5 second to pass through the target area.

Given equal spacing of 20m between cars; the number of cars which pass through the target area in that 1.0 second you are crossing is different for each lane.

Therefore...

In Lane A, the slow lane, cars are traveling at a rate of 20m/s, 20m apart from each other, passing through a 20m target area.  This means that in the 1.0 second it takes you to cross the target area, the maximum of number of cars which can pass through the target area during the time you are in the target area is one.

In Lane B, the fast lane, cars are traveling at a rate of 40m/s, 20m apart from each other, passing through a 20m target area.  This means that in the 1.0 second it takes you to cross the target area, the maximum number of cars which can pass through the target area during the time you are in the target area is two.

The faster cars (shells), despite equal spacing (ROF), due to their speed advantage (m/s) have a 100% greater chance (1 v 2) of impacting your car (plane) while crossing the target area (ideal range and deflection).

Does that not make sense or am I over-simplifying it?

[mtnman]

They won't have equal spacing in distance Saur, despite (and because) having an identical RoF.  The cars traveling at 1/2 the speed, will have 1/2 the distance between them.  The time will be equal, but not the distance.

Think of it this way, two guns are both firing one round every second (or ramps with cars, if you prefer).  Equal RoF (1 round/second).  Bullets are coming out at 2000 FPS from one gun, and 1000FPS from the other.  Fire them at precisely the same moment, and continue that process...

Precisely one second later, when the second bullets are fired, one of those initial bullets will have traveled 1000ft, while the other has traveled 2000ft.

So, while both guns are firing at 1 round/second, and each target is being hit at a rate of 1 round/second, the spacing between the two guns is different.

Where speed of flight really matters is in how arched the trajectory will be.  The slower round falls (assuming both guns are firing identical bullets) at precisely the same speed as the faster round, but it takes twice as long to arrive at the target, so appears to fall faster.

Back to the initial example of 1000 vs 2000 fps, firing both guns at a target at 2000ft, one round will take twice as long to arrive, so has twice as long to fall...  If the faster round hits 2ft low, the slower round would hit 4ft low.

Adjusting your convergence on nose-mounted guns is primarily about adjusting the trajectory, compared to effective range.  Effective range is reduced with a reduced velocity, because the increased arch of the trajectory makes hitting things harder.  Range-estimation is much more difficult, and much more critical as velocity is decreased.  Misjudging range with a slow projectile means you shoot over your target, or under it, with a small window of "just right".  A faster projectile has a flatter trajectory, for further, so is more forgiving of range-estimation errors.

[Lusche]

Therefore...

In Lane A, the slow lane, cars are traveling at a rate of 20m/s, 20m apart from each other, passing through a 20m target area.  This means that in the 1.0 second it takes you to cross the target area, the maximum of number of cars which can pass through the target area during the time you are in the target area is one.

In Lane B, the fast lane, cars are traveling at a rate of 40m/s, 20m apart from each other, passing through a 20m target area.  This means that in the 1.0 second it takes you to cross the target area, the maximum number of cars which can pass through the target area during the time you are in the target area is two.

The faster cars (shells), despite equal spacing (ROF), due to their speed advantage (m/s) have a 100% greater chance (1 v 2) of impacting your car (plane) while crossing the target area (ideal range and deflection).

Does that not make sense or am I over-simplifying it?

I see where you are coming from... but if you calculate this for a gun and airplane you will see  muzzle velocity makes no difference.

Target: (fighter sized)
Plane length 10m
Plane speed 150m/s (approximarety 338mph)

The plane takes 0.07 s to fully cross the gun's line of fire (plane length/plane speed)

Regardless of MV, every 0.1 s there is a round appearing at that point. As the plane's crossing time (0.07s) is shorter than this interval, it has a chance to get through before the next round arrives.

Don't look at the whole distance the round is traveling. That can be confusing. Just look at that target point, where the vectors of target & round are crossing. Every 1/(ROF) seconds, a bullet will appear there.

Take my "flashlight" example again. If you are flashing the light every 1 (or 0.1) second, there is a flash every 1 (or 0.1) seconds even at distance, though the "muzzle velocity" is 299 792 458 m/s

[Saurdaukar]

They won't have equal spacing in distance Saur, despite (and because) having an identical RoF.

THAT is what I'm missing... had not considered it.

Makes sense now, thanks.

[VonMessa]

For what it's worth, this is what works for me.....

I use the Revi16d gun sight.

I pull up and start aiming where the arrow is (in the attached modified pic).  Sometimes lower, depending on how many G's you are pulling.  I have convergence set at d400 for the 30mm (even though it's a single cannon, it will have an affect on how you aim and is a personal preference) .  I'll set mg's at d500, especially if I am flying with squaddies so I can spook a con off someone.   

Obviously it is best to pull as little lead as possible, but the one being pursued usually panics when they hear that 108 barking.  I have found that it is best to wait til you have a sure shot and squeeze off a few at a time.  If nothing else flying with only 65 cannon rounds will teach you to husband your ammo   



The 30mm rounds usually had 41 m (135 ft) of drop in the first 1,000 m (3,300 ft) of range.   

So.... in a perfect world in perfectly level flight, using a perfect angle of drop (not accounting for wind, or arc of drop as it would not fly at a perfect downward angle, or other adverse effects on ballistics) it would go like this.

d200 = 8' - 2" of drop

d300 = 12' - 3" of drop

d400 = 16' - 4" of drop

d600 = 24' - 6" of drop

the angle of descent is (2) degrees or roughly 4' of drop for every 100' of distance (again a perfect angle, not a true ballistics arc)









Taking into account that a B-24 Liberator is about 18' tall, you have to just about point the center of the pipper OVER the buffs to actually HIT them with taters at d400 

That's why I usually don't use the center of the pipper for aiming 30mm's

I have not accounted for rate of fire in any of the above, but without a doubt, it comes into play.

Hope this helps.

[PJ_Godzilla]

Back to the initial example of 1000 vs 2000 fps, firing both guns at a target at 2000ft, one round will take twice as long to arrive, so has twice as long to fall...  If the faster round hits 2ft low, the slower round would hit 4ft low.

This isn't correct since the motion is parabolic. The eqn. of motion for the shell is based off the fact that accel is constant. Even if you ignore aero effects on the shell, you've still got:

x=x(0)*v(0)*cos(theta)*t
y=y(0)+v(0)*sin(theta)*t-gt^2/2

t is time, v(0) is muzzle velocity, theta angle off target in a simple horizontal example, x horizontal, y vertical

what can be pretty easily shown is that, if they both hit the target, because you've got to change your initial angle theta to a higher trajectory (up to 45 degrees) for the lower-speed shell, the two shells actually have to travel different distances (more for the slower one, compounding the time disparity) - unless you're using some flat-trajectory assumption, in which case, that's great IF YOU'RE FIRING YOUR SHELLS IN THE ABSENCE OF GRAVITY. Even in the case you posit, where they both drop below the target, the slower shell still has to travel farther, even is you use a simple, triangular assumption to calculate the distance.

BORON MUST LIVE. ONLY MY FRIENDS WILL POINT OUT MY WEAKNESSES. THANK YOU COMRADE STALIN.