Wish List

[SMIDSY]

hay, E25280 listen up. there is anecdotal evidence from both sides in the vietnam war concerning this very thing. various MiGs tried to creep up on the 6 of B-52s armed with .50 cal tail guns while the MiGs had 20mm cannon with a higher muzzle velocity and longer range. however, due to the physics that HiTech so perfectly described, the MiGs were shot down by the tailguns of the B-52 before they could get in range.  nuff said.

also, in responce to the "but SMIDSY!! i still get blasted even when i attack from the side" argument, the B-17G model that we currently have ingame is equipped 12 .50 caliber machineguns. if you are at the plane's 9 or 3 there are a total of 9 of those guns that can hit you. everyone who knows how to fly a P-47 knows how lethal all those ma duces can be.

[FiLtH]

You guys have it all wrong. The reason some guys think bomber guns are too lethal is because they start at 1000 away and come straight in and sit at 400 away straight and level firing their guns.

[hitech]

Originally posted by E25280
Love the "hovering bullet" analogy.  

So I guess I need a physics lesson.  I tend to think of deceleration as a time-dependent variable.  Is deceleration due to drag purely dependent on distance traveled?  That is, a bullet fired at 2000 feet per second decelerates more rapidly than a bullet fired at 1000 feet per second?

Waiting patiently for my smackdown . . .

Until then, as once stated on Mythbusters;  "I reject your reality, and choose to substitute my own."

I watch that show also, anyway what your missing is deceleration (or more properly drag )  varies with the sqaure of vel.

The baisic equations forms are (combining constants all into BulletDragCO ).

Drag = Vel^2 * AirDensity * BulletDragCO.
and
Acceleration = Drag / BulletMass

So the end result is the drag is Zero in my case is Huge in the case you described.

Or more more precisly take the same muzzle vel in our sample case.

With a Total drag of 4 out the barrel if the gun was not moving.

In my case drag would be 0, in your case drag would be 16.




HiTech

[Saxman]

***Head explodes***

[viper215]

Originally posted by FiLtH
You guys have it all wrong. The reason some guys think bomber guns are too lethal is because they start at 1000 away and come straight in and sit at 400 away straight and level firing their guns.

RIGHT!!!! Come on guys 1 time a week someone complains about bomber guns...not includeing boxboy b/c i likes his other suggestions... but if your going to attack buffs you cant come in right from the back...get killed...curse...log off...come on bbs...and post about it. Every buff has its blind spot or a spot where it will be hard to shoot at a plane from. Ive done it before =)


<~V{Viper215}V~>

[E25280]

Originally posted by hitech

The baisic equations forms are (combining constants all into BulletDragCO ).

Drag = Vel^2 * AirDensity * BulletDragCO.
and
Acceleration = Drag / BulletMass

So the end result is the drag is Zero in my case is Huge in the case you described.

Or more more precisly take the same muzzle vel in our sample case.

With a Total drag of 4 out the barrel if the gun was not moving.

In my case drag would be 0, in your case drag would be 16.

HiTech

At the risk of losing another AH member . . . .

Originally posted by Saxman
***Head explodes***

I tried to do a "real world" example in AH land (a contradiction, I know).

Without a lot of time to browse, I quickly found one site (wikepedia, I believe) that listed the muzzle velocity of a WW2 Browning .50 Cal MG to be 3,050 feet per second, or about 2,080 mph.

I also assumed that the deceleration would not be linear, but would rather be an inverse curve (not sure if that is the proper term), which is a fancy way of saying that the rate of deceleration goes down as time goes on.  This is because as the bullet slows, drag resistance decreases, thus although the bullet is still slowing, it is slowing at a lesser rate.  Um, . . . clearest way to say is that drag slows the bullet the most immediately after it leaves the barrel.

Anyhoo, with this in mind I created a scenario where a 300mph bomber is flying 3000 feet in front of a fighter (D1000 in AH) also traveling 300mph (440 feet per second).  They both fire a single .50cal bullet.  What is the true difference in damage this single bullet can inflict?

One problem is that I have no idea how quickly a bullet will slow down IRL if fired from a stationary platform.  Tried to find a quick note on the internet, and patience ran out before the 3,000,000 pages relating to the drag of a bullet ran out.  So, I took a WAG at 300 feet per second per second at the muzzle.

Did a spreadsheet -- I'll spare the details.

Interesting was that I found the bullet fired from the fighter decelerates relative to the ground at an initial rate 1.8 times faster than the bullet fired backwards (3,490 feet per second vs 2,610 feet per second)

The approximate result is that the bullet fired from the bomber hits the fighter at a velocity of 2,843 feet per second after 1.02 seconds.  The fighter's bullet hits the bomber at 2,680 feet per second after 1.05 seconds.  The difference in velocity is 6.1% in favor of the bomber.

Redid the results if initial drag was 200ft and 100ft per second instead of 300.  Differences were smaller, of course (3.9% and 1.9% respectively), but still bomber-favorable.

Changed the speed of the bomber to 250MPH.  Difference in hitting power at the 300ft/sec^2 deceleration fell to 5.2%, but the hitting power of both went up.

Cut the distance between bomber and fighter, and the difference in damage goes down.

So, what's my point?  

I am totally wrong in my initial statement that the hitting power of a bomber's guns are not enhanced vs the fighter's guns because it fires backwards:o

However, if (granted, BIG IF) my 300ft/sec deceleration is anywhere near true, each bullet is only about 6% more effective at D1000, less as distance between target and shooter decreases.  Or, you get about an extra bullet impact effect for every 20 bullets that hit.

In other words, if you fire 20 rounds from each of your 6 .50 cals into a bomber, and he fires 20 rounds back, the reason he is flying on while you go crashing to earth a flaming wreck is more because he is firing 12 or more guns back at your much smaller single engine while you hit a rudder instead or perhaps took out one of his four engines.  It shouldn't be SOLELY because he has 6% faster bullets.

(For those of you wondering, NO, I am not Crumpp)

All the Best,                        

[E25280]

Seriously, though . . .

THANK YOU, Hitech, for taking time to teach me something I didn't know.  

[Spatula]

So by that rationale, attacking a bombing from the front aspect would be more effective, as the bomber's return fire wouldnt hit as fast as it would as if it was fired from the back downwind?

[E25280]

Originally posted by Spatula
So by that rationale, attacking a bombing from the front aspect would be more effective, as the bomber's return fire wouldnt hit as fast as it would as if it was fired from the back downwind?

No -- just that there would be no difference in the impact speed / effectiveness of either the fighter's or the bomber's bullets.

In my 3000 feet, single bullet, 300ft/sec WAG instant deceleration scenario, the following would be the impact speeds when both planes are traveling at 300mph:

Fighter flying behind bomber:

Fighter's bullet: 2,680 ft/sec impact at 1.05 seconds
Bomber's bullet:  2,843 ft/sec impact (6.1% higher than fighter) at 1.02 seconds.

Head On Shot, again fired at exactly 3000 feet distance.

Fighter's bullet:  3,645 ft/sec impact at 0.79 seconds
Bomber's bullet: 3,645 ft/sec impact at 0.79 seconds

This makes the Fighter's shot 36.0% faster at impact than when shooting from behind, but the bombers shot is also 28.2% deadlier than when you approach from behind.  

The survivability advantage, therefore, is simply the difficulty of the bomber hitting you when coming from the front.  Any "single bullet" is actually deadlier -- but the odds of and number of hits would be greatly reduced.

[hitech]

E25280: Intial deacceration on the 50cal would be in the 1400 FPSPS Range

HiTech

[E25280]

Originally posted by hitech
E25280: Intial deacceration on the 50cal would be in the 1400 FPSPS Range

HiTech

  My WAG was Way off And Generally useless.

Recalc (assuming the rest of my math isn't whacked):

Bomber and Fighter traveling 300mph, single bullet at 3000 feet range:

Bomber bullet impact: 2,231 feet/sec at 1.16 seconds
Fighter bullet impact:  1,586 feet/sec at 1.37 seconds
Difference:  +40.7%

Range 1500 feet:

Bomber bullet impact: 2,592 feet/sec at 0.54 seconds
Fighter bullet impact:  2,232 feet/sec at 0.58 seconds
Difference:  +16.1%

Notice the fighter bullet needs to start at 1500 feet to have the same power as the bomber bullet at 3000 feet?

Range 600 feet:

Bomber bullet impact: 2,859 feet/sec at 0.20 seconds
Fighter bullet impact:  2,702 feet/sec at 0.21 seconds
Difference:  +5.8%

Now, assume Bomber is at 240mph and Fighter is at 300mph, single bullet at 3000 feet range:

Bomber bullet impact: 2,287 feet/sec at 1.13 seconds
Fighter bullet impact:  1,719 feet/sec at 1.30 seconds
Difference:  +33.1%

Range 1500 feet:

Bomber bullet impact: 2,666 feet/sec at 0.52 seconds
Fighter bullet impact:  2,353 feet/sec at 0.55 seconds
Difference:  +13.3%

Range 600 feet:

Bomber bullet impact: 2,935 feet/sec at 0.20 seconds (but 8 feet through fighter)
Fighter bullet impact:  2,805 feet/sec at 0.20 seconds (but 5 feet through bomber)
Difference:  +4.6%

Spatula's HO question:

Single bullet HO at 3000 feet would impact both planes at 2,798 feet/sec at 0.91 seconds, 76.4% more power than the fighter bullet fired 3000 feet from behind and 25.4% more power than the bomber's return fire.

All "impact velocities" account for target movement.  That is, in the first case the bomber's bullet impact speed is actual ground speed of 1,791 feet per second plus fighter's speed of 440 feet per second (2,231 total), while the fighter's bullet impact speed is 2,036 feet per second ground speed minus the bombers speed of 440 feet per second (1,586 total).

By the way, the same wikepedia sight that gave me the muzzle velocity of 3,050 feet per second gave an "effective range" of 2000 meters, or 6,600 feet.  By my math and Hitech's initial deceleration, a bullet from a stationary platform would hit that target 6,600 feet away in 3.7 seconds with a velocity of about 1,128 feet per second, and still be "effective" (by whatever definition they use).

Restating the obvious, I WAS TOTALLY WRONG    and bow to the superiority of HiTech.

(Next, someone will point out I forgot to carry the two, and I'll have to calculate all over again . . . )

Once more, THANK YOU HiTech for teaching me.

[Boxboy]

What's amazing to me is how many B-17's we lost to fighters before we got fighter cover for them.  In one raid alone we lost 60 bombers and I find it VERY hard to believe that NONE of those were lost due a 6 oclock attack.

[frank3]

I hosted several heavy bomber missions in my room. A group of 4/5 bomber pilots (with drones, so 12/15 bombers in total) against a group of 4/3 attackers (usually FW-190's)

The outcome was VERY realistic, I believe Raptor calculated the loss percentage, and it came close to real life.
I don't think the .50's should be lowered in lethalty.

[SMIDSY]

E25280, you are arguing with minds that are able to cause your mind to explode. just like we did to Kenedy.

[Kurt]

Don't let how Hitech spells fool you... Anyone who programs flight sims is going to have a pretty firm grip on physics and especially lift/drag/power/gravity formulae.

It comes as no surprise to me that he's a little smarter than the average bear on this.