Gunsight convergence has been changed so that your bullets will strike high when the

[Elfie]

The reason I brough this up is from a recollection that the A10's guns can stop the plane in mid air if fired in a long enough burst. That caught my attention.

According to that F-4E pilot even the gun on the A-10 wont slow down the A-10 as it's being fired. Really wish I could recall what he said. The F-4E could carry as many as 5 gun pods, 2 under each wing and 1 under the belly,(plus the nose gun) I do recall him saying that even with 6 20mm gatling guns all firing at the high rof (6000 rpm) that it wouldnt affect the Phantom's speed.

I would suspect that the same principle that he explained would apply to the planes we fly as well. Our planes are lighter, but so are the guns, recoil and rof.


Gads, this is gonna keep me awake trying to recall what he said

[stantond]

Whoops,

My bad, didn't divide by 32.2 ft/sec^2 (lb to slugs)!  Math error.  The correct numbers (after dividing by 32.2) are 12608 ft-lb per round and 36 million ft-lb for the aircraft at 300 tas.  Expelling 2400 rounds in 30 seconds exchanges 30 million ft-lb of energy.  Bringing the total back to a 6 million ft-lb energy level for the aircraft giving a 179 tas after all the rounds have fired.


There, that's better.  Thanks for the math check!  My offline test showed a speed drop of 25 ias (from 300 originally).  

Apparently the 50 BMG rounds in AH don't quite have 12608 ft-lb of energy per round.  My calculations show they have 2400 ft-lb which is comparable to a 5.56 mm NATO round.   Quite deadly in the right hands!



Regards,

Malta

[stantond]

Nope, 2400 ft-lb is for a 7.62 NATO (308 Winchester),  not a 5.56.  My bad again!



Regards,

Malta

[stantond]

"Let's also check the slow down in speed.

Malta please use this example and apply your speed reduction formula:

A 200lb man is running 10 mph at a target firing a .45acp pistol which has a ME of 410 ft-lbs of energy. He fires 10 rounds in under 3 seconds.

How much should the man slow down?"


Ok... here goes:

Energy is conserved.  What that means to a person running while firing a 45 is that his body uses energy to absorb the recoil.  Lets go one step further and say the man has a 100 lb weight he is holding 3 feet from the ground and running with.  Will that slow him down?  Both answers depend on the physical condition of the person.   One thing is true in both cases, the man will use more energy.  If the person only has the energy to run 10 mph without firing the 45, then firing the 45 while running must slow him down because of conservation of energy.  

As an example, Mr. Grinch can easily fire the 45 and run after the Hoo's in Hooville.  However, little Mary Loo Hoo cannot both run and fire the 45 because she can barely run, much less shoot the 45 at the same time.




Regards,

Malta

[kevykev56]

Malta,

Not sure your calculations are still correct. Are you calculating that the thrust of the aircraft is still being generated by the prop while firing. And lets not forget newtons law's here. There would be alot of inerta built up in the forward motion of the aircraft. Also we have gravity to take into consideration. Is the aircraft level flight or is he in a dive. what alt is the aircraft at (air density). I could probably go on and on. But I will just leave that for you to think about.

Break out the calculator! its gona be a long night.

RHIN0

[stantond]

All I am doing is an energy balance.  Nothing more, nothing less.  If you want to include potential energy changes, feel free.  My flight tests were at 250 ft alt level flight in a 12000 lb (more or less) bird.  

2400 rounds of 50 caliber Browning machine gun ammuntion has a lot of energy!  This is an anit-vechicle round still used by the military.  All I am doing is looking at where that energy is going.

For a plane (here I am using an F4U) that does not accelerate well, not much energy is being 'supplied'.  There is enough energy to overcome drag forces and keep the plane airborn at any particular throttle setting (assuming you are flying).  If another negative force (recoil in this case) is included, the speed must change.  This change  does show up (which I am most impressed by)!  

However, just like converting kinetic to potential energy by changing altitude, changing the aircraft energy affects its speed.  Take an aircraft (again with the F4U) and have it go from 250 ft to 2750 ft alt in 30 sec.  An energy change has occurred and a change in speed has occurred.   Firing ammunition releases energy!  Now, 2400 0.1 lb bullets weight with a muzze velocity of 2850 ft/sec release 30 million ft-lb in a 30 second duration.  That is quite a bit of energy. In fact, the energy level is the same as going from 250 ft to 2750 ft in 30 seconds.   A drop in airspeed of 25 mph for that energy change does not agree with conservation of energy.





Regards,

Malta

p.s. BTW, there is no law stating AH has to conserve energy.

[kevykev56]

Malta,

Im not quite sure I know where you are going with this. Yes there is a great release of energy. However it is not a constant (time to release the energy/time to chamber and fire second round). You seem to be looking at these numbers in a pure sense. Nothing here is that way. All these factors play against one another. Yes that ammount of energy is being released but how much of the energy is going back into the guns via the gas ports to chamber the next round.

I guess I dont understand how you would be able to calculate how much of a reduction in aircraft  speed would be created without using the aircrafts kinetic energy.  

Are you saying that there isnt enough force being applied to the aircraft. Is it not slowing the planes down fast enough? Once again im not sure I am following your line of thought. Not to mention we have wandered way OT from the original thread.

Just curious to your mode of thinking. Nothing more nothing less.


RHIN0

[TimRas]

You should calculate recoil impulse, not energy, if you want to get some idea of the effect to airspeed. Hint:
Bullet mass * Bullet velocity * Rate of Fire, has the unit of force. You should also take account the effect of the propellant gases (see some rules of thumb from the web).

[stantond]

Ok,

I did not use conservation of momentum, which divides up how much energy is given to each part.  Since the plane weights 12000 lb and the bullet .1 lb, the bullet gets the lions share of the energy.  A momentum balance gives (2400)x.1 lb x 2850 ft/sec = 12000 lb x 57 ft/sec.  

In energy terms (since I started all this with energy) 1/2*12000*57^2/32.2= 605403 ft-lb of energy loss in the plane.  The plane's decrease in speed is 57 ft/sec or 38mph which is pretty close to what is seen offline.

So.... in the immortal words of Gilda Radner, "Never mind".  



Regards,

Malta

[hitech]

standond , the only thing you are not accounting for is thrust or loss of alt , adding to the total energy remaining.


HiTech

[JB82]

12 year old Scotch hu...you realy think I would share, that alone give a bottle away HT?  What do you think about Johnnie Walker Blue?

[eilif]

if you fire guns in il2 with some of the higher caliber guns you can actually lift the tail off the ground going in revers. talk about cheese ball. i think iv gone 50 mph like that.

[stantond]

This is my last post on this topic, really!!

I worked all this out on paper and without drinking beer, so I expect its correct.
Using a thrust of 1600 lb, the drop in airspeed is 25 mph as seen offline.  I worked backwards to get this number, but it seems reasonable.

The calculation uses the plane velocity from momentum conservation substituted into the energy equation.  Energy is then added due to excess horsepower (P) as the velocity drops (using HP=T*dv, =>  E=T*dv*t ).  


I am most impressed with the modeling.



Regards,

Malta