Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: Vinkman on March 18, 2013, 02:22:56 PM
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Do rounds fired from machinge guns slow down in game due to drag? If so does anyone have a table of measured data showing velocity at distance travelled for the guns in game?
Thank you
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Yes, no.
:)
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Is this what you are looking for?
(http://imageshack.us/a/img145/5593/apim8.gif)
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Is this what you are looking for?
(http://imageshack.us/a/img145/5593/apim8.gif)
Yes! thanks Buster. :salute
Any other gun calibers?
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HTC has modeled trajectory and velocity, yes.
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HTC has modeled trajectory and velocity, yes.
Trajectory yes (gravity has a big effect.) But I wasn't sure if they modelled drag, or assumed the speed doesn't change much to matter over bullet range in game (1500 yds).
I'm wondering what the terminal velocity of a MG round in game is. I believe they disapear at 1500 yrds. What I wanted to further understand was how much the damage value of a MG round changes with distance. This would be a function of it's speed. From buster's Table I can calculate the Drag and hence the terminal velocity. From that I could calculate the drop in damage as a function of the drop in Speed.
This goes back to my original question which, I now assume to mean the damage and velocity change over time and distance travelled.
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HTC has modeled trajectory and velocity, yes.
Wish they would add wind sheer from prop ect, and other wind forces,then players could not do the BS 1k shoots in AH.
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I now assume to mean the damage (...) change over time and distance travelled.
Against player objects (planes and GV) yes, against ground structures no.
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Against player objects (planes and GV) yes, against ground structures no.
Thanks Lusche. Have you done the calculations on MG damage/distance? I have been wondering about it for a long time with respect to the other (potential) advantage of cannons which is most of the damge is done by the explosion so they lose less of their damage effect over distance than MG do. Let's say at 100 ft it takes one 30mm, or 20 .50 cals to knock a wing off, at 1500yrds it still takes one 30mm but 40 .50cals.
Ever looked at it?
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No, because I don't have any means to test it.
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I would imagine that bullets follow the same physics in AH as GV guns. Put two tanks far enough apart while they're exchanging fire and hear one of them gripe about how many hit sprites his tiger caused on a M4 or an M4 driver laughing over how many times he's been hit without taking damage.
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I wonder if it can be tested with a p51 saddled on a b17 rear gunner and slowly coming forward as it fires?
You just compare the range of the rear guns vs the fighter pursuing.
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Wish they would add wind sheer from prop ect, and other wind forces,then players could not do the BS 1k shoots in AH.
They do disperse. With the sheer volume of rounds we fire, you're pretty much guaranteed to get a ridiculously long range kill every once in a while.
Vinkman- Just because you mentioned cannons having 'most of' their damage as chemical. If memory serves, cannon rounds do exactly the same damage at the muzzle as they do at range. Velocity doesn't affect cannon round damage in any way, only MG round damage. Although, typing that out just now makes me wonder, does a cannon round with the aircraft doing 0mph do the same amount of damage on a stationary target as a cannon round with the aircraft doing 300mph?
Wiley.
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Even a cannon round moving faster would help the damage done, most fuses had a delay so after penetration then the round would blow up.
Unless in the case of the round over penetrating and blowing up on the other side of the target, in this case just a really really big hole. :rock
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Even a cannon round moving faster would help the damage done, most fuses had a delay so after penetration then the round would blow up.
Unless in the case of the round over penetrating and blowing up on the other side of the target, in this case just a really really big hole. :rock
Not always. Some cannon warheads did not have TNT in them and were only meant to splinter apart upon any medium thicker than air. We can get in to the nitty-gritty all we want, but my guess is that HTC modeled the damage the same regardless if the impact is at 200 yards or 1000 yards for cannon HE rounds.
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Thanks Lusche. Have you done the calculations on MG damage/distance? I have been wondering about it for a long time with respect to the other (potential) advantage of cannons which is most of the damge is done by the explosion so they lose less of their damage effect over distance than MG do. Let's say at 100 ft it takes one 30mm, or 20 .50 cals to knock a wing off, at 1500yrds it still takes one 30mm but 40 .50cals.
Ever looked at it?
The controls in that kind of test would be very difficult to maintain. It almost isn't really worth testing because of the infinite amount of variables in the field that are present when considering the damage dealt to a player controlled object. Vs a static OBJ (town building, ammo bunker, fuel tank, barrack, etc) it is easy and that testing has proven that there is no difference in damage dealt be it 200 yards or 1000 yards. That is something I've directly tested using both AP and HE rounds.
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Velocity at 1800ft(600yds) drops to about 1915 ft\sec for the 50cal.
For all practicle concerns, the AAF determined that the fighter mounted version of the 50cal had an effecitve range to damage other aircraft and to still hold a practicle dispersion pattern was 2000ft(666yds). Ever wondered why the max convergence for fighters in the hanger is 650yds? Except for the Mk108, your dispersion patterning in all fighters is a 4Mil cone at all convergence distances 30cal, 50cal, 20mm.
4Mil @ 900ft (300yds) = 3-4ft
4Mil @ 2000ft(666yds) = 8-10ft
Now bounce this cone around by fractions of a degree shooting at a 666yd target and you get rounds everywhere. At 300yards half of everywhere. And so forth closer. Dispersion is your real issue trying to hit anything with 30 - 20mm past 400yds. The 50cal MG type with the longer barrel in the M16 would hold it's velocity enough to be effective at 1200yds if you can hit anything moving with the dispersion. Probably why 4 were ganged close together.
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Velocity at 1800ft(600yds) drops to about 1915 ft\sec for the 50cal.
For all practicle concerns, the AAF determined that the fighter mounted version of the 50cal had an effecitve range to damage other aircraft and to still hold a practicle dispersion pattern was 2000ft(666yds). Ever wondered why the max convergence for fighters in the hanger is 650yds? Except for the Mk108, your dispersion patterning in all fighters is a 4Mil cone at all convergence distances 30cal, 50cal, 20mm.
4Mil @ 900ft (300yds) = 3-4ft
4Mil @ 2000ft(666yds) = 8-10ft
Now bounce this cone around by fractions of a degree shooting at a 666yd target and you get rounds everywhere. At 300yards half of everywhere. And so forth closer. Dispersion is your real issue trying to hit anything with 30 - 20mm past 400yds. The 50cal MG type with the longer barrel in the M16 would hold it's velocity enough to be effective at 1200yds if you can hit anything moving with the dispersion. Probably why 4 were ganged close together.
Good info Buster. This thought was raised from another thread talking about the P-38 gun pack, and that it was great for killing buffs. Killing buffs is tough inside 1000 yrds because they kill you first. (just talking dead 6. I know.."never attack dead 6...blah...blah) But I wondered if a P-38 could hang back at 800-1000 and light up a wing tank without getting too shot up. All the guns in the nose would be an advantage because the dispersion is very low. But then I thought "maybe the rounds hit like duds at 1000 yrds so you can't take advantage of the guns being "clustered".
I've killed a lot of Buffs in a 410 with the 50mm by hanging back at 1500 yrds. The 50mm can reach them and still kill a buff with 1 hit. Again its the HE that kills the buff not the KE of the round. But 410 are SLOWWWWWWWWWWW and handle poorly. IT's not easy to catch 25-30k Bombers in it, and if fighter shows up you toast.
The 38 has great fire power, but I think it may drop significantly with distance. I want to calculate the KE difference but need to calculate the speed over the distance traveled to solve it. With your data and solve for the drag by comparing it to a no drag formula. With the coeficients and drag term solved for, I can calulate it at any distance after firing, then I can campare the KE at various differences.
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The controls in that kind of test would be very difficult to maintain. It almost isn't really worth testing because of the infinite amount of variables in the field that are present when considering the damage dealt to a player controlled object. Vs a static OBJ (town building, ammo bunker, fuel tank, barrack, etc) it is easy and that testing has proven that there is no difference in damage dealt be it 200 yards or 1000 yards. That is something I've directly tested using both AP and HE rounds.
Isn't the "hit" model in AH by area of the plane?
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Damage to ground targets and player controlled objects are 2 different systems.
For player controlled objects.
1. Drag is completely modeled , including altitude effects. A simple test shooting different directions from bombers will show this statement to be true.
2. Damage is modeled on a per round type.Both it's speed and explosive components can be modeled independently on a per bullet type basis.
3. Damage to ground structures are a very simple damage model Ordnance does type x does y based on 3 x values. There are basically 3 types of ground object soft, hard and armored.
Each ordnance has 3 damage values based on if it is hitting a soft, hard or armored structure.
The ordnance also do blast radius damage to near by structures based on the distance to the object.
HiTech
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Damage to ground targets and player controlled objects are 2 different systems.
For player controlled objects.
1. Drag is completely modeled , including altitude effects. A simple test shooting different directions from bombers will show this statement to be true.
2. Damage is modeled on a per round type.Both it's speed and explosive components can be modeled independently on a per bullet type basis.
3. Damage to ground structures are a very simple damage model Ordnance does type x does y based on 3 x values. There are basically 3 types of ground object soft, hard and armored.
Each ordnance has 3 damage values based on if it is hitting a soft, hard or armored structure.
The ordnance also do blast radius damage to near by structures based on the distance to the object.
HiTech
thanks HiTech :aok
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.50 cals lose 25% of their Kenetic engery at 400 yards, and 70% of their kenetic engery at 1000yrds. So at 1000yrds you have to hit with about twuce as many bullets as you do at 400yrds to do the same damage. That's a big difference.
Cannon engery loss would be less because the HE portion does not decrease with distance travelled.
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.50s fired from the tail gun of a bomber keep their energy better than the .50s fired from the pursuing P-51D. At the muzzle the bullets from the bomber's tail guns are going about 700mph slower and that means less E loss due to drag, plus the P-51 is running into the bomber's bullets whereas the bomber is running away from the P-51's bullets.
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.50s fired from the tail gun of a bomber keep their energy better than the .50s fired from the pursuing P-51D. At the muzzle the bullets from the bomber's tail guns are going about 700mph slower and that means less E loss due to drag, plus the P-51 is running into the bomber's bullets whereas the bomber is running away from the P-51's bullets.
They kept their energy better? :headscratch:
They start with less energy because they leave the gun 300mph slower, relative to the air. So they have less energy and travel less distance, relative to the point they were fired from. A persuing plane is running at the bullet at 300+ mph, so it will meet the bullet at the shorter distance, and it's forward velocity will add to the energy of the impact. I belive it's the same as if they are both stationary.
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Regarding the last two posts above this one involving the speed of the aircraft and its effect on the bullets.
The velocity on Bustrs chart >> 50cal = 2870 fps
The chart doesn't give a bullet weight but a quick search revealed ww2 50 cal ball ammo at 46 grams or converted to grains at 709
Using the standard formula for calculating foot pounds of energy we get the following
Bullet velocity 2870 at 709 grains =12964.72 ft-lbs
Bullet velocity 2870 plus speed of aircraft at 300 mph(440 fps)...actual bullet velocity of 3310 and a 709 grain bullet = 17244.69 ft-lbs
That is a difference of 4279.97 ft-lbs or about 25% more.
Of course this is assuming a situation where the bullet is fired strait with no g effects.
A bomber shooting a tail gun moving at 300 mph will have its bullet ft-lbs reduced by 25 %.
Gravity is already calculated in ft-lbs calculations. If you add in g forces or angle of attack...say the gun is pointed 45 deg of the tail of the bomber then the calculations will change.
If two planes fly directly at each other at the same speed firing the same gun and bullets then the ft-lbs will be the same for each plane. If one is flying slower than the other then the ft-lbs will change for each plane accordingly.
I submit that given average speeds of aircraft in the game and guessing at a closure speed depending on the angle of attack that the difference in ft-lbs one is gaining or loosing based on closure speed of BOTH aircraft that the effect of damage is less than 10 %. Then add in drop in velocity over distance and the ft-lbs becomes almost irrelevant in the damage model.
In the end one is only getting 1 to 3 % of additional or loss of ft-lbs of additional damage possibility....not very much and probably not worth considering.
But then one begins to think about it.....get going 500 mph in a dive on buffs at a 30 deg deflection angle head on....and add in the speed of the buff and the ft-lbs of your bullets start to look pretty good for getting the best bang for your buck.
Turn that around and start shooting at a plane accelerating away from you by 100 mph and distance is far the same goes...you aint damaging jack...LOL
just a thought.
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They kept their energy better? :headscratch:
They start with less energy because they leave the gun 300mph slower, relative to the air. So they have less energy and travel less distance, relative to the point they were fired from. A persuing plane is running at the bullet at 300+ mph, so it will meet the bullet at the shorter distance, and it's forward velocity will add to the energy of the impact. I belive it's the same as if they are both stationary.
Drag increases as speed increases. Therefore, relatively, the rounds fired from the tail gun lose the energy they start with at a slower rate than the rounds fired from the perusing P-51.
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Drag increases as speed increases. Therefore, relatively, the rounds fired from the tail gun lose the energy they start with at a slower rate than the rounds fired from the perusing P-51.
The tail gun bullets start with a negative value of velocity....less the fps the buff is flying. The p51 starts with a positive value.
Drag doesn't matter. It is just a constant used to calculate velocity...like gravity is a constant. Yes, drag is a coefficient of speed but drag is factored in to a bullets stated velocity. Drag is more of an aeronautical issue. Drag in bullets is not worth considering when calculating kinetic energy because it has already been considered based in the bullet design, its weight and powder load.
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The tail gun bullets start with a negative value of velocity....less the fps the buff is flying. The p51 starts with a positive value.
Drag doesn't matter. It is just a constant used to calculate velocity...like gravity is a constant. Yes, drag is a coefficient of speed but drag is factored in to a bullets stated velocity. Drag is more of an aeronautical issue. Drag in bullets is not worth considering when calculating kinetic energy because it has already been considered based in the bullet design, its weight and powder load.
If the two were shooting at a stationary target, yes. But in the case where they are shooting at each other, no. The P-51s bullets will lose more energy by the time they hit the B-17 than the B-17's bullets will lose by the time they hit the P-51. In this scenario each bullet hitting the P-51 has more energy than each bullet hitting the B-17.
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The tail gun bullets start with a negative value of velocity....less the fps the buff is flying. The p51 starts with a positive value.
Drag doesn't matter. It is just a constant used to calculate velocity...like gravity is a constant. Yes, drag is a coefficient of speed but drag is factored in to a bullets stated velocity. Drag is more of an aeronautical issue. Drag in bullets is not worth considering when calculating kinetic energy because it has already been considered based in the bullet design, its weight and powder load.
Think of the 0 speed case,The b17 drops a rock , the 51 will run into it, the 51 drops a rock, there is no way the b17 will ever run into the rock. The drag is what makes the rock from the b17 hit the 51.
The same condition exist when the bullet is moving.
HiTech
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Think of the 0 speed case,The b17 drops a rock , the 51 will run into it, the 51 drops a rock, there is no way the b17 will ever run into the rock. The drag is what makes the rock from the b17 hit the 51.
The same condition exist when the bullet is moving.
HiTech
When the b17 drops the rock its going the speed of the b17 to start with. Factor the speed, mass, gravity and air density and we can calculate ft-lbs for the rock. Obviously the ft-lbs of the rock will decrease as speed decreases relative to the target. Nothing else is changing. Only the speed of the rock.
Calculate the speed of the P51 and add that to the rocks speed then we can calculate the ft-lbs of the rock when it hits the P51.
In your case the rock starts of with 0 speed....so it must be falling due to gravity at a speed up to terminal velocity. The P51 must be moving at some amount of speed to hit the rock.
I don't see how drag is a factor here. Given an objects velocity, drag should have been already factored into the velocity to start with....as far as bullets go anyway.
Drag is a by product of atmosphere. In a vacume drag is not a factor.
My point is that when calculating ft-lbs(kinetic energy) of a bullet, drag is already included when the bullet velocity is calculated.
I may be totally misunderstanding your post though.
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Think of the 0 speed case,The b17 drops a rock , the 51 will run into it, the 51 drops a rock, there is no way the b17 will ever run into the rock. The drag is what makes the rock from the b17 hit the 51.
The same condition exist when the bullet is moving.
HiTech
you talking about a regular rock or skyrock? cause I swear he never misses my b17's no matter what the speed is.
semp
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Agent:
Let b = muzzle vel of bullet.
Let A = speed of plane.
Assume tail chase with both b17 and p51 travelling same speed.
let v1 = airspeed of b17 bullet
let v2 airspeed of p51 bullet
V1 = b - a
v2 = b + a
now the key to this is that drag is related to the square of vel.
Sinc V2 > V1 V2 will decelerate at a faster rate.
So after 1 sec of travel the delta of V2 will be greater then the delta of v1
Hence the builet from the b17 will travel a greater distance and retain a higher vel relative to the airplanes then the bullet from the p51.
HiTech
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Agent:
Let b = muzzle vel of bullet.
Let A = speed of plane.
Assume tail chase with both b17 and p51 travelling same speed.
let v1 = airspeed of b17 bullet
let v2 airspeed of p51 bullet
V1 = b - a
v2 = b + a
now the key to this is that drag is related to the square of vel.
Sinc V2 > V1 V2 will decelerate at a faster rate.
So after 1 sec of travel the delta of V2 will be greater then the delta of v1
Hence the builet from the b17 will travel a greater distance and retain a higher vel relative to the airplanes then the bullet from the p51.
HiTech
Pffft....I knew that....Geezus Agent. Dumbazz
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Agent:
Let b = muzzle vel of bullet.
Let A = speed of plane.
Assume tail chase with both b17 and p51 travelling same speed.
let v1 = airspeed of b17 bullet
let v2 airspeed of p51 bullet
V1 = b - a
v2 = b + a
now the key to this is that drag is related to the square of vel.
Sinc V2 > V1 V2 will decelerate at a faster rate.
So after 1 sec of travel the delta of V2 will be greater then the delta of v1
Hence the builet from the b17 will travel a greater distance and retain a higher vel relative to the airplanes then the bullet from the p51.
HiTech
It seems like we are looking at this problem from opposite sides of the equal sign...LOL
According to Newton:
"Every body persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces impressed on it."
I agree that drag is a factor. The issue is how much drag effects the bullet due to the speed of the airplane and how much this drag effects kinetic energy at the point of impact relative to the distance the bullet travels.
I submit that the degree of kinetic energy changed by drag is "insignificant" in terms of the damage model in this game...and in real life for that matter.
The fact that AH2 models this is awesome. It goes to show that this game is a notch above anything out there....hehe actually there's not really anything competing with AH2 but that's another issue.
In the formula you post I agree.
My case on the issue is that given actual closing speeds the factor of drag can almost be ignored.
Consider this:
I tried to write this in the form of a brain teaser.
There are two planes flying at the same altitude. Plane 1 is moving at 200 mph ( a little over cruising speed of B17). Plane 2 is following Plane 1 and is moving at 400 mph (reasonable attack speed in a p51)
Plane one is firing a gun from its tail exactly 180 deg from its heading of 0 deg. Plane two is flying a heading of 0 degrees and is on the same axis and heading as Plane 1. Both Planes are flying level.
Plane 1 and Plane 2 are firing a 50 cal bullet with a 709 grain weight and a power load which results in a bullet velocity of 2870 fps at the muzzle.
Both planes begin firing at each other at the exact same time with the exact same fire rate at a distance of 400 yards at the speeds stated above.
Assume any gravity value and any air density value as long as both are used for both Plane 1 and Plane 2
What will be the ft-lbs (kinetic energy) of the first bullet fired from Plane 1 when it hits Plane 2?
What will be the ft-lbs (kinetic energy) of the first bullet fired from Plane 2 when it his Plane 1?
Free beer to the first one to solve.
Bonus question:
Which bullet has more drag. Plane 1 or Plane 2?
We could re write this and say that both planes are moving at the same airspeed. In that case I do agree with you...in the fact that drag is effecting the bullet.
But, show me how drag effects kinetic energy and by how much? That is the question I've been seeking all along.
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Agent,
For AH bump the B-17's speed up to about 300mph and the range up to about 1000 yards. The difference becomes significant at that point.
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Agent : It is impossible to answer your question with out providing a bullet co. The drag is not an insignificant piece of the equation.
HiTech
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Agent : It is impossible to answer your question with out providing a bullet co. The drag is not an insignificant piece of the equation.
HiTech
Just for kicks say the bullet co is 0.61 (G1) ....just in case anyone wanted to actually solve.
About your formula example a few posts above.
Ok. So velocity is squared in the equations. So drag does increase with velocity. No issues there.
bullet A = 2000 fps
bullet B = 3000 fps
How is it possible for bullet B to have less kinetic energy than bullet A at any time during its trajectory?
It does not matter that it was fired from in front of or from behind a plane. We just have to correct the bullet velocity based on the speed of the plane.
I understand that the faster something goes the more drag it has and like wise the slower the less...hence a "trajectory"
If two shooters point at each other and fire the same bullet but one is firing at 2000 fps and the other at 3000 fps. How is it that the faster bullet ends up having less kinetic energy at the point of impact? (Remember you said that both planes were moving at the same speed)
PS - I know you don't have time to give a math education and that I am probably wrong and that I am currently the definition of your sig...LOL but I enjoy recreational mathematics and find this topic interesting. I may be more suited to dabble in something more elementary but I am enjoying this none the less :)
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Just for kicks say the bullet co is 0.61 (G1) ....just in case anyone wanted to actually solve.
About your formula example a few posts above.
Ok. So velocity is squared in the equations. So drag does increase with velocity. No issues there.
bullet A = 2000 fps
bullet B = 3000 fps
How is it possible for bullet B to have less kinetic energy than bullet A at any time during its trajectory?
It does not matter that it was fired from in front of or from behind a plane. We just have to correct the bullet velocity based on the speed of the plane.
I understand that the faster something goes the more drag it has and like wise the slower the less...hence a "trajectory"
If two shooters point at each other and fire the same bullet but one is firing at 2000 fps and the other at 3000 fps. How is it that the faster bullet ends up having less kinetic energy at the point of impact? (Remember you said that both planes were moving at the same speed)
PS - I know you don't have time to give a math education and that I am probably wrong and that I am currently the definition of your sig...LOL but I enjoy recreational mathematics and find this topic interesting. I may be more suited to dabble in something more elementary but I am enjoying this none the less :)
kinetic energy needs a frame of reference
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kinetic energy needs a frame of reference
Agent. What this means is the energy relative to the ground, the plane shooting the projectile, or the plane the projectile will hit. The faster projectile will have more energy if it hit a stationary target. But when hitting a target coming towards it, and another away from it, the one hitting the target from the front will have more total energy. And the one chasing the plane will have less.
HiTech
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Agent. What this means is the energy relative to the ground, the plane shooting the projectile, or the plane the projectile will hit. The faster projectile will have more energy if it hit a stationary target. But when hitting a target coming towards it, and another away from it, the one hitting the target from the front will have more total energy. And the one chasing the plane will have less.
HiTech
And the projectile hitting the target coming towards it will, functionally, have greater effective range due to the distance being closed and lower parasitic drag because of its lower velocity compared to the atmosphere it is traveling through.
If the B-17 fires at the 400mph P-51 at 1000 yards, the P-51 will move 586ft per second closer to the point from which the .50 was fired while the 300mph B-17 will move 440 feet per second further from the point at which the .50 was fired. In addition the higher parasitic drag on the .50 fired by the P-51 means that it loses velocity in relation to the B-17 faster than the .50 fired by the B-17 loses it in relation to the P-51. Additionally the B-17's 300mph retreat from the P-51's .50 slightly reduces the energy transfer of the hit while the P-51's 400mph rush to meet the B-17's .50 somewhat increases the energy transfer of the hit.