Bombs dont lose E

[juzz]

Please, go and find a high school physics textbook, read the section on projectile motion - and only then come back, when you might have some idea of what you are rambling on about.

[DA98]

Man, I wonder how did the USA put a man on the moon... Is the educational system REALLY that bad???  

[This message has been edited by DA98 (edited 02-19-2001).]

[TheWobble]

Ok well since you all think I am a total IDIOT,(ironic) i went ahead and wrote a question to a Mr.Harold Brochmann.  he is a retired professor and quite adept in the ares's of Fractals, Nonlinear Dynamics, Data Mining, in other words he knows his toejam.

I am sending him the following question:

If a WW2 bomber traveling around 220mph drops a bomb weighing around 500 pounds from an altitude of around 22,000 feet will the bomb land directly below the bomber that dropped it, or will it land behind it?
ALSO, would the lack of ANY drag or air resistance at all enable it to land DIRECTLY under the bomber?  reason?
 

when he sends me the reply I will post it here, I will ALSO post a TEMPORARY password to access my hotmail account so that you can look at the answer and see that I am not making the reply up.

that should settle it eh?

EDIT:
BTW, when am I going to see some chats graphs and formulas that disprove what I have posted, all I see so far is hot air.

[This message has been edited by TheWobble (edited 02-19-2001).]

[This message has been edited by TheWobble (edited 02-19-2001).]

[DA98]

OK, I have a better question:

"If a bomber drops a bomb weighting X lbs from Y feet and we IGNORE the bomb air resistance, will it land exactly under the bomber or slightly behind?"

I NEVER said that a bomb should fall DIRECTLY under the bomber, but it's due to air resistance, nothing else. Did you ever heard the phrase "movement is relative"?

[DA98]

OK, now I undestand your theory, and I must accept that it's correct... but there is a problem... if it's true... THE BOMB SHOULD IMPACT DIRECTLY AT THE DROP POINT!!!  

[This message has been edited by DA98 (edited 02-19-2001).]

[TheWobble]

 

movement is relative

Yes i know, but when the distance A (bomb) is traveling to get to a given point is longer than the the the path that B (bomber) is taking B will reach the destination befoe A will. hence the bomb landing behind the bomber.

I edited the question I seny him btw.

EDIT:  

if it's true... THE BOMB SHOULD IMPACT DIRECTLY AT THE DROP POINT!!!

LOL!  
only if it was a cotten ball  

The laws of physics do not allow it to land directly athe the point of release or for it to land directly under the bomber, but depending on the weight, or more importantly how aerodynamic the object is i could land almost anywhere inbetween thos 2 extremes.

BTW, this is the best argument I have ever been in, even if I end up being wrong, this was a blast(no pun intended)      



[This message has been edited by TheWobble (edited 02-20-2001).]

[DA98]

Just a little example to make you understand it: If you're inside a train that is travelling at 50 mph, and you drop an object from a 5 feets height to the floor, where will it fall? In that example the air resistance is eliminated in the horizontal axis because the air inside the train is inmobile respect you, the object, and the floor, so the only thing that can influence the impact point is your theory. Having said that, are you saying that the object will not fall at the exact same point where you dropped it?

I'm having a great time too  

[TheWobble]

Damn DA98, thats the best argument i have seen yet  

the only problems i have with it are.

A: train only going 50 (not very fast)
B:dropping it from 5 feet (extreamly low)


between the extreamly low height and the low speed even if you were trying to notice it it would be kinda hard to, a bomber going over 200mph at over 20,000 feet gives the effect alot of time to take effect, 5 feet in a slow train is barley enoigh to get the effect going. it would fall behind the drop point (in theory) but not by much

I really cant wait to get that guy's answer to this, its really got me thinkin,  

lol!, when I saw "train going 50mph" i had this horrid highschool math flashback  


lets apply the above ratio to this
(25,000/15840) at say 250mph (ballpark)

ok here goes

FIRST we must guess what the ratio would be without air resistance, so without it the bomb would go further..lets say 20% farther (modest guess)
15840x.2 = 3168
15840 + 3168 = 19008
New ratio 25,000/19008

ok  lets convert
25,000/5000 = 5 feet
19008/5000 = 3.8016

now to factor in speed
250/5 = 50mph
3.8016/5 = .76032 feet or 7 tenths of a foot.

actually thats more of a difference that I would have guessed, but still only about 8 inches from 5 feet.   I cannot use this as much proof because the number used for the orignal ratio are VERY un-precise, plus there was no given speed of the bomber, which is a required variable.  I wish I had more solid numbers to work with on this, not because i really wish to prove you wrong, but because this is so damn interesting, im having fun either way  

[Pepino]

I would say that, without air resistance, bomb would drop EXACTLY under the plane. Both are moving at the same speed, and air is the only force opposing. IIRC, Newton says that unless a force is opposing, or applying, everything is moving at a constant speed, forever. A bomb drop is a freefall movement in a fluid, where you have 3 forces acting. Gravity, that points right towards the center of the Earth; drag, generated by the speed of the bomb; and flotability, based on Archimedes' principle and neglectible in this case. So the trajectory of a freefall is a parabolic curve determined by the combined effect of initial speed (same as bomber) and gravity. Drag will certainly have an effect but given the fusiform shape of a bomb, its effect would not be great, and it will be quite easily predictable, since Drag would transform the constant speed into a uniformly deccelerated movement with regards to the X-axis of the freefall movement.

That's what I can recall without reading any book, so I welcome corrections.

OTOH, and sorry for not sticking to the thread, I think the main problem with buffs in MA is the lack of a more realistic Norden sight. You should have to spend a good time flying straight after it gives you an accurate reading. So no jinking buffs, and more navigational skills.

Cheers,

Pepe

[TheWobble]

Right on the head pepino, everything you said is correct, however remember that the bomb is moving AWAY from the bomber aswell as foreward, the bomber is moving foreward in a straight line, the bomb is moving foreward in a curve, therefore the bomb has to travel faster to maintain its evenness with the bomber becasue its takind a less direct path than the bomber is, ofcourse the bomb cannot gain any foreward speed so based of the fact that it has to travel farther than the bomber to reach the same point and  it is going the same speed as the bomber (which is unrealistic), it will reach the point after the bomber.
the old addage, the shortest distancebetween  2 points is a straight line, the bomber is going in a straight line, the bomb is not (downward curve) so it cant land directly under it.

[Pepino]

Wobble, the bomb has a combined movement. 2 axis. The vertical one is moving the bomb away from the buff. The horizontal one keeps the bomb steady with the buff. Let's say the buff flies at 250 kts. . Everything inside the buff is moving at 250 kts. Inside the plane, lift counters gravity, so vertical speed of everything inside equals zero (i'm not counting falling pencils from navigator's desk   ). Now you open doors, release the bomb. At this time, the bomb is still moving at 250 kts. on the X axis. Only thing that changes is that now, unattached to the plane, the bomb generates practically 0 lift force, so besides the 250 kts. in the X-axis, it's building speed on the vertical axis.  Given the 9'8 m/s^2 gravitational constant, each second the bomb is falling, it gains 9'8 m/s in vertical speed. The now freefalling bomb is moving on the path determined by the resultant of 2 vectors. One is horizontal and paralell to the Earth, the X-axis speed that, if we forget about drag, is constant at 250 kts. The other one is vertical, perpendicular to the Earth, and determines a uniformly accelerated movement. So, in the second zero, the bomb is moving 250 kts. on the horizontal, and 0 kts. on the vertical (flat trajectory). The first second  X-speed is 250 Kts., and Y-speed is 9'8 m/s. The second nr.2 X-speed is 250kts. and the Y-speed is 19'6 m/s. and so on. So the resultant for the freefalling bomb is a parabolic movement. This does not mean that the bomb will fall behing because it travels farther as, with regards to the plane, the bomb is not moving in the X-Axis (equal speeds) while it goes down.

So, if we do not put drag into the equation, relative X-position of the falling bomb and the buff, given the buff does not changes its course, speed or altitude, remains constant. If we introduce the drag, we alter the X-axis component of the freefall movement. Instead of a straight, constant speed movement, we have a straight, uniformly deccelerated one. The magnitude of the decceleration relates to the drag, and this aspect relates to the shape & mass of the bomb. Given the bomb is a fusiform object (close to the ideal aerodinamic shape, which is the one the drop of water adopts in a freefall), and very massive (thus neglecting the flotability - lift issues), I would say drag forces can be safely neglected without affecting hardly the "sim" side of this game.

So, as a conclusion, and correct me if I'm wrong, the bomb would fall exactly under the buff, if we negate drag's influence, and slightly behind the buff, if drag is modelled.

Cheers,

Pepe

[Pepino]

And, as Hans says, the norden would calculate the drift related to the drag, and compensate accordingly.

Again, IMHO, the only thing that is severely distortioned is the Norden ability to stay perfectly aligned. IIRC, Norden sights were dependent on gyroscopes, and these were VERY sensible to course changes. So if you move your heading even slightly, Norden lecture would be wrong, and ruin your aiming.

Cheers, and thks for an informative thread  

Pepe

[juzz]

Go to this website and click the "History of projectile motion" link at the bottom. There's even a game at the end where you drop water bombs from a building rooftop onto some poor sap!  

[TheWobble]

Still waiting on the reply from Mr. Brochmann, he's prolly out fishin or something   ..I wish I was  

EDIT: At least I have already won the battle so to speak, everybody keeps admitting theat they would only fall directly under the bomber if there was no drag, so even if i am TOTALLY wrong in my theory, the bombs still are not modeled correctly.  I know some say that there wasnt much drag on the bombs but.
A: 25,000 fett is a long way for it to take effect, at that distance even a small amount of force will creat a very visable result.
B:250mph creats alot of drag aswell (faster ya go the more drag ya have)
C: WW2 bombs were far from the aerodynamics bombs we have today, they were rather blunt ESPECALLY the lanc's cookie ( it looks like a hot water heater}



[This message has been edited by TheWobble (edited 02-20-2001).]

[pzvg]

Ah not to query the physics, but I had a talk with my sister (US Air Force Ord specialist) She said there are no differences between bomb cases from WW2 and today a Mk 82 is the same case as the Mk 41 used in 1940, most changes have been in the area of fusing, Also, she said forward motion of bombs is a definite factor, that's why modern A/C use CCIP (continuously computed impact point) sights to allow for that.
And last, on WWII bomber bomb dispersion, my mother cracked up on the nice straight lines comment, she says "yes very straight on film from way up there, covering 2 city blocks long and a block wide down here"
(My mother survived the bombing of Mannheim in WW2, she might not know physics but she does know a little about bombs)

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pzvg- "5 years and I still can't shoot"