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HotelBushranger
03-20-2006, 03:56 AM
Can somebody dig up from somewhere the mathematics behind bomb aiming? Surely in real life, they just didn't figure it out from 'Guess, Check and Improve' sorta methods. As I see it, its basic trig, with factors of bomb weight, aircraft velocity, and wind involved. Can anyone shed some light on this? It would make bomb aming a lot easier for me http://forums.ubi.com/images/smilies/sadeyes.gif

Megile_
03-20-2006, 04:14 AM
W = TF(2) where N = (00) + B

Results may vary

HotelBushranger
03-20-2006, 04:19 AM
http://forums.ubi.com/groupee_common/emoticons/icon_rolleyes.gif

russ.nl
03-20-2006, 04:21 AM
You mean with bomb site? Or on a fighter?

Capt.LoneRanger
03-20-2006, 04:44 AM
Nobody knows - except raaaid. I suggest you ask him over in the PF-forum. He's a grand-grand-great-nephew of Bernoulli and knows everything that is comparable to liquid physics.

He also invented this revolutionary drive, where you move forward with an increasing speed by letting a weight on a line circle around you. He has also prooven us all wrong and showed us, why jet-engines don't work in reality. http://forums.ubi.com/images/smilies/1072.gif

danjama
03-20-2006, 05:08 AM
Raaaid is why i dont go to the PF forum

Capt.LoneRanger
03-20-2006, 05:29 AM
Raaaid is why I **GO** to the PF forum.

It simply lights up your day. http://forums.ubi.com/images/smilies/34.gif

russ.nl
03-20-2006, 06:00 AM
Raaaid Rules http://forums.ubi.com/images/smilies/metal.gif http://forums.ubi.com/images/smilies/88.gif

HotelBushranger
03-20-2006, 06:23 AM
Lol on topic please! http://forums.ubi.com/images/smilies/88.gif

He's not such a bad bloke in real life, hadda chat with him on HL <screams like a female Beatles fan>

Oh, and, bomb site http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

LEBillfish
03-20-2006, 08:39 AM
HBR, I don't know but I believe there is much more involved then point and drop......Factors such as true air Speed, altitude, cross wind directions at various altitudes, temperatures, barometric pressures, bomb type on and on fall into it.

Now, if it is anything like "ballistics trjectories" the math is unbelievable.....In fact, it was for developing the tables for ground and ship artillery that computers were developed.........

However in the end, what would of been supplied to a bombadier my "guess" would be would be a simple chart much like we have here (IAS/TAS). Weather conditions that might affect it would have been determined before they left (if at all).....So in the end like we do they'd simply input altitude, true air speed, with perhaps an estimate of drift due to "known" cross wind direction & a weapon type modifier (if not already figured for them on the ground).

That's ALL a guess only, yet all those things affect a bomb drop......In the end probably the most important thing was keeping flight groups tight, and therefor covering a large enough area some of the bombs would hit the target.

(have a 50x50' foot building you want to hit?...flatten 40 acres....1/4mile sq.)

Tully__
03-20-2006, 03:19 PM
Bomb forward speed slows with time due to the horizontal component of drag. How quickly is slows depends on the horizontal component of the bomb's current air speed.
Bomb vertical speed increases due to gravity. How quickly it increased depends on the difference between gravity acting down and the vertical component of drag due to the bomb's current airspeed.

As altitude decreases, air density increases so drag will also increase.

Also affecting air density are current temp, humidity and barometric pressure.

Add to that varying wind speeds with altitude (already mentioned by Billfish) and you can see that it's not a simple calculation.

If my fluid dynamics training was a little less than 20 years old I might actually have a stab at providing a useful formula, but in addition to trigonometry it would involve integral calculus, specific gravity of air, humidity, wind speed, temperature, barometric pressure etc..
If there were serious windsheers involved (different layers of air travelling at different speeds or in different directions) you'd have to do the calculations for each layer of wind through which the bomb passed to be truly accurate.

In reality bombardiers had an "average" calculation either built into their bombsight or built into specialist slide rules/discs. The results were adjusted with fudge factors for "average" wind, temp, pressure etc (either from tables or by adjustment knobs on the sight itself) and they hoped for the best.

WTE_Galway
03-20-2006, 03:36 PM
Its why this thing

http://www.squadron13.com/B17/B17-21.html

was one of the biggest secrets of the war and had had to be removed from the plane every night and destroyed if the plane went down

AlGroover
03-20-2006, 05:05 PM
First thing is to get yourself a sliderule. Hate to think how much an original specialist one (like Tully says) would cost on ebay.

KrashanTopolova
03-20-2006, 05:22 PM
Hotelbushranger...you are obviously an Aussie...and a CBF...!

There could be at least one disturbing thing there http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

why use compicated maths...?

if you know the velocity of your aircraft and its height and you know the mass of the bomb you are carrying and there is a known constant acceleration of gravity which equals approx 9.82m/s...all you have to do is calculate the distance the bomb will travel under that acceleration downwards and under that constant velocity of the aircraft at the height at which it is dropped.

here are a few simple physics formulas:

speed = d/t
v = change in d/change in t

A better way is to practice...practice...practice...and forget about your free-running wheels because they don't come into this equation (very much).

DaimonSyrius
03-20-2006, 05:38 PM
Originally posted by KrashanTopolova:
why use compicated maths...?

if you know the velocity of your aircraft and its height and you know the mass of the bomb you are carrying and there is a known constant acceleration of gravity which equals approx 9.82m/s...all you have to do is calculate the distance the bomb will travel under that acceleration downwards and under that constant velocity of the aircraft at the height at which it is dropped.

here are a few simple physics formulas:

speed = d/t
v = change in d/change in t

A better way is to practice...practice...practice...and forget about your free-running wheels because they don't come into this equation (very much).
Hi KrashanTopolova,

And what about drag (air friction) building up, increasing with velocity squared and significantly modifying the way the bombs fall? That makes things quite different from what would otherwise be a frictionless free-fall.

And in this respect, what about air density changing significantly all the way along the fall, when it's a high-altitude level bombing run?

Those would be good reasons to use complicated maths, or to use sophisticated bombing sights... Which is what happened in actuality, I believe.

The simple free-fall equations only apply in that simple form when one disregards the effects of drag. When drag is taken into account, the concept, and the phenomenon, of terminal velocity has to be taken into account too, along with several others... and then things turn out to not be that simple.

About practice, yes that's always a good thing, but still, sights like the one in the link provided by Galway really helped a lot.

http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
S.

Tully__
03-20-2006, 05:53 PM
Please disregard Krashan, he seems to enjoy oversimplifying things to the point where the theory breaks down or overcomplicating things with the express purpose of frustrating other posters, though bomb ballistics in the game may actually be that simple and for low altitude bombing it would at least be a good approximation if you're not using parafrags.

DaimonSyrius
03-20-2006, 05:55 PM
I know the humourous ways of our resident conveyor-belt specialist, no problem http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif

Cheers,
S.

AlGroover
03-20-2006, 10:13 PM
You might find this article interesting. www.physicstoday.org/pt/vol-54/iss-8/p40.html (http://www.physicstoday.org/pt/vol-54/iss-8/p40.html)

HotelBushranger
03-21-2006, 04:10 AM
So, how did they figure out that you needed such an angle at such a height? Like I said before, it couldn't have been just guess check and improve could it?

WB_Outlaw
03-21-2006, 05:36 AM
Originally posted by HotelBushranger:
So, how did they figure out that you needed such an angle at such a height? Like I said before, it couldn't have been just guess check and improve could it?

High end bombsights of the day used an electro-mechanical computer to calculate the release point based on inputs provided by the bombardier. The best of these was the Norden bombsight and even it wasn't very good unless the conditions were perfect.

Pre-calculated tables were used to enhance accuracy when less sophisticated bomb sights were used. Without guided munitions it is much easier to build more aircraft, train more crews, drop more bombs, and suffer the losses than it is to increase accuracy with any type of sight. There are just too many variables on the way down from 30000 feet to bomb accurately.

--Outlaw.

HotelBushranger
03-21-2006, 06:02 AM
Faaaaaaaair enough

foxyboy1964
03-21-2006, 01:28 PM
Isnt it the case that most bombs dropped in WW2 missed the target by quite a wide margin? Wasn't it something like 2 out of 5 landed anywhere near the target? Maybe our inaccurate bombing is just about right.

StellarRat
03-21-2006, 03:19 PM
Yeah, they couldn't hit anything. That's why it took so many bombers and sometimes multiple missions to destroy or even damage targets. My Mom (she is German) said they blew up more fields and orchards than anything else. The British gave up on aiming and just tried to hit somewhere in the cities. This raised the hit probability, but believe or not they even missed cities quite often. They best bombing she saw was a single Mosquito at low level (tree top) that destroyed an entire factory in one run.

Akronnick
03-21-2006, 04:09 PM
Originally posted by WB_Outlaw:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by HotelBushranger:
So, how did they figure out that you needed such an angle at such a height? Like I said before, it couldn't have been just guess check and improve could it?

High end bombsights of the day used an electro-mechanical computer to calculate the release point based on inputs provided by the bombardier. The best of these was the Norden bombsight and even it wasn't very good unless the conditions were perfect.

Pre-calculated tables were used to enhance accuracy when less sophisticated bomb sights were used. Without guided munitions it is much easier to build more aircraft, train more crews, drop more bombs, and suffer the losses than it is to increase accuracy with any type of sight. There are just too many variables on the way down from 30000 feet to bomb accurately.

--Outlaw. </div></BLOCKQUOTE>

This is why modern PGMs use laser beams or GPS or Imaging Infra-red to guide them to the target. Even with highly sophisticasted CCRP, CCIP or DTOS systems can't control everything. The target could move, or the wind could change during the weapon's flight time. With Iron Bombs, once they are released they are on their own.

KrashanTopolova
03-21-2006, 04:12 PM
[QUOTE]Originally posted by DaimonSyrius:
Hi KrashanTopolova,

And what about drag (air friction) building up, increasing with velocity squared and significantly modifying the way the bombs fall? That makes things quite different from what would otherwise be a frictionless free-fall.

______________________________________________
air friction on a heavy falling object only reduces acceleration of that falling object to the point where it no longer exists and then a terminal velocity arises for the falling object (where the frictional force = the gravitational force producing the acceleration, thus once that happens the velocity is no longer changing and that means there is no more acceleration.)

The bomb falls predictably. I think you may have have been thinking of (insignificant) wind effects on the bomb that may or may not be able to divert the bomb from its trajectory.

The bomb in fact follows a normal curved flight as any other projectile under the influence of gravity. This is not appreciated by eye by the pilot/bomb aimer...it is appreciated by an observer on the ground.
_______________________________________________

And in this respect, what about air density changing significantly all the way along the fall, when it's a high-altitude level bombing run?
______________________________________________
the WWII bomb does not have flight (guiding) surfaces over the centre of gravity...hence air density does not affect the trajectory of this sort of falling missile.
_______________________________________________

Those would be good reasons to use complicated maths, or to use sophisticated bombing sights... Which is what happened in actuality, I believe.
_______________________________________________
they were sophisticated only in the sense that were frequently improved in their capability to utilise the simple physics involved.
_______________________________________________

The simple free-fall equations only apply in that simple form when one disregards the effects of drag. When drag is taken into account, the concept, and the phenomenon, of terminal velocity has to be taken into account too, along with several others... and then things turn out to not be that simple.
_______________________________________________
drag assists in producing the curved path of the trajectory...that's all
_______________________________________________

About practice, yes that's always a good thing, but still, sights like the one in the link provided by Galway really helped a lot.

http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
_______________________________________________
'...there are many who believe an Orator is an unpopular wind instrument...'
_______________________________________________

KrashanTopolova
03-21-2006, 04:14 PM
[QUOTE]Originally posted by Tully__:
Please disregard Krashan, he seems to enjoy oversimplifying things to the point where the theory breaks down or overcomplicating things with the express purpose of frustrating other posters, though bomb ballistics in the game may actually be that simple and for low altitude bombing it would at least be a good approximation if you're not using parafrags

_______________________________________________

'...I have listened to your humble opinion...and it certainly is...'
_______________________________________________

Stackhouse25th
03-21-2006, 04:16 PM
E=MC^2

With nukes bomb aiming is not required, wind is not a factor.

Tully__
03-22-2006, 12:56 AM
Originally posted by HotelBushranger:
So, how did they figure out that you needed such an angle at such a height? Like I said before, it couldn't have been just guess check and improve could it?
They started with the formula for an object falling free in a vacuum and added the result of wind tunnel testing on bombs to the formula.

The end result is not that complex but it looks considerably different to the simple newtonian formula for free falling in a vacuum and requires some quite advanced math to arrive at.

NonWonderDog
03-22-2006, 11:05 AM
It's a hell of a lot of calculus, actually. Nowadays they just do a bit of CFD on the bomb and plug numbers into the CCRP computers. Everything's done iteratively.

They can't be all that accurate thanks to wind currents and such, but rest assured that they do take air density, pressure, and temperature into account when dropping iron bombs today. Hell, they supposedly have the circuitry to take an unlimited number of winds aloft under the plane into account, if the target is somehow surrounded by weather balloons or something.


I might actually be capable of writing most of the program, now that I think about it. That's a scary thought, but I guess it's not all that complicated if you have access to the data you'd need.

DaimonSyrius
03-22-2006, 01:14 PM
Originally posted by KrashanTopolova:
**Utterly garbled post, committed (since 'composed' would be an overstatement) with a quoting style that seems as if it were purposefully conceived to impair critically the reader's comfort and, particularly, the attempts to reply to it; impairment that is achieved regardless of whether it was deliberate or just a spin-off from a peculiar aesthetic ability, or lack thereof**

KrashanTopolova,

What you are now saying about drag is quite fine and dandy (to put it very optimistically), but:

-Your initial, previous post in this thread (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/7531059424/r/1511022524#1511022524) ommitted any mention of drag or terminal velocity while you were pondering on how very very easy is to carry out level bombings at altitude (the subject of this thread) by simply using the very very simple maths involved in a couple of equations that would only need to use constant gravity and, as variables, the mass of the bomb, and the aircraft's velocity and altitude at which bombs are dropped.

-Even if you are now paying a bit more attention to drag and terminal velocity as factors that will influence the bomb's trajectory, making it different from what it would be in a frictionless free-fall, your contention that "The bomb falls predictably" is ludicrous, and were you to tell that to any of the gunners that dropped them during WWII, they would jump at your neck at once http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

-For one thing, if your predictability means to use just the four parameters you listed, and using just a couple of simple-math equations (which seemed to be your complete point), and then 'adjusting for some drag, some terminal velocity' (which you seem now to be acknowledging that will have a contribution in shaping the bombs trajectory), your 'adjustment factor' for the simple-maths you advocate would itself be variable; and would tomorrow be different from what it was today, not to mention season changes and location changes; and that's without taking winds into account at all; I'm just referring to athmospheric pressure, temperature, density.

-About your interpretation that air density will not affect the trajectory of the falling bomb: remember that you are, at the same time, acknowledging that air friction (depending itself on density) is what determines the fact that some terminal velocity will be reached (and its value will depend on air friction, hence on density). So if you are to compare (or 'predict' the difference in) how bombs fall when they're dropped at, say, 6.000m and 8.000m altitude, besides your simple-maths you'll need to account for the different density gradients along their different trajectories. When you throw in that bit about bombs not having "flight (guiding) surfaces" and you attempt to use that as the reason by which air density would not play any role at all in shaping the trajectory, you score almost as much close to physics target as you do when you ellaborate on your conveyor-belt poetry, I mean theory http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
S.

P.S.: And then, besides all of the above, winds enter the stage and would have to be accounted for somehow. Even if the bomb is only falling and not flying on airfoils, winds will of course play a quite significant role in the whole problem.

KrashanTopolova
03-22-2006, 03:47 PM
Originally posted by DaimonSyrius:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by KrashanTopolova:
**Utterly garbled post, committed (since 'composed' would be an overstatement) with a quoting style that seems as if it were purposefully conceived to impair critically the reader's comfort and, particularly, the attempts to reply to it; impairment that is achieved regardless of whether it was deliberate or just a spin-off from a peculiar aesthetic ability, or lack thereof**

KrashanTopolova,

What you are now saying about drag is quite fine and dandy (to put it very optimistically), but:

-Your initial, previous post in this thread (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/7531059424/r/1511022524#1511022524) ommitted any mention of drag or terminal velocity while you were pondering on how very very easy is to carry out level bombings at altitude (the subject of this thread) by simply using the very very simple maths involved in a couple of equations that would only need to use constant gravity and, as variables, the mass of the bomb, and the aircraft's velocity and altitude at which bombs are dropped.

-Even if you are now paying a bit more attention to drag and terminal velocity as factors that will influence the bomb's trajectory, making it different from what it would be in a frictionless free-fall, your contention that "The bomb falls predictably" is ludicrous, and were you to tell that to any of the gunners that dropped them during WWII, they would jump at your neck at once http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

-For one thing, if your predictability means to use just the four parameters you listed, and using just a couple of simple-math equations (which seemed to be your complete point), and then 'adjusting for some drag, some terminal velocity' (which you seem now to be acknowledging that will have a contribution in shaping the bombs trajectory), your 'adjustment factor' for the simple-maths you advocate would itself be variable; and would tomorrow be different from what it was today, not to mention season changes and location changes; and that's without taking winds into account at all; I'm just referring to athmospheric pressure, temperature, density.

-About your interpretation that air density will not affect the trajectory of the falling bomb: remember that you are, at the same time, acknowledging that air friction (depending itself on density) is what determines the fact that some terminal velocity will be reached (and its value will depend on air friction, hence on density). So if you are to compare (or 'predict' the difference in) how bombs fall when they're dropped at, say, 6.000m and 8.000m altitude, besides your simple-maths you'll need to account for the different density gradients along their different trajectories. When you throw in that bit about bombs not having "flight (guiding) surfaces" and you attempt to use that as the reason by which air density would not play any role at all in shaping the trajectory, you score almost as much close to physics target as you do when you ellaborate on your conveyor-belt poetry, I mean theory http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
S.

P.S.: And then, besides all of the above, winds enter the stage and would have to be accounted for somehow. Even if the bomb is only falling and not flying on airfoils, winds will of course play a quite significant role in the whole problem. </div></BLOCKQUOTE>

_____________________________________________
forget terminal velocity for a bomb...i never said it was a factor.
forget winds and other such like...they don't stop a bomb from getting into a target area with a significant level of accuracy...
forget conveyor belts...they don't work...!

...when I say the bomb *Falls* predictably i mean in the Newtonian sense...

eg.

a bomb dropped from an aircraft flying horizontally at (say) 120 m/s-1, reached the ground in 5.0 s.

from this all sorts of things can be computed...i will leave it to readers to work them out:

the vertical velocity of the bomb just before it strikes the ground target is:
Vv=u + at = 0.5 x 9.8 = 49m/s-1

the Total velocity of the bomb just before it strikes the ground is:
V=sqrt(120sq + 49sq) = 130 m/s-1
______________________________________________
'...Time is nature's way of stopping everything happening at once...'
______________________________________________

Boyington6
03-22-2006, 05:06 PM
Earlier in this thread, AlGroover recommended a Physics Today article. If you haven't read it yet, you should, as it answers a lot of questions you gents are posing. For instance:

"Early in the 21st Bomber Command's campaign, LeMay required B-29 crews to take bomb-strike photographs, to be used for competitive scoring of the various squadrons, groups, and wings. Unfortunately, taking bomb-strike photos required that the bomber fly straight and level for the duration of the bomb fall (typically 40 seconds for high-altitude releases). That, of course, increased the bomber's vulnerability to antiaircraft fire and fighter attack.

Happily, however, with a photograph of the bomb in relation to the ground only 10 seconds after release, we could use the laws of physics to calculate the precise impact point. That would spare the bomber and crew about 30 seconds of straight and level flight over the target. The problem was of double interest to the 509th Group. They needed to know where the atomic bomb would land but, to get away from its awesome shock wave, they needed to veer promptly after releasing the bomb. "

If this guy who was actually up there doing it says this, then, I guess they do fall predictably. But "predictable" back then meant a different thing than it does now doesn't it? http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

I think terms like "Daylight Precision Bombing" were as likely as not to have been cooked up for the folks at home, so they wouldn't think of things like Guernica.

DaimonSyrius
03-22-2006, 05:55 PM
http://www.airspacemag.com/ASM/Mag/Index/1995/FM/swpn.html


.../...
With production bombsights in hand, the Army Air Corps finally had an opportunity to rehearse daylight precision strategic bombing. Practice did not make perfect. Brigadier General H. H. "Hap" Arnold (later commander of the Army Air Forces) plotted the progress of one of his bomb groups in 1935: "From 15,000 feet, we began on the first day by placing eggs within 520 feet of the target, closing the gap to 480 feet at the end of seven days, to 300 feet at the end of 27 days, and placing its bombs regularly within 164 feet of a target no bigger than a woodshed at the end of 41 days. If airmen got to talking a little too confidently about tossing it right in the pickle barrel, our continued improvement in bombing with the Norden sight may explain why."

McFarland notes that close enough may count in horseshoes but not with bombs. A 2,000-pounder--the largest in Army inventory before the war--blew a crater 50 feet in diameter and had a maximum effective fragmentation range of 125 feet.

.../...

accuracy did improve with experience. However, real precision bombing was seldom achieved. Instead of a few sharpshooters, the Army Air Forces mustered vast armadas--up to 2,000 B-17 and B-24 bombers per mission--to darken the sky over Germany, bludgeoning the enemy by saturating target areas with 500-pound bombs. Except in the final month of bombing--April 1945--more than half the bombs dropped landed more than 1,000 feet from their targets. All told, less than one-third landed within 1,000 feet of the aiming point. Against Japan, the daylight precision strategic bombing record was only slightly better, prompting a shift toward area bombing, which finally turned the tide.

The whole article is of interest, IMO.

And by the way, it all doesn't sound as something to be easily done, with just a couple simple equations scribbled while whistling along nonchalantly http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif
(and crosswinds apparently had to be computed)

Cheers,
S.

DaimonSyrius
03-23-2006, 02:55 AM
Some more input about drag and ballistic trajectories. This figure and its corresponding text come from a good site explaining aircraft gun ballistics, but the principle of course applies to bombs if we consider them as being 'shot' with a 'muzzle velocity' equivalent to the bomber aircraft's velocity at the time of dropping.

Ballistics (http://www.geocities.com/CapeCanaveral/Hangar/8217/fgun/fgun-th.html)

http://www.infonegocio.com/daimon/img/Trajectory-drop.gif


In vacuum the projectile would retain its horizontal speed, and follow a parabolic curve (red). In air the projectile encounters drag, a speed-dependent force. The air molecules absorb part of the energy and momentum of the projectile, while the friction converts some of the energy to heat. In general this results in a trajectory that is more curved (blue), although a properly designed round might actually have some body lift, counteracting gravity. The line of sight is made to coincide as close as possible with the curve over the ranges expected in combat (green). Often the guns are given a slight upwards angle, to make the match easier. For guns in the wings this is convenient, because the wings have a positive angle of attack.

Notice the remark about 'properly designed' rounds (which completely lack airfoils, never mind the center of gravity http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif) having a finite amount of lift derived from its shape; wich further adds to decreasing simplicity.

This is of course related to:

Originally posted by KrashanTopolova:
bomb dropped from an aircraft flying horizontally at (say) 120 m/s-1
.../...
the Total velocity of the bomb just before it strikes the ground is:
V=sqrt(120sq + 49sq) = 130 m/s-1
where the horizontal velocity, after the long travel the bomb has almost completed, doesn't seem to account for drag at all, in this simple formulation that would be describing, rather, a bombing carried out in vacuum (hence, not on this planet) http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
S.

HotelBushranger
03-23-2006, 03:03 AM
Dangit http://forums.ubi.com/images/smilies/16x16_smiley-mad.gif

Was hoping for some calcs, instead of having to remember what angle at what speed/height <sigh>

El Turo
03-23-2006, 03:24 AM
I did read an article on WWII bombing that laid out exactly what Stack has quoted.. and went on to say that it took 2000 bombers to statistically guarantee a bomb landing within 100 feet of the intended target.

http://forums.ubi.com/images/smilies/35.gif

HotelBushranger
03-23-2006, 03:27 AM
Buggery!

KrashanTopolova
03-23-2006, 03:51 PM
Originally posted by DaimonSyrius:
Some more input about drag and ballistic trajectories. This figure and its corresponding text come from a good site explaining aircraft gun ballistics, but the principle of course applies to bombs if we consider them as being 'shot' with a 'muzzle velocity' equivalent to the bomber aircraft's velocity at the time of dropping.


<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">In vacuum the projectile would retain its horizontal speed, and follow a parabolic curve (red). In air the projectile encounters drag, a speed-dependent force. The air molecules absorb part of the energy and momentum of the projectile, while the friction converts some of the energy to heat. In general this results in a trajectory that is more curved (blue), although a properly designed round might actually have some body lift, counteracting gravity. The line of sight is made to coincide as close as possible with the curve over the ranges expected in combat (green). Often the guns are given a slight upwards angle, to make the match easier. For guns in the wings this is convenient, because the wings have a positive angle of attack.

Notice the remark about 'properly designed' rounds (which completely lack airfoils, never mind the center of gravity http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif) having a finite amount of lift derived from its shape; wich further adds to decreasing simplicity.

This is of course related to:

Originally posted by KrashanTopolova:
bomb dropped from an aircraft flying horizontally at (say) 120 m/s-1
.../...
the Total velocity of the bomb just before it strikes the ground is:
V=sqrt(120sq + 49sq) = 130 m/s-1
where the horizontal velocity, after the long travel the bomb has almost completed, doesn't seem to account for drag at all, in this simple formulation that would be describing, rather, a bombing carried out in vacuum (hence, not on this planet) http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

Cheers,
S. </div></BLOCKQUOTE>

_____________________________________________
by corollary: none of us experienced in il-2 need a bombsight when fast at low level...its all intuitive. Its only when the dimensions related to the fall of the projectile increases to a large factors that greater reliance needs to be placed on maths...
______________________________________________

DaimonSyrius
03-23-2006, 04:44 PM
Originally posted by KrashanTopolova:
by corollary: .../... Its only when the dimensions related to the fall of the projectile increases to a large factors that greater reliance needs to be placed on maths...

...which is what was being discussed here all along. Q.E.D.

Cheers,
S.

KrashanTopolova
03-24-2006, 05:19 PM
Originally posted by HotelBushranger:
Dangit http://forums.ubi.com/images/smilies/16x16_smiley-mad.gif

Was hoping for some calcs, instead of having to remember what angle at what speed/height <sigh>

___________________________________________
if you know how to transpose formulas...try these (but be careful, there is a lot to consider...its not necessarily just a case of plugging the numbers into these formulas):

where u = initial velocity (in this case u = 0)
a (g) = acceleration under gravity = 9/82m/s-1
t = time taken for bomb to impact ground
d = distance travelled by bomb

from v = u + at and from d = ut + 1/2a x tsq

v = -gt
d = - 1/2g x tsq

Set up a quick mission, mark your release point on the ground with a static object, note your speed at time of bomb release and determine how many seconds it takes the bomb to impact the ground...then do the calculations (don't forget the horizontal component of velocity under g.)

Note Well:
before using this information you must stand up from your computer...turn around 3 times on the spot with both hands in the air and swear that you will never reveal this information to anyone else...

Examples:
(note: friction is ignored by these equations)

1.
An aircraft flying at airspeed 180 kmh (50m per second)

i). How far horizontally from point of bomb release is ground impact?
=> 500m

ii). What is altitude of this aircraft at time of bomb release?
=> 500m

2.
u = 180 kmh (50ms-1)
t = 20s

i) What is the ground distance of target from point of bomb release?
=> 1 000m

ii) altitude of aircraft at bomb release?
=> 2 000m

3. Aircraft at 1 000m with airspeed of 300kmh

i) ground distance to target at point of bomb release?
=> 3 735m

May be good test of flight model's wind and friction modelling
_______________________________________________

KrashanTopolova
03-26-2006, 06:31 PM
I forgot...besides friction etc the above equations also do not take into account the motion of revolution of the earth...neither does IL-2 I suspect...

and g is rounded to 10ms-1 rather than 9.8ms-1

Z4K
03-27-2006, 02:42 AM
"tsq" has to be one of the most obscure ways of writing "t squared" I have ever seen.

Alternatives include:
t^2
t*t
(t squared)

All are clearer. Personally I tend to use a^b to mean "a raised to the power of b".
In your "example 1" (I think "example" in this context means "an example of Krashan's total failure to grasp the fundamentals of kinetics) you write the aircraft speed is 50m/sec and then list points i) and ii) both with values of 500m. Are these answers, or more information with which to solve the problem?

I can only assume they're answers.

In which case, "example (of why Krashan should have his internet connection terminated) 1", there's insufficient information to answer the problem. Either that, or ALL aircraft flying at 50m/s are at an altitude of 500m, and their bombs will land 500m downrange of where they're released.

It appears you've forgotten to put the time of 10 seconds.

In "example (of why Krashan doesn't, I hope, have a job where lack of critical thinking skills will ever put anyone in danger) 3":

Aircraft altitude 1000m:

S = ut + 0.5*g*t^2
1000 = 0 + 0.5*10*t^2
t^2 = 1000/5
t = 200^(0.5)
t ~= 14.1421sec

The bomb horizontal speed is 300km/hour (3.6km/hour = 1m/s):

S = ut + 0.5*a*t^2
S = (300/3.6)*(200^0.5) + 0
S ~= 83.3333*14.1421
S ~= 1179m

Where 3735 came from, I have no idea. I notice you've already edited the post at least once - athough apparently not to fix mistakes.

Kapteeni
03-27-2006, 08:33 AM
Originally posted by Capt.LoneRanger:
Raaaid is why I **GO** to the PF forum.

It simply lights up your day. http://forums.ubi.com/images/smilies/34.gif
http://forums.ubi.com/images/smilies/35.gif http://forums.ubi.com/images/smilies/351.gif

NonWonderDog
03-27-2006, 09:23 AM
You guys are REALLY oversimplifying this. The only practical way to solve this problem is iteratively; the calculus is really just too hard.

I don't have time to sit down and work all this out again, but I belive it's something like this:


You need a computer code to keep track of velocity in both x and y (it's easier that way), and set up loops with the following -- plus some more conditions.

V_x = V_x - 1/2 * rho * S_x * C_d * V_x*|V_x| * dt/m + V_wind;
V_y = V_y - 1/2 * rho * S_y * C_d * V_y*|V_y| *dt/m - g*dt + V_thermal;

"dt" is the time interval between updates. You have to keep track of altitude in this, too, in order to get the value of rho. So,

alt = alt_0 + V_y*dt;

where alt_0 is release altitude.

You can either use a lookup table or the barometric formula for rho or use this bit of code I picked up long ago, setting the appropriate T0, p0 and rho0:


function [T p rho] = StdAtpSI(h)

h1=11; h2=20; h3=32; a0=-6.5e-3; a2=1e-3; g=9.80665; mol=28.9644;
R0=8.31432; R=R0/mol*1e3;

T0=288.15; p0=1.01325e5; rho0=1.2250; T1=T0+a0*h1*1e3;
p1=p0*(T1/T0)^(-g/a0/R); rho1=rho0*(T1/T0)^(-g/a0/R-1); T2=T1;
p2=p1*exp(-g/R/T2*(h2-h1)*1e3); rho2=rho1*exp(-g/R/T2*(h2-h1)*1e3);

if h <= h1
T=T0+a0*h*1e3;
p=p0*(T/T0)^(-g/a0/R);
rho=rho0*(T/T0)^(-g/a0/R-1);
elseif h <=h2
T=T1;
p=p1*exp(-g/R/T*(h2-h1)*1e3);
rho=rho1*exp(-g/R/T*(h2-h1)*1e3);
elseif h <= h3
T=T2+a2*(h-h2)*1e3;
p=p2*(T/T2)^(-g/a2/R);
rho=rho2*(T/T2)^(-g/a2/R-1);
end



It's a far cry from high-school level projectile physics, and I'm certain that I left several things out. What I have here can still be solved with calculus, because I'm forgetting the lift created by the bomb casing (yes, it does exist). I might try to update this if I feel up to it.

KrashanTopolova
03-27-2006, 04:43 PM
Originally posted by Z4K:
"tsq" has to be one of the most obscure ways of writing "t squared" I have ever seen.

Alternatives include:
t^2
t*t
(t squared)

All are clearer. Personally I tend to use a^b to mean "a raised to the power of b".
In your "example 1" (I think "example" in this context means "an example of Krashan's total failure to grasp the fundamentals of kinetics) you write the aircraft speed is 50m/sec and then list points i) and ii) both with values of 500m. Are these answers, or more information with which to solve the problem?

I can only assume they're answers.

In which case, "example (of why Krashan should have his internet connection terminated) 1", there's insufficient information to answer the problem. Either that, or ALL aircraft flying at 50m/s are at an altitude of 500m, and their bombs will land 500m downrange of where they're released.

It appears you've forgotten to put the time of 10 seconds.

In "example (of why Krashan doesn't, I hope, have a job where lack of critical thinking skills will ever put anyone in danger) 3":

Aircraft altitude 1000m:

S = ut + 0.5*g*t^2
1000 = 0 + 0.5*10*t^2
t^2 = 1000/5
t = 200^(0.5)
t ~= 14.1421sec

The bomb horizontal speed is 300km/hour (3.6km/hour = 1m/s):

S = ut + 0.5*a*t^2
S = (300/3.6)*(200^0.5) + 0
S ~= 83.3333*14.1421
S ~= 1179m

Where 3735 came from, I have no idea. I notice you've already edited the post at least once - athough apparently not to fix mistakes.
_____________________________________________
I see my mistake...thank you.

I did it quickly at the computer with no references. My mistake must have arisen from miscontruing formulas in memory and thereby mixing up the transposition.

May you find yourself 1179m away on the ground from an online bomb-loaded aircraft flying at 300kmh-1 http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif
______________________________________________

WTE_Galway
03-27-2006, 09:17 PM
some nice person write me a little windows pop up applicaton that will do teh calcs for me please

Z4K
03-28-2006, 02:39 AM
NonWonderDog:

Don't get me wrong, I wasn't saying using basic idealised Newtonian mechanics would answer the question, just that Krashan had made mistakes.


This is the output of code (within Octave - like MatLab) I use for this sort of problem.

http://www.5images.com/uploads/1ab5869443.jpg

Each set of three lines represents a flight of 7 seconds (calculated at 0.001 sec intervals). Where a new line branches from the first, a new angle of attack has been set. Clumsy, but does the job (eventually - the above was about 5 to 10 minutes of work).

I've ignored lots of the variables you mention, because over a 50m flight and 8m drop, they're just not relevant.


Krashan:

The feeling's mutual http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

KrashanTopolova
03-28-2006, 08:04 PM
Originally posted by WTE_Galway:
some nice person write me a little windows pop up applicaton that will do teh calcs for me please

______________________________________________
You don't need a computer program...despite the math formulas its basically simple.

So in simple physics maths:

set up a mission and note your height (y) [say 1 000m) and initial velocity (u) [say 300 kmh airspeed at time of bomb release) => which in metres per seconds is 300 * 1000 to get metres and the answer to that is divided by 60 to get minutes and the answer to that is again divided by 60 to get (83.3)metres per second.

From here you don't even need the time taken for the bomb to impact the ground...you can work it out more accurately just from the above information.

Its best to work out each component of velocity (horizontal and vertical) separately instead of using one big formula.

The y component is: y = 1/2*a*tsq
where a = gravitational acceleration (9.8 ms-1) or a can be rounded up to the concept g (10m/s)
to find t (finding t is demonstrated in post by Z24).

after finding t, instead of using the rest of the formula used by Z24 simply use x = ut which gives you the ground distance covered by the parabolic fall of the bomb in that amount of t. (x (x co-ordinate) is in fact a different concept to s which is displacement (in physics terms))

anyway:

in that example u = 83.3 recur
and t was found to be 14. something)

=> x = ut => 83.3(14. something) = 1179m
(the unit seconds cancel out to leave just metres (m) in the answer)

With a cluster of bombs a hit is more likely than not in IL-2 or real life despite the simplification of the problem
___________________________________________

Tully__
03-28-2006, 09:00 PM
Originally posted by Z4K:
...I think "example" in this context means "an example of Krashan's total failure to grasp the fundamentals of kinetics...

Krashan understands physics better than most, he's deliberately messing it up to promote contentious discussion. If you're going to append labels to Krashan, I think something from the "Troll" category would be more appropriate than the "Ignorant" category http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

If he doesn't keep his posting down to a reasonable level of provocativeness he may find his frequency of posting is artificially curtailled... fortunately many of the members here are still finding it amusing rather than irritating so he may continue for a while. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

KrashanTopolova
03-29-2006, 04:39 PM
Consider it curtailed...see ya !