OutOnTheOP

Oh, ok. I concede; clearly your calibrated eyeball can tell the exact AOA and profile of the wings from a mere photograph. :megalol:

Oh, in that case, clearly the F-4 should outmaneuver the F-16; after all, the F-16 has a wing loading of 88.3 lb/ft^2, while the F-4 has a superior 78 lb/ft^2 (according to whatever arbitrary weight statuses wikipedia assumed). Does the Phantom II outperform the F-16?

False equivalence. Just because I am stating it makes it more combat effective, does not mean it is not ALSO more maneuverable. It has equal or superior TWR in combat loading, much superior TDR in combat loading, and you have not actually shown any comparisons of lift-versus-drag (the calculation that actually matters for sustained turns), because the "wing" on the F-35 is not the only lift-producing surface. Pilots that have flown it state that it turns as well as or superior to an F-16, and there is no hard evidence to counter those claims; wing-area-versus-assumed-(but always full fuel!)-combat-weight comparisons are entirely unconvincing.

Again, all that "shows" is that the nose appears to be at a higher angle in a static photo. Even assuming that the F-35 wasn't porpoising at the exact instant of the picture, that only speaks to fuselage angle, and not wing angle. All aircraft have a built-in AOA designed into the wing, so that the body sits level at a designed cruise airspeed. At lower than that speed, it will be nose-high. Higher than that speed, it will be nose-low. They intentionally set it to an airspeed selected deliberately during the design phase, so that the fuselage is most-level (IE, most aerodynamic) at a certain airspeed. SO. The nose may be higher, but that doesn't mean the wing's AOA actually is; you cannot inherently measure one from the other. It likely just means that the designers, realizing the F-35 will spend most of it's time at a higher cruise airspeed than the F-16 (no surprise here; the F-35, combat-loaded, is much less draggy), designed it with less built-in AOA in the wings. Wanna see what happens when an aircraft is designed for low-speed cruise, with a very high built-in AOA? Higher cruise speed makes the aircraft MORE combat effective, not LESS.

This depends on one's definition of "fast". It may not have the dash speeds of some of the '60s interceptors, but it has a much higher sustained speed. Exception for the SR71, which, well, isn't exactly a fighter, now is it? (no, it's not. Not even in YF12 guise) It's huge, for one. Besides, it's all fuel!

I was being sarcastic. Given that the F-35, with a comparable combat-range fuel load, has comparable or superior TWR, and VASTLY superior TDR, it is highly, HIGHLY likely that it can very easily maintain level flight. In fact, given it's TWR, it should be able to CLIMB out of 50 degree level flight with ease. I mean, if it has greater than 1:1 TWR, the only thing preventing it from hanging stationary on it's tail is lack of airflow past control surfaces, and possibly an engine stall from insufficient feed air.

Clearly, what he's trying to say is that, yeah, while the F-35 has control authority to maintain control at 50+degrees AOA, it OBVIOUSLY doesn't have the thrust-to-weight or thrust-to-drag to maintain level flight. :doh::lol::lol::lol:

The fuselage being at a higher angle nose-up only means that the aircraft was optimized for cruise at a higher airspeed. That, in turn, to me indicates that the F-35 is absurdly more aerodynamically efficient and has a far, far higher thrust-to-drag ratio. None of that points to the F-35 being inferior in maneuvering; only that it is capable of realistically sustaining much higher speeds.

You don't know exactly; you get it close. Like I said, it still drifts to level... you're just helping it out by getting it closer.

I *think* it is correct behavior: say you uncage on the ground, pointing north. The artificial horizon gyro "thinks" that it is level, but it is in fact tilted up (in the north) some 15-20 degrees, because the airplane is not at a level stance on the ground. If you then take off and turn to the east, the gyro will be tilted some 15-20 degrees to the right. As best I can tell, the artificial horizon is a ball with a slightly weighted bottom (to force it to level), and the gyro is just there to steady it and prevent it from tumbling. This means that if you continue to fly generally level, it will slowly settle into level. It also means that violent maneuvering will cause it to drift out of level. Both appear to be reflected in how the simulation seems to do it. If you're impatient, just go wings level and momentarily cage it, and it'll be good to go.

Agreed. I think it's a symptom of current film and television: everyone is trying to out-do the last guy. They're trying WAY too hard to be "edgy" and "push the envelope", and in BSG 2.0, it just got tiring. OK, we get it... it's supposed to be suspenseful that anyone could be a Cylon. But when you drag on for four seasons just beating up on that dead horse, it gets old. Perhaps consider some plot twists that aren't *exactly* the same every time. Oh, and people like for the good guys to win *sometimes*. When the plot is predicated on hopelessness and constant failure, the viewer base will just get weary of it. And lord help us, please, no more insanely archetyped characters like the 2003 Starbuck. Oh. My. God. Possibly the most annoying, cringe-inducing character I have ever, in all my television-watching life, been subjected to. ...and that's including the ones that are SUPPOSED to be annoying! I actually rate her worse than JarJar! All in all, I agree with the above poster: the characters in the series were a little *too* undisciplined and prone to bickering and infighting. I think that anyone who seriously believes that this is what would happen in an apocalypse situation, needs to reconsider: people will, in such situation, do what has the best odds of leading to their survival. In most cases (including BSG), it doesn't make sense for the characters to expend so much energy fighting each other, when they are better served acting in coordination. Also, it always irritated me that like three episodes in, they make a big deal about being down to their last explosive rifle bullets. You know... the kind of bullets that can actually hurt Centurions (and then they make a big deal about how, normal bullets just bounce off). I mean, if the main, existential threat to humanity- the threat that is the Colonial Fleet's very reason for existence- requires explosive bullets to hurt, then why, oh, why, did they not stockpile up primarily (or even exclusively) that kind? That's just D-U-M, dumb.

A-26B Invader with the 8-gun nose pack (14 .50 cal forward armament!). B-25H Mitchell with 75mm cannon and 8 forward .50 cal. Woe unto anything on ground or sea. 'nuff said.

And yet people are complaining the Mustang turns too good already. Sounds to me like you want the 75" Mustang... otherwise it'd just get boring, parking on their tail so effortlessly, huh?

Not true: at 1g of constant acceleration (9.8 meters per second per second), it would take just under a year to accelerate to the speed of light. 9.8 m/s/s * 60 seconds= 588 m/s in a minute 588*60= 35,280 m/s in an hour 35,288*24= 846,720 m/s in a day 846,720*365= 309,052,800 m/s after a year The speed of light is 299,792,458 meters per second. Sounds like a pretty reasonable timeframe, at a quite comfortable g-load to me!

No, it would not. It would make it an M4A3E8. All M4A3E8 had 76mm gun, all M4A3E8 had wet ammo stowage, and all M4A3E8 had HVSS. The (76) and (W) were only used for lower-block M4A3, to differentiate them from the M4A3 that had 75mm gun M3 (but wet stowage) (which would be M4A3(W)). You could refer to it as M4A3E8(76)(W)HVSS, but the additional equipment designators are an alternate to the E-series number, not in addition to it. Given the mouthful that is, it's easy to see why they became known simply as "easy eight" And come to think of it, I believe I made a mistake in designation, myself. I'm pretty sure ALL M4A3 with 76mm gun were designated M4A3(76) (because, if I remember correctly, all 76mm gun turreted M4A3 had wet stowage, so the W would be extraneous in that case, too) Also, they weren't bags, per se. The ammo was stored in the base of the turret (instead of in the sponsons above the tracks as earlier models), in a vertical "honeycomb" metal structure with a water-glycol mix filling the gaps between the individual cavities for the shells. As to the power of the 88L56... the 76L57 outperformed it in armour penetration with the ammunition in use by each side contemporaneously (76mm HVAP: 139mm at 500 m, 127mm at 1,000; 88mm APCBC: 110mm at 500m, 99mm at 1,000m). The 88L56 DID have an APCR (essentially the same construction as HVAP) round available which outperformed the 76mm HVAP, but tungsten was in such high demand in Germany for machine tools that almost none of the German APCR was produced or used... particularly later on. The US didn't issue TONS of HVAP, but by the time the M4A3E8 was in use, each tank would have had 2-5 rounds on hand (which the crew would reserve for the tougher targets like Tiger). Keep in mind, up until the end of the war, the majority of German tanks were the PzKw IV. The Sherman bested the IV in pretty much ever metric.

No, it is not. It is an M4A3(E8 ), the Easy Eight of late war (and Korea) fame. You can see the distinct running gear (the Horizontal Volute Spring Suspension and different return rollers) at 0:19 and 0:40 (oh... I should have looked past the first post or two before replying... would've saved much typing!) The A3 versions and later had wet ammo stowage, which fixed the infamous "ronson" effect (which was caused by ammo propellant cooking off, NOT by the gasoline fuel- and the Panzer IV statistically burned just as easily as the pre-A3 Shermans). It has a 76mm high velocity gun, M1; which had superior penetration to the 88L56 as used by Tiger I, particularly when the 76mm HVAP ammunition was utilized. The M1 had performance very similar to the British OQF 17 pounder- the 17lb had higher propellant capacity, but somewhat less effective ammunition (the 17lb APDS had significantly better penetration at very close ranges, but TERRIBLE accuracy) Armour on the Easy Eight was significantly improved, and while not quite up to the Tiger's all-round armour, was damn close. People mistakenly believe Sherman was a weakly armoured tank; it was not. Some of the (appears to be CG) tanks in other cuts appear to be M4A3(76)(W), as they have the older vertical volute spring suspension

This is true, cargo shells (rocket and tube artillery) tend to jettison their munitions at some couple hundred meters altitude, and would overfly their target significantly (unless at near max range, in which case their angle of fall is quite steep indeed). For either rocket or tube submunition rounds, the shift in center of gravity from the submunitions departing will destabilize the projectile, and it's pretty much impossible to know *exactly* where the casing is going to go (just a big elliptical pattern where it might go!). I'm not saying that lack of rocket airframes means it was NOT rockets, I am saying that since there are no carriers in the video (be they rocket, aerial, or tube), it's hard to assess if it was in fact from MRLs, or some other submunitions dispenser. Heck, I never even said that it wasn't MRL to begin with. The real point I was getting at is that the damage was clearly done by submunition warheads (of some kind)- rather than unitary HE rockets, which were the topic of the OP- and that the only possible unitary warhead crater in the film is the one left of the columns direction of march, which appears from the spill pattern was a high-angle impact, as it does not have the characteristic "V" shape of shrapnel patterning from a low-angle impact. It is, I suppose, possible that it was a unitary warhead rocket that either had a delay fuze, or which was fired at near-max range- but in that case, I *would* expect to see the propulsion portion of that rocket, as they tend to survive the detonation of the warhead in the nose surprisingly intact. Anyhow, even if that were the case, the unitary warhead one isn't what did the damage: it probably was not fired as part of the actual fire mission, but rather well ahead of time, as a registration round (to accurately "zero" the battery onto the crossroads)

Yes, I know there is such thing. However, you cannot say that with any certainty; at least from the footage I saw, I could not identify any rocket casings (or CBU canisters, or cargo shell carriers for ICM, or...). I'll admit I didn't watch the entire clip, though. I was a fire support officer (artillery spotter) for half a decade, including time in Iraq. I do know a thing or two about weaponeering and crater analysis. It is, you realize, quite possible to have a target hit by both ICM rockets AND conventional mortars. *edit* having gone back and watched the whole thing again, I still see no readily identifiable rocket casings. The only possible items that are the right dimensions both appear to be snorkel cases (normally mounted on the rear of the turrets), because they are both in the immediate vicinity of the rear of the tank turrets.

Clearly a submunition warhead; you can see numerous small craters from the submunitions around the area. The only large crater is to the right of frame (left of the tank's direction of motion), and from the blast pattern on it, that one came in at a VERY steep angle; either mortar or aerial munition, but almost certainly not a rocket. Could have been a mine, too. Quite obviously a pre-planned mission, too: the intersection was likely a registration point.

That pretty much sums it up, yes

Ok, I'll bite: the argument seems to keep coming up that the loss of the outboard section of the wing is less important, because the inboard portion produces a higher amount of lift per foot of span. This is true... however, how much of the relatively low lift produced by the tip portion of the span is due to the loss of efficiency due to spanwise flow? Keep in mind, if mid-wing BECOMES the tip (due to the loss of the outboard panels), it TOO loses a lot of lift to spanwise flow. How much, I don't know; I don't have those equations at hand.

At a rate of fire around 700-800 rounds per minute per gun (yes, the AN/M2, with it's lighter barrel, fires faster than the M2HB used on ground mounts), you get around 20 seconds of fire before the outer 4 guns run out of ammo, and then another 10-ish with the inboard guns after that. 20 seconds sounds like a lot, but undisciplined fire will run that out right quick. That said, if you set up a good firing position, it only takes a 2-3 second burst to kill a fighter, and I have, on multiple occasions, taken out 4-6 AI FW190 in a single loadout. That said, the AI planes, while a reasonably challenge to maneuver against, are pretty cooperative when it comes to guns defense (or lack thereof!); they fly a fairly smooth line that's easy to get a good tracking shot against. Human opponents tend to fly MUCH more unpredictable guns defense maneuvers, and your hit rate is likely to be far, far lower against human players (as long as they're aware of your presence- for which I REALLY hope they eliminate the "sound radar", as it makes it practically impossible to get into good firing position unnoticed, and really counteracts the value of the AN/APS-13 tail radar on the Mustang, which SHOULD be a huge advantage over other aircraft in SA, but at the moment is extraneous due to sound warning)

Did I ever say that? I said that the range of sweep on a cardboard/ paper airplane is whatever you want it to be (and, for that matter, you can use the ENTIRE wing surface as aileron by twisting it, and can do the same to use the horizontal stabilizers for roll control), and cannot be compared to a real airplane where the deflection angle and surface area of aileron is limited. No, I am not referring to any decision; I am pointing out that the loss of a wing is not likely to occur without also incurring other damages. Yes, I did think about that. It's why I asked the question; no one has addressed it. It is not necessary to be quite so antagonistic about it; how about if instead you simply post the math on how much force it would require? Also, as to "efficient ailerons"... they're not THAT much better than a P-51, because the P-51 rolls damn near as well as the FW190, despite having weight distribution WAY further toward the wingtips, and much higher roll inertia. No, the matters of faith IN science... but ONLY when the experiment is properly constructed. As yet, it has not been. A small-scale RC plane flying with 80% wing lost (but surplus thrust on the order of likely 2:1 TWR, and control authority vastly beyond that of a FW190) is not an equal comparison to a full-scale FW190. ...Particularly if roll control on that F-18 model is done with all-moving tailplane, in which case it hasn't even lost any control responsiveness, unlike a FW190 with a missing aileron. You can quit with the ad hominem inferences that I do not understand how science works. Your posts have shown that a paper airplane missing half its wing can be made to fly generally straight, so long as it has effectively unlimited control authority. It has not shown that a Dora, within it's control authority and distributions of lift and weight, can do the same with 2/3 it's wing missing, as shown in DCS in the original post. The fact remains that none of the real-life examples (despite other very impressive damage!) have been shown to be recovered with more than maybe 30-40% of the wing surface removed, while in DCS, the Dora appears flyable with as much as 60-70% removed. This does not align to what has been observed in reality, and is therefore open to questioning. To prove it could do so, you must show: 1) that the aileron on the remaining wing is capable of cancelling the lift produced by the portion of wing missing on the other side, less the force equal to the *weight* of the missing portion (so as to balance the lift on each side), and that 2) the aircraft is capable of remaining airborne with the lift force produced by what's left. As I see it, if you have lost 60% off one wing, you must then cancel out the lift on the same 60% of the other wing (again, it would in fact be somewhat under 60%, as you don't have to account for the weight of the now-missing portion). So... with 40% the effective lift, a Dora would be flying at something slightly over 100-120 pounds per square foot effective wing loading. That sounds REALLY high to me; we're talking half again higher loading than an F-16! It would be equivalent, wing-load/ AOA-wise, to flying a perpetual 3-ish G turn. I'm not entirely certain it would be sustainable (particularly with all the drag from the aileron at high deflection)

Agree 100% I too prefer flying combat missions (CAS/ strike/ SEAD/ CAP type, I mean), and in fact, not only find flying civilian aircraft boring, I find flying many combat profiles boring (15,000 AGL A-10 PGM runs? Snoooooooore. NOE any day of the week). However, there are clearly PLENTY of people that are interested in this type of flight, as evidenced by the long-term success of the MS Flight Simulator series and X-Plane (and similar). Attracting these people to DCS can be nothing but good: they may progressively become more interested in trying their hand at the high-performance combatants, and even if they never do, their dollars help fund the development of new theaters, larger maps, better ATC/ AWACS/ GCI procedures, better weather modelling, and ultimately, your favorite combat aircaft. Bring 'em on. ...just don't put my Strike Eagles and Sukhois on the back burner to do it :smilewink:

While I have no doubt that an aircraft can fly with significant portions of a wing missing *in theory*, the paper airplanes are entirely unconvincing to me, as are the RC aircraft, for a number of reasons: 1) RC aircraft generally have huge control authority, with vast sweeps of deflection available to their control surfaces. The paper airplanes have essentially infinite. You can push their ailerons and rudder through 90+ degrees of deflection. This is simply not true on a FW190 (or just about any other real, full-scale aircraft). It begs the question: how much wing can a FW lose before it runs out of control authority to counter it? 2) the RC and paper airplanes have parts deliberately, surgically removed. In combat, things are a bit different: if the wing of a metal-construction aircraft is violently removed, I would bet odds are pretty good that the twisted and crushed remains would bind up control rods, and the pilot would no longer be able to control even the remaining aileron. 3) even assuming that there was enough control authority, and that the controls remained unbound, how much stick force would be required, could the pilot put that much stick force in for an extended period, and would the aileron tolerate such high deflection angles at combat speeds (which the aircraft was presumably at when it lost the wing)? None of the "real-world" example photos yet shown have illustrated any more than perhaps 30% wing loss on a conventional tube-with-wings design (F-15 doesn't count; it has significant body lift that the WW2 designs just don't share). The FW190 photo posted earlier was particularly unimpressive; if you look closely, you'll see the picture was deliberately angled to ensure that perspective foreshortening made the wing look more damaged than it actually is; only the portion of wing outboard of mid-aileron is missing; perhaps two-three feet at most.