SwingKid

Gvardeyskoye also hosts a Russian Su-24 regiment. Where did the information about air defense sites come from? -SK

Ever since I read your sig, I knew that we would agree about everything. :beer: -SK

I think he means that if the dinosaur air force couldn't save them when it was happening then, we shouldn't expect USAF to save us from replacement now. Are you guys (the non-"environment whacko" types) for real, or just putting on an act? The elimination of Atlantic cod that destroyed industry in maritime Canada has pretty much exterminated such opinions from my country... Reading this is like going to a museum and seeing real, live dinosaurs still walking the earth - people with these opinions still exist?? I thought you guys were just fairy tales, made up by environmentalists to scare the rest of us at bedtime. :) "Mommy mommy, can I feed one?" :prop: -SK

I can confirm that yes, there is significant, measurable radiation, and yes, it ends up in your head. Whether it does any harm when it gets there is what the controversy is about, and is for a biologist or a doctor to tell you, that's not our specialty. The cell phone companies deny it, but having worked with them, we can also confirm that they neither know, nor want to know. Their strongest, truest and loudest argument is basically the same as StarForce when asked about ruining DVD-ROM players: "there is no data or evidence" - and they're quietly doing everything to keep it that way. I wouldn't trust StarForce with the health of my brain. -SK

"Thin-skinned?" I thought his post was more... :cheer3nc: I've been waiting to see that smiley used, like, forever! -SK

On a related topic, a recent (this month's?) issue of Combat Aircraft (or was it AFM?) had an article about the US "sacrificing" half of its B-52 fleet and dozens of other aircraft, in order to fund just 4 more F-22s. I was surprised by the amount of negative political opinion expressed in the article. It seemed like the author was going way beyond objectively reporting about military aviation, and almost venting frustration about "the world today." I wonder how some of the more optimistic posters here would react to that, considering that even I found it out of place. -SK

I can't speak for "environment whackos," but I can confirm that there are a remarkable number of my fellow microwave and antenna engineers who, like myself, refuse to own a mobile phone. We still encourage the rest of you to buy them though, and continue paying our salaries. They're perfectly safe. Really! ;) Speaking of which, have you upgraded your computer lately? -SK

If there isn't a problem, then why is the fuel price going up, and the proposed economic sanctions against Iran exclude oil sales? Money talks. -SK

I don't know about Black Shark, but Lock On and Flaming Cliffs are mainly GPU dependent for their speed. I think they also run better on Intel-based systems, compared to most other software. The reason why I can still run Flaming Cliffs on a 500 MHz Win98 machine is because I have a graphics card with 128 MB. -SK

You do realize, I can't possibly upgrade now... After holding out this long? That would be crazy!! Vista64 is practically on top of us! So close, so close to skipping ME, 2000, and XP completely... Just a little more time, and I've beat the system... :) -SK

We have a "beta forum?" :confused: ;) -SK

Wow - and after all that appreciation, he didn't even ask for what he wanted. What kind of flight simmer ARE you?! :shocking: -SK

Not quite - Flaming Cliffs runs fine on Windows 98/ME. Black Shark probably won't. :( -SK

They don't? ;) EDIT: seriously though, I don't think a dedicated PR person would be able to answer questions, that even programmers don't yet know the decisive answers.. -SK

:lol: :megalol: :lol: :poster_ban: -SK

I'm afraid Lock On has nothing to do with it. Simply, the Dash-1 manual seems to be - lying!? Take a look at the area circled in red in this distance-to-climb chart: If there was any weight for which the aircraft could climb vertically at 350 kts, there should be a horizontal section of the curve here, to indicate altitude gained for zero distance traveled. There isn't one. The best it says you can do is about 5000' per 1 nm horizontal distance, all the way down to sea level. That can't be right, can it? How on earth can a 30,000 lb F-15 experience more than 17,000 lbs drag at 350 kts? What am I missing? Anyway, Lock On's F-15C matches this chart fairly well, but the chart seems to describe an aircraft with only about 41000 lbf bench engine thrust. -SK

Ok, I just ran some more tests, and... I'm confused. I don't see any evidence, that Lock On's F-15C engines ever produce more than 41000 lbf thrust, if that. On the other hand, the time-to-height test matched very well. What's going on? -SK

Agreed, I forgot a unit coversion to divide by (9.8 N/kgf), my estimate should be the same as yours. Thanks for the correction. Good question. I admit I don't know what the exact reduction of thrust with altitude is but the applet you linked is very interesting! Now let's condsider 350 kts. The drag should be proportional to the square of the speed, so at 350 kts, the drag should be about 4.5 times what it was at 300 km/h - around 5100 lbs. Ok? So, let's take a 35000 lb F-15, add 5100 lbs drag = ~40000 lbs. So in maximum thrust with 47000 lbs, at what climb angle will the F-15 pilot be able to "maintain 350 KCAS," according to the instructions for a performance climb? It will be impossible - even flying vertically, he'd just accelerate with 7000 lbs excess thrust. In order to "maintain" such a speed, there must be some climb angle at which the F-15C will cease to accelerate, when it's flying 350 kts. So, I think that in addition to the engine thrust loss with altitude, I've just simply underestimate the Cd or aircraft area somehow, our new drag figure must be too low now. I admit I'm not sure myself, in fact I brought this topic up for discussion elsewhere recently convinced that the Eagle should accelerate in the vertical. I've only recently become more skeptical. Thanks for the help with the math, it's much appreciated. -SK

Ok, something's slightly off. If with the brakes out you were averaging 685 km/h, that's about 190 m/s (assumed TAS). That means you covered 190 * 45 = 8550 m slope. If the slope was 12 degrees (10 degree pitch + 2 degree AOA), then you should have descended only 8550 * sin(12) = 1778 m, instead of 2000 m. Assuming the 45 s and the speeds are correct, then asin(2000/8550) = 13.5 degree dive (10 deg. pitch-down + 3.5 degrees AOA?) With the brakes retracted, we have an average speed of about 766 km/h = 213 m/s, 42 seconds gives 8946 m slope, asin(2000/8946) = 12.9 degree dive (10 degrees pitch-don + 2.9 deg AOA?) Ok to find drag force without the brakes, let's see, engine thrust = (9.8 N/kgf) * (1900 kgf) = 18620 N gravity component = (9.8 N/kg) * (12000 kg) * sin(13) = 26454 N measured acceleration = (820-711 km/h)/(3.6 kms/mh)/(42 s) = 0.72 m/s^2 drag force = (18620 N) + (26454 N) - (12000 kg * 0.72 m/s^2) = 36434 N Cd = 36434 / (0.5 * 1.225 * 212.6^2 *8 ) = 0.165 So the body drag at 685.5 km/h should be: body drag = 0.165 * 0.5 * 1.225 * 190.4^2 *8 = 29310 N measured acceleration with airbrakes = (687-682 km/h)/(3.6 kms/mh)/(45 s) = 0.03 m/s^2 brake force = (0.03 m/s^2) * (12000 kg) + (29310 N) - (18620 N) - (26454 N) = 15404 N Cd(airbrakes) = 15404 / (0.5 * 1.225 * 190.4^2 *1.8 ) = 0.385 Not far off from the ~0.4 figure I calculated from my own measurement (thanks for the extra data). The airbrakes still seem to be only about as draggy (in shape) as a "boxy road car". Consider also these shapes, especially the "flat plate": http://www.grc.nasa.gov/WWW/K-12/airplane/shaped.html -SK

In order to get off the ground, an F-15C needs to be doing roughly 300 km/h. Very roughly speaking, we can estimate the drag by assuming a section-referenced coefficient Cd = 0.15, a section area 8 sq. m, to get: Fd = 0.15 * 0.5 * 1.225 * (300/3.6)^2 * 8 * 2.2 ~= 11000 lbf drag If the engines are producing 47000 lbf thrust to lift a 35000 lb aircraft AND overcome 11000 lbf drag, that only leaves about 1000 lbf excess thrust for acceleration. As the aircraft gains speed, the drag increases with the square of the speed, and as the aircraft increases altitude, the engine thrust decreases. So, while these are very rough numbers, you can see how any vertical acceleration is likely to be a very transient phenomenon, that only happens for a moment near the ground, at very low speed. For comparison, consider this video of an F-15C "Viking takeoff" from Boeing's site: http://www.boeing.com/defense-space/military/f15/video/15cvklbk.mov http://www.boeing.com/companyoffices/gallery/video/index.html Notice that he appears to be flying at a 45-degree climb angle, maintaining a constant speed, exactly according to the performance climb profile described at the top of the chart. EDIT: note, I think there is an error in my drag calculation somewhere, it seems too high, but the general idea is still true -SK

I don't doubt your word, but a flight sim as accurate as Lock On needs more precise numerical data. For example - how does Kula66 know the fuel load? The F-15C airshow routine and Viking takeoffs are reportedly done with full fuel. How long was the aircraft going vertical, to what altitude, at what speed, and how do we know it was accelerating? Note that the climb profiles shown above are for 350 KCAS. Whatever the precise details, vertical acceleration seems to be a very transient condition, that only happens near the ground imediately after take-off. -SK

Ok, so which part of this is in error? -SK

Oh sure. (a) there is no clear-cut plan, (b) even if there was, it would change all the time, © even if it didn't, such a statement would only stimulate many discussions, which would then (d) change the clear-cut plan, as it has every time before. Promises, promises. :) -SK

Some data is missing fror a proper Cd calculation, e.g. we don't know what was the AOA or the descent time (assuming 0 AOA it would seem about 60 seconds). So, my math from this example somehow resulted in a section-referenced brake Cd of 0.15, which is even lower than my previous estimate. I think Cd = 0.4 is a slightly more reliable measurement, and yes I still think it might be too low, based purely on the shape of the airbrakes. Coincidentally, I also think Lock On's A-10 airbrakes might be too slippery now, for the same reason, but I'm less sure about those because they have a different shape. Is it possible that ED had aerodynamic data available only about the less-effective old two-panel airbrakes of early Su-25s, and is using it to model the four-panel airbrakes of later aircraft? -SK

Ok, I took a look at the A-10A dive brake anyway... After a very rough calculation, it also seems to have a Cd of about 0.4. -SK