VanshilarNone of this has anything to do with the PAK-FA, or this thread.
I believe that’s about 70% of the F-35 thread. Why should this thread be any different?
VanshilarThe back half is all YF-23.. I just can’t “place the face”…..
Actually, the back half seems like McDonnell Douglas’s JSF entry while the front half seems like (heh) a chined version of the F-111.
VanshilarYou are so funny 😛
*Sigh* I guess rather than trying to defend or explain the basis for your calculations, you’re resorting to posting pictures of marine animals. In which case it’s easy to cherry-pick them:
http://www.baliberty.com/wp-content/uploads/2012/03/bali-raie-manta.jpg
http://cdn.pcwallart.com/images/wild-orca-underwater-wallpaper-1.jpg
But at any rate, since the discussion was about the F-35A’s range, there’s a couple of data points:
1. As posted earlier, the combat radius in air-to-air is 760 Nmi (1408 km), using internal fuel and (presumably) carrying 4 missiles. This means that the ferry range is going to be at least 1520 Nmi (2815 km), with whatever additional allowances (i.e. additional range) you want to make for the fact that the combat mission profiles usually have some hard maneuvering, some loiter time, may not be at optimum cruise altitude, and is carrying a weapons payload, when none of those would apply for a ferry flight. Any calculation which gives the F-35A’s ferry range as being less than this is simply wrong.
2. Consider the F-18. I choose this because the plane is considered relatively “draggy” and people argue that the F-35 is also relatively draggy. The US Navy fact sheet gives the original Hornet’s combat range (not combat radius) as 1089 Nmi (2003 km) clean (no external fuel tanks) plus 2 Sidewinders, while its ferry range with 3 330-gallon tanks as 1546 Nmi (2844 km). By comparison, Paralay’s spreadsheet calculations has the Hornet’s “clean” range at 1375 km (instead of the official 2003 km) and range with 3 fuel tanks as 2001 km (instead of the official 2844 km), or basically underestimating their range by about 30%.
3. The same US Navy spec sheet gives the Super Hornet as having a combat range (not radius) of 1275 Nmi (2346 km) clean with 2 Sidewinders, and a ferry range of 1660 Nmi (3054 km) with 2 Sidewinders and 3 480-gallon external fuel tanks.
4. The Super Hornet’s SAR gives its combat radius on an air interdiction mission (3 external fuel tanks, 2 Sidewinders, 4 1000-lb Mk 83 low drag bombs with low drag pylons) as 489 Nmi (906 km).
5. The F-35’s SAR gives its combat radius (the SAR itself doesn’t say, but it’s typically held to be on internal fuel only, 2 AMRAAM’s, and 2 2000 lb bombs) as 625 Nmi (1157 km).
6. Now here’s the thing. For the Super Hornet, apparently it has a ferry range of 1660 Nmi (3054 km) but once loaded up with 4000 lb of bombs and taking into account combat mission profile and maneuvers, etc., that drops to a 489 Nmi (906 km) combat radius, or about 29.5% of its ferry range. The ferry range of the F-35A isn’t given, but with 4000 lb of bombs (and 2 AMRAAM’s) its combat radius is given as 625 Nmi (1157 km). I’ll assume that the combat mission profiles for those two planes in their SAR’s are the same. Then, dividing the F-35’s combat radius by the same percentage (29.5%) yields 2122 Nmi (3929 km).
7. However, for the Super Hornet, when carrying its bombs they contribute not only additional weight (i.e. induced drag), but also additional parasitic drag. This does not apply to the F-35, which is carrying its bombs internally (and thus they only increase its induced drag but not its parasitic drag). Thus, the F-35 certainly will not lose as much of its range by carrying bombs internally as the Super Hornet does from external carriage. In other words, the combat radius for the F-35 is certainly going to higher than 29.5% of its ferry range. Therefore, the 2122 Nmi (3929 km) figure is a maximum.
8. Therefore, either one (or more) of the above assumptions is incorrect, or the F-35’s ferry range (on internal fuel) is going to be somewhere between 1520 Nmi (2815 km) and 2122 Nmi (3929 km). It’s likely not going to be very close to either of those figures; it won’t be close to 1520 Nmi (2815 km) because that assumed a combat mission profile with maneuvering, payload, etc., and it won’t be close to 2122 Nmi (3929 km) because the external carriage of bombs on the Super Hornet likely have a significant effect on drag. This is still a somewhat wide range of possible values, to be sure, but calculations which imply a ferry range outside of these values are going to be very suspect.
VanshilarIt is difficult to communicate through the Google translator. Download the table and look at the formula in the cells.
Um you’re the one that’s supposed to be able to back up and justify your calculations and why you did them, not force others to wade through your calculations with no explanation to help determine what you’re doing. But looking through it, there’s multiple mistakes:
1. The range (without external fuel tanks) seems to be calculated as:
410 * (cruising fineness) * (cruising speed (in kilometers per hour) / 1062, i.e. Mach number) / (fuel consumption, probably meaning specific fuel consumption) * fuel fraction term
with the fuel fraction term being:
((fuel weight) / (total weight)) / sqrt(1 – (fuel weight) / (total weight) )
You defined the cruising fineness as 0.91 * fineness, where fineness is the plane’s area when viewed from above (i.e. plane planform) divided by the plane’s area when viewed from the front. That’s not related to range at all (at least not directly). If you’re using this as something similar to the lift-to-drag ratio in the Breguet range equation (which seems to be what you’re trying to do, based on matching up the different parameters), it is very, very wrong, and is probably why you have the F-35 as having a very short range; your calculated fineness for it is around 8, with 10.9 for the F-22 and 12.4 for the PAK FA. No wonder why you’re calculating the PAK FA as having such a large range.
I’m also not sure what assumption you’re using (i.e. constant altitude, angle of attack, or whatever) for that fuel fraction term. I’m sure I could crack open some flight dynamics textbook again and look it up, but that’s the point: you’re the one that’s supposed to be explaining your assumptions, rather than getting other people to try to guess at what you were doing. But I don’t know offhand where it’s from.
Additionally, for the fuel consumption you have it as 0.65 for the F-35, 0.61 for the F-22, 0.62 for the PAK FA, and 0.67 for the Su-27. I’m not sure where you get your estimates for the PAK FA and the Su-27, but the F-35 has a higher bypass ratio than the F-22, so I don’t know why its fuel consumption would actually be worse, when higher bypass ratios typically mean lower specific fuel consumption.
2. The range (with external fuel tanks) seems to be calculated as:
472 * (cruising fineness) * (cruising speed (in kilometers per hour) / 1062, i.e. Mach number) / (fuel consumption, probably meaning specific fuel consumption) * fuel fraction term * correction term
with the fuel fraction term being the same as before, but using internal + external fuel weight (and adding the external fuel weight to the total weight), and with the correction term being:
(total weight without external fuel, i.e. empty + internal) / (total weight including the external fuel, i.e. empty + internal + external)
I’m not sure of the reason for the correction term, nor why you have 472 instead of 410 for the range with external tanks. But comparing these equations, you’re using the exact same “fineness” term, which means that you’re not taking into account the increased frontal area of the external fuel tanks, nor their increased drag (you’re not really taking lift and drag into account anyways, unless you’re using your definition of “fineness” to do it). IIRC something like 30-40% of the fuel carried by an external fuel tank is basically just used to offset their increased drag, and the remainder is what goes toward improving the plane’s range. Additionally, although more minor, external drop tanks and their associated pylons increase the weight of the aircraft, by roughly 15% of their fuel’s weight. So basically for your range with external fuel tanks, the planes effectively get free fuel without an increase in drag nor an increase in “empty” weight. Perhaps the correction term is there to balance it out, but I’m not sure how it works out.
As a side note, you gave the F-35’s range using the “internal fuel only” formula, with the others using the “internal + external fuel” formula, probably because we don’t know what’s the progress of fitting out the F-35 with external fuel tanks. But even so, you give the F-35 as 2269 km range on internal fuel, with the F-22 as 2511 km range, the PAK FA as 3502 km range, and the Su-27 as 2352 km range, all of these on internal fuel — despite the F-35 having a higher fuel fraction than all of the others. By your values the PAK FA has a fuel fraction of 0.388, compared with 0.327 for the F-22, 0.344 for the Su-27, and 0.403 for the F-35, and yet its range is about 1000 km more than all of them?
VanshilarKС-135 90700 kg of fuel, three times
KC-10 161480 kg of fuel, four times4 F-35A 5960 km (3 + 1, which is returned to the half way)
(3 + 4) * 4 = 28 refueling
(90700 + 161480): 2 = 126 090 kg: 28 = 4503 kg – 54% complete refueling
Wait. Are you saying that the tankers accompanying the F-35’s completely depleted themselves by the end of the final refueling? And that they themselves didn’t use up any fuel during the entire transit? Both assumptions are necessary if the tankers offloaded 4503 kg per refueling.
VanshilarSuppose that the plane refueled at 50% fuel in the tank. It turns out (8382 kg / 2): 745 km = 5.63 kg / km…
That’s a bad assumption. The purpose of these transits isn’t to show off the plane’s range. They’re to ensure that the plane can be sent safely across the Atlantic, or, in the event of some sort of malfunction, can be safely flown back to the U.S. So they basically keep the fuel at a high enough level such that at any point during the transit, if the plane receives no further fuel, it can still make it to an alternate base, including any reserves needed for weather, headwind, circling around for a second try at landing, etc. Additionally, the planes weren’t flying at cruise speed, i.e. the speed for optimal distance/fuel burn ratio, because they’re with tankers, which have a much lower cruise speed. So whatever range is calculated is simply a lower bound. IIRC when the F-35 first crossed the Atlantic, it was with a Typhoon — which also had 3 external fuel tanks. The F-35 refueled 7 times while the Typhoon refueled 8 times, but the Typhoon went an additional 500 miles or so after reaching the continental U.S. (i.e. plenty of alternate bases nearby, so can let fuel reserves go lower). The Typhoon’s ferry range is stated to be >3790 km on wikipedia; the F-35’s ferry range is probably going to be in the same ballpark.
VanshilarThe fineness of the aircraft – a big secret. It can be found in the wind tunnel.
I find that value in another way. Top view (m2) : Front view (m2) = fineness.
I’m confused. What do you mean by fineness? The length-to-width ratio or something else?
When you say top view and front view, do you mean the area of the plane when viewed from above and the area of the plane when viewed from the front, or something else? If this is the case, it doesn’t have a (simple and direct) relation to the lift/drag ratio or other parameters related to range. For example, if a plane were stretched (i.e. twice as long as before), the “fineness” that you’re calculating would double (and by the way it looks like you’re using it, the thrust required would be cut in half), when in reality the drag would increase due to having twice the wetted area.
VanshilarAerodynamic quality – the ratio of lift to the power of resistance, fineness
Okay I think there’s something lost in translation here.
By “power of resistance”, do you mean drag? As in lift-to-drag ratio? Or do you mean some mathematical power (i.e. exponent) of drag?
Fineness is actually just the ratio of a plane’s length and its maximum width.
VanshilarThe relative mass of fuel
Um I think you’re going to need to explain your calculations and your reasoning behind them in more detail. For example, based on wikipedia (I know, I know…) the empty weight of the Su-27 is 16380 kg, with 9400 kg fuel, so that’s 25780 kg. Yet you gave it 30500 kg (+4720 kg). For the F-22, it’s 19700 kg and 8200 kg, but you gave it 33906 kg (+6006 kg). For the F-35, it’s 13200 kg and 8390 kg, yet you gave it 28169 kg (+6579 kg). So the Su-27 is carrying a lot less load than the others? It’s unclear what you’re doing with your calculations or where they’re from. And the Su-27’s fuel consumption is 0.67 kg / kgs * hr while the F-35’s fuel consumption is 0.82 kg / kgs * hr? Is the F-35’s engine that bad at converting fuel to thrust, especially when it’s designed more for the subsonic regime, which is the speed that most fighter planes cruise at?
Vanshilari presume aerodynamic quality = drag,
russia got some off way of definitions.there’s no fookin way T-50 consume 0.62 compared to F-35 0.7 lest he meant per engine, but T-50 got 2 engines
I think it’s thrust specific fuel consumption. It doesn’t matter how many engines. At any rate, then, the claim is that the T-50’s engines are a bit over 10% more efficient than the F-35’s engine.
VanshilarI cant read Russian so how do you come up with these ratio ? I downloaded the table but everything inside seem like they are random number TBH
It’s just a score computed by taking the ratio of the compared plane’s spec in different areas with that of the Su-27 (i.e. the Su-27 is the normalizing plane), all multiplied together into a product, then take the 13th root (i.e. ^ (1/13) ). The different factors are:
Wing sweep angle (lower is better)
Wing aspect ratio (higher is better)
Wing loading (lower is better)
Fineness ratio? (higher is better)
Thrust-to-weight (higher is better)
Thrust vectoring max angle (higher is better)
Number of targets it can simultaneously track (higher is better)
Number of targets it can simultaneously attack (higher is better)
Number of hardpoints (higher is better)
RCS (lower is better)
Max g (higher is better)
Unknown (higher is better)
Range (higher is better)
The two that I don’t understand are the fourth one (удлинение планера), which is the length of the aircraft divided by the square root of its frontal cross-section area, which seems similar to the fineness ratio, and the next to last one (ускорение). I’m not quite sure how to parse that one. I also don’t understand why the fourth one is a ratio comparing to the Su-27’s удлинение планера (row 341) while the next to last one is compared to the Su-27’s удлинение фюзеляжа (row 354), nor why those values in row 341 should be different depending on the plane being compared to the Su-27 (row 341 is part of the section with Su-27’s stats, why is it different depending on column)? I’m sure there’s some simple explanation for them (i.e. wing thickness to chord ratio is probably there somewhere), it’s just not self-evident from the spreadsheet and Google translate.
At any rate, it’s just a numerical score, which means you’re assuming each of those things are equally important in a dogfight. For example, the V-22 Osprey probably fares pretty well under this scoring, since it has a low wing sweep angle, high wing aspect ratio, high thrust-to-weight ratio, and a high thrust vectoring max angle. (It would probably score pathetically bad in some of the other ones though.)
It also omits other factors that may be relevant to a dogfight. For example, why not max angle of attack, or max pitch rate, or max sustained turn rate, or even possibly max speed and max altitude? The first two are relevant for nose-pointing ability, while the latter two are relevant for high energy going into the merge (not to mention, implies something about the dragginess of the plane).
So in the end, it’s basically just a formula, using numbers, not “facts”.
VanshilarThere was Lt. Col Griffith stating
“What we can do in our airplane is get above the Mach with afterburner, and once you get it going … you can definitely pull the throttle back quite a bit and still maintain supersonic, so technically you’re pretty much at very, very min[imum] afterburner while you’re cruising,” Griffiths said. “So it really does have very good acceleration capabilities up in the air.” (Lt. Col. Griffiths, from a defensenews article)
The actual relevant quote about the F-35 supercruising is this one:
The F-35, while not technically a “supercruising” aircraft, can maintain Mach 1.2 for a dash of 150 miles without using fuel-gulping afterburners.
“Mach 1.2 is a good speed for you, according to the pilots,” O’Bryan said.
The high speed also allows the F-35 to impart more energy to a weapon such as a bomb or missile, meaning the aircraft will be able to “throw” such munitions farther than they could go on their own energy alone.
There is a major extension of the fighter’s range if speed is kept around Mach .9, O’Bryan went on, but he asserted that F-35 transonic performance is exceptional and goes “through the [Mach 1] number fairly easily.” The transonic area is “where you really operate.”
It also states that the F-35A can do 9 G’s with full fuel and missiles. (“Missiles”? Not “missiles and bombs”? Not “full internal load of missiles and bombs”? But that’s the quote.) Given that the interview was in 2012, it’s unclear if they had already flight tested the F-35’s capabilities or if it was designed or computer-simulated capability instead.
VanshilarHere is your 6th flying prototype. 😀
Must be a really fast plane based on the nose cone angle.
VanshilarThe figures does not seem reliable. F-15E with 3ET has 50% less combat radius than F-22 with 2ET. both practically identical for ferry ranges.
The F-15E has to carry its weapons externally, which adds a substantial amount of drag. It also has to go low for part of the mission, which also reduces its range.
VanshilarI dunno…I do note that the maintenance cost went up about 10% compared to earlier estimate. I presume that means spares and depot/factory level maintenance? It’s not too clear what the categories include.
Of course, actual cost of some prospective upgrades far in the future is extremely difficult to estimate.
It’s because the military decided to use each airframe for 32 years instead of 30, thus extending the life of the program out from year 2064 to the year 2070.
All of that is including inflation (yes, inflation out to year 2070). To give you an idea of how much inflation (plus other assumptions in the projections out to 2070) skews things, the document at the link says that in constant year 2012 dollars, using the airplanes for the longer time (1.6 million flight hours total) increases the overall operating and support costs by $45 billion, but if that hadn’t been done, it would have instead showed a decrease of $22 billion due to improvements in maintainability and sustainability. The kicker is that this plus other changes means the overall change to the cost of the program is an increase of $16 billion in constant year 2012 dollars, but due to inflation, this ends up being $95 billion in then-year dollars (i.e. inflated dollars). That’s how much the costs are skewed.
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