Part 2: Top-Level Analysis of F-35 Transonic Performance(Part 1 Here, Part 3 Here, Bonus Block 60 Comparison Here)
Therefore, I will try to base analyses upon relative differences in ‘sum totals’ and relative percentages or fractions to some baseline. We can’t get away from a minimal discussion of the nature of acceleration versus drag, but we can illustrate the things about drag and acceleration that we need to grasp without crunching too many of the numbers. Which is a good thing, because without the need to know, we don’t have the actual F-35 numbers (ex: drag values) available to us in the public domain for plugging into any calculations anyway.
First we’ll summarize the relevant things we now know-- or think we know-- about the F-35’s Transonic Acceleration KPP and overall performance. We will focus on the F-35A model in examining the relevant physics of accelerating through the transonic region and how they affect all aircraft and in particular the F-35A. We’ll then follow with a discussion examining just one likely ‘workaround’ that the F-35A might operationally employ (if it is even needed), and discuss the ‘tactical’ impacts and whether or not what may obviate or mitigate whatever the ‘popularly-perceived’ operational challenges the F-35 variants endure passing through the transonic region.
What we know or think we know
Information from ‘other’ sources.
Dr JENSEN: Air Vice Marshal Osley, in a previous hearing you responded to APA's criticism of the F35's aerodynamic performance and you said that it is inconsistent with years of detailed analysis undertaken by Defence, the JSF program office, Lockheed Martin and eight other partner nations. Given that the Director of Operational Test and Evaluation has indicated that the JSF program office, the JPO, has asked JROC to reduce the sustained turn and the acceleration performance essentially to exactly the numbers that APA was predicting years ago, what does that say about the detailed analysis by Defence, the JSF program office, Lockheed Martin and the eight partner nations?
Air Vice Marshal Osley: The points that the Director of Operational Test and Evaluation made there about the manoeuvrability, as you point out it was the sustained turn and the transonic acceleration. He pointed out that the targets that have been set for those parameters were not going to be met by the F35. The figure of I think it was 55 seconds for transonic acceleration, the F35 was going to take 63.9 seconds to do that. That is obviously at a certain altitude, I think it was 30,000 feet, and a range of mach 0.8 up to mach 1.2.
The point to make about those is that that acceleration by the F35 is in a combat configuration. If you look at the legacy aircraft and we talk about comparable performance, a legacy aeroplane would require weapons and, obviously, external fuel tanks to be in combat configuration.
An interesting factoid, one of the USMC test pilots mentioned this little tidbit—they have to use a modified Rutowski profile in order to get the F-35B and C up to Mach 1.6. Basically, you do one push over, unload the jet and accelerate, get up to 1.2, turn and repeat until you hit 1.4 Mach, turn and repeat till you hit Mach 1.6. It just barely gets there and barely has any gas left over afterwards. The kinematics are basically F/A-18C-like, though that was apparently exactly what was expected….
If the final speed is near the aircraft’s maximum speed, the large speed increase necessary renders the conventional method of using the peaks of the Ps curves useless. However, the energy method works well. Note in this example the optimum climb path includes an acceleration in a dive. This optimum energy climb path is also known as the Rutowski climb path, after its developer. The path (Figure 7.14) [at the link] consists of four segments to reach energy state E in minimum time. Segment AB represents a constant altitude acceleration from V = 0 to climb speed at state B. The subsonic climb segment follows a path similar to the one illustrated to the tropopause at state C. This subsonic climb is usually a nearly constant Mach number schedule. An ideal pushover or dive is carried out at constant Eh from C to D. The acceleration in the dive is actually part of the optimum climb path.
Deduced Transonic Acceleration KPP Times Analysis of the Variants
Why the differences, and why by ‘that’ much?
To help us investigate, we now need to take a look at the possible relevant differences. Those would be the differences between aerodynamic shapes, the engine installations, and relative weights: elements affecting the drag equation (Figure 3 from Part 1) which we now modify to remind us that the drag coefficient is but one element in the drag equation:
|Figure 8 (Numbering Continued From Part 1|
Using the F-35A Model weight as the basis, we find that over a range of mid-mission fuel loads, the F-35C weight, and therefore lift needed to be generated in level flight to be around 13-14% higher than the F-35A.
|Figure 9. Relative f-35 Variant Weights|
We can therefore conclude that the wing wave drag coefficient percentage contribution to the total drag coefficient that is due to lift for the F-35B is less than the F-35A (under the same flight conditions in the speed region we are looking at of course), and that the F-35C’s wing wave drag coefficient contribution due to lift is significantly higher than for the F-35A. Further, we can conclude there is little, if any, difference between the wing wave drag contribution due to volume for the F-35A and B because the F-35A and F-35B wings are identical in area: they have the same fixed length (chord), and span. We also know the area of the A and B wings are identical and their cross-sectional volumes very nearly so (more on this in the next section).
Cross-sectional area comparison
|Figure 10. Cross-sectional Are Differences|
The Impact of Lift Surface Area on Wave Drag
|Figure 11. Extended Vertical surfaces on C model not shown|
A Summary of Thrust vs. Drag
|Figure 12. Factor Differences between variants, F-35A Baseline|
Transonic Acceleration: What is it good for?
F-35A Transonic Acceleration Performance
How the F-35A new KPP standard stacks up against the F-16 Block 50/2If we examine the F-16’s ‘Combat Max AB’ transonic acceleration data (Ref #11, Table A8-12), and compare it to the F-35A’s newest transonic KPP time (63 seconds) we find the F-35A loaded with two AMRAAMs and two 2KLb JDAMs has better acceleration than the F-16C/D in 20 of the 30 possible weight/drag index combinations shown in the tables (weights from 20K lbs to 41K lbs, and Drag Indexes from 0 to 250).
|Figure 14: External Stores start adding drag quickly|
This is about as much as a Block 50/52 F-16C can weigh/carry without increasing the drag count:
|Figure 15. Max 'Clean' Block 50/52 weight.( Source Ref 11)|
|Figure 16: Empty Fuel Tanks and Store Stations Still Add Drag|
In short, even if the F-16 is running on fuel fumes carrying wingtip AMRAAMs and
LANTIRN pods, it can only carry just a little more internal fuel before the F-35A's latest transonic acceleration KPP standard can be said to be ‘better’ than an F-16C Blk50/52 in transonic acceleration.
|Figure 18: Only 2 AIM-9s greater load than F-35 Internal Load Out. It'll need those AIM-9s long before te F-35 will.|
Even the F-35B and F-35C transonic KPP times meet or beat a similarly-loaded but much lighter weight Block 50/52 F-16s:
|Figure 19: With comparable war loads: F-35A and F-35 B beats all & F-35C beats most F-16 configurations.|
The ‘meaning’ we can derive from the revised F-35A transonic dash KPP is this: It still represents the stated ‘F-16 Like’ performance goals and overall, it exceeds them. Other than that, everything else is guesswork.
BUT! The ubiquitous 'some' might ask....
‘What If’ Operational needs require the F-35A/B/C to get rid of all or part of those extra darned seconds enroute to Mach 1.2?
THAT topic we will address in closing the series in Part 3.