How did this F/A-18's execute a "low-boom maneuver"?

Question

The Engadget article NASA will publicly test quiet supersonic technology in November says:

The administration plans to conduct a series of public tests around the coastal city of Galveston in November. The F/A-18 Hornet aircraft at the heart of the tests will perform dive maneuvers that produce louder sonic booms out at sea, while quieter sonic "thumps" will take place over Galveston proper. After that, "at least" 500 local volunteers will provide feedback on what they heard, while audio sensors will provide more definitive noise readings.

and links to the NASA YouTube video NASA Social: Low-Boom Maneuver:

This footage is from a NASA Social event at Armstrong Flight Research Center during which the F/A-18 was flown to produce a regular sonic boom and then a low “boom” by performing the dive maneuver described in this story. The normal sonic double boom occurs at 0:43. The low “boom” occurs at 02:34. The rest of the footage includes a flyby, planes on the ground with NASA Social participants, a visit to the Ikhana UAS and a pilot signing.

Question: What would be a simple way to understand what a "low-boom maneuver" is, and how would an F/A-18 execute it? In this case would it require any modification of the Hornet's configuration?

Is producing the boom during a dive of particular relevance to the reduction in sound level, or is this just a convenient way to do this particular experiment? Since the sound level as well as the reduction of the level for the low-boom maneuver at ground level may have a directional dependence, I'm wondering if the direction of the dive needed to be carefully planned to be representative of how the reduction would work for civilian air travel boom reduction, where dives are (thankfully) less frequent.

Answer 1

I did some reading and this is what I could find out so far: https://www.nasa.gov/aero/nasa-prepares-to-go-public-with-quiet-supersonic-tech This article is similar to the Engagdet one but goes into slightly more detail(though not enough for a proper explanation) about the dive maneuver. However, Lockheed Martin is in the process of designing and building a low sonic boom demonstrator, slated to fly in 2021. The wikipedia page says that to prevent sonic booms from building up, the aircraft will have a long slender body and canards (thought the Concorde had these as well, so I'm not sure what the difference is). They will probably add canards to the F/A-18.

The dive maneuver happens in such a way that while the sonic booms directly underneath the aircraft are strengthened, there is a secondary pair of sonic booms created further ahead that are weaker, giving quiet thumps instead. Hope this helps.

Answer 2

I was doing some research when I came across this question and thought that maybe someone still cares about an answer. In fact both answers given so far are almost right but lack some background.

To elicit low sonic booms, commonly also described as sonic thumps, the F-18 first ascents to a height of a couple of thousand feet (40-50k). Once the height is reached the plane turns upside down and falls to a much lower height (15-20k ft), while breaking the mach border. The overall procedure is somewhat accurately depicted on the QSF 18 (Quite Sonic Flights) research crew patch from NASA:

https://www.google.com/url?sa=i&url=https%3A%2F%2Fnews.psu.edu%2Fstory%2F546964%2F2018%2F11%2F08%2Fresearch%2Fnasa-penn-state-survey-reactions-sonic-thumps&psig=AOvVaw2bFvYl9dBQZkQphIwsfwzs&ust=1586263795917000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCOi_vKHr0-gCFQAAAAAdAAAAABAr

I think that turning the plane around on it's way to the ground achieves the rear fins of the plane distorting the soundwaves on their way to the ground (in contrast to a flat belly of the usual plane, which reflects "hard" N-Waves (The sound signature of a conventional sonic boom on the ground roughly looks like the letter "N") to the ground (compare figure). https://www.google.com/url?sa=i&url=https%3A%2F%2Fvibrationdata.wordpress.com%2F2017%2F02%2F02%2Fquieter-supersonic-passenger-jets%2F&psig=AOvVaw36SVi23jWSoygdQMinXdiy&ust=1586264669269000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCPjr88Hu0-gCFQAAAAAdAAAAABAO

The pull up manoeuvre observed wasn't the actual supersonic transistion, but the exit from the manoeuvre alltogether. You can find a detailled description of the whole procedure in Sullivan et al.'s paper "Human Response to Low-Intensity Sonic Booms Heard Indoors and Outdoors".

I hope that helps. Kind regards and stay healthy

Answer 3

By flying straight up before exceeding Mach 1, to spread the boom more widely.

The commentary says: at 2:10, "pull up" and the smoke trail suddenly points upwards; "[2:16] subsonic below thirty thousand feet," "point zero nine Mach [for?]ty thousand [feet?], thump, copy one point zero nine Mach, more thumps, [2:39] one point zero eight Mach."

In the dive at 2:02 he's "supersonic" already before "he's pulling up." The dive is only to build up enough speed so that the subsequent climb starts fast enough to eventually exceed Mach 1.

Heard from the ground, what are the thumps at 2:30? Forty thousand feet up is eight miles, roughly 45 degrees above the horizon; so as many miles horizontal, say twelve miles total. The speed of sound is 4.5 seconds per mile, so the lag from video to audio is 54 seconds. 2:05 is when the dive is "supersonic right now." The thump at 1:36 shows that the horizontal approach to the dive was supersonic.

Faint thumps are at 3:01 and 3:15, thus generated at 2:07 and 2:21. 2:07 is just after ascent starts: maybe parts of the airframe exceeded Mach 1 in that high-G maneuver. 2:21 is mid-ascent, thus the cause of the intended low-boom maneuver, between the radio chatter that says subsonic at 2:16 and supersonic at 2:39.