NYT: GIMBAL Video of U.S. Navy Jet Encounter with Unknown Object

Getoffthisplanet

Active Member
Make the clouds as dark as they appear in the Gimbal video. There's a very low contrast between the tone of the clouds and the sky, but your example has a very high contrast.

Nice.

blurred_clouds_w_unsharp_mask_02.gif

I should probably stick something in there to represent the object, but in this case it would be difficult to see the results of the unsharp mask process on the clouds.
 

Getoffthisplanet

Active Member
So, why is Fravor out there saying there is a "force field" around the Gimbal object?

Again, shouldn't he be familiar with this or shouldn't someone at least have told him the aura is the result of image sharpening?
 

Agent K

Active Member
If the aura was caused by an unsharp mask process wouldn't/shouldn't Fravor be well familiar with its characteristics or would this sort of minutia be strictly under the purview of the WSO?

The WSO would know this for sure, and I'd expect the pilot to be familiar with it since it's the pilots and/or WSO's who request more image sharpening. How do single-seat F-18E pilots operate without a WSO? Are they trained more than the F-18F tandem-seat pilots?
 

igoddard

Active Member
Air speed is 241 Knots, 277mph, so in 10 seconds the jet would have travelled 0.77 miles.
Why is that conversation of knots to mph the conversion we should use?

That's the conversion of CAS to TAS-mph at sea level...


@ http://www.hochwarth.com/misc/AviationCalculator.html#CASMachTASEAS

But that does not match the ATFLIR's altitude of 25k ft, it also does not match the Mach number on the ATFLIR screen of 0.58. But if we enter an altitude of 25k ft, it matches the screen data but TAS-mph goes up to 403 mph.

Sorting this out is necessary to determine the circumference of the circle the Navy jet travels as well as the distance traveled. How can we model the situation correctly if we're using seal-level data for a jet at 25k feet?
 

Mick West

Administrator
Staff member
Why is that conversation of knots to mph the conversion we should use?
241 knots IS 277 mph, regardless of altitude. There's not question there. It's like converting from mph to km/h At the time I wrote that (Dec 2017) I was unaware that the HUD numbers were CAS not TAS (I only realized that when looking at the Go Fast video later), so I was not converting from CAS to TAS, I was converting from knots to mph. The result is incorrect because I should ALSO have converted from CAS to TAS.

The calculator you link to is doing TWO conversions. It's converting CAS to TAS, and it's converting knots to mph. You can make it just do the TAS -> CAS conversion. Like here I'm just doing it in knots.

Metabunk 2019-10-07 10-52-54.jpg
TAS (True airspeed) is what should be used, regardless of if it's in Knots, mph, fps, kph, or whatever. I'm not arguing with that at all. The problem here is when you say:

Why is that conversation of knots to mph the conversion we should use?

You seem to think that converting from knots to mph is some kind of conversion that changes with altitude. It's not. It's just multiplying by 1.15078.
 

igoddard

Active Member
TAS (True airspeed) is what should be used, regardless of if it's in Knots, mph, fps, kph, or whatever. I'm not arguing with that at all. The problem here is when you say:

So the correct TAS-mph to use in the circumference equation is 403 mph, not 277 mph?

You seem to think that converting from knots to mph is some kind of conversion that changes with altitude. It's not. It's just multiplying by 1.15078.
No, all I've thought is what I've shown, that CAS --> TAS outputs differ by altitude, and that's true whether it's CAS(knots) --> TAS(mph) or CAS(knots) --> TAS(knots).
 

Mick West

Administrator
Staff member
So the correct TAS-mph to use in the circumference equation is 403 mph, not 277 mph?
403 mph is the correct airspeed to use.

No, all I've thought is what I've shown, that CAS --> TAS outputs differ by altitude, and that's true whether it's CAS(knots) --> TAS(mph) or CAS(knots) --> TAS(knots).

Then there's no problem, it just wasn't really very clear when you said:
There seems to be an error in your calculations based on your assuming 241 Knots = 277 mph, which, however, is only true at sea level. At 25,000 ft altitude, 241 Knots = 403 mph (see my last reply above).
Which looked like you were only discussing Knots -> mph
 

gtoffo

Member
There seems to be an error in your calculations based on your assuming 241 Knots = 277 mph, which, however, is only true at sea level. At 25,000 ft altitude, 241 Knots = 403 mph (see my last reply above). Using the sea-level conversion for the Navy jet results in an underestimate of the circumference of the circle it travels, which you give as

((277/60)/60) * (360/1.4) = 19.8 miles

but which should instead be

((403/60)/60) * (360/1.4) = 28.8 miles

That larger circle may place the UFO further away. I'm not trying to be a nitpicking nag, just wanting to nail down exactly what the Navy jet's path was, because we can only know where or if the LOS angles (54˚L and 6˚R) intersect if we know the Navy jet's precise path. And it seems to me that it traveled about 3.4 miles around a circle with a circumference of about 29 miles.

There's another assumption in your above analysis that bugs me, which is: "If we take the target position as essentially fixed..." But what if it was not fixed? That assumption seems to set the UFO's position from the start, yet I want to find out what that position was, if possible. I suspect the best that can be done is to describe a range of possible flight-path scenarios as opposed to one definitive scenario. That the target was essentially fixed may be just one possible scenario.

I totally agree we could make some reasonable hypothesis. @Mick West could you show your mathematical model for the static UFO hypothesis updated with the correct turn radius? It would help visualise ranges I think.

The pilots say: "they are all going against the wind. The wind is 120 knots to the west". Could we use that to make another "moving UFO" scenario?
Could we use the brightness of the clouds to estimate where the sun is and therefore where "the west" is?
Given the brightest part of the clouds is at the bottom I would say the sun is probably behind the observer (or the top would be illuminated) so that would give us a general "south" as behind the observer and the left side of the screen would be west? So the apparent movement is somewhat confirmed by the pilot's statements. We should simulate the movement in that direction.

Also the 120kts will drift the F-18 sideways making its trajectory in real space more of a spiral rather than a circumference but I guess we can ignore that as the wind is uniform on both the object and the F-18.
 
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