Gimbal distance and Speed Range Estimates using Lines of Bearing and/or DCS

jarlrmai

Senior Member
I took a stab at the objects location and path based on the lines of bearing above, starting with the objects path and working backwards. I just eyeballed the lines in my drawing, keeping the last angle at 0 degrees, and the intersections at around 30 miles.

I'm assuming the object is another plane traveling about the same speed as the F-18 and in a constant direction, with the ATFLIR looking right down the tailpipe to see the bright IR glare from the engine.

1624306920544.png




1624306990758.png

For the object to appear to move from right to left over the clouds, as shown in the Gimbal video, the cloud tops need to be past the intersecting points.

There's another clue from the original video. The object appears to move quickly across the clouds at the beginning, and then slows to a stop at the very end. This means the space between lines of bearing (line of sight) should be larger where they intersect the clouds at first, and get smaller and smaller toward the end.

From my drawing, the object appears to travel about 0.4 NM in 30 seconds across the clouds. The ATFLIR would need to have a tiny field of view to match what we see in the video. Does anyone know the actual FOV when fully zoomed in?

Gimbal is NAR at 2.0 digital zoom so this is probably an FOV of ~0.35 with NAR at 1.0 being 0.7.
 

Cassi O

Active Member
Gimbal is NAR at 2.0 digital zoom so this is probably an FOV of ~0.35 with NAR at 1.0 being 0.7.

At the start of the video it takes about 2 - 2.5 seconds for a cloud top to move from one side of the frame to the other. In my drawing a 0.35 FOV doesn't really fit what we see in the video, but at least we're in the ballpark.

1624369942747.png
 

jarlrmai

Senior Member
At the start of the video it takes about 2 - 2.5 seconds for a cloud top to move from one side of the frame to the other. In my drawing a 0.35 FOV doesn't really fit what we see in the video, but at least we're in the ballpark.

1624369942747.png
Actually I might be wrong it might be 0.7 with 1.5 bring NAR 1.0
 

Cassi O

Active Member
I moved the objects path in my drawing to where the lines of bearing start to intersect, which fit better and makes sense since we can assume the hornet was turning to chase it.

1624399435870.png

I then played with Walter Bislin's Advance Earth Curvature Calculator to find the visible horizon formed by the cloud tops, which worked out to 140 miles (120 NM) away.. (Target 1 is the object, and Target 2 is the clouds, and the view angle is tilted down -2 degrees like ATFLIR in the video.) I adjusted the different distances and altitude until I found a view that fit the video.

http://walter.bislins.ch/bloge/inde...074.8503-59387.84-17162.8-4~0.0343-9-21-1~2-8

Pushing the clouds back to 120 NM helped the FOV fit the video's timing much better.

1624400093514.png
 

Mick West

Administrator
Staff member
I then played with Walter Bislin's Advance Earth Curvature Calculator to find the visible horizon formed by the cloud tops, which worked out to 140 miles (120 NM) away.. (Target 1 is the object, and Target 2 is the clouds, and the view angle is tilted down -2 degrees like ATFLIR in the video.) I adjusted the different distances and altitude until I found a view that fit the video.
Speaking of horizons. Is there possibly a visible ocean horizon between the object and the clouds? You can see this in the original, but here I've blurred it and then adjusted the levels to make it more apparent.

With standard refraction, the horizon would be down 2.5 degrees, consistent with the -2° displayed.

 

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Cassi O

Active Member
Speaking of horizons. Is there possibly a visible ocean horizon between the object and the clouds? You can see this in the original, but here I've blurred it and then adjusted the levels to make it more apparent.

With standard refraction, the horizon would be down 2.5 degrees, consistent with the -2° displayed.

Not according to Walter Bislin's calculator. At 35,000 ft, with standard refraction, and a 0.35 degree FOV, the ATFLIR will need to be pointing -3.1 degrees to see the ocean horizon. The calculated dip angle is 3.19609 degrees.
 

Cassi O

Active Member
So I just noticed the 25,000 number in the lower right side of original video. If this is the altitude then my altitude estimates were off. Should't affect the drawing, but would move the cloud horizon closer.

The Bislin still doesn't show the ocean horizon with a 0.35 FOV at 25,000 ft.
 

jarlrmai

Senior Member
So I just noticed the 25,000 number in the lower right side of original video. If this is the altitude then my altitude estimates were off. Should't affect the drawing, but would move the cloud horizon closer.

The Bislin still doesn't show the ocean horizon with a 0.35 FOV at 25,000 ft.
Yeah that number is 25000ft barometric altitude.
 

Cassi O

Active Member
Now that I have the f-18 altitude straightened out.

http://walter.bislins.ch/bloge/inde...23432-1~0.0065-12-1376.01556-1~34.53-81-1~2-8

1624465615907.png
1624465679558.png

Object is 35 miles away at 19,300 ft flying away from the F-18. It's well below commercial airliner's cruising altitude and is fleeing the airspace when a US fighter appears, (which any sane pilot would do). This scenario makes for a good UAP event.

Cloud horizon is 160 miles away at 10,500 ft, so the objects FOV and apparent path reasonably fits the video.
 
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jarlrmai

Senior Member
Now that I have the f-18 altitude straightened out.

http://walter.bislins.ch/bloge/inde...23432-1~0.0065-12-1376.01556-1~34.53-81-1~2-8

1624465615907.png
1624465679558.png

Object is 35 miles away at 19,300 ft flying away from the F-18. It's well below commercial airliner's cruising altitude and is fleeing the airspace when a US fighter appears, which any sane pilot would do. This scenario makes for a good UAP event.

Cloud horizon is 160 miles away at 10,500 ft, so the objects FOV and apparent path reasonably fits the video.

Does does your refraction calculator take into account the wavelength of IR used by the camera? If the even matters.
 
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Cassi O

Active Member
Does does your refraction calculator take into account the wavelength of IR used by the camera? If the even matters.
I don't think Walter's calculator has any adjustments for IR, but it does have an adjustable refraction variable.

Changing from zero to standard refraction doesn't make that much difference. It just means the object and cloud heights would need to be a little higher or lower to match the view.
 
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Cassi O

Active Member
Played with Walter Bislin's calculator some more today. If it's a cloud bank with nothing behind it, and not an overcast sky all the way to the horizon, then the clouds and object can be put at any distance, just by adjusting their relative altitudes.
 

Cassi O

Active Member
In the video the pilot said "They are all going against the wind. The wind is 120 knots to the west". If the object is going 120 knots as Graves said, against a 120 knot wind, it will have an air speed of 240 knots. 240 knots is the same speed Chris Lehto said the fighter that recorded the video was traveling and estimated the object was going as well.
 

Vizee

New Member
Speaking of horizons. Is there possibly a visible ocean horizon between the object and the clouds? You can see this in the original, but here I've blurred it and then adjusted the levels to make it more apparent.

With standard refraction, the horizon would be down 2.5 degrees, consistent with the -2° displayed.


Whoa.... I always looked at it like the clouds were very far away and went past the horizon. But now considering the horizon drop at 25,000ft (Wolfie6020 shows this on yt channel)... if that really is the ocean new numbers are coming into play...

Until now I haven't really been sure that 2X on the NAR FOV was actually 0.35 degrees. It didn't line up with my tests and I've been searching docs thinking 0.7 made more sense as the lowest available. But if what we're looking at in this video is twice as close as I thought it was it then the 2X digital zoom on 0.7 is making more sense.

Also, being closer makes much more sense regarding how bright object is ..

If that really is the ocean / horizon then I think we're actually much closer to a solution on this one.
 

Cassi O

Active Member
Whoa.... I always looked at it like the clouds were very far away and went past the horizon. But now considering the horizon drop at 25,000ft (Wolfie6020 shows this on yt channel)... if that really is the ocean new numbers are coming into play...
It's not the ocean. With a 2 degree down tilt at 25,000 feet, the FOV would need to be at lease 2.5 degrees with standard refraction to see the water.

http://walter.bislins.ch/bloge/inde...2-1~0.0065-12-1376.01556-1~34.53-81-1~2-1~2-7
 

Vizee

New Member
At 25k feet I got exactly a 2.8 degree drop to the horizon....

So with the target object centered at -2 degrees in the video at level flight I think It's reasonable to believe that the other -0.8 degrees below that is the ocean.
 
In the video the pilot said "They are all going against the wind. The wind is 120 knots to the west". If the object is going 120 knots as Graves said, against a 120 knot wind, it will have an air speed of 240 knots. 240 knots is the same speed Chris Lehto said the fighter that recorded the video was traveling and estimated the object was going as well.
I think when he says stationary it means it was still with respect to the jet, but flying against a wind of 120 knots, its TAS was exactly like the wind intensity.
 
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Cassi O

Active Member
At 25k feet I got exactly a 2.8 degree drop to the horizon....

So with the target object centered at -2 degrees in the video at level flight I think It's reasonable to believe that the other -0.8 degrees below that is the ocean.
With a 2 degree FOV, looking 2 degrees below eye level, the ocean is still 0.8 below the FOV.
 
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Cassi O

Active Member
I think when he says stationary it means it was still with respect to the jet, but flying against a wind of 120 knots, its TAS was exactly like the wind intensity.
To put it another way, with eye level centered in a 2 degree FOV, there's 1 degree between eye level and the bottom of the frame.
 

jarlrmai

Senior Member
With a 2 degree FOV, looking 2 degrees below eye level, the ocean is still 0.8 below the FOV.
For what it's worth the elevation angle is described in the sim documents as:

Lookpoint Elevation Indication - The FLIR's current elevation angle in degrees relative to the horizon is displayed here. Negative values are below the horizon and positive values are above.
 
That suggests a false radar target. Electronic warfare doesn't care about the wind.
Actually, I think the speed talk needs a more accurate explanation from Graves. We do not know if he was talking about IAS, most likely, or of the object's TAS. I have instead found this passage of the inteview very interesting:"Two of these videos, colloquially dubbed “Go Fast” and “Gimbal,” displayed one of the objects that had perplexed Graves and his fellow Red Rippers back in 2015...."
 

Mendel

Senior Member.
For what it's worth the elevation angle is described in the sim documents as:

Lookpoint Elevation Indication - The FLIR's current elevation angle in degrees relative to the horizon is displayed here. Negative values are below the horizon and positive values are above.
An aircraft's navigation systems are inertial, i.e. its idea of "level flight" is the same at any altitude. The attitude indicator's horizon is at 90° from the zenith no matter what altitude the aircraft is flying at, it is at "eye level".

If the actual horizon outside the aircraft is at -2.8° from eye level, that's what the artificial horizon would show if you pointed the aircraft at it.

A square camera picture with a 2° FOV pointed at -2° will cover the range from -1° to -3°.
 
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Hey guys. Some of you will recall that I did a Blender recreation of GOFAST. I'm now giving GIMBAL a similar treatment. Since it's a different approach from what others on this thread are doing, I hope it will turn up some new or otherwise interesting results.

As with last time, I'm analyzing only the last 12 seconds of the clip. Also as with last time, I tracked the bank angle (as I see others have done). Since GIMBAL's bank changes quite a lot, I decided to animate the turn radius so that it changes continuously. This is what that looks like in overhead view (jet is at right...you can barely make out the FLIR sightline). Each square is 1,000 feet.

Source: https://youtu.be/mYyDnPAYNNU

One thing I don't like is that the radius changes instantaneously with the bank angle. From my own flight-simulator experience, I know this isn't physically accurate; steering a plane is more like steering a boat than a car. Perhaps I should delay the appropriate keyframes by one second or so?

As with GOFAST, I have a camera on the sightline. I will set it to a FoV of 0.35 and try to get a handle on the size of the IR flare at various distances. I'd also like to nest two rotator-objects, representing the outer and inner gimbals, and try to reproduce the slightly chaotic glare rotation that we see near zero degrees.
 
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Cassi O

Active Member
A square camera picture with a 2° FOV pointed at -2° will cover the range from -1° to -3°.


Right. A 0.35° vertical FOV when tilted down 2° has a range from -1.825° to -2.175°. This is not enough to see a horizon that's -2.8° below eye level (2.67° with standard refraction) as seen from 25,000',
 

markus

Active Member
Right. A 0.35° vertical FOV when tilted down 2° has a range from -1.825° to -2.175°. This is not enough to see a horizon that's -2.8° below eye level (2.67° with standard refraction) as seen from 25,000',
Note that the numbers on the display are rounded, however, so 2 degrees could really be anywhere from 1.5 to 2.5.
 

Mick West

Administrator
Staff member
One thing I don't like is that the radius changes instantaneously with the bank angle. From my own flight-simulator experience, I know this isn't physically accurate; steering a plane is more like steering a boat than a car. Perhaps I should delay the appropriate keyframes by one second or so?
An F/A-18 is going to be more boat-like than your average plane. Can anyone figure out this delay in DCS?

It's probably not going to be a simple shift of the turn radius.

Let's say the bank angle change from 28° to 30° over 1 second. I think the turn radius will start to change at the same time, but will stabilize later - probably with some ease in/out - maybe over 2 seconds.
 

markus

Active Member
Note that the center of the turn circle is not a physical object and can accelerate extremely fast. Once the aircraft is flying with the desired bank angle, the turn circle is set; any 'delay' has to come from transient aerodynamic effects that may happen as the bank angle changes. Figuring those out in detail is probably quite difficult, but we can put an upper bound on them by noting that an air molecule just now hitting the leading edge of the wings has no idea where the wings were a second ago, so at the most we're looking at effects that happen during the typical time of travel of an air molecule across the wing.

With a true airspeed of around 360 knots, and assuming a wing chord length of about 4 meters (the chord length of an F-18 wing near the root) that gives a maximum delay of about 4 m / 360 kt = 22 ms. That's less than a single frame of 30 fps video, so even assuming the maximal conceivable effect, where nothing happens whatsoever to the lift vector throughout that entire period, this delay appears to be negligible.
 

Mick West

Administrator
Staff member
With a true airspeed of around 360 knots, and assuming a wing chord length of about 4 meters (the chord length of an F-18 wing near the root) that gives a maximum delay of about 4 m / 360 kt = 22 ms. That's less than a single frame of 30 fps video, so even assuming the maximal conceivable effect, where nothing happens whatsoever to the lift vector throughout that entire period, this delay appears to be negligible.
It would be good to verify either way - even if just in a flight sim. It's not simply a change in forces on a point mass, there's also angular momentum (although seems like a small change).
 

MclachlanM

Active Member
It would be good to verify either way - even if just in a flight sim. It's not simply a change in forces on a point mass, there's also angular momentum (although seems like a small change).
Markus is spot on here, the angular momentum doesn't really apply as the turn radius is the solution of a static problem rather than a dynamic one (i.e. it is based on instantaneous geometry alone):

1624827456103.png

One thing I don't like is that the radius changes instantaneously with the bank angle. From my own flight-simulator experience, I know this isn't physically accurate; steering a plane is more like steering a boat than a car. Perhaps I should delay the appropriate keyframes by one second or so?
An F/A-18 is going to be more boat-like than your average plane.
I'm not too sure what you meant by this so I might be talking about something completely different here.

I made the DCS AI fly the same 3 banks in different planes (A-10, F-18 & C-130) to see if they would end up in different spots and the considerable differences between the planes seemed to have no effect:
1624827736490.png


1624827820778.png
(After the third turn the AI does it's own thing which is why they deviate, the point is that constant altitude, speed and bank angle are the only things that effect the turns)
If there is no discernible delay between 45 and 60 degrees it's probably safe to assume it's instantaneous for 30/35 in the video.
 

Mick West

Administrator
Staff member
I'm not too sure what you meant by this so I might be talking about something completely different here.

I made the DCS AI fly the same 3 banks in different planes (A-10, F-18 & C-130) to see if they would end up in different spots and the considerable differences between the planes seemed to have no effect:
I think I'm kind of falling for the same error that Chris Lehto did. F/A-18s are denser planes than something like the C-130, so it seems like they would need more force to turn. But of course, all planes have exactly 1g upwards in level flight, and in a coordinated turn will still have 1g down, and the corresponding component in the horizontal direction.

(After the third turn the AI does it's own thing which is why they deviate, the point is that constant altitude, speed and bank angle are the only things that effect the turns)
If there is no discernible delay between 45 and 60 degrees it's probably safe to assume it's instantaneous for 30/35 in the video.
That's great info, thanks!
 

Vizee

New Member
What I really want to know about the Gimbal video is what the actual FOV is at "NAR 2.0". Every single document I read list the available FOVs as 6.0, 2.8, and 0.7. I've been digging as far as every search engine can take me for this info and personally I think these are the actual values available and the on screen data is simplified for the operators. 6.0 is the wide "WFOV".. "NAR 1.0" is 2.8... and "NAR 2.0" is just an indicator for 0.7. It's not literally 2X, but just meant to signal the zoom state.

If we can figure out the actual FOV on this system that'll really help nail down the possible movement. I'm currently left running two simulations with one being over 2X as fast of the other...
 

MclachlanM

Active Member
What I really want to know about the Gimbal video is what the actual FOV is at "NAR 2.0". Every single document I read list the available FOVs as 6.0, 2.8, and 0.7. I've been digging as far as every search engine can take me for this info and personally I think these are the actual values available and the on screen data is simplified for the operators. 6.0 is the wide "WFOV".. "NAR 1.0" is 2.8... and "NAR 2.0" is just an indicator for 0.7. It's not literally 2X, but just meant to signal the zoom state.

If we can figure out the actual FOV on this system that'll really help nail down the possible movement. I'm currently left running two simulations with one being over 2X as fast of the other...
1624867726387.png
You're missing MFOV which is 2.8, then NAR1.0 is 0.7. Both MFOV and NAR can zoom in to 2.0x. That way the system can do 6.0, 2.8, 1.4, 0.7, & 0.35 (WFOV, MFOV1.0, MFOV2.0, NAR1.0 & NAR2.0).

NAR2.0 is definelty 0.35 in DCS as well.
 

jarlrmai

Senior Member
As far as I can tell the MFOV and NAR modes can be digitally zoomed to half the FOV.

2.8 at 2.0 would be 1.4
So NAR is 0.7 and NAR 2.0 is 0.35
 
The 0.35° FOV puts significant constraints on the "actual size" of the glare, which presumably is larger than the object causing the glare.

In Blender, I set the camera's FOV to 0.35° and roughly recreated the appearance of the glare by putting a sphere at a distance:

Screen Shot 2021-06-28 at 11.13.47 AM.png

The glare occupies about 4.3% or 1/23rd of the width of the picture, which means an angular diameter of about 0.0152 degrees. Confirmed with a quick trig check, here are some possible distances and "actual sizes" of the glare:

1 NM - 1.5 feet
5 NM - 7.3 feet
10 NM - 14.5 feet
20 NM - 29 feet
40 NM - 58 feet
 
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XXL

New Member
Hi guys,

didn't read the whole thread, not sure if this helps or is already redundant:

I have manually created an image stitch of parts of the first 20 seconds / 600 frames. Stitching was done by aligning images so the highest / most distant clouds match. Closer clouds not always align perfectly due to some parallax.
Given we know the horizontal FOV we can derive an upper bound for the actual camera yaw rotation.

The movement of the reticle in the first 600 frames was measured to be 2344 pixels. Camera image width is 320 pixels. Having traveled 7.325 times the FOV and assuming a 0.35° FOV we end up at 2.56° in the first 20s.

In case you want to review or continue the stitch, or measure other frames, it is saved using Gimp layers here:
https://drive.google.com/drive/folders/1oqChbgBWltZDOeWQkUafno9MLLxD8js6?usp=sharing

Cheers
Axel

 
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