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

Mick West

Administrator
Staff member
[NOTE: This Thread is branched from: https://www.metabunk.org/threads/f-16-pilot-chris-lehto-analyses-gimbal-footage.11795/ and focuses on the calculation of the distance of the "Gimbal" object using lines of bearing, including validation and tests with the DCS flight simulator]


Chris Lehto's protractor diagram is quite similar to this:
Source: https://www.youtube.com/watch?v=1sHmuP_LIxI


From
https://www.metabunk.org/threads/ny...-navy-jet-encounter-with-unknown-object.9333/

Note I have a turn circle of 2.5 NM, he was 2.4, but that's essentially the same. The results end up being different, with a more distant non-moving point, and a narrower range of angles. I think the variable there would be the speed around the circle, which, looking back seems to be using CAS, not TAS.

Original calculations:

The easiest way to get some perspective on this is to note that the clouds never get any closer, even when the jet is heading directly for them at 240 kts.


It's difficult to wrap my head around. There's are very narrow field of view (0.75°). The object and the jet are both moving, and probably not in straight lines

Starting at frame 0 the clouds take 67 frames, or 2.23 seconds to cross the field of view, it's at a bit of an angle so that about 0.75° in 2 seconds, or 0.375°/sec
Starting at frame 400 it takes 103
600 - 758

Total cloud movement is about 6° to 7°. total camera rotation is 60° (54 to -6). So the camera rotates about 10x the rate the object is moving relative to the clouds, angularly.

In the first 300 frames (10 seconds), the heading changes from 54° to 40°, 14 degrees, or 1.4° per second. about 4.28 minutes for a full turn.

Air speed is 241 Knots, 277mph, so in 10 seconds the jet would have travelled 0.77 miles.

If we take the target position as essentially fixed (if it's far away), then the heading change is the actual turn rate of the jet and so would travel a circle of circumference 277/60/60*360/1.4 = 19.8 miles

Adding this all together in a VERY simple GeoGebra sim with a non-moving UFO seems to indicate the UFO is around 12-15 miles away



Here the circle is the path of the jet. The green line is the original line of sight to the UFO. The pink line is the Line of sight to the UFO, so the angle between them is the angular movement of the clouds behind the UFO. When the Jet moves though 60° the cloud angle moves about 6°

Notice the speed of movement of the pink line, it starts out moving smoothly, but then slows down and essentially stops as the Jet Heading (black arrow) crosses over it. Just like in the video.

This is making some gross simplifications about the turn rate and path of the jet, but I reckon it's in the ballpark.
 
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Alphadunk

Active Member
Without the rotation what exactly is unusual about the GIMBAL video?

Even if the object were rotating couldn't it just be another fighter jet doing a roll? The video itself has never been all that interesting despite the imagination of UFO fans. All the stuff that led up to the object being in the camera's sight is what makes it interesting but that's much harder to objectively analyze and draw conclusions from since we'll never have access to the complete dataset from the encounter.

His protractor diagram is quite similar to this:
Source: https://www.youtube.com/watch?v=1sHmuP_LIxI

Is your hypothesis the object isn't moving at all? I had considered the same but the pilots mention there is movement against the wind by the object(s). Although I'm not sure what criteria they were using to determine the trajectory and speed of the object.
 

DebunkMee

New Member
2 thoughts:

1. He claims that it can't be at the further, 9nm position because an airliner can't move that fast, but this doesn't rule out it being other fighter planes being at this (or greater) distances. So we need to figure out how detailed the fastest fighter plane would look like at the distance corresponding to wherever you'd draw the line given that top speed.
Namnlös.png


2. If the analysis above is roughly correct in terms of orientation, then how does it make any sense that the target creates an IR glare from the exhaust, since the camera is looking at the target directly from the target's left? If the engines of the target are not pointing towards the camera, what is creating the glare?
 

MarcusH

New Member
His protractor diagram is quite similar to this:
Source: https://www.youtube.com/watch?v=1sHmuP_LIxI


From
https://www.metabunk.org/threads/ny...-navy-jet-encounter-with-unknown-object.9333/

Note I have a turn circle of 2.5 NM, he was 2.4, but that's essentially the same. The results end up being different, with a more distant non-moving point, and a narrower range of angles. I think the variable there would be the speed around the circle, which, looking back seems to be using CAS, not TAS.

Original calculations:
Yes, Chris indicates a speed of 0.58 Mach or around 444 miles per hour as opposed to your figure of 277 miles per hour. Who is correct and does this sort out the discrepency?
 

TCarmickle

New Member
I may be totally wrong here but I thought mach does change with altitude in how it relates to spatial distance. Not that it changes the results of his math much though.
 

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Mick West

Administrator
Staff member
Yes, Chris indicates a speed of 0.58 Mach or around 444 miles per hour as opposed to your figure of 277 miles per hour. Who is correct and does this sort out the discrepency?

As I noted, the sim I did used CAS, which, for distance travelled is wrong.

Converting 241 CAS to TAS gives 350 Knots (402 mph)
https://aerotoolbox.com/airspeed-conversions/
2021-06-07_04-43-27.jpg
I've recreated Chris's diagram in Geogebra with a slider for Rate of Turn using the three points he used (PT1, PT2, PT3)

https://www.geogebra.org/classic/pxjp7p8u


This is fairly similar to his for a standard rate turn (3° per second), but diverges significantly for low rates of turn

2021-06-07_04-49-55.jpg

And below around 1.6 the intersections quickly veer to being infinitely far away.

So what's the actual rate of turn? It seems to be a function of true airspeed and bank angle. The bank angle varies from 26 to 35 degrees. It's about 26 over the first 10 seconds though
http://www.aviationwebdevelopment.com/samples/rateandradiusofturncalculator.aspx
2021-06-07_04-46-58.jpg

At that turn rate (1.52° per second), the LOB1 and LOB2 intersection is 72NM away2021-06-07_04-48-41.jpg

So is this turn rate calculation from TAS and bank angle correct?
 

Vizee

New Member
Yeah I find the actual turn rate to be an interesting part of the equation.

Here's some good reference from page 372 of the manual that you can find by googling "F18 EF 200 NATOPS".

TurnRatePage.png
 

FatPhil

Active Member
I may be totally wrong here but I thought mach does change with altitude in how it relates to spatial distance. Not that it changes the results of his math much though.

"Mach number changing with altitude" is fuzzy thinking. Compare with the gas law - does pressure change with temperature? That all depends on what else is varying. An object travelling with a fixed velocity can travel in a non-uniform medium with fixed velocity (inc. stationary) such that the vessel's Mach number changes with altitude, because of the medium's non-uniformity (temperature and pressure). Therefore any unqualified statement "Pilots use Mach number because it doesn't vary with altitude" (from your GIF, presumably a quote from the pilot?) is misleading at best, and downright wrong at worst. However, you can't fix the error in the statement just by flipping the "doesn't" into a diametrically opposite "does". "Can" would be better. It's complicated.

Of course, his short sentence actually contains three assertions, not just one. It's of the form "X use Y because Y has property Z". The above only addresses the "Y has property Z" aspect. If it doesn't hold, it calls into doubt the truth of the "because" clause too. I would guess, and this is pure speculation, that the real reason pilots use Mach number is because it's measurable and it's useful for awareness of what aerodynamic regime they are in.
 

BobArtwohl

New Member
I'm a bit confused. Wouldn't radar give the object's distance? Was this object was never captured on the F-18's radar? If not, why not? It seems to me that if I were investigating a UAP, the first thing I would try to do is get in on my radar. . .(forgive me if this has been answered elsewhere, I'm a newbie here.)
 

gtoffo

Active Member
3 observations:

  1. the object is not at "constant speed". It seems to be slowing down. So I think Chris's analysis would need to use a line that is rotated counterclockwise a little bit.
  2. the object does seem to be substantially bigger by the end of the video. We are getting closer to the object. This is consistent with point 1. A more counter clockwise trajectory would result in the object being closer at the end.
  3. Mick's analysis is based on an exact circle. While Chris considers that the angle of bank of the F-18 changes at some point (it increases the angle of bank). This is why in the video Mick you say his circle doesn't' seem a perfect circle. It deliberately isn't as the AOB is increased.
Question for Mick: given the fact you calculate such long distances isn't it impossible? The speed would be huge. Isn't that an indication there is some error in your calculations?
 

Mick West

Administrator
Staff member
Mick's analysis is based on an exact circle. While Chris considers that the angle of bank of the F-18 changes at some point (it increases the angle of bank). This is why in the video Mick you say his circle doesn't' seem a perfect circle. It deliberately isn't as the AOB is increased.
Yeah, he seems to have hand drawn it, so I don't think that's it.
 

Harabeck

New Member
Question for Mick: given the fact you calculate such long distances isn't it impossible? The speed would be huge. Isn't that an indication there is some error in your calculations?
At a more shallow rate of turn, the lines are more parallel, and so the apparent speed of object is slower.
 

Mick West

Administrator
Staff member
Question for Mick: given the fact you calculate such long distances isn't it impossible? The speed would be huge. Isn't that an indication there is some error in your calculations?
No, the huge distance is for an object that is NOT MOVING. A moving object would be even further away.
 

gtoffo

Active Member
No, the huge distance is for an object that is NOT MOVING. A moving object would be even further away.
I have your model open right now but I think I have missed something.

I can't see a point where all three lines converge as it would be necessary for a static object. There is no static solution I can find on your model.

What are the values for your proposed solution? What speed of the object and distance and what rate of turn you think is actually right?
 

TCarmickle

New Member
3 observations:

  1. the object is not at "constant speed". It seems to be slowing down. So I think Chris's analysis would need to use a line that is rotated counterclockwise a little bit.
  2. the object does seem to be substantially bigger by the end of the video. We are getting closer to the object. This is consistent with point 1. A more counter clockwise trajectory would result in the object being closer at the end.
  3. Mick's analysis is based on an exact circle. While Chris considers that the angle of bank of the F-18 changes at some point (it increases the angle of bank). This is why in the video Mick you say his circle doesn't' seem a perfect circle. It deliberately isn't as the AOB is increased.
Question for Mick: given the fact you calculate such long distances isn't it impossible? The speed would be huge. Isn't that an indication there is some error in your calculations?
I don't think that fact that he can calculate a longer distance indicates anything about the speed of the object. Because if the object is near the points of intersection then the object could be moving slow, far from intersection it may be moving fast and if right at intersection not moving at all.
 

Mick West

Administrator
Staff member
What are the values for your proposed solution? What speed of the object and distance and what rate of turn you think is actually right?
i think there's a wide range of possible solutions, but for me if the RoT is correct at around 1.5 to 1.7 then around 40+NM out, flying away (like along the red line)

To to get some sense of the increasing RoT, try at RoT about 1.55 and note where Intersect12 is, then try 1.65 and see Intersect13 is similar.

More detailed modeling could be done with the track of the bank angle (I think I extracted it earlier) but I don't have the time.
 

Vizee

New Member
I'm working on a "Gimbal UFO Explorer" program in Unity to show the math and let people play with all the variables. It's already been really helpful in determining what's going on here. I'll start a new thread for further details on this.

Prelim results show that this could definitely be an object flying at similar speeds to a passenger jet. There are a few key terms that play into Lehto's analysis like "convergence" and "aspect break" that reveal what he thinks he's seeing in the video and if he's wrong, or even off by a little bit in his conclusions/assumptions then it's throwing off his math. I think it's very possible we're seeing a real example of the limitations a skilled pilot might encounter when analyzing a flight solely through a moving camera with a ~2 degree field of view.


GimbalUFOExplorer.PNG
 

Daniel F

Member
Very nice analysis by Mick.
Christ ! He doesn’t hang about does he !

What is the significance of the standard 3 deg turn ?
He uses it to calculate the initial trajectory but then it changes due to a different bank angle. It’s not a true radius throughout as in Mick’s.

Is the manual not a guide as to an f18s maximum capability at a given speed and load ? A standard rate of turn seems more of an industry standard regardless of craft ?
I’m not sure I read that chart as what degrees of turn a pilot should do at any given speed. By their nature, they are performing dynamic movement dependant on a given situation.
 

jarlrmai

Senior Member
Very nice analysis by Mick.
Christ ! He doesn’t hang about does he !

What is the significance of the standard 3 deg turn ?
He uses it to calculate the initial trajectory but then it changes due to a different bank angle. It’s not a true radius throughout as in Mick’s.

Is the manual not a guide as to an f18s maximum capability at a given speed and load ? A standard rate of turn seems more of an industry standard regardless of craft ?
I’m not sure I read that chart as what degrees of turn a pilot should do at any given speed. By their nature, they are performing dynamic movement dependant on a given situation.
He was an F/16 pilot perhaps that jet does a tighter standard turn than the F/18.

Different planes have physical limits of the turns they can do at certain banks based on altitude/speed etc.

Don't confuse bank angle with turn angle.
 

MetaRoo

New Member
I watched Mick's video last night and it was one of the most impressive pieces of work I've seen on Mick's channel.
 

Daniel F

Member
He was an F/16 pilot perhaps that jet does a tighter standard turn than the F/18.

Different planes have physical limits of the turns they can do at certain banks based on altitude/speed etc.

Don't confuse bank angle with turn angle.
That’s kind of what what I mean. This chart is possibly the safe working limits for an f18, rather than standard expected turn. A fighters turn at any given time is dependent on a variety of variables obviously. In this case, needed to get behind the gimbal object.
As Mick said, the critical point is how Chris works out the first position turn radius. He doesn’t explain beyond why it’s in a standard rate of turn. I’m presuming he looks at bank and speed and knows it’s a 3 deg. Then at point two he makes a new radius due to the bank changing as bank angle or speed directly affects the turn radius.
 

jarlrmai

Senior Member
That’s kind of what what I mean. This chart is possibly the safe working limits for an f18, rather than standard expected turn. A fighters turn at any given time is dependent on a variety of variables obviously. In this case, needed to get behind the gimbal object.
As Mick said, the critical point is how Chris works out the first position turn radius. He doesn’t explain beyond why it’s in a standard rate of turn. I’m presuming he looks at bank and speed and knows it’s a 3 deg. Then at point two he makes a new radius due to the bank changing as bank angle or speed directly affects the turn radius.

No, we know from analysis of the artificial horizon marker on the footage the bank angle of the jet, knowing this bank angle gives us the turning circle.

Of course they could have turned faster if they banked more sharply or used more control surfaces, but the HUD shows a steady bank angle.

Given this bank angle and speed/altitude of the jet you can use the document to determine the turning circle.

Imagine trying to work out how much a car was turning, obviously impossible you only have speed, but if you have a video of the steering wheel with markers on it you could work it out with the specs of the car.
 

TCarmickle

New Member
I made a quick data sheet for some of the information in the gimbal video. I used GeoGebra to get the bank angles and the F18 turn capabilities sheet posted earlier in the discussion by Vizee to get the rate of turn using Mick Wests 350 value for true air speed (these values are obviously very approximate). Hopefully somebody can find this useful.

Turn Capabilities

F-18 Turn Capabilities.png
 

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FatPhil

Active Member
No, we know from analysis of the artificial horizon marker on the footage the bank angle of the jet, knowing this bank angle gives us the turning circle.

Of course they could have turned faster if they banked more sharply or used more control surfaces, but the HUD shows a steady bank angle.

Given this bank angle and speed/altitude of the jet you can use the document to determine the turning circle.

Imagine trying to work out how much a car was turning, obviously impossible you only have speed, but if you have a video of the steering wheel with markers on it you could work it out with the specs of the car.

Are there any clouds distant enough that their parallax will be negligible? If so, then each time a screenful of clouds passed across a screen width, the camera and plane together will have turned one field of view, and we know both the field of view and camera's angle. Of course, using the camera angle twice in a calculation might double the unknowns, but if the calculations drop out just the right way, the unknowns might cancel out.
 

Mick West

Administrator
Staff member
Are there any clouds distant enough that their parallax will be negligible? If so, then each time a screenful of clouds passed across a screen width, the camera and plane together will have turned one field of view, and we know both the field of view and camera's angle. Of course, using the camera angle twice in a calculation might double the unknowns, but if the calculations drop out just the right way, the unknowns might cancel out.
I think this is a valid approach, I had done it (and forgotten about it), here:

(THIS IS PROBABLY WRONG)
https://www.metabunk.org/goto/post?id=216629
The easiest way to get some perspective on this is to note that the clouds never get any closer, even when the jet is heading directly for them at 240 kts.


It's difficult to wrap my head around. There's are very narrow field of view (0.75°). The object and the jet are both moving, and probably not in straight lines

Starting at frame 0 the clouds take 67 frames, or 2.23 seconds to cross the field of view, it's at a bit of an angle so that about 0.75° in 2 seconds, or 0.375°/sec
Starting at frame 400 it takes 103
600 - 758

Total cloud movement is about 6° to 7°. total camera rotation is 60° (54 to -6). So the camera rotates about 10x the rate the object is moving relative to the clouds, angularly.

In the first 300 frames (10 seconds), the heading changes from 54° to 40°, 14 degrees, or 1.4° per second. about 4.28 minutes for a full turn.

Air speed is 241 Knots, 277mph, so in 10 seconds the jet would have travelled 0.77 miles.

If we take the target position as essentially fixed (if it's far away), then the heading change is the actual turn rate of the jet and so would travel a circle of circumference 277/60/60*360/1.4 = 19.8 miles

Adding this all together in a VERY simple GeoGebra sim with a non-moving UFO seems to indicate the UFO is around 12-15 miles away



Here the circle is the path of the jet. The green line is the original line of sight to the UFO. The pink line is the Line of sight to the UFO, so the angle between them is the angular movement of the clouds behind the UFO. When the Jet moves though 60° the cloud angle moves about 6°

Notice the speed of movement of the pink line, it starts out moving smoothly, but then slows down and essentially stops as the Jet Heading (black arrow) crosses over it. Just like in the video.

This is making some gross simplifications about the turn rate and path of the jet, but I reckon it's in the ballpark.

I don't have time to dig into this right now. But I not I'm using at FOV of 0.75, when I think it might actually be 0.35 (0.7 at 2x zoom). So probably worth revisiting.
 

jimmyslippin

New Member
I don't have time to dig into this right now. But I not I'm using at FOV of 0.75, when I think it might actually be 0.35 (0.7 at 2x zoom). So probably worth revisiting.

I'm curious as to where the 0.75° FOV comes from. When trying to find this figure to do some rough calculations regarding focal length, all I could find was this VR Simulations wiki (https://forums.vrsimulations.com/support/index.php/A/G_Advanced_Targeting_FLIR_(ATFLIR)) which gives the narrow FOV as 1.5° x 1.5°.
 

Mick West

Administrator
Staff member
I'm curious as to where the 0.75° FOV comes from. When trying to find this figure to do some rough calculations regarding focal length, all I could find was this VR Simulations wiki (https://forums.vrsimulations.com/support/index.php/A/G_Advanced_Targeting_FLIR_(ATFLIR)) which gives the narrow FOV as 1.5° x 1.5°.
That is where it comes from (2x zoom on 1.5 gives 0.75). However I now think it's 0.7 and 0.35, as that's what Raytheon and outer sources say, and it fits the videos better.
 

jackfrostvc

Active Member
@Mick West

In your calcs, what is your assumption for what the object is doing, speed, turn and direction wise?

Also, to me the wind is a factor, along with what the objects are doing.
I don't know if the concept that everything gets blown together by the wind, thus it makes no difference holds in all circumstances

To me , if say for example the object is flying into a 120 knot wind and remains geo stationary.
And the jet is flying perpendicular to the wind
Then you have a funky dynmaic going on. As the jet would then not just be going around in a circle relative to the object, but also be blown laterally (side ways) at the same time,
ie looking from above , it may seem like the jet almost flies straight, or a lot straighter than a circular motion the speed+heading change would indicate
 
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Mick West

Administrator
Staff member
Also, to me the wind is a factor, along with what the objects are doing.
The calculations are relative to the air mass - so that provides a single frame of reference (like being on a train). Wind might affect each one differently but the air-relative positions and velocity are independent of how that air is moving relative to the ground.
 

Mick West

Administrator
Staff member
I extracted the Gimbal bank angle over time by tracking the end-points of the horizon indicator


2021-06-09_08-45-17.jpg


Independently, Tero2021 created this:
Source: https://www.youtube.com/watch?v=zTuRheAI1LE

Their values are slightly higher than mine, but possibly more accurate as they are extracted from a linear fit of the horizon line, not just the ends - but it's essentially the same.
2021-06-09_08-55-29.jpg


They are simulating the path of the jet and lines of sight over time using this data. But it's not clear the math/algorithm that's used for that part.
 

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Mick West

Administrator
Staff member
Focusing on the intersection between point 1 and 2 in your simulation makes no sense to me. How could an object be at the intersection of line of bearings 1 and 2 but then Jump to the line of bearing of point 3? it makes no sense.
The object must be somewhere on each line of bearing, that's the only stipulation. It's probably flying away and far away, so the best fit is when the lines bearing are closer to parallel, and equally spaced.

We don't get that exactly here as the sim does not account for the change in Radius of Turn, but the intersection of LOB 1 and 2 indicated a region of possible lines between the LOBs that represent reasonable velocities. Further out, and they are too far apart (meaning the speed needed is too fast)
 

FatPhil

Active Member
Hello i'm a brand new member and really new to this whole thing so maybe this is a dumb point but here goes.

Mick, I think you missed one of the key points of Chris' video in your rebuttal. His point of reference for the location of the object is not the intersection between the camera views, but instead a trajectory of the object being recorded that places it at a location on each line at each time point. This has a few key assumptions, mainly that the object is going at a constant velocity without turning. That's why his representations for location of the object are places where the distance between each line of bearing is equal, not the intersections. Focusing on the intersection between point 1 and 2 in your simulation makes no sense to me. How could an object be at the intersection of line of bearings 1 and 2 but then Jump to the line of bearing of point 3? it makes no sense. Each line of bearing represents some location where the object must be at that point in time. Assuming it's moving (a decent assumption given its elevation) It takes 10 seconds to get from a point on line 1 to line 2 to line 3.

The error in estimating the position using the intersection of the bearings from times t and t+delta are small compared to the other errors that exist in the model, and don't change the results significantly. The intersection is best viewed as an estimate of the position of the object at time t+delta/2 - one line will be off by some amount, the other line will be off by some other amount the other way. Better schemes can be introduced when/if there's sufficiently reduced uncertainty in the inputs to warrant it, and right now there isn't. Pinning down the rate of turn is way more important, as it has far more significant numerical repercussions.
 

OneEleven

New Member
Lehto's response
Source: https://www.youtube.com/watch?v=1DTYqMqc3cU


He seems to resolve on a rate of turn higher than mine, but lower than 3.0

I don't have time to get into it this week. Maybe next week.
Chris seems to ramble a lot on irrelevant things and leaves thoughts unfinished. I wish he was a bit more focused. But maybe I'm just too stupid to get it.

It wasn't clear to me how the weight of the plane enters in reading that chart.

I also did not understand where he's coming up with those lines of intersection at the end. You could in principle put any line anywhere by assuming different speeds, no? Is he really just eyeballing based on the "constant size"? Isn't this going to be irrelevant if the object was a glare (flare?) from a far away object? Couldn't this assumption of his be biasing him towards assuming certain properties of the turning he admitted were guesswork at the beginning?

I also don't really understand the relevance of the standard rates of turn for airports for fighter jets doing exercise in the middle of the ocean. Surely none of those concerns matter under such circumstances?

This linear motion assumption could be a good way to analyze these videos (Gimbal and Go Fast): take agreed-upon trajectory of the plane and lines of sight, assume a constant speed object and find which linear motions are the most compatible with the data (instead of eyeballing). I could write such a software if there's any interest (taking positions and angles as input, not calculating the actual trajectory of the plane as I don't have the time to learn about this turn radius stuff). It would be a definitive statistical approach provided the position of the plane and the lines of sight are in agreement (which is where we're trying to reach).

It's also very sad to see highly trained pilots confidently using the Bernoulli principle as an explanation for lift. I wonder how many generations we need until that incorrect explanation finally dies. But that doesn't invalidate his points, it's just highlighting how pilots can be outdated in their physics too, so they are not authorities in the subject of physics of how planes actually fly. There's a significant amount of intuition-based skill, and intuition isn't objective or authoritative.
 
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