Tags:
  1. Mick West

    Mick West Administrator Staff Member

    You can't see if it's hotter or colder than the surrounding atmosphere as surrounding atmosphere does not show up on IR. The blob here is showing up as cooler than the ocean.

    It's cooler because it's higher, in cold air, and any source of heat is either hidden, insulated, or there is no source of heat. This could happed in different ways:

    1) Helium Balloon - no source of heat, so it's the same temp as the air at 13,000 feet.
    2) Bird - internal heat, but possibly insulated and masked by feathers which are cold, as they are in cold air
    3) Small plane - Engine heat, cabin heat, and exhaust heat. Unlikley to show up as cooler than the Ocean, but perhaps possible at a distance with the high wing hiding most of the heat sources, so on average it's cooler.
    Metabunk 2018-03-14 21-20-20.

    I'd lean towards that order for probability. I'm torn between bird and balloon as a balloon fits the temperature profile better, and the shape. But there's a lot more birds than balloons.
     
    • Agree Agree x 1
  2. Mick West

    Mick West Administrator Staff Member

    One point in favor of the weather balloon hypothesis is that they have radar reflectors.
    Metabunk 2018-03-14 21-27-14.
    But how does the RNG measurement actually work? I'd assume the tracking was image based, is there image based tracking and a radar based distance measurement? Or laser based? The radar reflectors are pretty shiny, so I'd assume the would work well with lasers.
    Metabunk 2018-03-14 21-30-33.
     
    • Agree Agree x 1
  3. Agent K

    Agent K Active Member

    Does the range come from the radar or from ATFLIR's laser rangefinder, since it's displayed when ATFLIR locks on to the target? I don't know if the radar would pick up a bird or a balloon.

    Edit: Of course the radar would pick up the radar reflectors carried by a balloon.
     
  4. Agent K

    Agent K Active Member

    ATFLIR has its own laser rangefinder, which should be able to range a bird or a balloon that it's tracking, but to detect it in the first place, it may be slaved to the radar, which is more likely to get a reflection off a weather balloon's radar reflector than off a bird.
     
  5. Agent K

    Agent K Active Member

    Unless it's heated by the sun, especially if the balloon is dark. But they're usually white.
     
    • Agree Agree x 1
  6. Mick West

    Mick West Administrator Staff Member

    A method just occurred to me - image tracking in Adobe After Effects
    Metabunk 2018-03-14 22-25-04.
    Metabunk 2018-03-14 22-25-37.

    You can then click on the feature center in the the timeline, Copy, then paste into Excel. Repeat for second feature center. Calculate angle. Done!
    Metabunk 2018-03-14 22-43-31.
    https://www.metabunk.org/attachments/go-fast-tilt-tracking-xlsx.32275/
     

    Attached Files:

    • Like Like x 4
  7. Mick West

    Mick West Administrator Staff Member

    There's clearly a bit of feature center drift there though, you can see it in the two images. However I think it's accurate to within 1°

    I'd be concerned about the accumulation of errors though, if you use a non-zero value when the plane is not really turning.
     
  8. Agent K

    Agent K Active Member

    How do you calculate the turn rate from the bank angle? Do you use the angle of bank formula?
     
  9. igoddard

    igoddard Active Member

    Well, I meant the surrounding environment. However, FLIR cameras can detect invisible warmer gases, and the atmosphere is gases.
    http://www.flirmedia.com/MMC/THG/Brochures/T820433/T820433_EN.pdf#page=4

    And isn't the temperature we're seeing on the ocean the reflection of the temperature of the sky, not the ocean itself? That's what someone said previously in this thread. Agent K posted this...

    [​IMG]

    There's more warmth in the sky near the horizon, and I suspect the thermal variation seen in the waves of the Go Fast video is caused by the slanted angles of the waves reflecting more of the warmer horizon than the cooler sky directly above. Looking at the horizon we're actually looking through a thicker volume of atmosphere than when we look straight up, and the atmosphere is warmer than outer space. So I suspect FLIR cameras do detect the temperature of the atmosphere when aimed at a sufficiently expansive volume of it.
     
    • Agree Agree x 1
  10. Agent K

    Agent K Active Member

    Does the horizon look warmer because it's actually warmer, or for the same reason it looks red during a total solar eclipse? The atmosphere scatters the blue light coming from the distant horizon, while the red and infrared light is more likely to reach you.
     
    • Like Like x 1
  11. Agent K

    Agent K Active Member

    Probably some of both. Climate scientists certainly use IR cameras to study sea surface temperature. It may also depend on wavelength, long-wave IR (LWIR) versus mid-wave (MWIR) which is what ATFLIR senses.
     
  12. Justin Shaw

    Justin Shaw New Member

    You can assume the pilot is making a "coordinated turn" which puts the acceleration vector (including gravity) orthogonal to the plane of the wings.

    Nice work @Mick West ! I will not be able to get to this till next week however. Open challenge to other in the meantime.
     
  13. jarlrmai

    jarlrmai New Member

    Planes can turn in multiple ways though, they can bank, or use rudders or both and fighter jets may have thrust vectoring (I don't think an F/A 18 does) also fly-by-wire means the plane chooses how it turns depending on conditions.

    My main issue here is that the video is released as is by TTSA and their subsequent analysis (https://coi.tothestarsacademy.com/2015-go-fast-footage/) says:

    'moving at high speed'

    The assumption is that this object is moving quickly, no mention of radar tracking data is made in this analysis, nor is a method for determining the speed of the object mentioned.

    It seems left open for the reader to arrive at the conclusion that the object in the video is anomalous because it is moving quickly (fighter jet speed) yet has no visible jet engine exhaust or wings etc.

    However any reasonable attempt to determine air/ground speed object based on data in the FLIR video which is all we have ends up with a figure well within the speed of a bird or an object carried by the wind.

    The only reference I can find to them countering this is a Facebook message saying that the speed of the object was measured on radar, but they never mention this at all in their analysis, they also never address the what must be very clear (outside of reasonable margins of error) flaws in the many mathematical methods that have been used to determine speed all of which are in the range that would make the object non anomalous, although they must be aware of them given they mention Mick West by name.

    Additionally it is now clear that the 'GO FAST' video is a cut from the same video as 'GIMBAL' no mention of this was made, and surely given the different parameters of the objects (as described by TTSA and apparent from the video, one hot one cold) this is worthy of mention surely 2 differently behaving UFO's within a short time period is a big thing.
     
    Last edited: Mar 15, 2018
    • Agree Agree x 1
  14. Getoffthisplanet

    Getoffthisplanet New Member

    Absolutely amazing.

    Any chance you can throw together a MaxScript CSV importer that would apply the rotation angle values to an object so one wouldn't have to manually rotate each of the 1040 frames? I'm kidding. Mostly.
     
  15. Mick West

    Mick West Administrator Staff Member

    I've not used 3Ds Max for a decade. But it would seem relatively straightforward to modify a script like this found here:
    http://makeitcg.com/maxscript-animation/1413/
    like:
    Code:
    with animate on
    (
       at time 1	   $needle.rotation.z_rotation = 0.92394394
       at time 2	   $needle.rotation.z_rotation = 0.99434054
       at time 3	 ... etc ...
       ...	
    )
    
    You could even generate this it in Excel with various columns and export CSV with spaces as the separators.

    I'm unfortunately a bit too busy book editing to look into it right now.
     
  16. Getoffthisplanet

    Getoffthisplanet New Member

    Very nice. I think I could hack away and get the script to work. The hard part, for me, would be the Excel.

    Short term, it would probably be faster to manually enter all the values in Max.

    Long term, if there are 2 dozen more of these videos coming down the pike, then...

    Anyways, appreciate that you took the time to look.
     
  17. James Thorpe

    James Thorpe New Member

    I have been working on this over at VORTEX.

    I also realised that we need to know the angle the aircraft turned during the 19 seconds it was locked onto to. I drew a 3D drawing assuming the aircraft flew straight and level, I did this to check the parallax only. But then realised this can tell us its turn angle. Bank angle and turn rate are two different things and the factors are many and varied why that would not equate.

    I found from my parallax ground track, that if the aircraft were flying straight and level at the end, the sea surface should be crossing the frame at an angle of 34 degrees. But in the video it is crossing at an angle of 55 degrees at the end and 63 degrees at the start. taking this back to the drawing I created a curve that matched these figures, so the ground track on the drawing matched the track on the video at both positions. I found that the best match was when the aircraft turned by 5 degrees. track.

    Putting that curve into Mick's graph, this places the object at almost the shortest distance from the original point. Meaning it was moving very slow at only 34 knots.
     
    Last edited: Mar 15, 2018
  18. James Thorpe

    James Thorpe New Member

    This is the location I got once the 5 degree aircraft turn was put into Mick's Geobra graph. It has only travelled 345 meters in the 19 seconds which is 35 knots. Track2.
     
    • Like Like x 1
  19. Justin Shaw

    Justin Shaw New Member

    Excellent work. Keep this analysis bookmarked. There will always be accumulation of errors, but this is the best method presented so far!

    I just wrote a script that converts the roll angle to acceleration assuming that the pilot is keeping a coordinated turn keeping the total acceleration (gravity + acceleration) orthogonal to the wings.

    if a, v, and p are acceleration, velocity, and position, then we can numerically integrate as follows:

    Let G = 9.8 m/s^2
    initialize v and p appropriatly then repeat the following steps:

    roll = next roll angle
    acc = -tan(roll) * G
    solve for a as a coordinated turn.
    v = v + a * dt
    p = p + v * dt

    I'll post the results as soon as I get them.
     
  20. Getoffthisplanet

    Getoffthisplanet New Member

    Couldn't have been a less painless process.



    Apologies if this is off topic.
     
    • Like Like x 2
  21. igoddard

    igoddard Active Member

    I stabilized one of the two non-track-locked segments (at half speed) to the waves so that the ocean is stable (with a bit of jitter, alas). As expected, the trajectory of the object is then found to be straight, denoted by its parallel track alongside a red straight line I overlaid (in contrast to the non-stabilized version I posted previously). In the original segment the object's trajectory has a slight curve, probably an artifact of camera motion.

    [​IMG]

    Next I'll stabilize the second non-track-locked segment, which shows a more curved path in the original footage.
     
    • Like Like x 2
  22. Justin Shaw

    Justin Shaw New Member

    @Mick West was kind enough to pull out the roll data from the artificial horizon.

    Here is the track with the roll information included (assuming coordinated turns as previously described).

    When I assume a crosswind of 100 kts the object speed is only about 50 kts.
    upload_2018-3-16_14-13-55.

    And zoomed in, in 3D the object motion looks pretty random.

    upload_2018-3-16_14-15-46.

    Here is the raw data:
    upload_2018-3-16_14-13-17.

    Try the code again @Mick West , it should all work for you except for the 3D plots.

    Code:
    from numpy import *
    from pylab import *
    from scipy.interpolate import *
    
    ################################################################################
    ## constants
    ################################################################################
    G = 9.8 ## m/s^2
    NMI = 1852.
    HOUR = 3600.
    DEG = pi / 180.
    KTS = NMI / HOUR
    FEET = FOOT = .3048
    
    ################################################################################
    ## raw data
    ################################################################################
    
    txt = 'goFastRoll.txt'
    
    roll_data = loadtxt(txt, skiprows=1)
    
    
    #		 t sec	az deg,  el deg, range nmi  v?
    az_data = array([
    	[12 + 10./30, -43],
    	[13 + 10./30, -44],
    	[13 + 20./30, -45],
    	[14 + 19./30, -46],
    	[15 + 13./30, -47],
    	[16 + 13./30, -48],
    	[18 +  0./30, -49],
    	[21 +  6./30, -50],
    	[23 +  6./30, -51],
    	[24 + 18./30, -52],
    	[26 +  6./30, -53],
    	[27 + 18./30, -54],
    	[29 +  6./30, -55],
    	[30 + 13./30, -56],
    	[31 + 19./30, -57],
    	[32 + 25./30, -58],
    	[33 +  0./30, -58]
    	])
    el_data = array([
    	[12 + 10./30, -26],
    	[13 + 19./30, -27],
    	[16 +  6./30, -28],
    	[18 + 25./30, -29],
    	[21 +  7./30, -30],
    	[23 + 13./30, -31],
    	[25 + 25./30, -32],
    	[28 +  1./30, -33],
    	[30 +  1./30, -34],
    	[32 +  7./30, -35],
    	[33 +  0./30, -35]
    	])
    rng_data = array([
    	[12 + 10./30, 4.4],
    	[13 + 10./30, 4.3],
    	[15 +  3./30, 4.2],
    	[17 +  4./30, 4.1],
    	[19 +  1./30, 4.0],
    	[20 + 29./30, 3.9],
    	[22 + 28./30, 3.8],
    	[24 + 28./30, 3.7],
    	[27 +  1./30, 3.6],
    	[29 + 13./30, 3.5],
    	[31 + 22./30, 3.4],
    	[33 +  0./30, 3.4],
    	])
    
    az = interp1d(az_data[:,0], az_data[:,1])
    el = interp1d(el_data[:,0], el_data[:,1])
    rng = interp1d(rng_data[:,0], rng_data[:,1])
    roll = interp1d(roll_data[:,0], -roll_data[:,1])
    
    dt = .1
    t = arange(15, 30, dt)
    N = len(t)
    
    ## align data to uniform time grid and standard units
    data = vstack([az(t) * DEG, el(t) * DEG, rng(t) * NMI, roll(t) * DEG]).T
    
    ################################################################################
    ## functions
    ################################################################################
    def get_accel(roll):
    	'''
    	assume a corrdinated turn so that accel vector (with gavity) is orthog to wings
    	'''
    	return -tan(roll) * G
    
    
    ################################################################################
    ## Numerical Integration of accel to get vel and pos
    ################################################################################
    for WIND_HDG in arange(225, 226, 45) * DEG:
    	# xoWIND_HDG = 90 * DEG
    	WIND_SPD = 100 * KTS
    	WIND_VEL = array([cos(WIND_HDG), -sin(WIND_HDG), 0]) * WIND_SPD
    	AIR_SPEED = 369 * KTS
    	pos = array([0, 0, 25000 * FEET])
    	vel = array([AIR_SPEED, 0, 0]) + WIND_VEL
    
    	poss = [] ### save track
    	vels = [] ### save vel
    	Xs = []
    	R = array([0, 0, 1]) ### unit up vector (assume minimal earth curvature over)
    	for i in range(N):
    		rho = data[i, 3] ## grab roll
    		I = vel - dot(R, vel) * R; I = I / linalg.norm(I) ## intrack vector
    		L = array([-I[1], I[0], 0])					   ## horizontal unit left vector
    
    		a = get_accel(rho) * L
    		dv = a * dt
    		vel = vel + dv
    		# vel = GND_SPEED * vel / linalg.norm(vel)
    		dp = vel * dt
    
    		pos = pos + dp
    		C = cross(R, I)
    		RIC_X = vstack([R, I, C]).T
    		RHO_X = array([[cos(rho), 0, -sin(rho)],
    					   [	   0, 1,		 0],
    					   [sin(rho), 0,  cos(rho)]])
    		X = RIC_X @ RHO_X ## include roll
    		X = RIC_X ### dont include roll
    		poss.append(pos)
    		vels.append(vel)
    		Xs.append(X)
    	poss = array(poss)
    	vels = array(vels)
    	Xs = array(Xs)
    	azs = data[:,0]
    	els = data[:,1]
    	rngs = data[:,2]
    
    	### compute 3d position of unknown object
    
    	x = cos(azs) * cos(els)
    	y = -sin(azs) * cos(els)
    	z = sin(els)
    
    	doas = vstack([z, x, y]).T
    	d_fixed = array([X @ d for X, d in zip(Xs, doas)])
    	gs = poss + d_fixed * rngs[:,newaxis]
    	figure(123);	plot(linalg.norm(abs(diff(gs, axis=0)), axis=1) / dt / KTS)
    
    	################################################################################
    	## PLOTS
    	################################################################################
    
    	figure(1)
    	ax = subplot(411)
    	plot(t, az(t))
    	plot(az_data[:,0], az_data[:,1])
    	ylabel('Az [deg]')
    
    	subplot(412)
    	plot(t, el(t))
    	plot(el_data[:,0], el_data[:,1])
    	ylabel('El [deg]')
    
    	subplot(413)
    	plot(t, rng(t))
    	plot(rng_data[:,0], rng_data[:,1])
    	ylabel('Rng [nmi]')
    
    	subplot(414)
    	plot(t, roll(t))
    	plot(roll_data[:,0], -roll_data[:,1])
    	ylabel('Roll [deg]')
    	xlabel('Time [sec]')
    
    
    	figure(2)
    	plot(-poss[:,1] / NMI, poss[:,0] / NMI)
    	plot(-gs[:,1] / NMI, gs[:,0] / NMI)
    	xlabel('<-- Port [nmi] Starboard-->')
    	ylabel('Intrack [nmi] -->')
    	axis('equal')
    
    
    	## try 3D plot... don't error out if libs not present
    	try:
    		import matplotlib as mpl
    		from mpl_toolkits.mplot3d import Axes3D
    		fig = plt.figure(3)
    		ax = fig.gca(projection='3d')
    		# ax.plot(-poss[:,1] / NMI, poss[:,0] / NMI, poss[:,2] / FEET)
    		ax.plot(-gs[:,1] / NMI, gs[:,0] / NMI, gs[:,2] / FEET)
    
    	except:
    		pass
    
    
    show()
    
    ROLL DATA, save as "goFastRoll.txt"
    
    Code:
    Time(sec)	ANGLE(deg)
    0.000000000	-1.527525442
    0.03336670003	-1.527525442
    0.06673340007	-1.529930785
    0.1001001001	-1.529727759
    0.1334668001	-1.529467672
    0.1668335002	-1.518095647
    0.2002002002	-1.53267112
    0.2335669002	-1.474672013
    0.2669336003	-1.442115169
    0.3003003003	-1.449033179
    0.3336670003	-1.448491076
    0.3670337004	-1.366693395
    0.4004004004	-1.362085868
    0.4337671004	-1.448560624
    0.4671338005	-1.366816098
    0.5005005005	-1.35875473
    0.5338672005	-1.347126873
    0.5672339006	-1.313522968
    0.6006006006	-1.316211803
    0.6339673006	-1.31669294
    0.6673340007	-1.319092193
    0.7007007007	-1.318687659
    0.7340674007	-1.310107293
    0.7674341008	-1.269931244
    0.8008008008	-1.271831854
    0.8341675008	-1.254297799
    0.8675342009	-1.252965901
    0.9009009009	-1.187814438
    0.9342676009	-1.18963841
    0.967634301	-1.194717652
    1.001001001	-1.097501285
    1.034367701	-1.091757519
    1.067734401	-1.08559533
    1.101101101	-1.085369117
    1.134467801	-1.06274827
    1.167834501	-0.9839292344
    1.201201201	-0.9869799868
    1.234567901	-0.9899353083
    1.267934601	-0.9880899279
    1.301301301	-0.9138886262
    1.334668001	-0.832264237
    1.368034701	-0.8366440038
    1.401401401	-0.8300653589
    1.434768101	-0.8177009379
    1.468134801	-0.8097928306
    1.501501502	-0.7414495604
    1.534868202	-0.7397942202
    1.568234902	-0.7470575709
    1.601601602	-0.8010169766
    1.634968302	-0.8052596287
    1.668335002	-0.8071316226
    1.701701702	-0.7127572193
    1.735068402	-0.6565412697
    1.768435102	-0.6482386193
    1.801801802	-0.5111619268
    1.835168502	-0.5426412843
    1.868535202	-0.545593627
    1.901901902	-0.555778022
    1.935268602	-0.4748472525
    1.968635302	-0.4755367486
    2.002002002	-0.4594720966
    2.035368702	-0.4087903385
    2.068735402	-0.4039990677
    2.102102102	-0.4086619848
    2.135468802	-0.4045519829
    2.168835502	-0.4076769789
    2.202202202	-0.4043038865
    2.235568902	-0.3952245231
    2.268935602	-0.3957967707
    2.302302302	-0.3478919869
    2.335669002	-0.3526225688
    2.369035702	-0.3477584524
    2.402402402	-0.2733829195
    2.435769102	-0.272165513
    2.469135802	-0.1613005246
    2.502502503	-0.1594394579
    2.535869203	-0.1652063862
    2.569235903	-0.1599757997
    2.602602603	-0.1588678392
    2.635969303	-0.06983149832
    2.669336003	-0.03879850644
    2.702702703	-0.03590305267
    2.736069403	-0.08649174158
    2.769436103	-0.08537901971
    2.802802803	-0.0076371359
    2.836169503	0.03037854839
    2.869536203	0.004731278231
    2.902902903	-0.001221433788
    2.936269603	0.008855984867
    2.969636303	0.007185573954
    3.003003003	0.01437383248
    3.036369703	0.02275023371
    3.069736403	0.04108559377
    3.103103103	0.05898131067
    3.136469803	0.07997317346
    3.169836503	0.04764238146
    3.203203203	0.0465891272
    3.236569903	0.04415099376
    3.269936603	0.04475849965
    3.303303303	0.126531145
    3.336670003	0.1336642461
    3.370036703	0.1342802926
    3.403403403	0.1313700651
    3.436770103	0.1332112978
    3.470136803	0.1307712464
    3.503503504	0.1330422158
    3.536870204	0.1646711233
    3.570236904	0.2614225177
    3.603603604	0.250964315
    3.636970304	0.2640974045
    3.670337004	0.2616513747
    3.703703704	0.2591527634
    3.737070404	0.2532082782
    3.770437104	0.2496449063
    3.803803804	0.2503271973
    3.837170504	0.2396128143
    3.870537204	0.2430830349
    3.903903904	0.3141675899
    3.937270604	0.2982442049
    3.970637304	0.3042371158
    4.004004004	0.288622051
    4.037370704	0.2924327798
    4.070737404	0.2763718307
    4.104104104	0.2816523015
    4.137470804	0.2872071261
    4.170837504	0.2925302875
    4.204204204	0.2902204131
    4.237570904	0.2871328248
    4.270937604	0.3006951125
    4.304304304	0.3147500142
    4.337671004	0.3223346498
    4.371037704	0.3270428544
    4.404404404	0.3302481266
    4.437771104	0.3224731411
    4.471137804	0.3186776727
    4.504504505	0.3176060991
    4.537871205	0.3284804441
    4.571237905	0.3303001286
    4.604604605	0.3230097479
    4.637971305	0.3372224496
    4.671338005	0.3360897312
    4.704704705	0.3276531444
    4.738071405	0.3275204786
    4.771438105	0.3264529126
    4.804804805	0.3289109136
    4.838171505	0.3313271274
    4.871538205	0.3348536606
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    • Like Like x 1
  23. Getoffthisplanet

    Getoffthisplanet New Member

    The blue represents the overhead trajectory of the jet across the entire 34 seconds/1129 frames?
     
  24. Justin Shaw

    Justin Shaw New Member

    No, is is only the time that the object was tracked.
     
  25. Getoffthisplanet

    Getoffthisplanet New Member

    But the blue does represent the trajectory of the jet from overhead?

    If so, would you be able to plot the entire clip and post it?
     
  26. Justin Shaw

    Justin Shaw New Member

    Not much to see here. This is the complete track from start of vid to end. upload_2018-3-16_18-21-30.
     
    • Useful Useful x 1
  27. Getoffthisplanet

    Getoffthisplanet New Member

    Great! Thanks.

    Now can I track that line in 3ds Max and use it as an animation trajectory. I think.

    I have zero aviation experience, so I'm confused and will sound pretty ignorant here, but why does the jet bank to the left in all these calculations when the artificial horizon banks to the right?

    And does anyone know, for a 3D recreation would I want to use the TAS or CAS?
     
  28. Mick West

    Mick West Administrator Staff Member

    Tilt your head to left, the horizon tilts down to the right. The plane is represented in the middle of the artificial horizon. It does not move because the screen is moving with the plane.

    It's easier to visualize this if you hold your cellphone camera at arms length and then tilt your head and camera together. Bonus points if you use a theodolite app.
     
  29. Getoffthisplanet

    Getoffthisplanet New Member

    Eureka. Got it now. App made all the difference.

    So in order to use the After Effects tracking data plugged into your Go Fast Tilt tracking.xlsx to drive the rotation of the jet in Max I would need to invert the values (or change the formula in Excel)?
     
  30. igoddard

    igoddard Active Member

    Here's the second segment with non-locked tracking at half speed stabilized to the ocean.

    [​IMG]


    Here's this same segment with overlaid path map not stabilized...

    [​IMG]

    Go Fast's true trajectory looks to be straight during both non-locked-tracking segments. There's some fine zig-zagness to the path here, but I suspect that's interlacing noise in the video. In these snippets I see no indication that the object is being consciously steered, it could be a wind-blown weather balloon looking to go faster than it is due to parallax.
     
    Last edited: Mar 16, 2018
    • Like Like x 1
  31. Tom Mellett

    Tom Mellett New Member

    Hello Mick,

    I’ve been conversing with Garry Nolan on the TTSA Facebook group page and he has raised some issues and questions about the conclusions of the altitude of the object being so high ca, 13,000 feet. He states that the object is moving along about 100 feet above the ocean surface. I told him I would copy his comment for you here:

     
  32. Mick West

    Mick West Administrator Staff Member

    He can state it all he wants but he really need to show some reason why he thinks that.

    Why not?

    And where is the focal plane exactly? Nothing really appears to be in particularly sharp focus.
     
    Last edited: Mar 16, 2018
  33. switch70

    switch70 New Member

    The part of the video referenced above, "We got very clever at dual-using", is not actually in the YouTube clip provided, although I have seen it. Elizondo is referring to his time as program director when he says, "We got very clever at dual using..." The statement is referring to the programs ability to use resources and gather data. I do not believe it has anything to do with his time at TTSA and in any way makes reference to the "GO FAST" video and how they were obtained or released..
     
  34. Mick West

    Mick West Administrator Staff Member

    What part?
     
  35. Z.W. Wolf

    Z.W. Wolf Senior Member


    What Gary Nolan seems to be saying is that if the object were at 12,500 feet altitude, both it and the surface of the sea could not be in focus at the same time, because the object would be too close to the camera and the sea would be too far. So, because they are both in focus, the object must be at a lower altitude.

    Let's find out.

    Let's start at the beginning with an example of how naive people can get into trouble with the subject at hand. I've seen a certain confusion about focus over and over among FE believers. A good example:



    The video author first focuses his camera on the star Sirius and then switches his view to the moon... which is still in focus! The implication is that they are at the same distance, otherwise they would not both be in focus. I've seen this argument extended. Stars that are said to be in our galaxy and distant galaxies are both in focus in the same astrophotograph ... which in the minds of FE believers would be impossible. It's as if they expect to keep adjusting the focus on a camera or telescope infinitely. It's as if they've never seen the the infinity setting on a focus ring. (Which of course they haven't.)

    If we are talking about cameras, the simplest explanation would be to say that the moon and Sirius are both far enough away that both would both be in focus if the lens were set to infinity. You could take a picture of the moon rising over a mountain, and the mountain, the moon and Sirius would all be in focus with the lens set to infinity.

    But what about the case in question? Would the surface of the sea and the object at 12,500 feet of altitude be both far enough away from the camera so that both would be in focus if the camera was focused on the surface of the sea?

    Let's get more technical...

    First, scupper the term "focus plane." It doesn't really mean what people seem to think it does. It's just confusing the issue. Let's talk about "hyperfocal distance."


    From Wikipedia:
    https://en.wikipedia.org/wiki/Hyperfocal_distance

    The second definition is easier to understand, and I'm going to go with it.

    The simplest way to think about this is, how far away does something have to be from the camera for you to just twist the focus ring all the way (to infinity)? It's different for different lenses. The hyperfocal distance for wide angle lenses is close and for telephoto lenses it's far.

    It can cause some problems when you're shooting landscapes and you want to get that tree in the foreground and the mountains both in focus. You may not be able to do that with the lens you want to use. But you can be sure that you can get both the clouds and the mountain (and the rising moon) in focus at the same time. Doesn't matter even if you are using a powerful telephoto. They're all far enough away to be beyond the hyperfocal distance.

    Getting even more technical, let's figure the hyperfocal distance for various 35 mm telephoto lenses.

    We can use this formula (from the same Wiki article):

    The "circle of confusion limit":
    https://en.wikipedia.org/wiki/Circle_of_confusion
    It's figured partially by what is acceptable, so there's no one absolute way to calculate it. If you really want to understand it there is plenty of info in the article and elsewhere on the Net. I'm going to use 0.03, just because that's what I'm used to in 35 mm photography.

    Even though the formula is hyper-technically for the first definition, I'm going to use it as if it were for the second... because in the real world it doesn't matter.

    Just to get a feeling for this: Using an f16 setting I've calculated the hyperfocal distance for various lenses in the 35 mm format:

    100 mm = 68 feet (2 power magnification)

    500 mm = 1,706 feet (10 power magnification)

    1000 mm = 6,835 feet (20 power magnification)

    2000 mm = 27,340 feet (40 power magnification)

    You can use another traditional fudge factor and say "everything past half that distance from the camera will be in 'acceptable' focus if you set the focus to that distance." In other words, definition one from above. But that's an old photographer's trick, and I'm confident that the auto-focus on the camera in question doesn't use that trick on distant objects. It just focuses on the most "important" object.

    But, let's get one thing clear. The lens in the footage in question would not necessarily be set at infinity. If the surface of the sea were not as far away as the hyperfocal distance of the lens/camera that would be another question to chew on.

    Regarding the footage in question, if anyone can find the focal length of the lens, and the size of the sensor we could figure the circle of confusion limit and the hyperfocal distance.

    To clarify, or at least re-cap:

    What is the calculated distance between the camera and the object? Is it 4.4 nautical miles - 26,700 feet?

    And the further distance between the object and the surface of the sea?


    Another question: Is the "object" really in good focus?

    Or acceptable focus? Or poor focus?

    When considering that question we have to think about resolution also.
     
    Last edited: Mar 17, 2018
    • Informative Informative x 4
  36. Kaen

    Kaen Member

    Maybe Garry Nolan can explain how an object viewed from a shallow angle of 26 degrees from the jet can be both just above sea level AND at the same distance from the jet as the sea surface right below the jet. I would like to see him present a drawing where he explains this.

    The only 'assumptions' made in the calculations are about the wind speed and the precise curve of the jet while it is banking.The rest is based on hard data in the ATFLIR display. What hard data does Garry Nolan have to support his statements?
     
    Last edited: Mar 17, 2018
    • Like Like x 1
  37. Agent K

    Agent K Active Member

    It's in the YouTube clip before the part that I skipped to when I posted it here. I gave the context of the statement. If Elizondo was cleverly dual-using the program's funding, he may have cleverly dual-used the release of the videos.
     
    • Like Like x 1
  38. Agent K

    Agent K Active Member

    Are we sure that the background is water and not clouds like in the Gimbal video? Looks like water to me.
     
  39. Agent K

    Agent K Active Member

    The dot at the top left of the video indicates the direction towards North. You could use it as an absolute reference.

    Edit: On second thought, it probably indicates the sensor azimuth relative to the platform.
     
    Last edited: Mar 17, 2018
    • Like Like x 1
  40. Mick West

    Mick West Administrator Staff Member

    I thought that was just a visual indication of the direction of the camera relative to the plane?

    Might be useful to track that in After Effects. It's finer grained than the ° numbers, and could tell us something about the ground tracking mode which confuses people.
     
    • Like Like x 1
    • Agree Agree x 1