Sphere, Acorn, Metallic Blimp - Three iPhone Photos From an F-18 via Mystery Wire

gtoffo

Member
Ive seen the Batman balloon listed as being 33in = 86cm, but exactly what dimension this relates to is unknown, (link)

It would be interesting to do the calculations the other way around, ie, if we know the size of the Batman balloon, can we calculate the speed of the aircraft using the photos and metadata, and is it within the flight envelope of the F/A-18E...?
If the balloon was 1/3 of the size I've calculated you would need to scale down the whole "triangle" and that would give you a third of the distance (so very very close) and the aircraft would need to go 1/3 of the speed and would be stalling even with full afterburner.

Even at half the speed I used in my lower estimate the plane would be at the minimum controllable speed for sea level and probably would be stalling at that altitude even with full after burner and full flaps.

No intercept would be done at stall speed with full after burner. Jets like to fly fast especially at high altitudes and afterburners are very inefficient.

300/350knots (what I used in my lower bound) is probably the minimum given the altitude. Maybe 250 is the extreme minimum but highly unlikely in this scenario.
 

CeruleanBlu

Senior Member.
While doing calculations on the speed of an object based on known sizes it's best to remember that when it comes to mylar balloons they can be created at home, in almost any conceivable size and shape. Here's a demonstration video of designing, cutting, sealing, filling and releasing into the sky a balloon of the video makers own design.

There are few limits to the size, shape, color and release location of an unidentified object that could be created by anyone at any time.

DIY Type Balloons
 

jarlrmai

Active Member
Using the calculator above and given we know the full camera specs so by measuring pixels we can calculate:

The size given a distance or the distance given a size, however we do not know either of the variables so we don't gain much here alone.

We can however measure approx jet speed by assuming a fixed position object and noting the increase in apparent size over the gap between 2 pictures (0.374s)

The object goes from 20 pixels tall to 32 pixels tall in 0.374s

If we assume a 1m object that's 163m to 101m in 0.374 seconds = 322kts
If we assume a 5m object that's 815m to 509m in 0.374 seconds = 1590kts

I would suggest this puts bounds on the object size of 0.5m (approaching stall speed) and 3m (very fast cruise speed) given they were making passes to take photos and get a closer look on the slower and thus smaller end.
 

DavidB66

Active Member
https://www.scantips.com/lights/subjectdistance.html

This is the best calculator to use in my opinion

Image is 4032x3024 4/3 3.99mm 28mm equiv

Object is 20pixels in the one image I have
What image are you referring to? An iPhone 8 camera screen is said to have a standard pixel count of 1334 x 750, giving a h:w or w:h ratio of about 1:1.78. Presumably if it is redisplayed in some other format the pixel count might change, but the ratio should not, if the only change is a proportionate increase or decrease in pixels . 4032 x 3024 gives a ratio of 1:1.33, which is significantly different from 1:78 - the ratio is increased by about a third. Does that imply we are looking at a cropped image?

If the object image is 20 pixels wide in an image with total width of 4032, the ratio of object to total is 1:202, which (much to my surprise!) is very similar to the result I got from my stone-age method of using a ruler. So I think we are agreed that the width of the object is about 1/200 the width of the overall image.

There remains a question about the field of view. I think you assumed a horizontal fov of 65.5 degrees. I assumed 30 degrees. Is there any way of determining the correct value?
 

gtoffo

Member
Using the calculator above and given we know the full camera specs so by measuring pixels we can calculate:

The size given a distance or the distance given a size, however we do not know either of the variables so we don't gain much here alone.

We can however measure approx jet speed by assuming a fixed position object and noting the increase in apparent size over the gap between 2 pictures (0.374s)

The object goes from 20 pixels tall to 32 pixels tall in 0.374s

If we assume a 1m object that's 163m to 101m in 0.374 seconds = 322kts
If we assume a 5m object that's 815m to 509m in 0.374 seconds = 1590kts

I would suggest this puts bounds on the object size of 0.5m (approaching stall speed) and 3m (very fast cruise speed) given they were making passes to take photos and get a closer look on the slower and thus smaller end.

We are in similar ballparks. Just a note: 161knots is lower than the stall speed of the aircraft at that altitude for sure.

The F-18 stalls at around 200 knots with no flaps and minimum load at sea level.
 
What image are you referring to? An iPhone 8 camera screen is said to have a standard pixel count of 1334 x 750, giving a h:w or w:h ratio of about 1:1.78. Presumably if it is redisplayed in some other format the pixel count might change, but the ratio should not, if the only change is a proportionate increase or decrease in pixels . 4032 x 3024 gives a ratio of 1:1.33, which is significantly different from 1:78 - the ratio is increased by about a third. Does that imply we are looking at a cropped image?
The screen aspect ratio doesn't have any bearing on the camera's sensor, which is indeed 4:3. The lens has a 28mm focal length (35mm equivalent, actual sensor size is 4mm), which gives a 65.5º horizontal angle of view.
 

gtoffo

Member
Extreme approximation of the calculations here...

Using FOV of 53° for the camera (not sure it's the right one tried finding an exact value but I failed...in any case this is such an approximation that I think it shouldn't introduce such a big error).
I calculated number of horizontal pixels between an approximate centerline for the aircraft and the object and I see: 10,15° in the first one and 15,06° in the second one.

Using the most probable speeds above and assuming the object is stationary we have two triangles:
Screen Shot 2021-04-08 at 13.45.04.pngScreen Shot 2021-04-08 at 13.45.24.png

This would indicate a range between 203 and 336 meters in the second image. The faster the F-18 the longer the range.

I see 30 pixels of width in the second image so at that range using the same approach I would estimate:
@203 meters: 1.40 meters wide
@336 meters: 2.31 meters wide

Obviously several factors are not being taken into account and this is a big approximation (and FOV could be very wrong. Anyone know the exact FOV for iPhone 8?) but the result is pretty close to an expected intercept range and the size of something not impossible to find in the sky.

Conclusion: the batman mylar ballon seems to be sold in 45x70cm formats. I argued in the past this size was almost impossible to spot and intercept in the air with an F-18. I think the calculations above (if correct...big IF... please double check) show the object should be 3X-5X as wide and we can therefore exclude a party balloon candidate.
The screen aspect ratio doesn't have any bearing on the camera's sensor, which is indeed 4:3. The lens has a 28mm focal length (35mm equivalent, actual sensor size is 4mm), which gives a 65.5º horizontal angle of view.
My calculations above are not as dependent upon FOV as I thought. Basically changing the FOV yields minimal changes in the results.
The main factor is the F-18s speed.

however I made a mistake in my previous calculation.
To calculate the size I used the distance of the first picture with the apparent pixel size of the second picture.

Rerunning the 350 knots simulation with 65° FOV yields:

Screen Shot 2021-04-09 at 13.43.59.png

And using the correct a=137 meters of distance (I used the b=202 meters in my past calculation) we get an estimated minimum size of 1.16 meters.
Screen Shot 2021-04-09 at 13.49.50.png

At higher speeds we get bigger sizes. I think this is basically as low as we can go. If we had GPS coordinates we could get the full picture.

p.s. new idea: could we estimate the altitude given the curvature of the earth observed in the pictures?
 

DavidB66

Active Member
The screen aspect ratio doesn't have any bearing on the camera's sensor, which is indeed 4:3. The lens has a 28mm focal length (35mm equivalent, actual sensor size is 4mm), which gives a 65.5º horizontal angle of view.
I know next to nothing about cameras, but surely the fov depends on the zoom factor, if any? If I understand correctly, the iPhone 8 has an optical zoom of x2 (by switching to a different lens) and a digital zoom of up to x10.
 

jarlrmai

Active Member
only the iPhone 8 plus has a 2x optical zoom, as far as I know this is not a plus, unless the EXIF for a plus still puts iPhone 8 in the camera model field.

Trying to judge the angles from the iPhone in hand in a moving banking jet seems less accurate than just doing the increase in size in pixels from a known camera sensor, which is fully known quantity.

I get between 0.5m and 3m but 3m is only if the jet is doing 1000 kts which seems really unlikely, I still think the most likely range is between 0.6 and 1.5m which given margins of error still allows a balloon, anything bigger than 3m has the jet performing at or above it's max speed.

It seems more likely to be in the range of 0.8 to 1.5 meters which given error margins for pixels etc is still in the range of a balloon.
 

gtoffo

Member
My calculations above are not as dependent upon FOV as I thought. Basically changing the FOV yields minimal changes in the results.
The main factor is the F-18s speed.

however I made a mistake in my previous calculation.
To calculate the size I used the distance of the first picture with the apparent pixel size of the second picture.

Rerunning the 350 knots simulation with 65° FOV yields:

Screen Shot 2021-04-09 at 13.43.59.png

And using the correct a=137 meters of distance (I used the b=202 meters in my past calculation) we get an estimated minimum size of 1.16 meters.
Screen Shot 2021-04-09 at 13.49.50.png

At higher speeds we get bigger sizes. I think this is basically as low as we can go. If we had GPS coordinates we could get the full picture.

p.s. new idea: could we estimate the altitude given the curvature of the earth observed in the pictures?

Another issue with those calculations: we are using knots which is dependent upon the air density. What we ideally would need is the TAS=true air speed of the jet to calculate its movement.

The higher you go the less dense the air is and the indicated airspeed will drop. So 350knots indicated air speed at 10k feet of altitude is approximately 420 knots of True air speed. The 350 knots true speed I used would actually correspond to 290 indicated at 10k feet (see: http://www.csgnetwork.com/tasinfocalc.html)

This complicates matters and makes my estimation above even more conservative. 1 meter is probably the absolute minimum size of the object. It is probably larger.
 

gtoffo

Member
only the iPhone 8 plus has a 2x optical zoom, as far as I know this is not a plus, unless the EXIF for a plus still puts iPhone 8 in the camera model field.

Trying to judge the angles from the iPhone in hand in a moving banking jet seems less accurate than just doing the increase in size in pixels from a known camera sensor, which is fully known quantity.

I get between 0.5m and 3m but 3m is only if the jet is doing 1000 kts which seems really unlikely, I still think the most likely range is between 0.6 and 1.5m which given margins of error still allows a balloon, anything bigger than 3m has the jet performing at or above it's max speed.

It seems more likely to be in the range of 0.8 to 1.5 meters which given error margins for pixels etc is still in the range of a balloon.
See my comment above. You are using true air speed while the jet flies with the available air and indicated air speed. 1000 knots true air speed (800knots indicated at 10k feet) is not out of the ordinary for an F-18 that can reach mach 1.8+

Consider maximum-range cruise airspeed is around 0.85 Mach http://krepelka.com/fsweb/learningcenter/aircraft/flightnotesboeingfa18hornet.htm

When manoeuvring it wouldn't be strange to be flying transonic however we can set mach 1 as a limit as the f-18 won't supercruise (break the sound barrier without afterburners) and it is unlikely they were making an intercept with afterburners.

Altitude is a big factor to determine probable speed. Can we estimate it with the earth's curvature?
 

gtoffo

Member
Given the object is stationary the speed I am calculating is ground speed relative the stationary object.
Yes me too.

The problem is the speed assumptions you are making are based on the flight envelope of the F-18. The F-18 doesn't care what the ground is doing. It only takes into account how much air is under its wings and entering its engine.

Hence the difference between TAS and IAS.

The higher you go the faster you need to fly but the slower it "feels" like you are going.

For example if you are at 85k feet you need to be doing Mach 3+ and you need to be in a pretty capable machine to keep your engines running and enough air under your wings (this was top alt and speed for the SR-71).
Indicated airspeed will be extremely low (almost stall speed) as there is almost no air entering your pitot tube. But the ground is moving pretty fast....

Also let's not forget that the higher you go the more wind you can have. That would complicate things a bit if the "Acorn" was indeed stationary.

Wind would actually be the easiest way to determine if this was a balloon or not. It would take the Navy 1 second to determine that this was a balloon at such altitudes given a radar contact establishing its position. If it doesn't follow the wind exactly... can't be a passive balloon.
 

jarlrmai

Active Member
I understand the way wind affects the ability of a plane to fly, perhaps we can look up some data for the day they were taken.

How would they establish the radar position if it were small balloon not reflecting radar?

This goes back to Micks statement in the OP

Article:
The Task Force reports noted that the objects were able to remain stationary in high winds, with no movement, beyond the capability of known balloons or drones.
Source: https://www.mysterywire.com/ufo/new-uap-photographs/

For me this one line makes the whole story seem ridiculous. Stationary relative to what? And how was this measured. High altitude air movement is not very turbulent, so balloons would appear still.
 

deirdre

Senior Member.
ok my photoshop transform skill are lacking big time. i was trying to 'deflate' balloon bits but dont really know how to get the tail to hang right without messing up the arm fin. anyway....
1617980656847.png

original balloon below
 

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only the iPhone 8 plus has a 2x optical zoom, as far as I know this is not a plus, unless the EXIF for a plus still puts iPhone 8 in the camera model field.
It's not a Plus so only the single lens. To rule out digital zoom, this would show up in EXIF as both "Digital Zoom Ratio" (which is absent from these leaked photos) and would affect the "Focal Length in 35mm Film" figure too (these are Apple's display labels, I'm not sure what fields they correspond to in the metadata itself).

This is a photo from an iPhone with a 26mm lens with what was displayed as "5x" zoom in the camera app:

And 26 * 5.1 ~= 132

Here's the corresponding data from The Debrief's photo:

So it's pretty safe to say that there was no zoom on these UAP photos and they're all 28mm equiv.

edit: According to the developer documenation, the EXIF field for digital zoom has been available since iOS 4, circa 2010. It's just absent when there's no digital zoom applied.
 
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gtoffo

Member
To me the horizon looks curved (although clouds are present).

I found something that could be interesting: https://pubmed.ncbi.nlm.nih.gov/19037349/
Visual daytime observations show that the minimum altitude at which curvature of the horizon can be detected is at or slightly below 35,000 ft, providing that the field of view is wide (60 degrees ) and nearly cloud free. The high-elevation horizon is almost as sharp as the sea-level horizon, but its contrast is less than 10% that of the sea-level horizon. Photographs purporting to show the curvature of the Earth are always suspect because virtually all camera lenses project an image that suffers from barrel distortion. To accurately assess curvature from a photograph, the horizon must be placed precisely in the center of the image, i.e., on the optical axis.
 

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