5 replies to this topic

[Marcin_78]

I've taken a break from astronomy, but apparently my brain has to stay busy.

 

The first picture below is a screenshot of my spreadsheet with useful formulas for plane-spotting (with my comments). Farther down below there are pictures explaining some formulas for the crucial steps of calculations.

 

Click to enlarge.

 

 

 

 

Edited by Marcin_78, 27 June 2023 - 04:51 AM.

[TOMDEY]

Yep. Other interesting computations are resolvability with your scope and wavefront defocus (relative to infinity) because of the finite range. Applies equally well to the moon, satellites, birds, bats and gnats. I'm in the habit of deriving this stuff ~on the fly~ in my head each time. For the closer stuff, you can actually use your telescope as a rangefinder... especially interesting when determining the distance to the atmospheric "thermals" that are degrading resolution --- often much closer than one would imagine.   Tom

[TOMDEY]

Oh! Coincidentally, a formation of refueling tankers will be flying right over my rural home here about six hours from now. I'll have my Zeiss 20x60 S binocular on the ready and waiting. That bino is wonderful for seeing and identifying planes "flying at altitude". My daughter frequently was up front on the receiving end when they were refueling. Inherently dangerous maneuver involving a lot of skill and the proper temperament and mindset.    Tom

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[Marcin_78]

I had been troubled by my own remarks I had made in another topic:

 

(...) From my own experience I can say that any plane within the smaller circle (20km radius from my location) is great. And any plane outside the bigger circle (40km radius) is usually invisible or very weak because it's very low above the horizon, so there is lots of air in the way. (...)

 

(...) The visibility was great and the range at which I could clearly see a plane's contrails was remarkable. There were 2 cases when I was really surprised that I could clearly see the contrails of a plane that was AT LEAST 60km away. At such distance a plane is only (around) 10 degrees above the horizon (the precise angle depends also on its altitude), so there is lots of air in the way. (...)

 

I had been somewhat troubled by these (relatively small) distances because I had known that some people claimed that they spotted contrails of planes that were over 300 km away! How could that be, when I was impressed already by distance a little more than 60 km? Here's what I found.

 

I will describe things in this order (more or less in the order of importance):

 

1. Air clarity/transparency.
2. Plane's altitude.
3. Observer's (eye) altitude.
4. Atmospheric refraction.
5. Clear view to the horizon.

 

 
1. Air clarity/transparency.

 

The proper info in a table form, based on the STCW (International Convention on Standards of Training, Certification and Watchkeeping for Seafarers) was easy to find on the Polish Wikipedia site for visibility (widzialność) – it was described as “horizontal visibility scale”. On the English Wikipedia site for visibility there was nothing concrete (no table/scale) but I did find something similar, described as “Meteorological optical range table”, here:

 

 

Both scales use the unit nautical mile, which is equal to 1.852 km (for comparison a mile on land is equal to 1.609 km). I combined the last/best 3 levels of visibility on both scales (when there was a slight difference I used the higher value):

 

Visibility / Weather (Meteorological Optical Range):

 
7) Good / Clear – from 10km to 20km.
8) Very good / Very clear – from 20km to 52km.
9) Exceptional / Exceptionally clear – over 52km.

 

To me it was quite surprising that good visibility is already at ONLY 10 km! Very good visibility is at only 20 km and any visibility over 52 km is exceptional! So my value of 60 km doesn't look so bad after all.

 

I have to point out that Meteorological Optical Range (MOR) is “the length of path in the atmosphere required to reduce the luminous flux in a collimated beam from an incandescent lamp to 5% of its original value”. I think it means that a very bright object can “beat” the distances given above and it probably happens with sunlit contrails of a distant plane around sunset (or sunrise), when the sky/background is relatively dark.

 

I have to also point out that on some Internet sites you can find some things that are clearly wrong, for example the (wrong) info that the atmospheric visibility is less than 300 km, which is misleading without any additional info. The Wikipedia site for visibility states that “AT SEA LEVEL (…) in the cleanest possible atmosphere, visibility is limited to about 296 km.”, so this is purely theoretical number because AT SEA LEVEL you can't see anything at such a huge distance. Some Internet sites repeat this value without any explanation/additional info, as if this number could be used for any practical purpose. No, it can't.

 

 
2. Plane's altitude.

 

Plane's altitude is usually much higher than observer's (eye) altitude, so it's much more important by definition. In order not to confuse things I will keep all the calculations in the present point for the observer's altitude of 2 meters. Please notice that this (small) observer's altitude is “local” (it's NOT altitude above the sea level), because I assume that the terrain all around the observer is roughly at the same level.

 

Max visible distance in km = 3.57 * square root(observer's altitude in meters) + 3.57 * square root(plane altitude in meters).

 

The observer's (eye) altitude of 2 meters alone gives distance (to horizon) of (only) 5 km:
3.57 * square root (2) = 5.0

 

I prepared a cheat sheet with different plane's altitudes (with a bonus of the highest mountain on the Earth):
 

 

I marked with the red color the most important thing – there are diminishing results when increasing a plane's altitude, because of the Earth's curvature. Subsequent increases of plane's altitude by 2000 meters result in smaller and smaller increases of the max visible distance (in km).

 

There is a catch, however – if there were no atmosphere, all these planes would “visible” right at the horizon, so it would be actually impossible to see them because an object has to stick out above the horizon in order to be visible/discernible.

 

Obviously the max visible distance works both ways – when an observer can see a plane (or a mountain) then the plane (or the mountain) can “see” the observer. It's remarkable that from the highest mountain on the Earth (Everest) you can see a 2-meter-high object/person that is “only” 341 km away!

 

 

3. Observer's altitude.

 

Observer's (eye) altitude adds to the effect of plane's altitude, but it's usually much smaller than plane's altitude, so it's much less important by definition.

 

I prepared a cheat sheet with different observer's altitudes (with a bonus of the longest photographed line of sight between two mountains):

 

 

I marked with the red color the most important thing – there are diminishing results when increasing an observer's altitude, because of the Earth's curvature. Subsequent increases of observer's altitude by 500 meters result in smaller and smaller increases of the max visible distance (in km).

 

The catch is the same – if there were no atmosphere, all these planes would “visible” right at the horizon, so it would be actually impossible to see them because an object has to stick out above the horizon in order to be visible/discernible.

 

It's worth to note that the longest photographed line of sight on surface of the Earth is “only” 483.5km. That photographed line of sight was between two very high mountains (Aconcagua photographed from Champaqui) that were “spaced” perfectly – just below the max visible distance! You can see the photograph (and also a video) here:
https://dalekiewidok...cord-andes.html

 

 
4. Atmospheric refraction.

 

All the values for atmospheric refraction were given for bearable temperature of 0 degrees Celsius and max (best) available air pressure, according the online calculator at this site: http://www.jgiesen.d...ract/index.html

 

Atmospheric refraction may give some impressive results very close to the horizon, “lifting” an object from below to above the horizon like this:
 

 

By the way: this visual calculator shows perfectly why the sun appears flattened close to the horizon – it’s because there are significant differences in the atmospheric refraction very close to the horizon.

 

The higher above the horizon the weaker the effect of atmospheric refraction. When an object is just 2.0 degrees above the horizon the difference is only 0.304 degree (the object appears higher by that much, so at. 2.304 degrees above the horizon):
 

 

Such a small difference is hard to notice with the naked eye – here’s how it looks in broader scale:

 

 

I prepared a cheat sheet for atmospheric refraction at different angles above the horizon:
 

 

I can't post any more pictures here, so I will continue in the next post.

[Marcin_78]

5. Clear view to the horizon.

 

Even a very small difference in altitude above the horizon is very important when it occurs very low above the horizon because at such a low angle you approach the max visible distance for a particular plane AND at the same time you approach the max atmospheric refraction, which improves the distance even more.

 

A very small angle doesn't mean much on its own – the most important thing is still the plane's altitude:

 

 

With the orange color I marked the values that are SLIGHTLY incorrect because my formula is an approximation based on a rectangular triangle, so the angle of 0 degrees is calculated at infinity instead of at the max visible distance that takes into account the Earth's curvature. For each plane I added a column where I (proportionally) corrected the error (from the approximation), so the angle at the max visible distance is 0 degrees.

 

The table above doesn’t take into account the atmospheric refraction, so I prepared the FINAL TABLE, combined with a visual example based on a real photo. I can't see the horizon from my window (nor from my balcony), so I estimated the horizon in a way not to “hurt” the results, which means that it may be marked a little higher than it is in reality. Then I determined particular degrees above the horizon by using my 8x42 porro binoculars that have the field of view of exactly 8 degrees. Click to enlarge!

 

 

So, theoretically I could spot contrails of a plane that is 300 km away myself, but the plane would have to be very high (12km) AND it would appear extremely low above the horizon, barely over the building in the very far distance – nearly impossible to spot. It looks to me that spotting contrails of a plane at the distance of 200 km would already be a superb achievement.

 

The most important thing – if a person on an Internet forum claims to have photographed contrails of a plane at more than 300 km and shows a photo where a plane is relatively high above the horizon then there was SURELY a mistake during the plane’s identification. Such a distant plane would have to be almost touching the horizon.

 

By the way, here’s a very fun topic, with some real photos of plane contrails:
https://www.metabunk...isible-at.5628/

 

This hobby (looking for most distant plane contrails) is yet another hobby that complements astronomy perfectly:

 

1. Good weather during night – astronomy/astrophotography.
2. Decent weather during day + a close plane – planespotting with binoculars or a telescope, possibly combined with an astro camera (relatively good amount of details either way).
3. Decent (but preferably good) weather during day + a distant plane – planespotting without any observing equipment, possibly combined with taking smartphone pictures AND taking screenshots of the plane’s position shown on an Internet site like globe.adsbexchange.com.
4. Bad weather during day – recording timelapses of moving clouds, possibly with an astro camera.

 

Clear skies!

Edited by Marcin_78, 24 May 2024 - 09:32 AM.

[Marcin_78]

On 19 January 2025 I captured, among others, three pictures that were very important for theoretical reasons – they allowed me to check the precision of my formulas:

 
1. Size of a plane from precise data about the plane, its altitude and its position.

2. Distance to a plane from angle above the horizon.
3. Size of a plane from rough data about the plane, its altitude and its position.

 

In this post I will describe only the point #1 and two more posts will follow.

 

1. Size of a plane from precise data about the plane, its altitude and its position.

 

This is the captured picture (cropped) :
 

 

As you can see the plane was still relatively big when it disappear behind a nearby apartment building. I instantly took my smartphone and took a screenshot of the plane's position and other info (I was using the site globe.adsbexchange.com for preparing for coming planes – checking from which side a plane should appear, so I had the site running constantly):

 

 

So, I got 3 important things (the third one thanks to Google Maps) about the plane:

1) it was an Airbus a321, so its length was 44.5m long,
2) it was flying at the WGS84 altitude of 36425 ft,
3) it was around 33km away (horizontally).

 

If anybody is interested:
 

 

The last thing I needed was the resolution in arcs/pixel. For imaging I was using the camera ZWO ASI 715MC with a very small sensor pixel size (1.45 µm) and a GUIDESCOPE 40/160 (achromatic doublet):
 

 

So, the resolution of my equipment combination was 1.87 arcs/pixel – I added to my spreadsheet a line for the size of a plane using this resolution in arcs/pixel.
 

 

Let's check the actual size of the plane:
 

 

According to my final calculations the size of the plane should be 141 pixels while the actual size of the plane on the picture is 143 pixels. The precision of my calculations is 98.6%, so it's fantastic!!!

 

To be continued … (the next 2 points, but I don't know when)