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Possible revision of our vision range/"twinkle in the sky" feats

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Armorchompy

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Hey guys, I'm not really the type of guy who usually cares about making wiki-wide revisions, but this one's been bugging me for a while, so I thought I should. Currently, we calculate things being sent flying across the horizon or doing the ole "twinkle in the sky" gag using a distance crossed of 20 km, because that's what we consider to be the "vision range". That's something I've been a bit suspicious about. I mean, it's a bit silly to assume that it's 20 km for everything. You can see clouds from much further away than you can see an insect. So I did a bit of research and have found results, but keep this in mind, I'm not exactly an expert on what I'm about to say, hence the "possible" in the title. I'm decently confident in what I'm about to say, but am also very willing to reconsider.

Anyhow, I looked it up and there's a couple things of note, the biggest one of them is that is that it's obviously way harder to see in the dark, but I'll get to that in a second. The second biggest is that typically your vision range is determined by your height due to the Earth's curvature, being less than 3 km in most circumstances.

That's all well and good but it does depend on the actual size of whatever's in question, and straight-up doesn't matter if that something is in the air. For that purpose, I found this handy site that says that the Angular Resolution limit (UNDER DAYLIGHT) for most people is 5 x 10^-4 rad, while for people with extremely acute vision under perfect circumstances (can probably be used for Enhanced Senses) it's 2 x 10^-4 rad. The theoretical maximum for any eye of human size (often ignored by fiction, obviously) is 1.22 x 10^-4 rad.

Now, what does this mean calculation-wise? Well the formula to get distance out of that is simply Angular Resolution x Distance = Minimum Size (Of things you can see), which in turn becomes Size / Angular Resolution = Maximum Distance. When it comes to a man-sized object, we get a maximum vision distance of 1.8 meters / (5 x 10^-4) = 3600 meters, which, yeah, makes more sense than 20000. This does actually imply an increase in vision range regarding massive objects, but I think that's fine. Famously, you can see the Great Wall of China from space (and less famously, a whole lot of other man-made things), and while I couldn't find a source that I deemed reliable enough to post here (mostly stuff like travel and pop culture websites, alongside answers on Quora and r/askscience), clouds can be seen from several times more than 20 km away.

So, that's the revision, simple enough. Feats using the 20 km thing should preferably recalculated, starting with the one in References for Common Feats. I think the biggest impact of this revision is that our "Creating a Storm" calculation uses that assumed range of 20 km, but given that I couldn't find a number for that one, it should preferably maybe wait a bit (You can't just calculate the range in question using the formula I gave because planet curvature does end up playing a part).

EDIT: Nope they're fine, as far as this thread is concerned at least

Most of the "twinkle in the sky" feats I mentioned will probably end up with a hefty downgrade, including my poor baby God Hand. But, yeah, I guess that's all I got. By the way I'm not sure if I should or even can specify this but I'm not opposed to non-staff members weighing in, any opinion can potentially be valuable as far as I'm concerned. That said please don't post if it's just stuff like "oh no/yes, [verse] will be downgraded/upgraded!"
 
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I haven't re-derived the 3,600 meter number; I'll hope that someone else does that.

Conceptually, this sounds like a good change. I always wondered what an actually good range for "not visible to the human eye" was; having that would help out for some other weird feats I have.

I'd like us to have the numbers for nighttime as well, if you can pull those out from somewhere.

A different distance would make more sense for storm calcs, since clouds are notably not human-sized. I'd hope that there's some actual basis for our 20km assumption there, but I don't have it myself.
 
In that case if that's the max visibility than they're probably fine to leave untouched then.
 
Storm feats I reckon were based on maximum visibility based on the horizon stuff in wikipedia, @DontTalkDT can elaborate on that prolly.

EDIT: Found it.

https://en.wikipedia.org/wiki/Horizon
The only usage of 20 km I see there is in an arbitrary example; showing that for a viewer standing 10m above sea level, their horizon would be 11.3 kilometers away, and an object 20km away would only have its portions that are at least six meters above sea level visible. This seems to be far from the limit we'd expect, especially for clouds hundreds of meters in the sky.

Should we extend their method of estimation to clouds, or something?
 
Same, might be best to just wait on that for a bit.
 
Same, might be best to just wait on that for a bit.
I will say the method I showed here probably doesn't work with clouds cause planet curvature ends up mattering for it, otherwise the radius would be like half the planet lol
Ugh I don’t wanna revise all my cloud calcs for the billionth time
too bad : ) (it might not end up happening rn)
 
I will say the method I showed here probably doesn't work with clouds cause planet curvature ends up mattering for it, otherwise the radius would be like half the planet lol
Right; I was meaning the method explained in the Wikipedia page, which seems pretty much purely based on planet curvature.
 
Storm feats I reckon were based on maximum visibility based on the horizon stuff in wikipedia, @DontTalkDT can elaborate on that prolly.

EDIT: Found it.

2560px-Zimba_Panorama.jpg

Nope, they were based on visibility. I would have to look for the thread where it was decided, but basically the air is usually somewhat foggy (or, perhaps more accurately, light is always scattered by something) even when it's clear. Which is why usually seeing things way past the horizon (in cloud cases that's theoretically easily 200+ km) is not practically feasible. Make no mistake, it's possible. As the cloud calculations page mentions, on an extraordinarily clear day you could see as far as 280 km through regular atmosphere (thin air from up high is different). However, that's the exception, not the rule. 20km is the practical low-end for what is considered clear conditions.

Needless to say I disagree with any notion of ignoring such things and massively inflate cloud feats by using far beyond-horizon distances.


That said, visibility was a consideration for cloud feats as those are massive objects. For smaller objects that become invisible sooner, other visibility considerations are welcome. (Let me note that minimum angular resolution an eye can take is likely an overestimation, given both scattering of light and the observer likely having no time to properly focus in many circumstances)

Also, you likely can't see the Great Wall of China from space.
 
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I looked it up myself a long time ago. The 20 km range is only usable for the most ideal of conditions and the range should minimize at 5 km. We can see larger stuff further away, like an airplane that's 32000 feet away from us, or a bunch of tiny islands from the ISS 254 miles away: https://www.bbc.co.uk/bitesize/articles/z69yd6f

My preferred method of going for the twinkle in the sky feat, as it typically represents someone being completely out of sight, is the distance needed to reach the bare minimum angular size visible, which is generally agreed to be 1 arc-minute (1/60 of a degree).
 
Nope, they were based on visibility. I would have to look for the thread where it was decided, but basically the air is usually somewhat foggy (or, perhaps more accurately, light is always scattered by something) even when it's clear. Which is why usually seeing things way past the horizon (in cloud cases that's theoretically easily 200+ km) is not practically feasible. Make no mistake, it's possible. As the cloud calculations page mentions, on an extraordinarily clear day you could see as far as 280 km through regular atmosphere (thin air from up high is different). However, that's the exception, not the rule. 20km is the practical low-end for what is considered clear conditions.
Gotcha, doesn't seem like something that this thread should affect, I'm not sure about this being always true for truly massive things such as clouds but if there was a thread it's probably fine as an estimate.
Needless to say I disagree with any notion of ignoring such things and massively inflate cloud feats by using far beyond-horizon distances.
I did address such a thing in OP in and in the comments, and I do agree with this.
For smaller objects that become invisible sooner, other visibility considerations are welcome.
Should I consider this an agreement or just a general statement?
(Let me note that minimum angular resolution and eye can take is likely an overestimation, given both scattering of light and the observer likely having no time to properly focus in many circumstances)
I agree (mostly, the observer bit doesn't really work when it's not really done from a character's POV but just from an impartial camera) but I still think it's a much fairer method of estimation than just assuming 20 km, especially for smaller objects (There was a feat I calculated with the 20 km method that gave 9-A for hitting a tiny punching bag into the horizon, and recalculating it with my method gives a much saner baseline 9-B rating).

Maybe we should add in an addendum that if calculating distance this way, the result should not surpass 20 km?
Huh, TIL. Regardless, you get what I meant with those examples, right?
 
My preferred method of going for the twinkle in the sky feat, as it typically represents someone being completely out of sight, is the distance needed to reach the bare minimum angular size visible, which is generally agreed to be 1 arc-minute (1/60 of a degree).
What's the reasoning behind the "general agreement"? My method is the same but with research stats
 
What's the reasoning behind the "general agreement"? My method is the same but with research stats
It's one of those general rule of thumb deals, but if you need sourced arguments, I got two.:
https://landsat.gsfc.nasa.gov/wp-content/uploads/2013/06/RS_32.pdf

https://www.researchgate.net/figure...of-one-line-pair-per-arcminute_fig2_328353906

Plus Wikipedia cited the figure from an ophthalmology book (you know, a book from someone who knows their sh*t around eyes): Yanoff, Myron; Duker, Jay S. (2009). Ophthalmology 3rd Edition. MOSBY Elsevier. p. 54. ISBN 978-0444511416.
 
Should I consider this an agreement or just a general statement?
Just a general statement. I would need to take a closer look at everything, but have no time to invest until after the 15th this month.

I agree (mostly, the observer bit doesn't really work when it's not really done from a character's POV but just from an impartial camera) but I still think it's a much fairer method of estimation than just assuming 20 km, especially for smaller objects (There was a feat I calculated with the 20 km method that gave 9-A for hitting a tiny punching bag into the horizon, and recalculating it with my method gives a much saner baseline 9-B rating).
Fairer estimation, sure.

What the camera is concerned: I assume you mean when you scale such a scene from some anime or cartoon. If you have such a feat and infer that it becomes not visible from the visuals, what you first need to consider is that there is a limited resolution. Something smaller than 1px would not be drawn even if it were still within what could be seen by angular resolution. Hence, in such cases, it's likely better to determine the distance by scaling from the last frame where they could be seen.

I think an angular resolution method would be more viable for cases where the point is that the characters are not supposed to be able to see them anymore, not the "camera".
 
It's one of those general rule of thumb deals, but if you need sourced arguments, I got two.:
https://landsat.gsfc.nasa.gov/wp-content/uploads/2013/06/RS_32.pdf
This one is just a math problem so I wouldn't really use it, they tend to simplify stuff.
This one is "under ideal conditions" so it's really just the exact same as mine in terms of context.
Plus Wikipedia cited the figure from an ophthalmology book (you know, a book from someone who knows their sh*t around eyes): Yanoff, Myron; Duker, Jay S. (2009). Ophthalmology 3rd Edition. MOSBY Elsevier. p. 54. ISBN 978-0444511416.
This one seems fair though.

How would you calculate using this value, to be sure?
What the camera is concerned: I assume you mean when you scale such a scene from some anime or cartoon. If you have such a feat and infer that it becomes not visible from the visuals, what you first need to consider is that there is a limited resolution. Something smaller than 1px would not be drawn even if it were still within what could be seen by angular resolution. Hence, in such cases, it's likely better to determine the distance by scaling from the last frame where they could be seen.
I disagree, I think the twinkle is meant to indicate that the character is no longer in view in-verse. There's also videogame cases in which the model/sprite straight-up just disappears no matter how much you turn up the resolution.
 
I usually just use the angular size calculator to do it. If you want a formula... Well, basically it's working this backwards: https://en.wikipedia.org/wiki/Visual_angle

For some reason, it's in radians, which... Honestly, who the hell uses radians? They don't even sell protractors like that.
Fair enough I actually did have one with both though

I may be getting the numbers very wrong here, but this gives me a result of 6188.2 meters for a human sized target. Is this right?
Something smaller than 1px would not be drawn even if it were still within what could be seen by angular resolution. Hence, in such cases, it's likely better to determine the distance by scaling from the last frame where they could be seen.
I looked into this and I'm actually not sure this would end up playing a role. Assuming 1080px of screen height (which admittedly isn't going to be the resolution of older movies/animation) and 0.5px for the size of the character in question (which imo is better than 1px, because even if the object was slightly less than one full px, by occupying most of that pixel it would still "fill it"), then the distance for a human-sized character ends up being 1.8 / 0.5 * 1080 = 3888 meters of distance, which is more than you'd get from my method. This is without considering higher resolutions which are very common nowadays, most movies nowadays are 4096 x 2160 and basically any videogame can be cranked as high as your PC can handle.
 
Well like, that's the same method, that being "yours", when I say "mine" I mean the one I detail in OP, that gives 3600 meters.
 
I disagree, I think the twinkle is meant to indicate that the character is no longer in view in-verse. There's also videogame cases in which the model/sprite straight-up just disappears no matter how much you turn up the resolution.
I don't really think there's that much intention behind it. As a casual observer, I always just interpreted that sort of thing as a comedic flourish for something going a long way away. Until people started using it in a battleboarding context, I never assumed that it meant they were now invisible.

Video game cases are also weird due to draw distances.
 
I don't really think there's that much intention behind it. As a casual observer, I always just interpreted that sort of thing as a comedic flourish for something going a long way away. Until people started using it in a battleboarding context, I never assumed that it meant they were now invisible.

Video game cases are also weird due to draw distances.
That's usually not an issue in cutscenes, not to mention they can just shrink the model rather than pushing it that far away. But fair enough on the rest, still I don't think it really matters with modern resolutions given that it ends up higher than 3600 meters, especially if you use my logic of 0.5 px being better than 1 as an assumption.
 
Disagree with changing the 20 km value, it's already about as big as most storms in real life, using that formula Armorchompy provided will give us inflated results for the most part, as even being more than 10 meters up into the air gives higher results than 20 kilometers.

This means we can just find any shot high into the sky and use it to storm, since when was POV in any series limited to being 1.8 meters off the ground?
 
Disagree with changing the 20 km value, it's already about as big as most storms in real life, using that formula Armorchompy provided will give us inflated results for the most part, as even being more than 10 meters up into the air gives higher results than 20 kilometers.

This means we can just find any shot high into the sky and use it to storm, since when was POV in any series limited to being 1.8 meters off the ground?
Uh, we've talked about clouds, that's not something this affects, it's for how far you can see smaller things. My formula is also just not something that includes the viewer's height so I'm not sure you've understood it properly, the 1.8 meters is the size of the object in question.
 
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So lemme get this straight, this Vision Range doesn't affect storm clouds anymore?
 
So lemme get this straight, this Vision Range doesn't affect storm clouds anymore?
Yeah 20 km is still the (approximated) vision limit for most things, including storms, it's just that smaller objects, like people, cannot be seen from that far away, and that's where the angular resolution formula comes into play.
 
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