Planet Curvature Page Additions and Questions

[Firestorm808]

Planet Curvature Scaling

For many feats of very high destructive power scaling from measuring sticks like humans, trees or even mountains becomes very difficult, because they are too small compared to the phenomenon we are scaling. For such feats scaling from the size of the planet itself can be of interest. However...

vsbattles.fandom.com

When Not To Use Curvature Scaling

I made a calc recently that angsizes the Earth (not fully visible in the panel) to find out how far the point of view was away. @KLOL506 commented to suggest I use curvature scaling, to which I replied "you only do that when you need to scale something near/behind the horizon AND you intend to...

vsbattles.com

This discussion is somewhat related to the linked thread. My thread's original purpose was to add more details to the reference page, but I encountered an issue.

General Example and Diagram​

The basic premise of the reference page is that not all of the planet is viewable from a distance. I figured we could add similar diagrams of those below to show the point.

Using a unit circle, I made rough estimates of the maximum view.

1.00 Diameter away from Surface = 94% Max Diameter Viewable
0.75 Diameter away from Surface = 91% Max Diameter Viewable
0.50 Diameter away from Surface = 87% Max Diameter Viewable
0.25 Diameter away from Surface = 75% Max Diameter Viewable
0.10 Diameter away from Surface = 54% Max Diameter Viewable
and so on.

My observation of the trendline is approx. 16.2+17.4*ln(x) where x = %
Let me know if there is an existing formula for this sort of thing.

Real-World Example for Partial Planet View​

With this in mind, I wanted us to add a real-life example to help portray the page's point. Below is a photo of Earth from the ISS.

The ISS orbits at 422 km or 0.033 or 3.3% Original Diameter.

From the above formula estimate, about 37% of the maximum diameter should be visible.

Earth's real diameter is 12742 km.

In comparison to the original image, the extended diameter of Earth is 5317 px.

The Panel Height is 694 px.

Based on the current formula on the reference page, the Corrected Earth Diameter is 2335.0 km or 18.3% of the Original Diameter.

Now we have a difference of about 18%.

Am I missing something? Is it because of the original FOV angle of the photo?

 

[Firestorm808]

@SeijiSetto @DontTalkDT @KLOL506

 

[KLOL506]

This is more up DT's alley, he's the one who made the rules, so...

 

[DontTalkDT]

Probably the original FOV of the image. Angsizing is always janky. We should have the thread where I derived the planet curvature formula somewhere. If you find it you can check the math. Totally not having a hard time finding it myself.
Edit: Found it! It was a page lost in the forum move, but one archive actually had it. - Turns out my derivation there provides yet a third percentage formula for you! Aren't cha lucky.

As for your % viewable diameter formula: We can derive one using distance to horizon formulas as a base.
As wikipedia gives us that the angle between the two lines is γ = arccos(R/(R+h)) in radians.
The angle at the observer O is then pi-γ.
Now imagine a triangle that has as corners O and two points at opposite sides of the horizon H1 and H2. For the lengths of the sides we have OH1 = OH2 = distance to horizon.
That distance is given (again, by wikipedia) as sqrt(h(2R+h)).
Since this is a isosceles triangle we have that the remaining two angles are pi-(pi-γ)/2 = γ/2 in size.
Using law of sines we get for the remaining side H1H2 / sin(pi-γ) = OH1 / sin(γ/2) <=> H1H2 = OH1 / sin(γ/2) * sin(pi-γ) = sqrt(h(2R+h)) / sin(arccos(R/(R+h))) * sin (pi - arccos(R/(R+h))).
H1H2 / earth diameter is then your formula.

A different, perhaps more elegant way, is to use that we know that curved surface on the earth to horizon s is given as s = R* arccos(R/(R+h)). Then we can use circular segment formulas to solve to the side wikipedia calls c. We need to be careful: s in the horizon case would be 2s for us here, as we se s far in both directions.
θ = 2s/R = 2*arccos(R/(R+h))
and then
c = 2*R*sin(arccos(R/(R+h))).
Now sin(arccos(x)) = sqrt(1-x^2) and therefore c = 2*R*sqrt(1-R^2/(R+h)^2)
c / earth diameter is then your formula. Or specifically 2*R*sqrt(1-R^2/(R+h)^2) / (2R) = sqrt(1-R^2/(R+h)^2). Just take * 100 for percent and all.

Both formulas should be the same... but the second one has a lesser chance of me having screwed up in the math so use that one lol

So for instance for 0.1*diameter away we get sqrt(1-6371^2/(6371+0.2*6371)^2) = 0.5527707983925667 = 55.277% of the whole diameter.

 

[KLOL506]

So uh... is there anything else left to do here?

 

[Firestorm808]

@DontTalkDT Do you have any concerns in adding supplementary diagrams and formulas to the reference page?

Is there perhaps a better real life example for us to showcase the application?

 

[DontTalkDT]

I don't mind diagrams. For formulas... depends on which formulas?
What real life examples is concerned: Not that I know of. But fictional examples work just fine ¯\(ツ)/¯

 

[Firestorm808]

I don't mind diagrams. For formulas... depends on which formulas?
What real life examples is concerned: Not that I know of. But fictional examples work just fine ¯\(ツ)/¯

Thanks. I'll try to get some examples prepared for review to be added.