Formula for Extremely high Large Sizes? (Galaxy-size or more)

[StrymULTRA]

As the title says. I won't hide the context, as it's about Gravity Falls.

Essentially, everything that can be seen about this feat is basically in this blog, so reading it is fundamental. What is the bad part of it, obviously, is the fact that it was downplayed to just 4-A, when humans are muuuuch denser than galaxies.

I made the calc for this CRT, but I'll just report it here for simplicity:



1st image:

• Galaxy's diameter = 1706.075 px = 9.5e17 km
• Ford's chin to forehead = 259 px = 1.4421992e17 km

2nd image:

• Ford's chin to forehead = 454.58 px = 1.4421992e17 km
• Ford's full height = 1588 px = 6.55594912e19 km = 6.55594912e22 m

So, let's find the weight using the measurements of large size calcs:

(6.55594912e22/1.71)^3*62 = 3.49389687e69 kg (Aka Universal LS lol)

Ford's volume = 3.49389687e69 / 985 = 3.54710342e66 m^3

I did the main things, getting height, volume and mass, but there's a problem.

According to the Large Size page this should give me 3-B given Ford exceeds the 1.03458×10^20 m of height needed but I dunno the formula to use to get that result so... help, coz if I do the GPE of: 9.81 * 3.49389687e69 * (6.55594912e22 / 2) I get 1.12352999e93 Joules which is 3-A, so something is off here.

 

[Agnaa]

Yeah idk what's going on there. Using their methods explained on the page, their supposed minimum for 3-A would actually land at 1.6e113 Joules, well above the minimum for 3-A.

 

[StrymULTRA]

Yeah idk what's going on there. Using their methods explained on the page, their supposed minimum for 3-A would actually land at 1.6e113 Joules, well above the minimum for 3-A.

Can you ping the ones who know better about Large Size stuff? Seems a pretty big contradiction.

 

[StrymULTRA]

Actually I have found this from @DontTalkDT, with the formula being:

• E = (6.674*10^-11)5.972*10^24((Height / 1.595)^3 * 62) * (1/6371000 - 1/(6371000 + Height/2)) for women
• E = (6.674*10^-11)5.972*10^24((Height / 1.71)^3 * 62) * (1/6371000 - 1/(6371000 + Height/2)) for men

Now, if I put it for Ford I get:

• (6.674e-11) * (5.972e24) * ((6.55594912e22/1.71)^3 * 62) * (1/6371000 - 1/(6371000 + 6.55594912e22/2)) = 2.18579022e77 Joules / 2.18579022 TenaKiloFOE (which is 3-B, rightfully)

Yeah idk what's going on there. Using their methods explained on the page, their supposed minimum for 3-A would actually land at 1.6e113 Joules, well above the minimum for 3-A.

So the mystery is solved. I dunno why the formula wasn't added to the page, so it could be a problem.

 

[Agnaa]

Oh right, doh.

It's using generalised GPE, not GPE for objects on Earth.