Calc Horizon Forbidden West calcs, part 2

1. Erend destroys a cavern wall


Erend breaks through a cavern wall to continue the pursuit of Asera. Aloy is 5'6'', or 1.6764m.


907 pixels = 1.6764m
1 pixel = 1.6764m/907 = 0.00184829107m
0.00184829107m X 919 = 1.69857949m
260 pixels = 1.69857949m
1 pixel = 1.69857949m/260 = 0.00653299804m
0.00653299804m X 219 = 1.43072657m
0.00653299804m X 336 = 2.19508734m
0.00653299804m X 63 = 0.411578877m
0.00653299804m X 72 = 0.470375859m
0.00653299804m X 83 = 0.542238837m
0.00653299804m X 60 = 0.391979882m
0.00653299804m X 121 = 0.790492763m
0.00653299804m X 78 = 0.509573847m
0.00653299804m X 93 = 0.607568818m
60 + 106 + 78 = 244 pixels
0.00653299804m X 244 = 1.59405152m
0.00653299804m X 111 = 0.725162782m

We'll calculate each part of the hole as a different shape, such as rectangles, right-angled triangles and an irregular trapezium.

A = lh
= 2.19508734 X 1.43072657
= 3.14056978m^2

A = lh
= 0.411578877 X 0.470375859
= 0.193596768m^2

A = ab/2
= 0.542238837 X 0.391979882/2
= 0.106273358m^2

A = ab/2
= 0.790492763 X 0.411578877/2
= 0.162675062m^2

A = ab/2
= 0.509573847 X 0.607568818/2
= 0.15480059m^2

For the irregular trapezoid, we'll enter our values (1.59405152m for base, 0.725162782m for top and 0.542238837m for height) into this calculator to get an area of 0.6288m^2. Now to add all the areas up for our total area.

A = 3.14056978 + 0.193596768 + 0.106273358 + 0.162675062 + 0.6288
= 4.23191497m^2

Next we'll need the width of the section of the wall destroyed, then we'll be able to find our energy.
0.00653299804m X 59 = 0.385446884m

V = 4.23191497 X 0.385446884
= 1.63117844m^3

The fragments are flung everywhere, so I think violent fragmentation would be justified (violent fragmentation of rock being 69000000 joules/m^3).

E = 1.63117844 X 69000000
= 112551312 joules
= 0.0269004091778 tons of TNT

Plus in the background we see some pretty large rocks move, so chances are this could be even higher.

2. Corruptors destroys obelisks


Some corruptors who were unaffected by the Faro Plague destroy some obelisks, one with missiles, another with a tailspin.
637 pixels = 1.6764m
1 pixel = 1.6764m/637 = 0.00263171115m
0.00263171115m X 511 = 1.3448044m
0.00263171115m X 372 = 0.978996548m
148 pixels = 1.3448044m
1 pixel = 1.3448044m/148 = 0.00908651622m
0.00908651622m X 93 = 0.845046008m
0.00908651622m X 74 = 0.6724022m
0.00908651622m X 560 = 5.08844908m
0.00908651622m X 100 = 0.908651622m
127 pixels = 1.3448044m
1 pixel = 1.3448044m/127 = 0.010589011m
0.010589011m X 113 = 1.19655824m
0.010589011m X 88 = 0.931832968m

Volume of the obelisk as a trapezoidal prism and a rectangular pyramid.

V = 1/2(b1+b2)hl
= 1/2 X (0.6724022 + 0.845046008) X 5.08844908 X 0.931832968
= 3.59755453m^3

V = 0.908651622 X 0.908651622 X 0.908651622/3
= 0.250075395

V = 0.250075395 + 3.59755453
= 3.84762993m^3

We see some of the fragments being flung out while others fall, so we'll go with a low end of fragmentation and a high end of violent fragmentation.

(Low end)
E = 3.84762993 X 8000000
= 30781039.4 joules
= 0.00735684498088 tons of TNT

(High end)
E = 3.84762993 X 69000000
= 265486465 joules
= 0.0634527880019 tons of TNT

3. Tilda saves Aloy


Tilda saves Aloy from the other members of Far Zenith by unleashing a burst of energy, which results in a whiteout where even Gerard and Erik can't see. Tilda even says it was an overload of the senses, accompanied by an energy charge (and notes that both Gerard and Erik were affected, although their shields protected them).

Once again, we'll be scaling off Aloy.
413 pixels = 1.6764m
1 pixel = 1.6764m/413 = 0.0040590799m
0.0040590799m X 421 = 1.70887264m

Now we'll need the distance to the Far Zeniths when they first appear, as Gerard doesn't approach like Tilda or Erik, but he's still affected by the burst.

object degree size = 2*atan(Object_Size/(Panel_Height/tan(70/2)))

Click to expand...

= 2*atan(100/(1417/tan(70/2)))
= 0.0668508468 rad
= 3.830271378525146542 degrees

Enter this and the height through the angscaler and we get a distance of 25.553m to Tilda and the other Far Zeniths. With all of that, we can finally get our energy via luminousity.

M(star) is the apparent magnitude of the light source
m= is the apparent magnitude of the sun (-26.73)
L(star) is the Luminosity
L= Luminosity of the sun (3.486*10^26)
d= Distance to the sun from earth (0.000004731537734207877 parsecs)
d(star)= distance to the light source

Click to expand...

As it's a complete whiteout, I think it's safe to use the luminousity of the Sun (that being 100,000 lux). The conversion factor from lux to apparent magnitude is -2.5 log I - 14.2.

L = -2.5 log I - 14.2
= -2.5 log 100,000 - 14.2
= -26.7

Our formula is as follows...

-26.7 = -26.73 - 2.5log((L/3.846*10^26)(146000000000/(25.553))^2)

Let's go through this step by step.

(146000000000/(25.553))^2) = 3.26453945e19
-26.7 - -26.73 = 3.26453945e19/((L/3.846*10^26)
0.03 = -2.5Log(3.26453945e19/((L/3.846*10^26))
0.03/-2.5 = (3.26453945e19/((L/3.846*10^26))
10^(-0.012) = (3.26453945e19/((L/3.846*10^26))
0.972747224 = (3.26453945e19/((L/3.846*10^26))
0.972747224 X (3.864*10e26) = 3.26453945e19
3.75869527e27/3.26453945e19 = 115137076 joules = 0.0275184216061 tons of TNT

As one final aside, this is the energy output per second, so let's also get the amount of time she shone for.
Timeframe is 6 seconds and 21 frames.

T = 1s/24
= 41.6666667ms X 21
= 0.875000001s + 6s
= 6.87500s

E = 115137076 X 6.87500
= 791567398 joules
= 0.189189148662 tons of TNT