Explosion Calc Question

[Ninkan12]

Anyone know how I would start calcing this?

Feat: Destroying rocks here


Runs through a Creeper explosion (the explosion is 0.00212307737 Tons of TNT but idk how to do the cross section of a giant pig)


And dies to a giant explosion

(the one in the next clip is a different hog)

 

[Oiguana2701]

Runs through a Creeper explosion (the explosion is 0.00212307737 Tons of TNT but idk how to do the cross section of a giant pig)

If you can find me the height of this pig and a 3d model of him, I can calc that in Blender.

 

[Ninkan12]

If you can find me the height of this pig and a 3d model of him, I can calc that in Blender.

Heights easy, where the hell do I find a model of a movie character-

 

[Oiguana2701]

Heights easy, where the hell do I find a model of a movie character-

There's a lot of sites where people can post models or fan-models that are extremely close to the real models. For instance, look at this Destoroyah 3d model. Obviously, it won't be 100% accurate, but it'll be more accurate than anything we can calculate just using pictures.

 

[Ninkan12]

There's a lot of sites where people can post models or fan-models that are extremely close to the real models. For instance, look at this Destoroyah 3d model. Obviously, it won't be 100% accurate, but it'll be more accurate than anything we can calculate just using pictures.

https://makerworld.com/en/models/1317209-minecraft-movie-the-great-hog#profileId-1352593

Is this anything? There's an STL download on the page

 

[Oiguana2701]

https://makerworld.com/en/models/1317209-minecraft-movie-the-great-hog#profileId-1352593

Is this anything? There's an STL download on the page

Worked! Now, I just need his height.

 

[Ninkan12]

Worked! Now, I just need his height.

I'll run a quick calc on it - thankfully it's easy cause a block in the movie is 8ft^3

 

[Ninkan12]

Worked! Now, I just need his height.

Its really hard to get this guy next to a block and I was only able to get it with him crouched so its not really accurate

Feet per pixel = 2 ft ÷ 50.01 px ≈ 0.039992 ft/px

Value = 274 px × 0.039992 ft/px ≈ 10.96 ft

 

[Ninkan12]

I might be able to compare the head size with this to get a better reading

 

[Oiguana2701]

Its really hard to get this guy next to a block and I was only able to get it with him crouched so its not really accurate

Feet per pixel = 2 ft ÷ 50.01 px ≈ 0.039992 ft/px

Value = 274 px × 0.039992 ft/px ≈ 10.96 ft

Do the torches ever stand next to another object? If so, pixel scale a torch, use that shot to find the Great Hog's head's size and pixel scale its entire body using its head

 

[Ninkan12]

Do the torches ever stand next to another object? If so, pixel scale a torch, use that shot to find the Great Hog's head's size and pixel scale its entire body using its head

Not a redstone one, I should be able to find just his head closer to a block so I can measure on the horizontal plane

 

[Ninkan12]

Do the torches ever stand next to another object? If so, pixel scale a torch, use that shot to find the Great Hog's head's size and pixel scale its entire body using its head

Finally

Length = (385.58 px / 100.32 px) × 2 ft
Length ≈ 3.8435 × 2 ft
Length ≈ 7.69 ft

 

[Oiguana2701]

Height: 2.34299808612552 meters
I added his dimensions to Blender:

And got this surface area:

Area: 18.1544 m^2.
Both pictures were uploaded to vsbw.

 

[Ninkan12]

Height: 2.34299808612552 meters
I added his dimensions to Blender:

And got this surface area:

Area: 18.1544 m^2.
Both pictures were uploaded to vsbw.

Thank you so much!



Does it matter that this is the most he was in it?

 

[Oiguana2701]

Thank you so much!



Does it matter that this is the most he was in it?

We only use around 40% of the surface area.

 

[Ninkan12]

We only use around 40% of the surface area.

exposed area = 0.4 × 18.1544 = 7.26176 m²
surface area of explosion sphere = 4π × (2.9782428115)² ≈ 111.5 m²
energy intensity at radius = explosion yield ÷ surface area of explosion sphere
= 0.00212307737 ÷ 111.5 ≈ 1.90 × 10⁻⁵ tons of TNT per square meter
energy absorbed by character = energy intensity × exposed area
= (1.90 × 10⁻⁵ tons of TNT per square meter) × 7.26176 m²
≈ 1.38 × 10⁻⁴ tons of TNT
energy absorbed in joules = 1.38 × 10⁻⁴ × 4.184 × 10⁹
≈ 5.77 × 10⁵ joules
final durability result = ≈ 1.38 × 10⁻⁴ tons of TNT ≈ 577,000 joules


Something like this?

 

[Oiguana2701]

exposed area = 0.4 × 18.1544 = 7.26176 m²
surface area of explosion sphere = 4π × (2.9782428115)² ≈ 111.5 m²
energy intensity at radius = explosion yield ÷ surface area of explosion sphere
= 0.00212307737 ÷ 111.5 ≈ 1.90 × 10⁻⁵ tons of TNT per square meter
energy absorbed by character = energy intensity × exposed area
= (1.90 × 10⁻⁵ tons of TNT per square meter) × 7.26176 m²
≈ 1.38 × 10⁻⁴ tons of TNT
energy absorbed in joules = 1.38 × 10⁻⁴ × 4.184 × 10⁹
≈ 5.77 × 10⁵ joules
final durability result = ≈ 1.38 × 10⁻⁴ tons of TNT ≈ 577,000 joules


Something like this?

How far away is Great Hog from the epicenter? Because it feels like you may have done something wrong, though, granted, I'm not a CGM and haven't done many of those calculations.

 

[Ninkan12]

How far away is Great Hog from the epicenter? Because it feels like you may have done something wrong, though, granted, I'm not a CGM and haven't done many of those calculations.

I think I've accidentally put him at the max distance of ~3m

 

[Ninkan12]

He isn't more than like a foot away from the epicenter

exposed area = 0.4 × 18.1544 = 7.26176 m²
distance to epicenter = r < 0.3 m
surface area of explosion sphere = 4π × r²
= 4π × (0.3)² (upper bound) ≈ 1.130973 m²
→ surface area of explosion sphere < 1.130973 m²
energy intensity at radius = explosion yield ÷ surface area of explosion sphere
= 0.00212307737 ÷ (4πr²)
→ energy intensity > 0.00212307737 ÷ 1.130973
≈ 1.88 × 10⁻³ tons of TNT per square meter
energy absorbed by character = energy intensity × exposed area
= (1.88 × 10⁻³ tons of TNT per square meter) × 7.26176 m²
→ energy absorbed by character > 1.37 × 10⁻² tons of TNT
energy absorbed in joules = 1.37 × 10⁻² × 4.184 × 10⁹
≈ 5.70 × 10⁷ joules
final durability result = > 1.37 × 10⁻² tons of TNT ≈ > 5.70 × 10⁷ joules

 

[Ninkan12]

He isn't more than like a foot away from the epicenter

exposed area = 0.4 × 18.1544 = 7.26176 m²
distance to epicenter = r < 0.3 m
surface area of explosion sphere = 4π × r²
= 4π × (0.3)² (upper bound) ≈ 1.130973 m²
→ surface area of explosion sphere < 1.130973 m²
energy intensity at radius = explosion yield ÷ surface area of explosion sphere
= 0.00212307737 ÷ (4πr²)
→ energy intensity > 0.00212307737 ÷ 1.130973
≈ 1.88 × 10⁻³ tons of TNT per square meter
energy absorbed by character = energy intensity × exposed area
= (1.88 × 10⁻³ tons of TNT per square meter) × 7.26176 m²
→ energy absorbed by character > 1.37 × 10⁻² tons of TNT
energy absorbed in joules = 1.37 × 10⁻² × 4.184 × 10⁹
≈ 5.70 × 10⁷ joules
final durability result = > 1.37 × 10⁻² tons of TNT ≈ > 5.70 × 10⁷ joules

But does this even make sense because how can the dura be more than the AP of the explosion?

 

[Oiguana2701]

But does this even make sense because how can the dura be more than the AP of the explosion?

I don't think this is necessarily wrong, but it is an interesting case. It may be better to just post it and see what a CGM would say.

 

[Ninkan12]

I don't think this is necessarily wrong, but it is an interesting case. It may be better to just post it and see what a CGM would say.

I've messaged a few regardless to see if they have any guidance on any of the other feats

 

[Ninkan12]

bump

 

[Ninkan12]

bump

 

[Ninkan12]

bump

 

[Ninkan12]

bump