For a series where the premise is that they die horribly in almost every episode, this series really does have a ton of busted feats. Anyways, be wary of blood & gore.
1. Lumpy blows up the Moon
3:36
Lumpy activates his Christmas lights, which are so overpowered they not only vaporize several bystanders, but it lances out and blows up the Moon. The distance to the Moon is 384,400km.
Timeframe is 2 frames.
T = 1s/24
= 41.6666667ms X 2
= 83.3333334ms
= 384,400km/83.3333334ms
= 4.6128e9/299792458
= 15.3866446 C
Next we get the energy for exploding the Moon. The diameter of the Moon is 3,475km and has a mass of 7.35e22kg.
108 pixels = 3,475km
1 pixel = 3,475km/108 = 32.1759259km
Timeframe is 1 second and 15 frames.
32.1759259km X 18 = 579.166666km
32.1759259km X 43 = 1383.56481km
32.1759259km X 122 = 3925.46296km
32.1759259km X 98 = 3153.24074km
32.1759259km X 70 = 2252.31481km
32.1759259km X 73 = 2348.84259km
32.1759259km X 69 = 2220.13889km
32.1759259km X 94 = 3024.53703km
Let's find our average speed.
L = 579.166666 + 1383.56481 + 2220.13889 + 2252.31481 + 2348.84259 + 3024.53703 + 3153.24074 + 3925.46296
= 18887.268496/8
= 2360.90856km
T = 41.6666667ms X 15
= 0.625000001s + 1s
= 1.62500s
T = 2360.90856km/1.62500s
= 1452866.81m/s
KE = (0.5)mv^2
= (0.5) X 7.35e22 X 1452866.81^2
= 7.75727073e34 joules
= 18.540322012428298137 yottatons
2. Lumpy & the Mole aren't vaporized
From the above video at 4:01.
Both Lumpy and the Mole aren't vaporized by the energy of Lumpy's lightshow, while others such as Toothy and several Generic Tree Friends are vaporized instantly and Cub explodes (Lumpy is a tall character, so this would apply to him, although as his eyes melt out, it's unknown if he actually does survive or not; the Mole on the otherhand only has a burnt thumb). Thankfully, we already have a level of energy for vaporizing someone; 3 gigajoules, or 0.717017 tons of TNT. Of course, this is our low end. For our high end, see above.
3. The Demon generates clouds
0:49
Pop accidentally summons the Demon when reading the wrong book to Cub, which brings clouds around their house (and a hail of dead birds). It's hard to find the perfect volume, but we'll low ball a bit to account for the abnormal shape of the ring, and we'll scale the average height of nimbostratus clouds, that being 500m to 5000m.
Timeframe is 2.44 seconds.
= 1.02701259 rad
= 58.843486913945583 degrees
Put that through the angcaler and we have a diameter of 443.29m.
857 pixels = 443.29m
443.29m/2 = 221.64500m
1 pixel = 443.29m/857 = 0.517257876m
0.517257876m X 46 = 23.7938623m
0.517257876m X 110 pixels = 56.8983664m
56.8983664m/2 = 28.4491832m
Volume as a torus (obviously).
V = (πr^2)(2πR)
= (π X 28.4491832^2)(2 X π X 221.64500)
= 3541011.06m^3
The weight of air is 1.003kg.
M = 3541011.06 X 1.003kg
= 3551634.09kg
T = 23.7938623m/2.44s
= 9.75158291m/s
KE = (0.5)mv^2
= (0.5) X 3551634.09 X 9.75158291^2
= 168868426 joules
There are no clouds around before Pop started reading the wrong book, so we can also get the energy to form these clouds.
Density of the water droplets a cumulus cloud is around 0.5 grams per m^3 (not the air as a whole, but the water droplets of the cloud).
M = 3541011.06 X 0.5g
= 1770505.53g
Condensation of water vapour to liquid is 2260 joules per gram.
E = 1770505.53 X 2260
= 4.0013425e9 joules
Finally, to add both values together for our total energy.
E = 168868426 + 4.0013425e9
= 4170210926 joules
= 0.996704332218 tons of TNT
3 . Giggles cuts down the beanstalk
3:09
Giggles cuts down the beanstalk (in a parody/retelling of Jack & the Beanstalk). This causes giant Lumpy's castle to be shaken apart. Giant Lumpy's castle is not attached to the beanstalk, so that means Giggles would have to hit the beanstalk with enough force to dislodge a large part of Lumpy's castle. The castle looks smaller on the outside than it actually is on the inside (thanks to cartoony visuals), so we'll need to find the minimum dimensions of the interior, scaling off Giant Lumpy and the regular sized Happy Tree Friends. Once again, the average height of a Happy Tree Friends character is 1.11353712m.
60 pixels = 1.11353712m
1 pixel = 1.11353712m/60 = 0.018558952m
0.018558952m X 527 = 9.7805677m
0.018558952m X 45 = 0.83515284m
0.018558952m X 668 = 12.3973799m
0.018558952m X 442 = 8.20305678m
524 pixels = 8.20305678m
1 pixel = 8.20305678m/524 = 0.0156546885m
0.0156546885m X 1269 = 19.8657997m
81 pixels = 1.11353712m
1 pixel = 1.11353712m/81 = 0.0137473719m
0.0137473719m X 1273 = 17.5004044m
144 pixels = 1.11353712m
1 pixel = 1.11353712m/144 = 0.00773289667m
0.00773289667m X 1272 = 9.83624456m
W = 19.8657997m + 17.5004044m + 9.83624456m
= 47.2024487m
That is the minimum diameter of the main portion of the castle. Now, let's find out how big the different segments of the castle are.
439 pixels = 47.2024487m
47.2024487m/2 = 23.6012243m
1 pixel = 47.2024487m/439 = 0.107522662m
0.107522662m X 352 = 37.847977m
0.107522662m X 16 = 1.72036259m
23.6012243m - 1.72036259m = 21.8808617m
0.107522662m X 99 = 10.6447435m
0.107522662m X 70 = 7.52658634m
7.52658634m/2 = 3.76329317m
3.76329317m - 1.72036259m = 2.04293058m
0.107522662m X 94 = 10.1071302m
0.107522662m X 175 = 18.8164659m
0.107522662m X 133 = 14.300514m
14.300514m/2 = 7.150257m
7.150257m - 2.04293058m = 5.10732642m
0.107522662m X 174 = 18.7089432m
0.107522662m X 105 = 11.2898795m
11.2898795m/2 = 5.64493975m
5.64493975m - 2.04293058m = 3.60200917m
0.107522662m X 47 = 5.05356511m
0.107522662m X 58 = 6.2363144m
Volume of all the segments as cylinders, except for the battlements, which will be rectangular prisms. We'll also take away the hollow parts of each section via the thickness of the wall (also as a cylinder).
V = πr^2h
= π X 23.6012243^2 X 37.847977
= 66231.0452m^3
V = πr^2h
= π X 21.8808617^2 X 37.847977
= 56927.4049m^3
V = 66231.0452 - 56927.4049
= 9303.6403m^3
V = πr^2h
= π X 3.76329317^2 X 10.6447435
= 473.610343m^3
V = πr^2h
= π X 2.04293058^2 X 10.6447435
= 139.570069m^3
V = 473.610343 - 139.570069
= 334.040274m^3
V = πr^2h
= π X 3.76329317^2 X 10.1071302
= 449.690629m^3
V = πr^2h
= π X 2.04293058^2 X 10.1071302
= 132.521075m^3
V = 449.690629 - 132.521075
= 317.169554m^3
V = πr^2h
= π X 7.150257^2 X 18.8164659
= 3022.2559m^3
V = πr^2h
= π X 5.10732642^2 X 18.8164659
= 1541.96729m^3
V = 3022.2559 - 1541.96729
= 1480.28861m^3
V = πr^2h
= π X 5.64493975^2 X 18.7089432
= 1872.91363m^3
V = πr^2h
= π X 3.60200917^2 X 18.7089432
= 762.585876m^3
V = 1872.91363 - 762.585876
= 1110.32775m^3
That last one is for the left-most turret which is dislodged. The battlements are next, and as there are 5 of them, we'll times the final result by 5.
V = lhw
= 5.05356511 X 6.2363144 X 1.72036259
= 271.091476m^3
Keep that in mind for later, but first we shall add all the volumes together.
V = 9303.6403 + 334.040274 + 317.169554 + 1480.28861 + 1110.32775 + 271.091476
= 12816.558m^3
Although the scaling isn't perfect, please not that this is not taking into account the walls, ceilings, rooftops or stairs, so the final volume should ultimately be even higher. Medieval castles were made out of a few rock types, limestone (2560kg per m^3), sandstone (2323kg per m^3) and granite (2725kg per m^3).
M = 2323 + 2560 + 2725
= 7608/3
= 2536kg
M = 12816.558 X 2536
= 32502791.1kg
At long last, let's get our distance moved, our timeframe and our sweet kinetic energy!
226 pixels = 37.847977m
1 pixel = 37.847977m/226 = 0.167468925m
0.167468925m X 427 = 71.509231m
Timeframe is 1 frame.
0.167468925m X 423 = 70.8393553m
71.509231m - 70.8393553m = 0.6698757m
T = 0.6698757m/41.6666667ms
= 16.0770168m/s
KE = (0.5)mv^2
= (0.5) X 32502791.1 x 16.0770168^2
= 4.20050583e9 joules
= 1.00394498805 tons of TNT
But that's not all...the castle was shook from a large distance away, and far away from the epicenter, so why not go for a high end, and find the source of the energy at the epicentre with inverse-square law? Let's use the above distance to the clouds of 500m.
4.20050583e9=S/(4 X π X 500^2)
S=4.20050583e9 x (4 X π X 500^2)
S=1.31962783e16 joules = 3.1539862093690249623 megatons
Shiiiiiiiit that's way higher than I thought it would be. I asked a friend for advice about this, and they also suggested using a shockwave calculator. Using the MIT shockwave calculator (for a distance of 0.5km for "peak overpressure:10 psi") we get a yield of 3.1 kilotons.
Final Results
Lumpy's Christmas lights lances the Moon = 15.387 C
Lumpy's Christmas lights destroys the Moon = 18.540 yottatons
Lumpy & the Mole resist being vaporized = >0.717 tons of TNT
The Demon forms clouds = 0.997 tons of TNT
Giant Lumpy's height = 9.781m
Giggles cuts down the beanstalk (low end) = 1.004 tons of TNT
Giggles cuts down the beanstalk (mid end) = 3.1 kilotons
Giggles cuts down the beanstalk (high end) = 3.154 megatons
1. Lumpy blows up the Moon
3:36
T = 1s/24
= 41.6666667ms X 2
= 83.3333334ms
= 384,400km/83.3333334ms
= 4.6128e9/299792458
= 15.3866446 C
Next we get the energy for exploding the Moon. The diameter of the Moon is 3,475km and has a mass of 7.35e22kg.
1 pixel = 3,475km/108 = 32.1759259km
32.1759259km X 18 = 579.166666km
32.1759259km X 43 = 1383.56481km
32.1759259km X 122 = 3925.46296km
32.1759259km X 98 = 3153.24074km
32.1759259km X 70 = 2252.31481km
32.1759259km X 73 = 2348.84259km
32.1759259km X 69 = 2220.13889km
32.1759259km X 94 = 3024.53703km
Let's find our average speed.
L = 579.166666 + 1383.56481 + 2220.13889 + 2252.31481 + 2348.84259 + 3024.53703 + 3153.24074 + 3925.46296
= 18887.268496/8
= 2360.90856km
T = 41.6666667ms X 15
= 0.625000001s + 1s
= 1.62500s
T = 2360.90856km/1.62500s
= 1452866.81m/s
KE = (0.5)mv^2
= (0.5) X 7.35e22 X 1452866.81^2
= 7.75727073e34 joules
= 18.540322012428298137 yottatons
2. Lumpy & the Mole aren't vaporized
From the above video at 4:01.
3. The Demon generates clouds
0:49
Pop accidentally summons the Demon when reading the wrong book to Cub, which brings clouds around their house (and a hail of dead birds). It's hard to find the perfect volume, but we'll low ball a bit to account for the abnormal shape of the ring, and we'll scale the average height of nimbostratus clouds, that being 500m to 5000m.
= 2*atan(857/(720/tan(70/2)))
= 1.02701259 rad
= 58.843486913945583 degrees
Put that through the angcaler and we have a diameter of 443.29m.
857 pixels = 443.29m
443.29m/2 = 221.64500m
1 pixel = 443.29m/857 = 0.517257876m
0.517257876m X 46 = 23.7938623m
0.517257876m X 110 pixels = 56.8983664m
56.8983664m/2 = 28.4491832m
Volume as a torus (obviously).
V = (πr^2)(2πR)
= (π X 28.4491832^2)(2 X π X 221.64500)
= 3541011.06m^3
The weight of air is 1.003kg.
M = 3541011.06 X 1.003kg
= 3551634.09kg
T = 23.7938623m/2.44s
= 9.75158291m/s
KE = (0.5)mv^2
= (0.5) X 3551634.09 X 9.75158291^2
= 168868426 joules
There are no clouds around before Pop started reading the wrong book, so we can also get the energy to form these clouds.
M = 3541011.06 X 0.5g
= 1770505.53g
Condensation of water vapour to liquid is 2260 joules per gram.
E = 1770505.53 X 2260
= 4.0013425e9 joules
Finally, to add both values together for our total energy.
E = 168868426 + 4.0013425e9
= 4170210926 joules
= 0.996704332218 tons of TNT
3 . Giggles cuts down the beanstalk
3:09
Giggles cuts down the beanstalk (in a parody/retelling of Jack & the Beanstalk). This causes giant Lumpy's castle to be shaken apart. Giant Lumpy's castle is not attached to the beanstalk, so that means Giggles would have to hit the beanstalk with enough force to dislodge a large part of Lumpy's castle. The castle looks smaller on the outside than it actually is on the inside (thanks to cartoony visuals), so we'll need to find the minimum dimensions of the interior, scaling off Giant Lumpy and the regular sized Happy Tree Friends. Once again, the average height of a Happy Tree Friends character is 1.11353712m.
1 pixel = 1.11353712m/60 = 0.018558952m
0.018558952m X 527 = 9.7805677m
0.018558952m X 45 = 0.83515284m
0.018558952m X 668 = 12.3973799m
0.018558952m X 442 = 8.20305678m
1 pixel = 8.20305678m/524 = 0.0156546885m
0.0156546885m X 1269 = 19.8657997m
1 pixel = 1.11353712m/81 = 0.0137473719m
0.0137473719m X 1273 = 17.5004044m
1 pixel = 1.11353712m/144 = 0.00773289667m
0.00773289667m X 1272 = 9.83624456m
W = 19.8657997m + 17.5004044m + 9.83624456m
= 47.2024487m
That is the minimum diameter of the main portion of the castle. Now, let's find out how big the different segments of the castle are.
47.2024487m/2 = 23.6012243m
1 pixel = 47.2024487m/439 = 0.107522662m
0.107522662m X 352 = 37.847977m
0.107522662m X 16 = 1.72036259m
23.6012243m - 1.72036259m = 21.8808617m
0.107522662m X 99 = 10.6447435m
0.107522662m X 70 = 7.52658634m
7.52658634m/2 = 3.76329317m
3.76329317m - 1.72036259m = 2.04293058m
0.107522662m X 94 = 10.1071302m
0.107522662m X 175 = 18.8164659m
0.107522662m X 133 = 14.300514m
14.300514m/2 = 7.150257m
7.150257m - 2.04293058m = 5.10732642m
0.107522662m X 174 = 18.7089432m
0.107522662m X 105 = 11.2898795m
11.2898795m/2 = 5.64493975m
5.64493975m - 2.04293058m = 3.60200917m
0.107522662m X 47 = 5.05356511m
0.107522662m X 58 = 6.2363144m
Volume of all the segments as cylinders, except for the battlements, which will be rectangular prisms. We'll also take away the hollow parts of each section via the thickness of the wall (also as a cylinder).
V = πr^2h
= π X 23.6012243^2 X 37.847977
= 66231.0452m^3
V = πr^2h
= π X 21.8808617^2 X 37.847977
= 56927.4049m^3
V = 66231.0452 - 56927.4049
= 9303.6403m^3
V = πr^2h
= π X 3.76329317^2 X 10.6447435
= 473.610343m^3
V = πr^2h
= π X 2.04293058^2 X 10.6447435
= 139.570069m^3
V = 473.610343 - 139.570069
= 334.040274m^3
V = πr^2h
= π X 3.76329317^2 X 10.1071302
= 449.690629m^3
V = πr^2h
= π X 2.04293058^2 X 10.1071302
= 132.521075m^3
V = 449.690629 - 132.521075
= 317.169554m^3
V = πr^2h
= π X 7.150257^2 X 18.8164659
= 3022.2559m^3
V = πr^2h
= π X 5.10732642^2 X 18.8164659
= 1541.96729m^3
V = 3022.2559 - 1541.96729
= 1480.28861m^3
V = πr^2h
= π X 5.64493975^2 X 18.7089432
= 1872.91363m^3
V = πr^2h
= π X 3.60200917^2 X 18.7089432
= 762.585876m^3
V = 1872.91363 - 762.585876
= 1110.32775m^3
That last one is for the left-most turret which is dislodged. The battlements are next, and as there are 5 of them, we'll times the final result by 5.
V = lhw
= 5.05356511 X 6.2363144 X 1.72036259
= 271.091476m^3
Keep that in mind for later, but first we shall add all the volumes together.
V = 9303.6403 + 334.040274 + 317.169554 + 1480.28861 + 1110.32775 + 271.091476
= 12816.558m^3
Although the scaling isn't perfect, please not that this is not taking into account the walls, ceilings, rooftops or stairs, so the final volume should ultimately be even higher. Medieval castles were made out of a few rock types, limestone (2560kg per m^3), sandstone (2323kg per m^3) and granite (2725kg per m^3).
M = 2323 + 2560 + 2725
= 7608/3
= 2536kg
M = 12816.558 X 2536
= 32502791.1kg
At long last, let's get our distance moved, our timeframe and our sweet kinetic energy!
1 pixel = 37.847977m/226 = 0.167468925m
0.167468925m X 427 = 71.509231m
0.167468925m X 423 = 70.8393553m
71.509231m - 70.8393553m = 0.6698757m
T = 0.6698757m/41.6666667ms
= 16.0770168m/s
KE = (0.5)mv^2
= (0.5) X 32502791.1 x 16.0770168^2
= 4.20050583e9 joules
= 1.00394498805 tons of TNT
But that's not all...the castle was shook from a large distance away, and far away from the epicenter, so why not go for a high end, and find the source of the energy at the epicentre with inverse-square law? Let's use the above distance to the clouds of 500m.
4.20050583e9=S/(4 X π X 500^2)
S=4.20050583e9 x (4 X π X 500^2)
S=1.31962783e16 joules = 3.1539862093690249623 megatons
Shiiiiiiiit that's way higher than I thought it would be. I asked a friend for advice about this, and they also suggested using a shockwave calculator. Using the MIT shockwave calculator (for a distance of 0.5km for "peak overpressure:10 psi") we get a yield of 3.1 kilotons.
Final Results
Lumpy's Christmas lights lances the Moon = 15.387 C
Lumpy's Christmas lights destroys the Moon = 18.540 yottatons
Lumpy & the Mole resist being vaporized = >0.717 tons of TNT
The Demon forms clouds = 0.997 tons of TNT
Giant Lumpy's height = 9.781m
Giggles cuts down the beanstalk (low end) = 1.004 tons of TNT
Giggles cuts down the beanstalk (mid end) = 3.1 kilotons
Giggles cuts down the beanstalk (high end) = 3.154 megatons
