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I was looking at Homelander's (TV series) profile and I think his durability is wrong. These are my opinions and analysis, so anyone feel free to disagree if they find any errors.
Durability is justified by this calculation. He struck me as incorrect in many ways.
1. Distance from Homelander to the explosion seems incorrect.
At the beginning of the calculation, 1 pixel is stipulated at 0.87354731707 inches.
To paraphrase the author of the calculation:
"Homelander’s knees are 0.6 cm tall or 22.677165354 pixels. Knowing 22.577165354 pixels equals 1.64351851852 feet or 19.7222222222 inches, this means 1 pixel equals 0.87354731707 inches."
In the end, those 0.87354731707 inches are used in the formula I = P/A as the distance from Homelander to the explosion. Again, paraphrasing:
"Then we use I = P/A, or I = 43069.457229 / (4π((0.87354731707 )^2))."
The feat is here. Pause at 1:41. Homelander cannot be 0.87354731707 inches (about 2.2 cm) away from the explosion. At 1:32 his arm is extended towards the object that exploded and there was a man between him and the object.
I couldn't find the average size of a single arm, but I'm smaller than Homelander (I'm 172 cm) and my arm is 65 cm from wrist to shoulder (I tried to imitate the Homelander position and measured my arm with a tape measure and got 55 cm; sorry, the best I could).
I know, it's not exactly a good source to use my measurements, but I imagine most people have an arm that is close in size (and longer than the 2.2 cm of the original calculation).
Not only that, but Homelander took a step back in 1:41. Accordingly, the average step distance is 0.415 x the height. Homelander is 180 cm. So, 74.7 cm.
I find it very difficult to estimate the final distance from Homelander in the 1:41 frame to the bottom of the object (and I don't particularly trust the angsize calculations), but it seems fair to add up the arm and stride distance I provided, concluding a distance of 1,322 meters. It seems to make sense to me looking at that frame.
2. The estimated smoke stack size problem.
According to the original calculation, a smoke stack of 275 meters was used as a base.
Pause at 1:42. There was a building in front of the smoke stack. The building appears to be half the size of the smoke stack (and that's considering the smoke stack was at the back, so it could be smaller from perspective). That building would have to be about 137.5 meters to cover so much of the smoke stack. Assuming that non-residential buildings average 4 meters high per floor, the building would have to be about 36 floors.
Now pause at 00:01.
We see that the buildings are clearly much smaller. We even see a small smoke stack (the size of a low-rise building) on the left (I don't think it's the one used in the original calculation, but it shows that there were small smoke stacks there).
It seems clear to me that that smoke stack was much smaller than 275 meters and that building in front of it looked small (in view of the buildings shown at 00:01 in the same video I quoted earlier).
I'm going to use a 3-story building (which should be about 12 meters high) as the base for the one in front of the calculation smoke stack. If the smoke stack was twice as tall, it would be 24 meters.
Taking advantage of this, I think the author of the calculation made a mistake at one point.
To paraphrase: "Now we're using the size equation, but now with 60 pixels (The explosion's size) instead of 66.
275 * 273/[66*2*tan(70deg/2)] = 812.259179 meters. The explosion covers the whole screen, so we'll use this as our blast radius as well."
It was said that the 66 pixels would be replaced by 60, but the 66 was repeated.
I'm going to reuse the formula using the 60 pixels of the explosion that the author of the original calculation got and replace the 275 meters of the smoke stack with the estimated 24 meters. It seems to me that the author also multiplied 275 meters by 273 pixels from the image he posted, but in the image the panel has 260.15999999999997 pixels. I will use the 260.15999999999997 pixels as well.
24 * 260.15999999999997/[60*2*tan(70deg/2)] = 74.3093970868 meters.
74.3093970868^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 = 32.9774240045 Tons.
Now using Homelander's distance for the explosion I estimated above. I = P/A, or I = 43069.457229 / (4π((1.322)^2)) =1.50156436769 Ton of TNT.
Homelander is 180 cm and 75 kg. According to this, the frontal surface area of his body would be 0.97 m².
1.50156436769 x 0.97 = 1.45651743666 Ton of TNT.
This is Building Level+, far below the Small Town Level proposed by the original calculation.
3. Consistency
The original calculation comes to a result of about 43 kilotons of TNT. Fat Man (the Nagasaki nuclear bomb) was 21 kilotons. It doesn't seem to make sense to me that an explosion from what appears to me to be a gas tank would be more powerful than a nuclear bomb. A woman who was in a building near the explosion survived without major injuries (a person would not survive only a few dozen meters from a nuclear explosion, right?). Even the walls of the surrounding buildings did not completely fall.
Soldier Boy was shown to be able to injure Homelander in season 3 of The Boys and using his energy, he did it with a building (pause at 1:49)
It seems consistent with Homelander if this feat really is Building Level+ and not Small Town Level. In fact, both feats look like Building Level+ to me.
Building Level seems much more logical than Small Town Level.
4. Conclusion
Basically, it's my opinions of why I think Homelander's durability should be reduced to Building Level+ (if we consider that the formulas and calculations made by the original calculation are correct, which was not my purpose to comment here). Some estimates I've made (like assuming Homelander took an average step when he clearly walked away from the man after splitting him with heat vision and the size of his outstretched arm) aren't such accurate evidence, but I believe it's something around that.
Anyway, I think Building Level+ is consistent.
Durability is justified by this calculation. He struck me as incorrect in many ways.
1. Distance from Homelander to the explosion seems incorrect.
At the beginning of the calculation, 1 pixel is stipulated at 0.87354731707 inches.
To paraphrase the author of the calculation:
"Homelander’s knees are 0.6 cm tall or 22.677165354 pixels. Knowing 22.577165354 pixels equals 1.64351851852 feet or 19.7222222222 inches, this means 1 pixel equals 0.87354731707 inches."
In the end, those 0.87354731707 inches are used in the formula I = P/A as the distance from Homelander to the explosion. Again, paraphrasing:
"Then we use I = P/A, or I = 43069.457229 / (4π((0.87354731707 )^2))."
The feat is here. Pause at 1:41. Homelander cannot be 0.87354731707 inches (about 2.2 cm) away from the explosion. At 1:32 his arm is extended towards the object that exploded and there was a man between him and the object.
I couldn't find the average size of a single arm, but I'm smaller than Homelander (I'm 172 cm) and my arm is 65 cm from wrist to shoulder (I tried to imitate the Homelander position and measured my arm with a tape measure and got 55 cm; sorry, the best I could).
I know, it's not exactly a good source to use my measurements, but I imagine most people have an arm that is close in size (and longer than the 2.2 cm of the original calculation).
Not only that, but Homelander took a step back in 1:41. Accordingly, the average step distance is 0.415 x the height. Homelander is 180 cm. So, 74.7 cm.
I find it very difficult to estimate the final distance from Homelander in the 1:41 frame to the bottom of the object (and I don't particularly trust the angsize calculations), but it seems fair to add up the arm and stride distance I provided, concluding a distance of 1,322 meters. It seems to make sense to me looking at that frame.
2. The estimated smoke stack size problem.
According to the original calculation, a smoke stack of 275 meters was used as a base.
Pause at 1:42. There was a building in front of the smoke stack. The building appears to be half the size of the smoke stack (and that's considering the smoke stack was at the back, so it could be smaller from perspective). That building would have to be about 137.5 meters to cover so much of the smoke stack. Assuming that non-residential buildings average 4 meters high per floor, the building would have to be about 36 floors.
Now pause at 00:01.
We see that the buildings are clearly much smaller. We even see a small smoke stack (the size of a low-rise building) on the left (I don't think it's the one used in the original calculation, but it shows that there were small smoke stacks there).
It seems clear to me that that smoke stack was much smaller than 275 meters and that building in front of it looked small (in view of the buildings shown at 00:01 in the same video I quoted earlier).
I'm going to use a 3-story building (which should be about 12 meters high) as the base for the one in front of the calculation smoke stack. If the smoke stack was twice as tall, it would be 24 meters.
Taking advantage of this, I think the author of the calculation made a mistake at one point.
To paraphrase: "Now we're using the size equation, but now with 60 pixels (The explosion's size) instead of 66.
275 * 273/[66*2*tan(70deg/2)] = 812.259179 meters. The explosion covers the whole screen, so we'll use this as our blast radius as well."
It was said that the 66 pixels would be replaced by 60, but the 66 was repeated.
I'm going to reuse the formula using the 60 pixels of the explosion that the author of the original calculation got and replace the 275 meters of the smoke stack with the estimated 24 meters. It seems to me that the author also multiplied 275 meters by 273 pixels from the image he posted, but in the image the panel has 260.15999999999997 pixels. I will use the 260.15999999999997 pixels as well.
24 * 260.15999999999997/[60*2*tan(70deg/2)] = 74.3093970868 meters.
74.3093970868^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2 = 32.9774240045 Tons.
Now using Homelander's distance for the explosion I estimated above. I = P/A, or I = 43069.457229 / (4π((1.322)^2)) =1.50156436769 Ton of TNT.
Homelander is 180 cm and 75 kg. According to this, the frontal surface area of his body would be 0.97 m².
1.50156436769 x 0.97 = 1.45651743666 Ton of TNT.
This is Building Level+, far below the Small Town Level proposed by the original calculation.
3. Consistency
The original calculation comes to a result of about 43 kilotons of TNT. Fat Man (the Nagasaki nuclear bomb) was 21 kilotons. It doesn't seem to make sense to me that an explosion from what appears to me to be a gas tank would be more powerful than a nuclear bomb. A woman who was in a building near the explosion survived without major injuries (a person would not survive only a few dozen meters from a nuclear explosion, right?). Even the walls of the surrounding buildings did not completely fall.
Soldier Boy was shown to be able to injure Homelander in season 3 of The Boys and using his energy, he did it with a building (pause at 1:49)
It seems consistent with Homelander if this feat really is Building Level+ and not Small Town Level. In fact, both feats look like Building Level+ to me.
Building Level seems much more logical than Small Town Level.
4. Conclusion
Basically, it's my opinions of why I think Homelander's durability should be reduced to Building Level+ (if we consider that the formulas and calculations made by the original calculation are correct, which was not my purpose to comment here). Some estimates I've made (like assuming Homelander took an average step when he clearly walked away from the man after splitting him with heat vision and the size of his outstretched arm) aren't such accurate evidence, but I believe it's something around that.
Anyway, I think Building Level+ is consistent.
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