My Hero Academia: The Final Smash Calculation

[TheRustyOne]

Alright, time to bite the bullet and discuss what calculation we're going to use for Izuku's Final Smash during the Final War Arc.

In My Hero Academia Chapter 423-424, Izuku throws a powerful punch that splits apart the storm clouds above him.

The most biggest effect of this Smash is that it reached the USA and changed the weather. What should've been a week of stormy weather became a week of clear skies.

The above scan takes place one week later and Meryl says they're still expecting strong winds, which is confirmed to have been caused by Izuku.

The air pressure Izuku generated from near Mount Fuji, went all the way across the ocean to the USA.

And even though one week passed, they were still expecting strong winds.

I'm currently aware of two calculations that I believe are usable for this feat.

Izuku's Smash Clears the Storm Over Japan: 226.68 Teratons of TNT (Large Country level) to 453.37 Teratons of TNT (Large Country level+)

Izuku's Smash Clears Weather From Japan to America: 21.18 Petatons of TNT to 42.37 Petatons of TNT (Multi-Continent level)

Both of these calculations are similar. The first calculates the storm over Japan being dispersed.

While the second one calculates the entire area that was affected by Izuku's attack. I believe using the distance from Japan to America is alright.

Since we have no reason to believe the shockwave reached America yet also slowed down enough that it doesn't effect our results.

One Week of Wind: 908.886 Teratons of TNT (Continent level)

I'm aware of my calculation here, but I don't think it compares to the two calculations above, which are far more sensible. Though you're free to make your own opinions. I've been informed it could be possible to add this one together with Therefir's calculation, though I'm not sure how that all works.

This thread is not just about discussing these calculations. But to discuss any alternative calculations, assuming you think they're accurate.

Note: This is a discussion on what calculation will be accepted. Character profiles or scaling is not to be discussed here. No profiles will be changed from this thread regardless of what is accepted. Depending on what is accepted, a CRT will be made in the future to discuss character scaling and everything it entails.

 

[TheRustyOne]

@Damage3245 @Therefir @CloverDragon03 @Dalesean027 @DemonGodMitchAubin

 

[Damage3245]

Izuku's Smash Clears Weather From Japan to America: 21.18 Petatons of TNT to 42.37 Petatons of TNT (Multi-Continent level)

I don't believe that this one would be the most accurate way of calcing it.

The storm is not shown to have reached the United States by the time that Izuku throws his punch that clears it, so Izuku's shockwave doesn't need to have affected a volume of air as large as that in such a short span of time to have produced the result of a week of sunny skies.

One Week of Wind: 908.886 Teratons of TNT (Continent level)

This one I don't think is quite right either. Strong winds still being expected a week after the incident doesn't mean it is solely the energy of Izuku's punch driving those winds as it's not a closed-energy system where he is the only input. There's the whole weather system of the rest of the world and Izuku's punch may have messed up the patterns significantly but that wouldn't mean he's the only contributing factor for why it would be windy - nor does it mean that just because strong winds are expected means that there has been non-stop constant high winds every second of the day across the entire country.

Izuku's Smash Clears the Storm Over Japan: 226.68 Teratons of TNT (Large Country level) to 453.37 Teratons of TNT (Large Country level+)

I think this first calc is the best one for now for the feat but this is just at as first glance and I'll double-check the contents of the calc myself soon just to verify that everything is alright.

 

[Dalesean027]

WATASHI WA KITA!!!
Oh im late... I'll check after work

 

[CloverDragon03]

Meryl says that "the gust that blew that day came in the wake of a great battle." Wouldn't that mean there was a gust of wind that reached from Deku all the way to America? My thought process is that this statement implies America felt a gust of wind from that fight (specifically from Deku's final punch, given the flashback), and Meryl is explaining what caused it

 

[Dalesean027]

Meryl says that "the gust that blew that day came in the wake of a great battle." Wouldn't that mean there was a gust of wind that reached from Deku all the way to America? My thought process is that this statement implies America felt a gust of wind from that fight (specifically from Deku's final punch, given the flashback), and Meryl is explaining what caused it

Yes America did get directly affected by the shockwaves which are still even after a week causing strong winds even in America

 

[CloverDragon03]

Yeah, I'm primarily wondering if the fact that the shockwave reached America could be useful for something

 

[TheRustyOne]

The storm is not shown to have reached the United States by the time that Izuku throws his punch that clears it, so Izuku's shockwave doesn't need to have affected a volume of air as large as that in such a short span of time to have produced the result of a week of sunny skies.

That's not what I'm calculating. You should take a closer look. Izuku produced a shockwave that dispersed the storm and we know that shockwave reached America. So the total volume of air effected should be from Izuku to America. Unless we have reason to believe Izuku's air pressure changed size/shape and slowed down?

The reasoning is the exact same. It's a blast of air being disperse from a center point. In the case of the first calculation it only measures the storm's area. But we know the shockwave reached America so I'm including that area as well. Which increases the results.

The calculator used to get weight can be used for just air as well. I'm not saying he blew away a storm of that size.

Meryl says that "the gust that blew that day came in the wake of a great battle." Wouldn't that mean there was a gust of wind that reached from Deku all the way to America? My thought process is that this statement implies America felt a gust of wind from that fight (specifically from Deku's final punch, given the flashback), and Meryl is explaining what caused it

Yes, the shockwave Izuku produced here was large enough to reach America.

The second calculation in the OP is taking that into account.

 

[57Dev]

Finally, took long enough.

Without a doubt Izuku did affect the weather of USA and by extension the whole planet. The Multi continental calc itself is good, but I wonder if such a massive Upgrade will be seen as outlier by the low ballers.

 

[Damage3245]

That's not what I'm calculating. You should take a closer look. Izuku produced a shockwave that dispersed the storm and we know that shockwave reached America. So the total volume of air effected should be from Izuku to America. Unless we have reason to believe Izuku's air pressure changed size/shape and slowed down?

Well, a shockwave would naturally slow down from its initial velocity as it spreads out and affects more mass, no?

I can't think of a reason why the mass of air at the edge of the calculated radius would move with the same velocity as the air closest to Izuku at the start.

 

[57Dev]

Well, a shockwave would naturally slow down from its initial velocity as it spreads out and affects more mass, no?

I can't think if a reason why the mass of air at the edge of the calculated radius would move with the same velocity as the air closest to Izuku at the start.

Good point

 

[TheRustyOne]

Well, a shockwave would naturally slow down from its initial velocity as it spreads out and affects more mass, no?

I can't think of a reason why the mass of air at the edge of the calculated radius would move with the same velocity as the air closest to Izuku at the start.

Yes, but the difference is minimal. It doesn't change anything because of the sheer speed it's already moving at and the fact we know strong winds still exist one week later.

In the third calc in the OP, which I agree is unusable. The strong wind speed I assume is 12.51712 m/s, which is a strong breeze. I'll use that to explain.

The initial speed is 29640.80 m/s and the final speed is 12.51712 m/s. It's been one week so it took one week for the wind to slow down to 12.51712 m/s.

Using the acceleration calculator, we can find the deceleration speed. Which is a value of -0.04899 m/s^2.

Meaning after seven minutes, the speed of the shockwave would be 29620.2 m/s. Which is a decrease less than 0.07%. The final results aren't even changed by this.

However, there are external factors effecting the shockwave as we said. So that final wind speed might not be accurate. Let's low ball and assume only 1% of that speed is caused by Izuku's shockwave, which contradicts what is stated in universe but I'm doing this to make it clear.

This means the final wind speed is actually 0.1251712 m/s, which means the deceleration is now -0.04901 m/s^2. This doesn't change the results at all.

The reason I'm alright with doing this is because there are still strong winds reaching America even after one week had passed. Meaning the deceleration was not significant.

Also, not all of the air is moving at initial speed, the 1/12 KE formula already takes into account the air moving at the edge is slower than air moving in the center.

 

[Damage3245]

I'll see if I can reword the issue I have with that version of the calc to better to explain myself. I don't think what I said came across well because I didn't mention anything about it slowing down from 29640.80 m/s to 12.51712 m/s over the course of the full week.

EDIT: I think I know how to address it, I just need time to do it tomorrow.

 

[TheRustyOne]

I don't think what I said came across well because I didn't mention anything about it slowing down from 29640.80 m/s to 12.51712 m/s over the course of the full week.

I feel like you're misunderstanding me, maybe I'm not being clear myself.

I'm the one who mentioned that difference, I don't understand how you reach that conclusion. That wasn't about you mentioning anything, I was explaining myself.

The initial speed of the shockwave 29640.80 m/s, with the possible "final" speed being either 12.51712 m/s or 0.1251712 m/s.

I brought this up to show you how the speed doesn't change by any serious value in the timeframe needed for the shockwave to reach the USA. If the wind speed caused by Izuku is 0.1251712 m/s after one week, that means the deceleration of that wind from the external factors was small.

My calculation and Therefir's are exactly the same. The only difference is the size of the area being measured. The storm was stated to be unprecedented in size, so they scaled it as being comparable to Typhoon Tip. While I used the distance from Mount Fuji to the end of America as my size.

Reason being? We're clearly told the shockwave Izuku generated reached all of America, and I've already explained why the speed didn't slow down significantly. Because after one week we're told the wind generated by Izuku is still strong, which is still being felt in America.

Meaning there is no noticeable deceleration from the initial speed by the time it reaches America. Which is less than 7 minutes, even taking the deceleration into account.

 

[Damage3245]

I feel like you're misunderstanding me, maybe I'm not being clear myself.

I'm the one who mentioned that difference, I don't understand how you reach that conclusion. That wasn't about you mentioning anything, I was explaining myself.

Gotcha, I just wasn't sure why you'd be doing that.

I'll get my next post up later this morning.

 

[Damage3245]

Okay, so when I mentioned the shockwave slowing down earlier in the thread, I was not referring to the idea of it decelerating as a gradual consistent rate from an initial speed of 29640.80 m/s to a final speed is 12.51712 m/s over the course of one week. I get where you're coming from with your explanation in this post and you'd be right that it is neglible deceleration over such a long time frame.

But that's not what I'm getting at. In the calc there is this section:

This is the mass of all the air in the effected range, now we need to find the speed of the shockwave Izuku made to get KE.

As we can see here, the shockwave from his punch instantly reached the top of the cloud, so fast that his fist didn't even change position by the time we see the clear sky.

Izuku wasn't that far above ground level.

Distance From Izuku To The Top of The Cloud = 21336 + 8304.80 = 29640.80 m

Minimum timeframe we can use until the anime arrives would be 1 second.

This is not the timeframe needed to clear the storm or for the shockwave reach the USA.

This is just the speed of the shockwave Izuku produced that is pushing all of this air.

Speed = 29640.80/1 = 29640.80 m/s

So, you establish that the force of Izuku's punch can move an initial mass of air upwards at a speed of 29640.8 m/s.

If this was just some kind of energy bullet unimpeded with air resistance or being slowed down by anything around it, then sure, in a vacuum it could continue indefinitely moving at 29640.8 m/s so long as no other forces act on it.

But using this velocity for the entire mass of air in a volume with a radius from Mt. Fuji to the edge of the United States? There's an issue with this.

Because mass being affected is not constant here. That initial speed of 29640.8 meters per second? That velocity isn't being found for a mass of air weighing 1210897470122445800 kg. It's found only for the air actually above Izuku in the initial volume affected by his punch.

As the shockwave then spreads outward, it impacts more air. It affects a larger and larger mass as it spreads out. The shockwave slows down, yes because the energy is being emitted in all directions, but because it is affecting more mass too.

To give a visual example, the people recording this video are hit by the "shockwave" of the volcano so they are within the "radius of the shockwave" just like how the edge of the United States is within the radius of your calc. But you wouldn't use the initial velocity of the volcanic eruption there and say that the air where the people recording were at was moving that quickly.

The 1/12 KE formula that you mentioned doesn't fully take this into account for this scenario because in that formula, what you do is find the speed that the entire mass of air travels at and use that in the K.E. formula with the mass of the entire air. That formula only takes into account the energy spreading outwards, not about it affecting more and more mass.

The 29640.8 m/s speed value you've used is for a significantly smaller mass of air.

To put it another way:

You've got Izuku initially affecting X kg mass of air at a calculated speed of 29640.8 m/s.

Izuku's shockwave then keeps going, affecting more and more mass until at least 1210897470122445800 kg mass of air is affected.

But that end amount of mass is far greater than X.... So unless more and more kinetic energy is being inputted into the scenario over time, why would velocity stay constant when mass has increased?

If I'm misunderstanding the physics of what is going on here I apologize and will be happy to be corrected but at first glance something doesn't appear to be adding up about it.

Meaning there is no noticeable deceleration from the initial speed by the time it reaches America.

This isn't proven to be the case just by strong winds still existing in the US. When a shockwave dissipates, it can create ongoing turbulence and shifts in atmospheric pressure that persist long after the initial impact. Energy doesn't just disappear; the residual energy can circulate in the atmosphere, creating wind patterns that last a long time.

It doesn't need to have impacted the United States with Mach 86 winds in order to still have an effect a week later. (Plus, if the initial winds were that high hitting America, I'm pretty sure they'd have torn down every single structure on the Western coast)

 

[Therefir]

Isn't it specifically stated in the chapter that it was the gust of wind that Deku created the one thing that caused the strong winds over the US and not some sort of changes in atmospheric pressure? I feel you're reaching with things that were never stated or explained in the manga, Damage.

I'm not saying it's impossible, but the reporter in the chapter was clearly trying to imply that it was a gust of wind created during the final battle the same one that will be hitting the US over the week.

As for which calculations seem best to me, I still like the approach of using both storm dispersion and the one week of strongs winds.

 

[Damage3245]

Isn't it specifically stated in the chapter that it was the gust of wind that Deku created the one thing that caused the strong winds over the US and not some sort of changes in atmospheric pressure? I feel you're reaching with things that were never stated or explained in the manga, Damage.

You misunderstand me, the change in atmospheric pressures / residual energy in the atmosphere is the after-effects of Deku's punch.

That's besides the main point of my post above.

 

[Dalesean027]

You've got Izuku initially affecting X kg mass of air at a calculated speed of 29640.8 m/s.

Izuku's shockwave then keeps going, affecting more and more mass until at least 1210897470122445800 kg mass of air is affected.

But that end amount of mass is far greater than X.... So unless more and more kinetic energy is being inputted into the scenario over time, why would velocity stay constant when mass has increased?

So basically you're wanting Rusty to do this?

Spoiler

To describe how clouds can affect wind speeds, we can use the concept of drag force. The drag force (F_d) can be calculated using the formula:

F_d = 0.5 * ρ * v^2 * C_d * A

Where:

• F_d = drag force
• ρ = air density
• v = wind speed
• C_d = drag coefficient (which can vary based on the shape and type of clouds)
• A = cross-sectional area

This formula helps to understand how clouds can create resistance against wind flow, thus slowing it down. Keep in mind that the actual impact of clouds on wind speed can be influenced by many other atmospheric factors.

Edit: at work so I couldn't say more but Ik this is drag but the only way I can see him taking into account what you've said is by either using this or just switching the timeframe of deceleration of the shockwave from 1 week to just the one day to account for the increasing mass as the shockwave spreads which wouldn't be as accurate

 

[Damage3245]

So basically you're wanting Rusty to do this?

Spoiler

To describe how clouds can affect wind speeds, we can use the concept of drag force. The drag force (F_d) can be calculated using the formula:

F_d = 0.5 * ρ * v^2 * C_d * A

Where:

• F_d = drag force
• ρ = air density
• v = wind speed
• C_d = drag coefficient (which can vary based on the shape and type of clouds)
• A = cross-sectional area

This formula helps to understand how clouds can create resistance against wind flow, thus slowing it down. Keep in mind that the actual impact of clouds on wind speed can be influenced by many other atmospheric factors.

Edit: at work so I couldn't say more but Ik this is drag but the only way I can see him taking into account what you've said is by either using this or just switching the timeframe of deceleration of the shockwave from 1 week to just the one day to account for the increasing mass as the shockwave spreads which wouldn't be as accurate

Switching the timeframe of deceleration from a week to the day wouldn't solve the issue at all.

I'm not sure about that drag force method yet; perhaps I'll understand it better seeing it in action.

 

[Dalesean027]

Switching the timeframe of deceleration from a week to the day wouldn't solve the issue at all.

I didn't think so either but had to suggest it

I'm not sure about that drag force method yet; perhaps I'll understand it better seeing it in action.

Spoiler: Example:

• Air density (ρ) = 1.225 kg/m^3 (typical value at sea level)
• Velocity (v) = 30 m/s
• Drag coefficient (C_d) = 0.47 (typical value for a smooth sphere)
• Cross-sectional area (A) = 0.1 m^2

Now, we can plug these values into the drag force formula:

F_d = 0.5 * ρ * v^2 * C_d * A

Now substituting the values:

F_d = 0.5 * 1.225 kg/m^3 * (30 m/s)^2 * 0.47 * 0.1 m^2

Calculating step-by-step:

1. Calculate v^2:
(30 m/s)^2 = 900 m^2/s^2

2. Multiply ρ and v^2:
1.225 kg/m^3 * 900 m^2/s^2 = 1102.5 kg*m/s^2 (which is also Newtons)

3. Now multiply by C_d and A:
1102.5 N * 0.47 * 0.1 m^2 = 51.2675 N

4. Finally, multiply by 0.5:
F_d = 0.5 * 51.2675 N = 25.63375 N

So, the drag force acting against the object moving at 30 m/s is approximately 25.63 N. This drag force would slow down the object as it moves through the air.

F = m * a


Let's assume the mass (m) of the object is 10 kg. The drag force (F_d) we calculated is 25.63 N. Since the drag force acts in the opposite direction to the motion, it will cause a deceleration.

Using Newton's second law:

a = F / m

Substituting the values:

a = 25.63 N / 10 kg = 2.563 m/s^2

This means the object would decelerate at a rate of 2.563 m/s^2 due to the drag force.

To determine how much the speed decreases over a certain time period, we can use the kinematic equation:

v_f = v_i - a * t

Where:

• v_f is the final velocity.
• v_i is the initial velocity (30 m/s in this case).
• a is the deceleration (2.563 m/s^2).
• t is the time.

For example, if we want to know the speed after 5 seconds:

v_f = 30 m/s - 2.563 m/s^2 * 5 s
v_f = 30 m/s - 12.815 m/s
v_f = 17.185 m/s

So, after 5 seconds, the speed of the object would decrease to approximately 17.19 m/s due to the drag force.

We know due to Rusty's math it takes like 6 minutes for the shockwave to reach the US so we'd use that as our timeframe for for that final portion

 

[Dalesean027]

Unless either Rusty or someone else here gets to it first when I'm off work and at home in a few hours I can take his numbers and run them through this to get better results

 

[Damage3245]

We know due to KTs math it takes like 6 minutes for the shockwave to reach the US so we'd use that as our timeframe for for that final portion

KT's math?

 

[Dalesean027]

KT's math?

whoops like I said at work, definitely a typo lol

 

[Dalesean027]

Air density (ρ) = 0.5258 kg/m^3
Velocity (v) = 29640.80 m/s
Drag coefficient (C_d) = 0.42 (typical value for a half sphere)
Cross-sectional area (A) = 67428900m^2 (using a half sphere)

Now, we can plug these values into the drag force formula:

F_d = 0.5 * ρ * v^2 * C_d * A:
0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 0.42* 67428900m^2 = 6.541326e15 N

a = F / m:
6.541326e15 N / 1210897470122445800kg = 0.00540204779m/s^2

v_f = v_i - a * t:
29640.80m/s - 0.00540204779m/s^2 * 362.04s = 29638.8442m/s

1/12*1210897470122445800*29638.8442^2 = 8.8643859e+25 Joules or 21.186391 Petatons of TNT

Honestly someone can check my work but it looks like the actual difference is negligible at best here so Rusty's method basically is still good to go unless my math is off here

Edit: here's a version using a cylinder for this instead of a half sphere

Spoiler: Cylinder Version

Air density (ρ) = 0.5258 kg/m^3
Velocity (v) = 29640.80 m/s
Drag coefficient (C_d) = 0.82 ~ 1.15
Cross-sectional area (A) = 67428900m^2 (using a half circle)

0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 0.82 * 67428900m^2 = 1.27711603e16N
0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 1.15 * 67428900m^2 = 1.79107736e16N


a = F / m:
1.27711603e16N / 1210897470122445800kg = 0.0105468552m/s^2
1.79107736e16N / 1210897470122445800kg = 0.0147913213m/s^2

v_f = v_i - a * t:
29640.80m/s - 0.0105468552m/s^2 * 362.04s = 29636.9816m/s
29640.80m/s - 0.0147913213m/s^2 * 362.04s = 29635.445m/s

1/12*1210897470122445800*29636.9816^2 = 8.8632718e+25 Joules or 21.18372 Petatons of TNT
1/12*1210897470122445800*29635.445^2 = 8.8623527e+25 Joules or 21.181531 Petatons of TNT

 

[Damage3245]

I'll get back to you on that point because I think I've spotted an issue.

 

[CloverDragon03]

Air density (ρ) = 0.5258 kg/m^3
Velocity (v) = 29640.80 m/s
Drag coefficient (C_d) = 0.42 (typical value for a half sphere)
Cross-sectional area (A) = 67428900m^2 (using a half sphere)

Now, we can plug these values into the drag force formula:

F_d = 0.5 * ρ * v^2 * C_d * A:
0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 0.42* 67428900m^2 = 6.541326e15 N

a = F / m:
6.541326e15 N / 1210897470122445800kg = 0.00540204779m/s^2

v_f = v_i - a * t:
29640.80m/s - 0.00540204779m/s^2 * 362.04s = 29638.8442m/s

1/12*1210897470122445800*29638.8442^2 = 8.8643859e+25 Joules or 21.186391 Petatons of TNT

Honestly someone can check my work but it looks like the actual difference is negligible at best here so Rusty's method basically is still good to go unless my math is off here

Are you calculating the air movement or the cloud movement? If it's the latter, I'd say use a cylinder instead of a half sphere, since we tend to use cylindrical shapes for cloud movement

 

[Dalesean027]

Are you calculating the air movement or the cloud movement? If it's the latter, I'd say use a cylinder instead of a half sphere, since we tend to use cylindrical shapes for cloud movement

air since i'm doing drag for the speed of the wind/shockwave i'm just taking the numbers from Rusty's calc with the weight and speed

 

[CloverDragon03]

Also, the kinematic equation is v_f = v_i + a*t

You did subtraction instead of addition

 

[Dalesean027]

Also, the kinematic equation is v_f = v_i + a*t

You did subtraction instead of addition

well yeah tho this is finding how the drag affects it tho so adding would be making it faster not slower due to drag?

 

[CloverDragon03]

Oh wait yeah, acceleration would then be negative here, my b

 

[Dalesean027]

all good I've made some adjustments for a cylinder as well

 

[CloverDragon03]

Your cross-sectional area is still that of a half circle. Are the cross-sectional areas for those two shapes the same?

 

[Dalesean027]

Your cross-sectional area is still that of a half circle. Are the cross-sectional areas for those two shapes the same?

yup I mean slice a cylinder in half and the cross sectional area would just the base which is still a circle, a half sphere would have the same cross section since the base is what that would be

 

[CloverDragon03]

Anyways here's my scuffed take on it. Since it's pure air, I might be doing things a bit different:

Volume = 3.6181125e+14 * 21336 = 7.7196048e+18 meters^3

Density = 0.5258 kg/m^3

Mass = 7.7196048e+18 * 0.5258 = 4.0589682e+18 kg

The air moved out in a cylindrical motion so I'll use a cylinder.

Drag Coefficient = 1.12 (between 1.08 and 1.16 from what I found)

Cross-Sectional Area = 67428900 meters^2

0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 1.12 * 67428900m^2 = 1.7443536e+16 N

a = 1.7443536e+16 / 4.0589682e+18 = 0.0042975296 m/s^2 (It's negative because it's deceleration)

v_f = v_i + a*t

v_f = 29640.80 - (0.0042975296 * 362.04) = 29639.24412 m/s

KE = (1/12) * 4.0589682e+18 * 29639.24412^2 = 2.9714515e+26 Joules or 71.01 petatons (High 6-A)

Chat I might've freaked it

 

[Dalesean027]

Anyways here's my scuffed take on it. Since it's pure air, I might be doing things a bit different:

Volume = 3.6181125e+14 * 21336 = 7.7196048e+18 meters^3

Density = 0.5258 kg/m^3

Mass = 7.7196048e+18 * 0.5258 = 4.0589682e+18 kg

The air moved out in a cylindrical motion so I'll use a cylinder.

Drag Coefficient = 1.12 (between 1.08 and 1.16 from what I found)

Cross-Sectional Area = 67428900 meters^2

0.5 * 0.5258 kg/m^3 * (29640.80 m/s)^2 * 1.12 * 67428900m^2 = 1.7443536e+16 N

a = 1.7443536e+16 / 4.0589682e+18 = 0.0042975296 m/s^2 (It's negative because it's deceleration)

v_f = v_i + a*t

v_f = 29640.80 - (0.0042975296 * 362.04) = 29639.24412 m/s

KE = (1/12) * 4.0589682e+18 * 29639.24412^2 = 2.9714515e+26 Joules or 71.01 petatons (High 6-A)

Chat I might've freaked it

at a quick glanec through (I'm cooking so actually a quick glance) I don't see any mistakes right off hand

 

[Damage3245]

at a quick glanec through (I'm cooking so actually a quick glance) I don't see any mistakes right off hand

Okay, so regarding this:

We know due to Rusty's math it takes like 6 minutes for the shockwave to reach the US so we'd use that as our timeframe for for that final portion

The main reason why there's only negligible differences in both yours and Clover's calculations from the original calculation in the OP is because that timeframe is found by assuming that the shockwave travelled at constant speed from Mt. Fuji to the United States:

Distance / Velocity = Time

10731640 m / 29640.8 m/s = 6.03427258 minutes = 362.04 seconds

I'm sure you can see why assuming constant speed wouldn't be appropriate for finding the timeframe for this given the issues I raised with assuming constant speed for the kinetic energy in the first place.

Moreover, for your cross-sectional area, I'm not sure how exactly you got that value of 67428900 meters^2 for that area, but wouldn't the area be changing and growing as the shockwave expands? Your calculation seems to assume that for the full assumed timeframe of 362.04 seconds, the area of the shockwave was a circle measuring only 67428900 meters^2.

 

[Dalesean027]

Moreover, for your cross-sectional area, I'm not sure how exactly you got that value of 67428900 meters^2 for that area, but wouldn't the area be changing and growing as the shockwave expands? Your calculation seems to assume that for the full assumed timeframe of 362.04 seconds, the area of the shockwave was a circle measuring only 67428900 meters^2.

Yes the shockwave in its totality reaches from mount fuji to America so using the cross sectional area of what would be omnidirectional shockwave thhat reaches America is how we get that total area affected.

The main reason why there's only negligible differences in both yours and Clover's calculations from the original calculation in the OP is because that timeframe is found by assuming that the shockwave travelled at constant speed from Mt. Fuji to the United States:

Distance / Velocity = Time

10731640 m / 29640.8 m/s = 6.03427258 minutes = 362.04 seconds

I'm sure you can see why assuming constant speed wouldn't be appropriate for finding the timeframe for this given the issues I raised with assuming constant speed for the kinetic energy in the first place.

That's still fine using said speed isn't wrong aa that would be our initial speed in the equation he only thing needes to change would be the timeframe for how long we wanna say it took to reach America but the obviously if you don't want a constant speed then we'd assume so other arbitrary value so what are you wanting to work with 10 minutes instead? 15 minutes, 30, An hour? I can't see it taking longer than that considering the effects and energy at work

 

[Damage3245]

Yes the shockwave in its totality reaches from mount fuji to America so using the cross sectional area of what would be omnidirectional shockwave thhat reaches America is how we get that total area affected.

Could you explain how you calculated the area for that? I didn't get that part.

That's still fine using said speed isn't wrong aa that would be our initial speed in the equation he only thing needes to change would be the timeframe for how long we wanna say it took to reach America but the obviously if you don't want a constant speed then we'd assume so other arbitrary value so what are you wanting to work with 10 minutes instead? 15 minutes, 30, An hour? I can't see it taking longer than that considering the effects and energy at work

I don't know which timeframe which would be best yet but we can't use the current timeframe.

 

[Dalesean027]

Could you explain how you calculated the area for that? I didn't get that part.

The shockwave reaches america, both the cross sectional area of a half sphere and cylinder are a circle and both would be the same value (a shockwaves at ground level expands as a half sphere, cloud formations expand in a cylinder hence the two versions) its the total distance of the shockwaves because doing otherwise for the initial would be incorrect here for how this calc works

I don't know which timeframe which would be best yet but we can't use the current timeframe

Hence my saying we have to do something arbitrary then since we're already taking into account drag even using something like a large meteor impacting earth the shockwaves would only take a little over an hour to travel the earth since this is pretty comprable I can't see it being some insane amount of time I'm going an hour for this one tops