This page is to show all the calcs for any feats that includes energy attacks and explosions.
Potential Energy Calculation
This is for Calcs that involve slicing and cutting feats or lifting objects (E (Joules) = mgh), there are a few things in order for you to know to calc this type of feat:
- h = first you would need the distance of which whatever the person cuts that separates the target (meters)
- m = The mass of the target in (kg)
- g = Then multiple both of those with the Gravitational Constant (9.8 m/s^2) to get your results in joules.
This page is to calc the amount of energy is use to destroy something.
- First you must find the volume of said energy attack in radius and depth, if it is spherical, then just radius to find (cm^3).
- Then go for Comprehensive Destruction (See below) to find how powerful the energy attack is.
- multiply both to get the result in joules.
Gravitational Binding Energy
Based on the OBD wiki:
Want to find out how much energy it takes to blow up a planet? This is the way to do it. Gravitational Binding Energy is defined as the amount of energy it would take to scatter the mass of a gravitationally bound body to the point that its own gravity will not pull it back together again. There are precise calculations for this via integration, but a good approximation can be achieved with the following formula:
Gravitational Binding Energy
Where U = GBE, M = the mass of the body in question, r = its radius, and G = the gravitational constant (6.67408x10^-11).
Ignoring this formula often leads to vast underestimates of the energy required to destroy astronomical objects (for example, some people assume it scales linearly with mass or volume).
Please note that this formula only works on objects that are mostly held together by their own gravity (meaning: large objects in space such as asteroids, moons, planets, stars, etc.) It also doesn't work on black holes, for obvious reasons.
List of approximate GBE values for various objects:
-Earth's moon (Luna): 1.24e29j
-Earth: 2.24e32j (calculated with a more accurate method than the above formula)
-The sun (Sol): 6.87e41j
-Average neutron star: 5.23e46j
This applies for Implosions of planets as well if said planet actually is destroyed and cannot be bound by gravity anymore.
for Stars it is a different matter:
U is GBE in joules, G is the gravitational constant of 6.67408x10^-11, M is mass in kilograms, r is radius in meters, and n is the polytropic value attributed to the type of star. While this formula is not perfectly accurate, it is widely applicable, and is still within the acceptable margin for error.
Inverse Square Law
NOTE: Inverse Law is a little complicated and requires careful consideratio, otherwise we do not use it for every calc.
What is Inverse Square Law?
Wikipedia describes Inverse Square Law as this:
|“||In physics, an inverse-square law is any physical law stating that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity.||„|
A simplified formula for this is area of the larger object or explosion/area of the object x the initial value
It can be used for many different subjects:
Example 1: Finding mass from a size change
A character can lift a sphere that weighs 3000 kilograms. This sphere is half a meter in radius. Now, this sphere is 10 meters in diameter, and he lifts it. How much does he lift?
- First we find the area of the two spheres. The area of the sphere originally was 3.14159 square meters, and the upsized sphere is 1256.64 square meters.
- Then we apply the formula ‘’new area of the object/original area of the object x the initial mass of the object’’
- 1256.64 square meters/3.14159 square meters * 3000 kilograms
- Plugging it in, the person lifted a 1,200,003.82 kilogram sphere, which is Class M
Example 2: Finding energy based on something destroyed within an explosion
A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?
- First we use the known value for vaporization onto the brick, which is 25700 joules per cubic centimeter. It requires 27.4798773 megajoules to vaporize said brick.
- When dealing with ground explosions, one should use a hemisphere as a basis for the explosion’s shape. The area of a hemisphere with a radius of 5 meters is 157.08 square meters
- 157.08 square meters/0.07116953508 square meters * 27.4798773 megajoules
- If we plug in these numbers, it results in 60.651501 gigajoules, or 14.496 tons of tnt, which is City Block level
Example 3: Finding the durability of a character given an explosion via Intensity
In this case the example shown in the The Explosion Yield Calculations page shall be utilized
- The amount of energy that hits a target if r meters away from an explosion is E=I*CA, where I is calculated as described above and CA is the area of the cross section of the target (the cross section orthogonal to the direction the explosion expands into, to be specific) and E is the energy the target is hit by.
- CA is approximately 0.68 m^2 for a grown human. It can also be estimated as half of the bodies surface area calculated using this, but that is a slight overestimation.
EX. (Cell Solar Kamehameha)
An omnidirectional explosion of 7 kilotons of TNT occurs, and a human 30 meters away from the epicenter endures the explosion. How high is the durability of the human?
- First, we set P = 7000 Tons of TNT, as that is the yield of said explosion.
- Second, we set the radius, or r = 30m
- Third, we find the value of I, or the intensity of the explosion at a specific distance.
- I = (7000 Tons of TNT) / 4π((30m)^2)
- This means at 30 metres away from the epicenter of the explosion, the shockwave is hitting with an intensity of I = 0.619 Tons of TNT per m^2.
- CA = 0.68 m^2 for a human.
- So I*CA = 0.619 Tons of TNT per m^2 * 0.68 m^2 = 0.42092 Tons of TNT = Energy.
- The character can withstand a 0.42092 Tons of TNT blast, meaning Building level durability.
Why should we use Inverse Square Law
One thing commonly forgotten when dealing with attack potency-related feats is the area in which the attack or movement or energy has compared to a target. For example, the energy you exert in your footsteps in theory is more than enough to kill a very small ant. However, the area of the foot compared to the ant’s surface area is so large in disparity, the ant does not take the full brunt of the force, and may possibly live. However, if one presses a finger on an ant, it would be crushed under the force as the area ratio is not nearly as large. This is why stepping on a flea, obviously much weaker than a beetle, would most likely not kill it while it would kill a beetle.
Furthermore, a character who endures an explosion may not necessarily have taken its full yield. Given the consistency of a writer, it is possible he could have survived it due to inverse-square law making him endure a far lesser yield of energy, especially if he endured such explosion from a long distance.
When Should we use Inverse Square Law?
Inverse Square Law has a numerous amount of applications, here is a small list of some uses:
- In finding the mass of objects
- Finding the resultant yield of an explosion
- Finding one’s durability from an explosion
- Newton’s Law of Universal Gravitation
- Coulomb's Law
There are many more applications for this law.
Dealing with explosions and Inverse Square Law in fiction
When a character in fiction is to be hit by an explosion, we must keep in mind that at times he may not endure its full brunt, realistically. We must apply this formula at times, to see how much he truly survived. If such yield proves more consistent with his showings, then it would be preferred to use such number given from inverse square law. In terms of feats dealing with a character enduring an explosion, we use this formula.
Area of the explosion/area of the character x the explosion's current energy
Inverse Square Law and our Attack Potency Chart
It should be noted that our values pertaining to Attack Potency when dealing with 4-B to 3-A use Inverse Square Law to find their results, which lead to numbers far greater than such things’ mass energy output or gravitational binding energy output. It is because of this that we don’t stringently require calculations or defined yeilds for a character who creates many stars, splits a galaxy in half, causes an explosion half the size of the galaxy, or similar feats. If they were to be calculated, it would be most likely that they result with numbers either within the 4-B or 4-A range, which runs blatantly against the blatant intent of several feats at this scale.
this Calculation is for any form of explosion that is done in space and includes both an area and celestial bodies, here is the formula needed:
4*U*(Er/Br)^2 = E
- U = GBE (joules)
- Er = Radius of Explosion (meters)
- Br = Radius of Celestial Body origin (meters)
And this link measures the sound pressure in distance
This is used for typically nuke like explosions that have air blast effects over a certain radius, this is to find the area (near-total fatalities) that an explosion should cover to reach certain AP tier.
use this equation to find the total yield of said Airburst explosion.
Y = ((x/0.28)^3)
R = Y^(1/3)*0.28
- Y = Kilotons
- x/R = radius of explosion in km
- 0.28 = adimentional constant
For getting Megatons, divide the results from above calc by 1000 (not necessary really),
Using the formula from the explosion yield calculations page, R = Y^(1/3)*0.28, where Y is the yield in kilotons and R the radius in km, so that:
- Yield*2 ==> Radius*2^(1/3) = Radius*1.26
- Yield*5 ==> Radius*5^(1/3) = Radius*1.71
- Yield*10 ==> Radius*10^(1/3) = Radius*2.1544
When you get said result for yield, multiply it by .45 as only 40-50% of the total energy of the explosion is from the blast. (Unless it is a nuclear explosion or proof of radiation based or result, disregard this point)
Here are a few links to help with finding explosion feats.
- http://www.stardestroyer.net/Empire/Science/Nuke.html (Use near-total-fatalities part)
Or use this to find the ground Level explosions for the results:
W = R^3*((27136*P + 8649)^(1/2)/13568 - 93/13568)^2
- W is yield in tons of TNT
- R is radius in meters (from between airblast and fireball in nuke calc link above)
- P is pressure of the shockwave in bars (the standard overpressure is 20 psi or 1.37895 bars.)
You can use the same formula above for smaller explosions as well, just replace a few things
- x in meters
- Y in grams of TNT
This section is about what would be considered Radiation and when it would be appropriate to disregard the requirement to divide by half the result from the formulas above, it must follow most of the criteria:
- The Energy in question is labeled as “Radiation” or "Nuclear"
- The Energy comes from sources that would cause radiation such as Natural Light sources (Sun, Moon), Radioactive material, etc..
- The Energy causes similar effects as radiation would:
- Damages atoms of living beings
- Causes physical contaminations
- Is considered poisonous/sick
- Causes a malignant growth to grow or a cancer
- Cause Some form of mutation.
- Either it would be highly concentrated or repeated exposure to such radiation overtime
- Radiolysis mentioned
NOTE: This could be used for some verses for their power system if they follow the requirements
Matter-energy conversion is something that is often assumed for processes in calculations, but rarely justified, unless it is explicitly stated to be the case or be supported by information that leads to the usage of this calculation.
One of the most important reasons why it is not a common equation to use is that it simply produces unrealistic values in virtually all cases. The energy required to do so is so ridiculously high that it is almost never realistic by any means.
Matter-energy conversion happens quite often in nature. For example any chemical reaction that requires energy to run will convert the energy required by the reaction into a very slight additional mass. For example, photosynthesis which takes
However, the most common equation for this is this (E=mc^2)
Here is a link to a calculator.
This is a way of getting feats or powers calculated via for Attack Potency, Destructive Capacity and for TNT Equivalence. Here are a few links that would help get started for that sort of thing.
For other miscellaneous means of Calculations such as what you would find on OBD or VSBattle Wiki, here are a means of what you would need to find to make those calcs.
Use the visuals of the Game, Comic, Manga, Movie/TV Show to help with the calcs. Or the description of the feat (Unless it is considered Hyperbolic or if the Lore is separate from the media or are different versions from one another).
Some with certain measurements such as Height, Width and Depth, multiply them together to get the Volume, which ever unit you get, convert them to cm^3, then multiple it by how ever powerful the means of destruction is to get joules, then convert it to which ever TNT unit to find the stat. here is the links for that:
Comprehensive Destruction is a way of measuring the volume of how precise and concentrated an attack is and equates to how much energy equivalence there is to destroy something. Here is the methods of destruction in order from least to most destructive:
- Minor Fragmentation = 8 j/cm³
- Fragmentation = 69 j/cm³
- Violent Fragmentation = 120 j/cm³
- Pulverization = 214.35 j/cm³
- Cutting = 120 (j/cc)
- Vaporization = 25700 j/cm³
- Liquification = 12957.175 (j/cc)
- Atomization = 30852.2j/cc
- Sub-atomization = 5.403E13 (j/cc)
We apply this to the Feat Measurements.
Destroying the Target
This is to talk about the destruction of a target rather than a massive area of effect explosion types, this can be done by finding the Volume (cm^3) of the target, and the method of destruction shown above.
You can find the Volume by finding the mass and density of the target and divide them with mass/density.
This website should help with that: https://www.calculatorsoup.com/calculators/physics/density.php
This calc is to determine the energy of a given lightning bolt feat, which would require a few things:
- The surface area of which the lightning bolt struck at (meters squared) for a circular area.
- Then multiply it by the density of lightning which is 955e+6 A/m^2.
- Then multiply both to get the Amps.
- Then you plug in the information to the Electrical Work Relationship which is "P = work done per unit time = (Q*V)/t = I*V"
- P = Power = Watts
- I = Amps = A
- V = Volts = 10^9 estimate voltage of a lightning bolt
- t = Time = seconds
- Multiply both Amps and Volts to get the Watts.
- Then multiply by number of seconds, which the average lightning bolt remains is .2 seconds.