Potential Energy Calculations (2024)

This page is to show all the calcs for any feats that includes energy attacks and explosions.

Contents

  • 1 Potential Energy Calculation
  • 2 Inertia Calculation
  • 3 Nuclear Binding Energy
  • 4 Rotational Kinetic Energy
  • 5 Energy Attack
  • 6 Gravitational Binding Energy
  • 7 Inverse Square Law
    • 7.1 What is Inverse Square Law?
      • 7.1.1 Example 1: Finding mass from a size change
      • 7.1.2 Example 2: Finding energy based on something destroyed within an explosion
      • 7.1.3 Example 3: Finding the durability of a character given an explosion via Intensity
      • 7.1.4 Example 3: Finding the energy of an attack that ended up destroying something at a distance
    • 7.2 Why should we use Inverse Square Law
    • 7.3 When Should we use Inverse Square Law?
    • 7.4 Dealing with explosions and Inverse Square Law in fiction
    • 7.5 Inverse Square Law and our Attack Potency Chart
    • 7.6 Simplified
  • 8 Explosion/Airblast Yield
    • 8.1 Radiation Requirements
  • 9 Mass-Energy Equivalence
  • 10 TNT
  • 11 Comprehensive Destruction
  • 12 Destroying the Target
  • 13 Lightning Energy
  • 14 Heat of Combustion
  • 15 Release of Gravitational Energy via Black Holes

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.

Inertia Calculation[]

This section will talk about on finding the Mass Moment of Inertia.

MoI depends on what volume you are using, such as a cylinder, which you would need to find the Radius (cm) and the height/depth (cm) and the mass of said object (kg), plug them together with (Iz= (mass * radius * 2)/2) and find the Mass moment of inertia about Z axis.

Nuclear Binding Energy[]

This section talks about finding the energy of mass of objects at an atomic or particle level,

  1. Find the mass of the object (Ibs)
  2. Find the atomic mass of the material said object is made from and convert to grams.
  3. Divide pounds by atomic grams to find # of atoms.
  4. for particles, multiple by 56 which are the number of particles around an atom.
  5. Then find the Electronvolt (MeV) of the material needed.
  6. Multiply both to get the overall energy (Joules)

Rotational Kinetic Energy[]

This section talks about finding the Kinetic energy of rotating objects.

Taking what you have for Angular Velocity with Frequency (Rotations per Minute) and plugging it into this along with the found MOI (Mass Moment of Intertia) to find how much energy (Joules). and plug them into the formula here (E rotational = (1/2) I (moment of inertia around the axis of rotation=angular velocity) * ω(angular velocity)^2).

Energy Attack[]

This section 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

Potential Energy Calculations (1)

Where U = GBE = (3*G*M^2)/(5*r), M = the mass of the body in question kg, r = its radius meters, and G = the gravitational constant (6.67408x10^-11).

For Stars it is different, while the equation is similar however = E = (3*G*M^2)/r(5-n)

  • While the factors are the same, where as "n" is the polytropic value attributed to the type of star.

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

SD.net's Planetary Parameter Calculator gives a simplified approach to finding a celestial body's GBE with diameter or gravity mods, for FXSolver , gives one control over the Mass and Radius mods.

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:

Potential Energy Calculations (2)

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 thepolytropic valueattributed 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

EX. (Cell Solar Kamehameha)

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.

Example 3: Finding the energy of an attack that ended up destroying something at a distance[]

The use of this where two people fought that their power would end up destroying a distant place relative to what would be in their area.

  • You would need to divide the area where the explosion/shockwave was (meters squared) by the location's surface area (meters squared) and then multiply by energy (joules)

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.

Ex. (Thor Taking a Neutron Star)

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:

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.

Simplified[]

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

Explosion/Airblast Yield[]

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.

Near-Total Fatalities is all large above-ground structures are destroyed within that radius, as well as causing 100% fatalities to anyone that is affected within the radius due to sheer pressure alone.

use this equation to find the total yield of said Airburst explosion .

  • Thermal Radiation radius (3rd degree burns) = ((x)^2.439)/400 = Y
  • Air blast radius (Near-Total Fatalities) = Y = ((x/0.28)^3) or R = Y^(1/3)*0.28.
  • Air blast radius (widespread destruction) = ((x)^3)/400 = Y
  • Ionizing radiation radius (500 rem) = ((x)^5.26)/400 = Y
    • Y = Yield in Kilotons or Megatons divided with by 1000
    • x/R = radius of explosion in km
    • 0.28 = adimentional constant
    • D = duration (in seconds)

For Fireball Radius Explosions: you would need to follow this calculation;

  • Fireball duration = (x/4.5)^2.22 = D
  • Fireball radius (minimum) = ((x/(27.3))^2.5)/1000 = Y
  • Fireball radius (airburst) = ((x/33.33)^2.5)/1000 = Y
  • Fireball radius (ground-contact airburst) = ((x/43.93)^2.5)/1000 = Y

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

Since the calculator didn't give results below 1 kiloton, the above would be useful; using these lists we will find the required area.

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.

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 (divide by 1000 to get Kilotons)
  • R is radius in meters (from between airblast and fireball in nukesecrecy calc link above or Near-Total Fatalities in Star Destroyer link)
  • P is pressure of the shockwave in bars (the standard overpressure is 20 psi or1.37895 bars.)

For more information, follow these two links.

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

Radiation Requirements[]

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/sickening
    • 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

You can also use certain Bomb craters to estimate a calculated explosion yield from a known explosion yield in real life.

or if you have the diameter, you can use this Crater Diagram to calc the explosion size from the NUKEMAP website.

Mass-Energy Equivalence[]

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.

TNT[]

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.

  • Height
  • Width
  • Area
  • Density
  • TimeFrame
  • Depth
  • Speed
  • Force
  • Energy
  • Mass

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:

  • Length, Depth, Volume, Angle Measurement
    • Comprehensive Destruction
    • Technique Speed

Comprehensive Destruction[]

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

Lightning Energy[]

This calc is to determine the energy of a given lightning bolt feat, which would require a few things:

  1. The surface area of which the lightning bolt struck at (meters squared) for a circular area.
  2. Then multiply it by the density of lightning which is 955e+6 A/m^2.
  3. Then multiply both to get the Amps.
  4. Then you plug in the information to the Electrical Work Relationship which is "P = work done per unit time = (Q*V)/t = I*V"
  5. Multiply both Amps and Volts to get the Watts.
  6. Then multiply by number of seconds, which the average lightning bolt remains is .2 seconds.

Heat of Combustion[]

Heat of Combustion is the heating value of a substance, usually a liquids and gas and solids, is the amount of heat released during the combustion of a specified amount of it. Here is the calculation of the method.

For when you find the value of the substance in question (g/mol) you would then multiply the mass of the same substance to get the number of moles (mols), where you will then multiply by the Kilojoules per Mole (kj/mol) to find how much Kilojoules there are. then you can convert it to Joules (1 KJ = 1000 Joules)

Release of Gravitational Energy via Black Holes[]

Here we will be looking at how Gravitational Energy is released, we will be using the Black Hole Merger as an example.GW150914, the subject of OP's post, was a black hole merger involving black holes with pre-merger masses of 36 and 29 solar masses. The resultant black hole has a mass of 62 solar masses. The before/after difference:

  • (36 + 29) - 62 = 3 solar masses

was radiated as gravitational waves during the final moments of the merger. Using the mass energy equivalence equation, E = mc2, where,

  • m = 3 x (2 x 10^30 kg)
  • c = 3 x 108 m/s

yields

  • E = 5.4 x 10^47 joules.

A release of such energy within the final 150 milliseconds of the merger:

  • 5.4 x 10^47 joules / .15 s = 3.6 x 10^48 watts
Potential Energy Calculations (2024)

FAQs

Potential Energy Calculations? ›

Multiply the mass of the object ( m ) and the height above the reference level ( h ) by the acceleration g to find the potential energy: E = m · g · h .

How do you calculate potential energy? ›

The most common type of potential energy (U) is gravitational potential energy, which is calculated based on the mass of the object (m), the gravitational acceleration constant (g), and the height above the ground (h). The potential energy formula is U=mgh.

How do you calculate potential energy strain? ›

Strain Energy Formula
  1. Form 1: The strain energy formula is equal to half the product of the compression factor (δ) and force(F) applied to the body.
  2. Thus, StrainEnergy,U=F×δ2.
  3. δ = compression factor.
  4. Form 2: The strain energy can be expressed in terms of stress (σ), the volume(V) of the body and Young's modulus (E) as,
Dec 8, 2022

What is the formula for total potential energy? ›

The total potential energy (Π) of a system is the sum of the strain energy (U) and the work (W) done by the external loads, and it is expressed as(1)Π=U+W,where W = − ∑Piqi, and U = U(εij).

What is the formula for potential? ›

The electric potential formula is: V = W q . Also, electric potential or voltage can be calculated by using the equation: V = E ∗ r or the formula: V = R ∗ I depending on the information given in the excercise.

How to solve PE without height? ›

If you dont know the height, then you cannot calculate the GPE directly. The alternative is to consider some process where the GPE is converted to some other form of energy and you measure that. example- ball rolls down a hill- measure its speed and mass and calculate the KE of the ball - (KE =1/2mv^2).

How do you calculate energy potential in electricity? ›

To calculate the electric potential energy of a system, use the equation U_e = k(q_1q_2)/r. In this equation, k stands for the Coulomb constant, q_1 and q_2 are two charges, and r is the distance between them.

What is the formula for energy? ›

The equation developed by Albert Einstein, which is usually given as E = mc2, showing that, when the energy of a body changes by an amount E (no matter what form the energy takes), the mass (m) of the body will change by an amount equal to E/c2.

How to calculate potential energy in a spring? ›

Formula of Potential Energy of A Spring
  1. P.E = ½ k x2
  2. Example 1: Determine the potential energy of a spring whose spring constant is 200 N/m and the displacement is 0.8 m.
  3. Solution:
  4. P.E = 64 J.
  5. Example 2: The spring constant and displacement of a stretched string is 100 N/m and 0.5 m respectively. ...
  6. Solution:
  7. P.E = 12.5 J.

What is the formula for true potential energy? ›

Potential energy
Common symbolsPE, U, or V
SI unitjoule (J)
Derivations from other quantitiesU = m ⋅ g ⋅ h (gravitational) U = 1⁄2 ⋅ k ⋅ x2 (elastic) U = 1⁄2 ⋅ C ⋅ V2 (electric) U = −m ⋅ B (magnetic) U =
1 more row

How to derive potential energy formula? ›

Let's consider an object taken to a height 'h' from the surface of the earth. When, h<<R, then, R + h = R and g = GM/R2. On substituting this in the above equation we get, Gravitational Potential Energy ΔU = mgh.

How to calculate total potential? ›

Step 1: Determine the distances r1 and r2 from each point charge to the location where the electric potential is to be found. Step 2: Apply the formula V = k Q r for both charges to calculate the potential due to each charge at the desired location. Step 3: Find the sum of the potentials of charges 1 and 2.

How to calculate potential energy? ›

Multiply the mass of the object ( m ) and the height above the reference level ( h ) by the acceleration g to find the potential energy: E = m · g · h .

How to calculate loss of potential energy? ›

The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy. This can be written in equation form as -ΔPEg = ΔKE. Using the equations for PEg and KE, we can solve for the final speed v, which is the desired quantity. mg|h|=12mv2.

What is an example of a potential energy? ›

Four common examples of potential energy include a skydiver waiting to jump from an airplane, a rubber band that has been stretched out between two fingers, water sitting behind a dam, and the energy in a battery. All four examples are examples of stored energy that has the potential to do work upon its release.

What is the formula for potential energy charge? ›

1) The equation for electric potential is U e = k q 1 q 2 r . The first step is to write the knowns. Note that the distance between charges is measured from the center of each charge. 2) Plug the values from 1) and the value for the Coulomb constant into the electric potential energy equation.

What is the formula for kinetic and potential energy? ›

Difference Between Kinetic Energy and Potential Energy
Kinetic energyPotential energy
The formula used to calculate kinetic energy is K E = 1 2 m v 2The formula used to calculate potential energy is m g h
Vibrational energy is an example of kinetic energyGravitational potential energy is an example of potential energy
1 more row
Feb 20, 2021

What is the potential energy of a rock that weighs 150 newtons? ›

Answer: The potential energy of the rock is 30,000 J.

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