What does my physics book mean? What values do these represent?

In summary: Since I'm not sure what Joules is (and I'm not sure if it's the same as J), I'm going to go with 5 J.So then taking that number and plugging it into the equation...5.537 * 8.025 = 47.1 J.So, the armor stops the bullet with a force of 47.1 J.
  • #1
Stradigos
9
0
I bought physics for game programmers today to study ballistic impact, but I'm confused as to where a few numbers are coming from. I'm basically looking to understand what the values represent since the book does a bad job at telling me.

I get that Ek = .5mv.

In talking about the F-Formula, developed by Dr. L. Thompson (neither of which I can find anything about) "The equation was presented as a ratio between the plate thickness and the diameter of the projectile."

That equation: t/d = 0.0623 (((mv)^2)/((d^3)(F^2)))cos^2 angle

I realize that the above was just an example of the equation in the book, but .0623 is pissing me off. I have no idea where they get this from. Probably some number relating to steel armor since that's what he's talking about... so I'll continue...

"The coefficient F is a measure of the penetration resistance of the armor"

That equation: F = 1.8288((t/d) - 0.45)(angle^2 + 2000) + 12192.

Can someone tell me what those numbers are? I don't care where they got them from, but what are they suppose to be? I need to know so I can plug my own in...

Apparently it's also possibly to "rearrange the Thompson formula so that it is in the terms of the minimum projectile energy, Ek, necessary to penetrate the armor.

Ek = .5mv^2 = 8.025((td^2)(F^2))/(cos^2 angle)

Again... another imaginary number. What is this 8.025? What should go there?

It says the right hand side of the last equation there tells which factors influence steel armor pen, so I'm guessing the 8.025 is relating to that, but what factor? Thickness?

To further confuse me, it seems like he then tries to take everything he's taught me so far and then simplify it.

Assuming the impact is head on and uses the second to last equation I provided to compute the F coefficient:

F = 1.8288((.01m/.009m) - .45)(2000) + 12192 = 14610

So then taking that number

.5mv^2 = 8.025 * .01 * .009^2 * 14610^2 = 1387.5 J.

It footnotes and says a 9mm bullet has a mass of .0082 kg and a muzzle velocity of 440 m/s. The bullet only does 794 J, so the armor stops the bullet.

I just need someone to tell me what those numbers represent. Much thanks!

Sorry for the lack of units, but that's literally all the book tells me. It doesn't provide units and the text doesn't indicate it. I think kg would be appropriate though.
 
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  • #2
They kind of look like empirical formulae.

There is a relatively simple model you can use for penetration test on hard materials. It might converge to the above results if you work it out. The idea is simple. It takes a certain amount of energy to create a crack. It's the energy of the bonds that keep the material together. So the total energy required to penetrate hard armor is proportional to the area of the surface where separation occurred.

A bullet will typically punch out a cone. You can estimate the geometry of the cone by taking the ratio of the speed of sound in the material to the speed of the bullet. That will be equal to the difference of radii (base radius - bullet radius) to the thickness of material.

Finally, you need to know amount of energy per surface area required. You can estimate it from stress intensity factor or energy release rate.
 
  • #3
Stradigos said:
In talking about the F-Formula, developed by Dr. L. Thompson (neither of which I can find anything about) "The equation was presented as a ratio between the plate thickness and the diameter of the projectile."

Did you try Googling "Thompson F-formula impact". The first hit I got was http://www.combinedfleet.com/formula.htm
 
  • #4
Just wanted to post up and say that I've decided to go with a pretty generalized equation based off the Thompson F-Formula for my purposes (a game, keep in mind).

This website seemed to really explain it well: navweaps.com/index_tech/tech-033.htm

Typically, the equation is T = (K)[(0.5)(W/g)V^2]^p

T is the thickness of the plate armor barely penetrated.

K = "catch-all" that changes with projectile nose shape, projectile size, projectile damage, definition of "penetration," plate type, and obliquity angle of impact)

W = Weight g = Gravity (Note, you don't need to do W/g since that's already figured in with kg, if you use kg).

V = Velocity

p = something I don't quiet understand, but am going to ignore for my purposes.

K was the tricky one for me. I took that list and basically started crossing off things I didn't want to factor in for the battle system. I cross off nose shape, definition of penetration (since T is enough for me) and plate type. That left me with size and damage of the projectile.

Size and damage I'm going to create my own rating called Hardness with.

In my other thread about combat design, I figured that a sword with a mass of .997 and velocity of 3.333 m/s does about 5.537 Joules of damage (yes, I'm going to call joules damage for the sake of the battle engine unless I can come up with something better).

If I want my armor to withstand that just up to the point of being penetrating, then my armor rating has to be = Projectile Joules I want to protect against * Toughness. Toughness will basically be the durability, and is the amount of beating the armor can take before breaking. I think I'm going to keep it a 0 to 1 range.

Let's say the armor has taken a couple hits but hasn't been penetrated and the toughness is now at 60%.

Against the incoming sword, 5.537 * .6 = 3.222

5.537 > 3.222, so 2.214 damage goes through and the armor rating is reduced .6/2.214 = .27 points.

This isn't official yet, but that's what I'm thinking of, as sloppy as it is.
 
  • #5


I understand your frustration with the lack of clarity in the book. Let me try to break down the equations and values for you.

First, the equation Ek = 0.5mv^2 represents the kinetic energy of an object, where m is the mass and v is the velocity. This equation is commonly used in physics to calculate the energy of a moving object.

Next, the F-Formula was developed by Dr. L. Thompson and is used to calculate the penetration resistance of armor. The value 0.0623 in the equation represents a constant that is specific to steel armor. It is derived from experimental data and is used to determine the thickness of the armor necessary to resist a projectile.

The coefficient F in the second equation is also related to the penetration resistance of the armor. It takes into account the thickness and diameter of the projectile, as well as the angle at which it impacts the armor. The numbers in the equation, such as 1.8288 and 0.45, are again constants derived from experimental data.

The value 8.025 in the rearranged Thompson formula represents a combination of factors that influence the penetration of steel armor. These factors include the thickness and diameter of the projectile, as well as the angle of impact.

In the final example, the value 14610 is the calculated F coefficient for a specific scenario. The value 1387.5 J represents the minimum kinetic energy necessary for a projectile to penetrate the armor, calculated using the F coefficient and the projectile's mass and velocity.

The footnote about the 9mm bullet is just an example to illustrate the use of the equations and values. It is not directly related to the previous calculations.

Overall, the numbers in these equations represent constants and factors that have been determined through experimentation and are used to calculate the penetration resistance of steel armor. I hope this helps clarify the meaning and significance of these values for you.
 

1. What is the meaning of the symbols and equations in my physics book?

The symbols and equations in your physics book represent the fundamental principles and relationships of the physical world. They are used to describe and predict the behavior of matter and energy.

2. How do I interpret the values given in my physics book?

The values in your physics book represent different physical quantities, such as mass, velocity, and force. They are often measured in standard units, such as meters, kilograms, and seconds, and can be used to solve problems and make predictions in various scenarios.

3. Why are there different units for the same physical quantity in my physics book?

Different units are used to measure the same physical quantity in order to accommodate different scales and systems of measurement. For example, length can be measured in meters, feet, or kilometers, depending on the context and preference.

4. How can I apply the concepts in my physics book to real-life situations?

The concepts in your physics book are applicable to many real-life situations, such as understanding the motion of objects, the behavior of fluids and gases, and the principles of electricity and magnetism. By understanding these concepts, you can better analyze and explain various phenomena in the world around us.

5. How can I use my physics book to improve my problem-solving skills?

Physics involves a lot of problem-solving, and your physics book is a valuable resource for developing your skills in this area. By studying the concepts, equations, and examples in your book, you can learn how to approach and solve problems using logical and systematic methods.

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