# B Finding impact depth

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1. Apr 11, 2017

### JoeSalerno

I'm trying to find the impact or penetration depth of a projectile in the (reasonably) most realistic way possible. I assume this has many factors including velocity, the densities of both materials, shapes of both materials, the medium of which the projectile travels through, gravity, etc. Because I know this could be infinitely complicated by variables, I will be assuming the air will be standard temperature pressure. This should ease finding the air resistance on the projectile. Depending on the necessary equations, I may end up omitting other variables. Thanks in advance to anyone who can help.

2. Apr 11, 2017

### Baluncore

Is it it an asteroid hitting the Earth, or maybe a hailstone landing on snow?
Rather than asking an all-encompassing question, could you narrow it down a bit.

What range of mass and velocity does the projectile have on impact?
What material is the target? Liquid or solid? Structured or isotropic?

3. Apr 11, 2017

### JoeSalerno

To be more specific, this is for a .950 JDJ bullet impacting 4340 steel. To find the velocity of the projectile on impact, I would need to find out what effect the air has on it. We can assume the starting speed is 670 m/s and the weight is 233 grams. The material that will be impacted is 4340 steel, so it is a hardened metal alloy. I assume shape matters for aerodynamic reasons, so the projectile is 24.13mm wide and 50mm tall. For simplicity, we can assume it is a "basic" pointed bullet, the closest shape I could find to match it would be an ogive on top of a small cylinder. Hopefully this should be enough to get started, let me know if this isn't sufficient.

4. Apr 11, 2017

### Khashishi

The air resistance depends how long it travels through the air.

5. Apr 11, 2017

### JoeSalerno

We c
We can assume 100 Meters for simplicity.

6. Apr 11, 2017

### Baluncore

I think you can assume that bullet will still be travelling at 670 m/s after 100m.
There are ballistics tables for every bullet ever sold. The ballistics table or program will give you accurate velocity data.

7. Apr 11, 2017

### JoeSalerno

if we were to assume that the impact velocity was 670 m/s, what would be the next step?

8. Apr 11, 2017

### Baluncore

Probably find out what the projectile energy is on impact.
Then what mass of the target material could be heated with that energy to the temperature where it ceases to be strong.
As a worst case estimate you could work out the depth of a 1" hole having that mass of steel.

9. Apr 11, 2017

### rootone

Well it definitely helps to have your projectile made up a least partly from depleted Uranium,
but there are, em, 'issues' with that approach.

10. Apr 12, 2017

### JoeSalerno

So, I need to make some major unit corrections. The projectile is 77mm long, 25.4mm wide, and weighs (get ready for it, this is why it's theoretical) 46,238 grams. We will also give it an unreasonable speed of 914 m/s. When you say that I need the energy on impact, would that be E=(1/2)*mass*velocity^2 ?

11. Apr 12, 2017

### Baluncore

Yes. E=(1/2)*m*v2
I think your estimate of projectile mass at 46 kg is on the high side by a factor of 100. I would expect it to be below 1 kg.

77mm long
diam = 25.4 * 0.95 = 24.13 mm
radius = 12.065 mm
area = 457.3 mm2
max vol = 35,212 mm3 = 35.212 cc, assuming cylinder
Lead density = 11.34
max mass = 399.3 gram

Last edited: Apr 12, 2017
12. Apr 13, 2017

### JoeSalerno

I tried getting as accurate of a volume as basic geometry could get me by putting the volumes of a frustrum, cylinder, and paraboloid. This gave me 24,994mm^3. The kicker however is the material. I looked at a list of some of the densest materials that were maybe semi-feasable (yet ridiculous) and decided on a tungsten alloy that's 18.5g/cm^3 (I realize that it would take a massive amount of propellant to get this thing going 3000 ft/s). I just multiplied 18.5 by 2,499.4 and got that ridiculous weight. Would the result of the energy equation be in joules?

13. Apr 13, 2017

### JoeSalerno

I just realized that the way you went from cubic mm to cubic centimeters was by multiplying by 1000. This is probably because of my ignorance, but is that because we are working in cubic instead of regular length, because I multiplied by 10.

14. Apr 13, 2017

### Comeback City

Yes

15. Apr 13, 2017

### Baluncore

Yes, the System International is good that way.

No, I divided by 1000. You must go very carefully when using mm or cm rather than metres.
There are 10 mm in 1 cm. So there are 1000 mm3 in 1 cm3

16. Apr 13, 2017

### JoeSalerno

Also, for the formula, is the mass in g or kg, and is the speed m/s?

17. Apr 13, 2017

### Baluncore

The System International, SI, uses the MKSA = metres, kilogram, second, ampere units. One litre of water weighs 1 kg.
Convert everything to meters per second, kilograms and joules.
https://en.wikipedia.org/wiki/International_System_of_Units

18. Apr 13, 2017

### JoeSalerno

19. Apr 13, 2017

### Baluncore

Whoops, you entered your text inside the quote.
The data you will need on your plate alloy will be thermal capacity and the temperature at which it will lose it's strength.

Rapid compression of metal causes an instant temperature rise with shear and flow. You can suddenly deform metals that could never be deformed slowly when cold with the same force.
There are two competing effects. Firstly, you could punch a conical hole clean through a plate by delivering energy to a circular area on the surface and shearing the metal. Secondly you could heat a volume of a solid block of material so that it flows out to leave a hole in the alloy.

20. Apr 14, 2017

### JoeSalerno

Sorry about accidentally putting that last post in the quote. So, by thermal capacity I think you're saying specific heat. I found that's 486 joules/kg kelvin. Using the units you mentioned earlier there weren't any thermal ones so I assumed Kelvin would work. I'm sorry if I'm being a bit dense, but what would you call the value "of which a material loses its strength"? Is this just a general number i.e. Around 2000 Fahrenheit? For this specific steel the only thermal specs given are the melting point (2600 Fahrenheit), mean coefficient of thermal expansion (6.6), and the thermal conductivity (21).

21. May 27, 2017

### JoeSalerno

What exactly do you mean "the temperature at which it will loose it's strength"?
Also, sorry for abandoning this forum for over a month, got real busy and forgot about it.

22. May 27, 2017

### Baluncore

As the temperature of a metal is increased it reaches a point where the material becomes softer and more malleable.
My guess for steel would be about 500°C. You can probably find data on young's modulus and the yield point at different temperatures.

23. May 28, 2017

### JoeSalerno

I found a website that lists a bunch of the properties of 4340 steel, and some of the charts are a bit confusing. For the modulus of elasticity (young's modulus), it says 10^3 N/mm^2 is 165 at 500 Celsius. The yield strength in N/mm^2 is about 1050 at 500 Celsius (I'm estimating off a graph).

24. May 28, 2017

### Baluncore

Well found. Having numbers is all very useful, except you have not listed for comparison what the parameters are at 20°C. Do you have a link to the site so I can see what the graph looks like ?

25. May 28, 2017

### JoeSalerno

http://www.ssm.co.nz/sitefiles/4340.pdfhttp://www.ssm.co.nz/sitefiles/4340.pdf This is where I got my information. The graph for young's modulus doesn't go that low, but there is a bit on the yield strength at 20 Celsius. It is however That chart is kind of confusing as it mentions diameter, but no unit of measure.